3. Fixed point approach to the solution of differential equations

Next, we will show a differential equation which solving by fixed point theorem in suitable spaces.

#### 3.1. Ordinary differential equation

Lemma 3.1. ref. [18] u is a solution of u<sup>0</sup> ðÞ¼ t f tð Þ ; u tð Þ satisfying the initial condition u tð Þ¼ <sup>0</sup> u<sup>0</sup> if and only if u tðÞ¼ <sup>u</sup><sup>0</sup> <sup>þ</sup> <sup>Ð</sup><sup>t</sup> <sup>t</sup><sup>0</sup> f sð Þ ; u sð Þ ds.

Proof. Suppose that u is a solution of u<sup>0</sup> ðÞ¼ t f tð Þ ; u tð Þ defined on an interval I and satisfying u tð Þ¼ <sup>0</sup> u0. We integrate both sides of the equation u<sup>0</sup> ðÞ¼ t f tð Þ ; u tð Þ from t<sup>0</sup> to t, where t is in I

$$\int\_{t\_0}^{t} \mu'(s)ds = \int\_{t\_0}^{t} f(s, \mu(s))ds$$

$$\mu(t) - \mu(t\_0) = \int\_{t\_0}^{t} f(s, \mu(s))ds.$$

Since u tð Þ¼ <sup>0</sup> u0, we have

$$
\mu(t) = \mu\_0 + \int\_{t\_0}^t f(\mathbf{s}, \mu(\mathbf{s})) d\mathbf{s}, \quad t \in I. \tag{3.1}
$$

We will show that, conversely, any function which satisfies this integral equation satisfies both the differential equation and the initial condition. Suppose that u is a function defined on an interval I and satisfies (3.1). Setting t ¼ t<sup>0</sup> yields u tð Þ¼ <sup>0</sup> u0, so that u satisfies the initial

condition. Next, we note that an integral is always a continuous function, so that a solution of (3.1) is automatically continuous. Since both u and f are continuous, it follows that the integrand f sð Þ ; u sð Þ is continuous. We may therefore apply the fundamental theorem of calculus to

The contraction mapping theorem may by used to prove the existence and uniqueness of the initial problem for ordinary differential equations. We consider a first-order of ODEs for a

The function f tð Þ ; u tð Þ also take value in <sup>R</sup><sup>n</sup> and is assumed to be a continuous function of <sup>t</sup> and

Definition 3.2. Suppose that <sup>f</sup> : <sup>I</sup> � <sup>R</sup><sup>n</sup> ! <sup>R</sup><sup>n</sup> where <sup>I</sup> is on interval in <sup>R</sup>. We say that f tð Þ ; u tð Þ is a globally Lipschitz continuous function of u uniformly in t if there is a constant C > 0 such that

> ðt t0

By the fundamental theorem of calculus, a continuous solution of (3.5) is a continuously

u ¼ Tu

ðt t0

Theorem 3.3. ref. [19] Suppose that <sup>f</sup> : <sup>I</sup> � <sup>R</sup><sup>n</sup> ! <sup>R</sup><sup>n</sup> where <sup>I</sup> is on interval in <sup>R</sup> and <sup>t</sup><sup>0</sup> is a point in the interior of I. If f tð Þ ; u , is a continuous function of ð Þ t; u and a globally Lipschitz continuous function of <sup>u</sup> uniformly in <sup>t</sup> on <sup>I</sup> � <sup>R</sup><sup>n</sup>, then there is a unique continuously differentiable

Proof. We will show that T is a contraction on the space of continuous function defined on a

Suppose that u, v : ½ �! <sup>t</sup>0; <sup>t</sup><sup>0</sup> <sup>þ</sup> <sup>δ</sup> <sup>R</sup><sup>n</sup> are two continuous function. Then, form (3.4), (3.5) we

f sð Þ ; u sð Þ ds:

u0

ðÞ¼ <sup>t</sup> f tð Þ ; u tð Þ . □

http://dx.doi.org/10.5772/intechopen.74560

13

ðÞ¼ t f tð Þ ; u tð Þ (3.2)

Fixed Point Theory Approach to Existence of Solutions with Differential Equations

u tð Þ¼ <sup>0</sup> u0: (3.3)

∥f tð Þ� ; u f tð Þ ; v ∥ ≤ C∥u � v∥ (3.4)

f sð Þ ; u sð Þ ds: (3.5)

(3.1) and conclude that u is differentiable, and that is u<sup>0</sup>

a Lipschitz continuous function of u on suitable domain.

The initial value problem can be reformulated as an integral equation.

u tðÞ¼ u<sup>0</sup> þ

differentiable solution of (3.2). Eq. (3.5) may by written as fixed point equation.

Tu tðÞ¼ u<sup>0</sup> þ

function u tð Þ that take value in <sup>R</sup><sup>n</sup>

for all x, y∈ R<sup>n</sup> and all t∈ I.

for the map T defined by

estimate,

function <sup>u</sup> : <sup>I</sup> ! <sup>R</sup><sup>n</sup> that satisfies (3.2).

time interval t<sup>0</sup> ⩽ t⩽ t<sup>0</sup> þ δ, for sufficiently small δ.

condition. Next, we note that an integral is always a continuous function, so that a solution of (3.1) is automatically continuous. Since both u and f are continuous, it follows that the integrand f sð Þ ; u sð Þ is continuous. We may therefore apply the fundamental theorem of calculus to (3.1) and conclude that u is differentiable, and that is u<sup>0</sup> ðÞ¼ <sup>t</sup> f tð Þ ; u tð Þ . □

The contraction mapping theorem may by used to prove the existence and uniqueness of the initial problem for ordinary differential equations. We consider a first-order of ODEs for a function u tð Þ that take value in <sup>R</sup><sup>n</sup>

$$u'(t) = f(t, u(t))\tag{3.2}$$

$$
\mu(t\_0) = \mu\_0. \tag{3.3}
$$

The function f tð Þ ; u tð Þ also take value in <sup>R</sup><sup>n</sup> and is assumed to be a continuous function of <sup>t</sup> and a Lipschitz continuous function of u on suitable domain.

Definition 3.2. Suppose that <sup>f</sup> : <sup>I</sup> � <sup>R</sup><sup>n</sup> ! <sup>R</sup><sup>n</sup> where <sup>I</sup> is on interval in <sup>R</sup>. We say that f tð Þ ; u tð Þ is a globally Lipschitz continuous function of u uniformly in t if there is a constant C > 0 such that

$$\|f(t,u) - f(t,v)\| \le C \|u - v\|\tag{3.4}$$

for all x, y∈ R<sup>n</sup> and all t∈ I.

We denote C Tð Þ ;R ≔f g x ∈ D : Tx ¼ Rx the set of all coincidence points of two self-mappings T

Theorem 2.32. ref. [17] Let ð Þ X; ω be a regular modular metric space with a graph Gω: Assume that D ¼ V Gð Þ <sup>ω</sup> is a nonempty ω-bounded subset of X<sup>ω</sup> and the pair ð Þ D; G<sup>ω</sup> has property (A) and also satisfy ΔM-condition. Let R, T : D ! D be two self mappings satisfying the following

Proof. See in [17]. □

Next, we will show a differential equation which solving by fixed point theorem in suitable

ðt t0

ðt t0 f sð Þ ; u sð Þ ds

f sð Þ ; u sð Þ ds:

ðÞ¼ t f tð Þ ; u tð Þ satisfying the initial condition u tð Þ¼ <sup>0</sup> u<sup>0</sup>

ðÞ¼ t f tð Þ ; u tð Þ defined on an interval I and satisfying

ðÞ¼ t f tð Þ ; u tð Þ from t<sup>0</sup> to t, where t is in I

f sð Þ ; u sð Þ ds, t∈ I: (3.1)

3. Fixed point approach to the solution of differential equations

<sup>t</sup><sup>0</sup> f sð Þ ; u sð Þ ds.

ðt t0 u0 ð Þs ds ¼

u tðÞ¼ u<sup>0</sup> þ

u tð Þ� u tð Þ¼ <sup>0</sup>

ðt t0

We will show that, conversely, any function which satisfies this integral equation satisfies both the differential equation and the initial condition. Suppose that u is a function defined on an interval I and satisfies (3.1). Setting t ¼ t<sup>0</sup> yields u tð Þ¼ <sup>0</sup> u0, so that u satisfies the initial

and R, defined on D.

i. there exists x<sup>0</sup> ∈ D such that ð Þ Rx0; Tx<sup>0</sup> ∈ E Gð Þ <sup>ω</sup> ,

ii. T is an F-Gω-contraction w.r.t R,

12 Differential Equations - Theory and Current Research

3.1. Ordinary differential equation

if and only if u tðÞ¼ <sup>u</sup><sup>0</sup> <sup>þ</sup> <sup>Ð</sup><sup>t</sup>

Since u tð Þ¼ <sup>0</sup> u0, we have

Lemma 3.1. ref. [18] u is a solution of u<sup>0</sup>

Proof. Suppose that u is a solution of u<sup>0</sup>

u tð Þ¼ <sup>0</sup> u0. We integrate both sides of the equation u<sup>0</sup>

conditions:

iii. T Dð Þ⊆R Dð Þ,

Then C Rð Þ ; T 6¼ Ø.

spaces.

iv. R Dð Þ is ω complete.

The initial value problem can be reformulated as an integral equation.

$$
\mu(t) = \mu\_0 + \int\_{t\_0}^t f(s, \mu(s)) ds. \tag{3.5}
$$

By the fundamental theorem of calculus, a continuous solution of (3.5) is a continuously differentiable solution of (3.2). Eq. (3.5) may by written as fixed point equation.

$$u = Tu$$

for the map T defined by

$$T\mathfrak{u}(t) = \mathfrak{u}\_0 + \int\_{t\_0}^t f(\mathbf{s}, \mathfrak{u}(\mathbf{s}))d\mathbf{s}.$$

Theorem 3.3. ref. [19] Suppose that <sup>f</sup> : <sup>I</sup> � <sup>R</sup><sup>n</sup> ! <sup>R</sup><sup>n</sup> where <sup>I</sup> is on interval in <sup>R</sup> and <sup>t</sup><sup>0</sup> is a point in the interior of I. If f tð Þ ; u , is a continuous function of ð Þ t; u and a globally Lipschitz continuous function of <sup>u</sup> uniformly in <sup>t</sup> on <sup>I</sup> � <sup>R</sup><sup>n</sup>, then there is a unique continuously differentiable function <sup>u</sup> : <sup>I</sup> ! <sup>R</sup><sup>n</sup> that satisfies (3.2).

Proof. We will show that T is a contraction on the space of continuous function defined on a time interval t<sup>0</sup> ⩽ t⩽ t<sup>0</sup> þ δ, for sufficiently small δ.

Suppose that u, v : ½ �! <sup>t</sup>0; <sup>t</sup><sup>0</sup> <sup>þ</sup> <sup>δ</sup> <sup>R</sup><sup>n</sup> are two continuous function. Then, form (3.4), (3.5) we estimate,

$$\begin{split} |Tu - Tv|\_{\boldsymbol{\alpha}} &= \sup\_{t\_0 \leqslant t \leqslant t\_0 + \delta} |Tu(t) - Tv(t)| \\ &= \sup\_{t\_0 \leqslant t \leqslant t\_0 + \delta} |\int\_{t\_0}^t f(s, u(s)) - f(s, v(s))ds| \\ &\leq \sup\_{t\_0 \leqslant t \leqslant t\_0 + \delta} \int\_{t\_0}^t |f(s, u(s)) - f(s, v(s))| ds \\ &\leq \sup\_{t\_0 \leqslant t \leqslant t\_0 + \delta} \mathbb{C} |u(s) - v(s)| \int\_{t\_0}^t ds \\ &\leq \mathbb{C} \delta |u - v|\_{\boldsymbol{\alpha}}. \end{split}$$

If follow that if δ ≤ <sup>1</sup> <sup>c</sup> then T is contraction on C t ð Þ ½ � <sup>0</sup>; t<sup>0</sup> þ δ . Therefore, there is a unique solution <sup>u</sup> : ½ �! <sup>t</sup>0; <sup>t</sup><sup>0</sup> <sup>þ</sup> <sup>δ</sup> <sup>R</sup><sup>n</sup>.

Let f xð Þ ; y be a continuous real-valued function on ½ �� a; b ½ � c; d . The Cauchy initial value problem is to find a continuous differentiable function y on ½ � a; b satisfying the differential equation

$$\frac{dy}{dx} = f(x, y), \quad y(x\_0) = y\_0. \tag{3.6}$$

<sup>C</sup><sup>≔</sup> <sup>y</sup><sup>∈</sup> C x½ � <sup>0</sup> � <sup>h</sup>; <sup>x</sup><sup>0</sup> <sup>þ</sup> <sup>h</sup> : <sup>j</sup>y xð Þ� <sup>y</sup>0<sup>j</sup> <sup>≤</sup> Mh � �:

Then C is a closed subset of the complete metric space C x½ � <sup>0</sup> � h; x<sup>0</sup> þ h and hence C is complete. Note T : C ! C is a contraction mapping. Indeed, for x∈ ½ � x<sup>0</sup> � h; x<sup>0</sup> þ h and two contin-

<sup>≤</sup> <sup>∣</sup><sup>x</sup> � <sup>x</sup>0<sup>∣</sup> sup <sup>s</sup>∈½ � <sup>x</sup>0�h;x0þ<sup>h</sup>

Therefore, T has a unique fixed point implying that the problem (3.6) has a unique fixed point.

Now, we consider the existence of solution for the second order nonlinear boundary value

x00ðÞ¼ t k t; x tð Þ; x<sup>0</sup> ð Þ ð Þt , t ∈½ � 0; Λ , Λ > 0,

x tð Þ¼ <sup>2</sup> x2, t1, t<sup>2</sup> ∈½ � 0; Λ;

ð Þ t<sup>2</sup> � t ð Þ s � t<sup>1</sup> t<sup>2</sup> � t<sup>1</sup>

ð Þ t<sup>2</sup> � s ð Þ t � t<sup>1</sup> t<sup>2</sup> � t<sup>1</sup>

and βð Þt satisfies β<sup>00</sup> ¼ 0, βð Þ¼ t<sup>1</sup> x1, βð Þ¼ t<sup>2</sup> x2: Let us recall some properties of G tð Þ ;s , precisely

<sup>∣</sup>G tð Þ ;<sup>s</sup> <sup>∣</sup>ds <sup>≤</sup> ð Þ <sup>t</sup><sup>2</sup> � <sup>t</sup><sup>1</sup>

<sup>∣</sup>Gtð Þ <sup>t</sup>;<sup>s</sup> <sup>∣</sup>ds <sup>≤</sup> ð Þ <sup>t</sup><sup>2</sup> � <sup>t</sup><sup>1</sup>

where k : ½ �� 0; Λ W Xð Þ� W Xð Þ! W Xð Þ is a continuous function. This problem is equivalent

� � � f x; <sup>y</sup><sup>2</sup>

� �dt∥

Fixed Point Theory Approach to Existence of Solutions with Differential Equations

http://dx.doi.org/10.5772/intechopen.74560

k∣y1ð Þ� s y2ð Þs ∣

G tð Þ ;s k s; x sð Þ; x<sup>0</sup> ð Þ ð Þs ds þ βð Þt , t∈ ½ � 0; Λ , (3.9)

if t<sup>1</sup> ≤ s ≤ t ≤ t2,

if t<sup>1</sup> ≤ t ≤ s ≤ t2,

2

8

<sup>2</sup> :

(3.8)

15

Ð x <sup>x</sup><sup>0</sup> f x; y<sup>1</sup>

≤ kh∥y<sup>1</sup> � y2∥:

uous functions y1, y<sup>2</sup> ∈C, we have

3.2. Ordinary fuzzy differential equation

problem:

we have

and

to the integral equation

∥Ty<sup>1</sup> � Ty2∥ ¼ ∥

x tð Þ¼ <sup>1</sup> x1,

8 ><

>:

x tðÞ¼ <sup>ð</sup><sup>t</sup><sup>2</sup>

where the Green's function G is given by

t1

8 >><

>>:

ðt2 t1

ðt2 t1

G tð Þ¼ ;s

Consider the Banach space C a½ � ; b of continuous real-valued functions with supremum norm defined by ∥y∥ ¼ supf g y xð Þj : x ∈½ � a; b :

Integrating (3.6), that yield an integral equation

$$y(\mathbf{x}) = y\_0 + \int\_{x\_0}^{\mathbf{x}} f(t, y(t)) dt. \tag{3.7}$$

The problem (3.6) is equivalent the problem solving the integral Eq. (3.7).

We define an integral operator T : C a½ �! ; b C a½ � ; b by

$$Ty(x) = y\_0 + \int\_{x\_0}^{x} f(t, y(t)) dt.$$

Therefore, a solution of Cauchy initial value problem (3.6) corresponds with a fixed point of T. One may easily check that if T is contraction, then the problem (3.6) has a unique solution.

Theorem 3.4. ref. [20] Let f xð Þ ; y be a continuous function of Dom fð Þ¼ ½ �� a; b ½ � c; d such that f is Lipschitzian with respect to y, i.e., there exists k > 0 such that

$$|f(\mathbf{x}, u) - f(\mathbf{x}, v)| \le k|u - v| \text{ for all } u, v \in [c, d] \text{ and } \text{ for } \mathbf{x} \in [a, b].$$

Suppose x0; y<sup>0</sup> � � <sup>∈</sup>intð Þ Dom fð Þ : Then for sufficiently small <sup>h</sup> <sup>&</sup>gt; <sup>0</sup>, there exists a unique solution of the problem (3.6).

Proof. Let M ¼ supf g jf xð Þj ; y : x; y∈ Dom fð Þ and choose h > 0 such that

$$\mathcal{C} \coloneqq \left\{ y \in \mathbb{C}[\mathbf{x}\_0 - h, \mathbf{x}\_0 + h] \, : \, |y(\mathbf{x}) - y\_0| \le Mh \right\} \dots$$

Then C is a closed subset of the complete metric space C x½ � <sup>0</sup> � h; x<sup>0</sup> þ h and hence C is complete. Note T : C ! C is a contraction mapping. Indeed, for x∈ ½ � x<sup>0</sup> � h; x<sup>0</sup> þ h and two continuous functions y1, y<sup>2</sup> ∈C, we have

$$\begin{aligned} \|Ty\_1 - Ty\_2\| &= \|\int\_{\mathbf{x}\_0}^{\mathbf{x}} f(\mathbf{x}, y\_1) - f(\mathbf{x}, y\_2)dt\| \\ &\le \|\mathbf{x} - \mathbf{x}\_0\| \sup\_{s \in [\mathbf{x}\_0 - \mathbf{h}, \mathbf{x}\_0 + h]} k |y\_1(s) - y\_2(s)| \\ &\le \|h\| \|y\_1 - y\_2\|. \end{aligned}$$

Therefore, T has a unique fixed point implying that the problem (3.6) has a unique fixed point.

### 3.2. Ordinary fuzzy differential equation

Now, we consider the existence of solution for the second order nonlinear boundary value problem:

$$\begin{cases} \mathbf{x}''(t) = k(t, \mathbf{x}(t), \mathbf{x}'(t)), & t \in [0, \Lambda], \quad \Lambda > 0, \\\mathbf{x}(t\_1) = \mathbf{x}\_1, & \\\mathbf{x}(t\_2) = \mathbf{x}\_2, & t\_1, t\_2 \in [0, \Lambda, ] \end{cases} \tag{3.8}$$

where k : ½ �� 0; Λ W Xð Þ� W Xð Þ! W Xð Þ is a continuous function. This problem is equivalent to the integral equation

$$\mathbf{x}(t) = \int\_{t\_1}^{t\_2} \mathbf{G}(t, s) k(\mathbf{s}, \mathbf{x}(s), \mathbf{x}'(s)) ds + \boldsymbol{\beta}(t), \quad t \in [0, \Lambda]. \tag{3.9}$$

where the Green's function G is given by

$$G(t,s) = \begin{cases} \frac{(t\_2 - t)(s - t\_1)}{t\_2 - t\_1} & \text{if } \ t\_1 \le s \le t \le t\_{2\prime} \\\frac{(t\_2 - s)(t - t\_1)}{t\_2 - t\_1} & \text{if } \ t\_1 \le t \le s \le t\_{2\prime} \end{cases}$$

and βð Þt satisfies β<sup>00</sup> ¼ 0, βð Þ¼ t<sup>1</sup> x1, βð Þ¼ t<sup>2</sup> x2: Let us recall some properties of G tð Þ ;s , precisely we have

$$\int\_{t\_1}^{t\_2} |G(t, s)| ds \le \frac{\left(t\_2 - t\_1\right)^2}{8}$$

and

jTu � Tvj

14 Differential Equations - Theory and Current Research

If follow that if δ ≤ <sup>1</sup>

<sup>u</sup> : ½ �! <sup>t</sup>0; <sup>t</sup><sup>0</sup> <sup>þ</sup> <sup>δ</sup> <sup>R</sup><sup>n</sup>.

defined by ∥y∥ ¼ supf g y xð Þj : x ∈½ � a; b :

Integrating (3.6), that yield an integral equation

We define an integral operator T : C a½ �! ; b C a½ � ; b by

is Lipschitzian with respect to y, i.e., there exists k > 0 such that

Proof. Let M ¼ supf g jf xð Þj ; y : x; y∈ Dom fð Þ and choose h > 0 such that

equation

Suppose x0; y<sup>0</sup>

of the problem (3.6).

<sup>∞</sup> <sup>¼</sup> sup <sup>t</sup><sup>0</sup> <sup>⩽</sup> <sup>t</sup>⩽t0þ<sup>δ</sup>

<sup>¼</sup> sup <sup>t</sup><sup>0</sup> <sup>⩽</sup> <sup>t</sup>⩽t0þ<sup>δ</sup>

≤ sup t<sup>0</sup> ⩽ t⩽t0þδ

≤ sup t<sup>0</sup> ⩽ t⩽t0þδ

dy

y xð Þ¼ y<sup>0</sup> þ

Ty xð Þ¼ y<sup>0</sup> þ

The problem (3.6) is equivalent the problem solving the integral Eq. (3.7).

≤ Cδju � vj

j ðt t0

ðt t0

∞:

Let f xð Þ ; y be a continuous real-valued function on ½ �� a; b ½ � c; d . The Cauchy initial value problem is to find a continuous differentiable function y on ½ � a; b satisfying the differential

Consider the Banach space C a½ � ; b of continuous real-valued functions with supremum norm

ðx x0

> ðx x0

Therefore, a solution of Cauchy initial value problem (3.6) corresponds with a fixed point of T. One may easily check that if T is contraction, then the problem (3.6) has a unique solution.

Theorem 3.4. ref. [20] Let f xð Þ ; y be a continuous function of Dom fð Þ¼ ½ �� a; b ½ � c; d such that f

∣f xð Þ� ; u f xð Þ ; v ∣ ≤ k∣u � v∣ for all u, v∈ ½ � c; d and for x∈ ½ � a; b :

� � <sup>∈</sup>intð Þ Dom fð Þ : Then for sufficiently small <sup>h</sup> <sup>&</sup>gt; <sup>0</sup>, there exists a unique solution

f tð Þ ; y tð Þ dt:

jTu tð Þ� Tv tð Þj

Cju sð Þ� v sð Þj

f sð Þ� ; u sð Þ f sð Þ ; v sð Þ dsj

jf sð Þ� ; u sð Þ f sð Þj ; v sð Þ ds

ðt t0 ds

<sup>c</sup> then T is contraction on C t ð Þ ½ � <sup>0</sup>; t<sup>0</sup> þ δ . Therefore, there is a unique solution

dx <sup>¼</sup> f xð Þ ; <sup>y</sup> , yxð Þ¼ <sup>0</sup> <sup>y</sup>0: (3.6)

f tð Þ ; y tð Þ dt: (3.7)

$$\int\_{t\_1}^{t\_2} |G\_t(t, s)| ds \le \frac{(t\_2 - t\_1)}{2}.$$

If necessary, for a more detailed explanation of the background of the problem, the reader can refer to the reference [21, 22]. Here, we will prove our results, by establishing the existence of a common fixed point for pair of integral operators defined as

$$T\_i(\mathbf{x})(t) = \int\_{t\_1}^{t\_2} G(t, \mathbf{s}) k\_i(\mathbf{s}, \mathbf{x}(\mathbf{s}), \mathbf{x}'(\mathbf{s})) d\mathbf{s} + \beta(t), \quad t \in [0, \Lambda], \quad i \in \{1, 2\} \tag{3.10}$$

In [23], it is proved that ð Þ C;≼ satisfies the following condition:

view of condition (r), hypothesis (i)-(iii) of Corollary 2.16 hold true.

ð t1

≤ D xð Þ ; y

<sup>≤</sup> ð Þ <sup>t</sup><sup>2</sup> � <sup>t</sup><sup>1</sup>

ðt2 t1

2 <sup>8</sup> D xð Þ ; <sup>y</sup>

t2∣Gtð Þk t;s k<sup>1</sup> s; x sð Þ; x

∣Gtð Þ t;s ∣ds

2 <sup>8</sup> <sup>þ</sup> <sup>δ</sup>

!

ðt2 t1

<sup>2</sup> D xð Þ ; <sup>y</sup> :

Consequently, in view of hypothesis (d), the contractive condition (5) is satisfied with

Now, we consider the boundary value problem for second order differential equation

2 <sup>8</sup> <sup>þ</sup> <sup>δ</sup>

Therefore, Corollary 2.16 applied to T<sup>1</sup> and T2, which have common fixed point x<sup>∗</sup> ∈ C, that is, x<sup>∗</sup> is a common solution of (3.9). □

Next, for all comparable x, y∈ C, From hypothesis (c) we obtain successively

n∈ℕ∪ f g0 .

and

<sup>∣</sup>ð Þ <sup>T</sup>1ð Þ<sup>x</sup> <sup>0</sup>

common fixed point of T<sup>1</sup> and T2.

∣T1ð Þx ð Þ� t T2ð Þy ð Þt ∣ ≤

ð Þ� <sup>t</sup> ð Þ <sup>T</sup>2ð Þ<sup>y</sup> <sup>0</sup>

From (3.12) and (3.13), we obtain easily

3.3. Second-order differential equation

ð Þt ∣ ≤ ð t1

≤ D xð Þ ; y

<sup>≤</sup> ð Þ <sup>t</sup><sup>2</sup> � <sup>t</sup><sup>1</sup>

D Tð Þ <sup>1</sup>x; <sup>T</sup>2<sup>y</sup> <sup>≤</sup> <sup>γ</sup> ð Þ <sup>t</sup><sup>2</sup> � <sup>t</sup><sup>1</sup>

<sup>q</sup> <sup>¼</sup> <sup>γ</sup> ð Þ <sup>t</sup><sup>2</sup> � <sup>t</sup><sup>1</sup>

(r) for every nondecreasing sequence f g xn in C convergent to some x∈ C, we have xn≼x for all

Let T1, T<sup>2</sup> : C ! C be two integral operators defined by (3.10); clearly, T1, T � 2 are well defined since k1, k2, and β are continuous functions. Now, x<sup>∗</sup> is a solution of (3.9) if and only if x<sup>∗</sup> is a

By hypothesis (a), T1, T<sup>2</sup> are increasing and, by hypothesis (b), x0≼T1ð Þ x<sup>0</sup> : Consequently, in

∣G tð Þ ;s ∣ds

t2∣G tð Þk ;s k<sup>1</sup> s; x sð Þ; x<sup>0</sup> ð Þ� ð Þs k<sup>1</sup> s; y sð Þ; y<sup>0</sup> ð Þ ð Þs ∣ds

Fixed Point Theory Approach to Existence of Solutions with Differential Equations

http://dx.doi.org/10.5772/intechopen.74560

0 ð Þs � � � <sup>k</sup><sup>1</sup> <sup>s</sup>; y sð Þ; <sup>y</sup>

> ð Þ t<sup>2</sup> � t<sup>1</sup> 2

ð Þ t<sup>2</sup> � t<sup>1</sup> <sup>2</sup> <sup>&</sup>lt; <sup>1</sup>: D xð Þ ; y :

0 ð Þs � �∣ds (3.12)

17

(3.13)

where <sup>k</sup>1, k<sup>2</sup> <sup>∈</sup>Cð Þ ½ �� <sup>0</sup>; <sup>Λ</sup> W Xð Þ� W Xð Þ; W Xð Þ , x<sup>∈</sup> <sup>C</sup><sup>1</sup> ð Þ ½ � 0; Λ ; W Xð Þ , and β∈ Cð Þ ½ � 0; Λ ; W Xð Þ :

Theorem 3.5 ref. [6] Assume that the following conditions are satisfied:


$$\mathbf{x}\_0(t) \le \int\_{t\_1}^{t\_2} G(t, s) k\_1 \left( t, \mathbf{x}\_0(s), \mathbf{x}\_0'(s) \right) ds + \beta(t) \lambda$$

where t1, t<sup>2</sup> ∈½ � 0; Λ ,

iii. there exist constants γ, δ > 0 such that, for all t∈ ½ � 0; Λ , we have

$$|k\_1(t, \mathbf{x}(t), \mathbf{x}'(t)) - k\_2(t, y(t), y'(t))| \le \gamma |\mathbf{x}(t) - y(t)| + \delta |\mathbf{x}'(t) - y'(t)|$$

for all comparable x, y∈C<sup>1</sup> ð Þ ½ � 0; Λ ; W Xð Þ ,

iv. for γ, δ > 0 and t1, t<sup>2</sup> ∈½ � 0; Λ we have

$$
\gamma \frac{\left(t\_2 - t\_1\right)^2}{8} + \delta \frac{\left(t\_2 - t\_1\right)}{2} < 1,
$$

v. if x, y∈ C<sup>1</sup> ð Þ ½ � 0; Λ ; W Xð Þ are comparable, then every u∈ð Þ T1x <sup>1</sup> and every v∈ ð Þ T2y <sup>1</sup> are comparable.

Then the pair of nonlinear integral equations

$$\mathbf{x}(t) = \int\_{t\_1}^{t\_2} G(t, s) k\_i(s, \mathbf{x}(s), \mathbf{x}'(s)) ds + \beta(t) \quad t \in [0, \Lambda], \quad i \in \{1, 2\} \tag{3.11}$$

has a common solution in C<sup>1</sup> ð Þ ½ � t1; t<sup>2</sup> ; W Xð Þ :

Proof. Consider <sup>C</sup> <sup>¼</sup> <sup>C</sup><sup>1</sup> ð Þ ½ � t1; t<sup>2</sup> ; W Xð Þ with the metric

$$D(\mathbf{x}, y) = \max\_{t\_1 \le t \le t\_2} \left( \gamma |\mathbf{x}(t) - y(t)| + \delta |\mathbf{x}'(t) - y'(t)| \right).$$

The ð Þ C; D is a complete metric space, which can also be equipped with the partial ordering given by

$$\forall \mathbf{x}, \mathbf{y} \in \mathcal{C}, \quad \Leftrightarrow \mathbf{x}(t) \leq \mathbf{y}(t) \quad \text{for all} \ t \in [0, \Lambda].$$

In [23], it is proved that ð Þ C;≼ satisfies the following condition:

(r) for every nondecreasing sequence f g xn in C convergent to some x∈ C, we have xn≼x for all n∈ℕ∪ f g0 .

Let T1, T<sup>2</sup> : C ! C be two integral operators defined by (3.10); clearly, T1, T � 2 are well defined since k1, k2, and β are continuous functions. Now, x<sup>∗</sup> is a solution of (3.9) if and only if x<sup>∗</sup> is a common fixed point of T<sup>1</sup> and T2.

By hypothesis (a), T1, T<sup>2</sup> are increasing and, by hypothesis (b), x0≼T1ð Þ x<sup>0</sup> : Consequently, in view of condition (r), hypothesis (i)-(iii) of Corollary 2.16 hold true.

Next, for all comparable x, y∈ C, From hypothesis (c) we obtain successively

$$\begin{split} |T\_1(\mathbf{x})(t) - T\_2(\mathbf{y})(t)| &\leq \int\_{t\_1} t\_2 |G(t, s)| |k\_1(s, \mathbf{x}(s), \mathbf{x}'(s)) - k\_1(s, \mathbf{y}(s), \mathbf{y}'(s))| ds \\ &\leq D(\mathbf{x}, y) \int\_{t\_1}^{t\_2} |G(t, s)| ds \\ &\leq \frac{(t\_2 - t\_1)^2}{8} D(\mathbf{x}, y) \end{split} \tag{3.12}$$

and

If necessary, for a more detailed explanation of the background of the problem, the reader can refer to the reference [21, 22]. Here, we will prove our results, by establishing the existence of a

i. k1, k<sup>2</sup> : ½ �� 0; Λ W Xð Þ� W Xð Þ! W Xð Þ are increasing in its second and third variables,

ð Þ ½ � 0; Λ ; W Xð Þ such that, for all t∈ ½ � 0; Λ , we have

G tð Þ ;s k<sup>1</sup> t; x0ð Þs ; x<sup>0</sup>

∣k<sup>1</sup> t; x tð Þ; x<sup>0</sup> ð Þ� ð Þt k<sup>2</sup> t; y tð Þ; y<sup>0</sup> ð Þ ð Þt ∣ ≤ γ∣x tð Þ� y tð Þ∣ þ δ∣x<sup>0</sup>

2 <sup>8</sup> <sup>þ</sup> <sup>δ</sup>

ð Þ t<sup>2</sup> � t<sup>1</sup> <sup>2</sup> <sup>&</sup>lt; <sup>1</sup>,

γjx tð Þ� y tð Þj þ δjx<sup>0</sup>

The ð Þ C; D is a complete metric space, which can also be equipped with the partial ordering

x, y∈C, ⇔ x tð Þ ≤ y tð Þ for all t ∈½ � 0; Λ :

ðÞ� t y<sup>0</sup> ð Þ ð Þj t :

ð Þ ½ � 0; Λ ; W Xð Þ are comparable, then every u∈ð Þ T1x <sup>1</sup> and every v∈ ð Þ T2y <sup>1</sup> are compa-

G tð Þ ;s ki s; x sð Þ; x<sup>0</sup> ð Þ ð Þs ds þ βð Þt t∈½ � 0; Λ , i∈ f g 1; 2 (3.11)

ð Þ ½ � 0; Λ ; W Xð Þ ,

<sup>γ</sup> ð Þ <sup>t</sup><sup>2</sup> � <sup>t</sup><sup>1</sup>

ð Þ ½ � t1; t<sup>2</sup> ; W Xð Þ :

t<sup>1</sup> ≤ t ≤ t<sup>2</sup>

D xð Þ¼ ; y max

ð Þ ½ � t1; t<sup>2</sup> ; W Xð Þ with the metric

<sup>0</sup>ð Þ<sup>s</sup> � �ds <sup>þ</sup> <sup>β</sup>ð Þ<sup>t</sup> ,

G tð Þ ;s ki s; x sð Þ; x<sup>0</sup> ð Þ ð Þs ds þ βð Þt , t∈ ½ � 0; Λ , i∈ f g 1; 2 (3.10)

ð Þ ½ � 0; Λ ; W Xð Þ , and β∈ Cð Þ ½ � 0; Λ ; W Xð Þ :

ð Þ� t y<sup>0</sup>

ð Þt ∣

common fixed point for pair of integral operators defined as

Theorem 3.5 ref. [6] Assume that the following conditions are satisfied:

ðt2 t1

iii. there exist constants γ, δ > 0 such that, for all t∈ ½ � 0; Λ , we have

ðt2 t1

where <sup>k</sup>1, k<sup>2</sup> <sup>∈</sup>Cð Þ ½ �� <sup>0</sup>; <sup>Λ</sup> W Xð Þ� W Xð Þ; W Xð Þ , x<sup>∈</sup> <sup>C</sup><sup>1</sup>

x0ð Þt ≤

Tið Þx ðÞ¼ t

16 Differential Equations - Theory and Current Research

ii. there exists x<sup>0</sup> ∈ C<sup>1</sup>

v. if x, y∈ C<sup>1</sup>

rable.

where t1, t<sup>2</sup> ∈½ � 0; Λ ,

for all comparable x, y∈C<sup>1</sup>

iv. for γ, δ > 0 and t1, t<sup>2</sup> ∈½ � 0; Λ we have

Then the pair of nonlinear integral equations

x tðÞ¼

has a common solution in C<sup>1</sup>

Proof. Consider <sup>C</sup> <sup>¼</sup> <sup>C</sup><sup>1</sup>

given by

ðt2 t1

$$\begin{split} |\left(T\_1(\mathbf{x})\right)'(t) - \left(T\_2(y)\right)'(t)| \leq & \int\_{t\_1} t\_2 |G\_l(t,s)| \|k\_1\left(s, \mathbf{x}(s), \mathbf{x}'(s)\right) - k\_1\left(s, y(s), y'(s)\right)\| ds \\ \leq & D(\mathbf{x}, y) \int\_{t\_1}^{t\_2} |G\_l(t,s)| ds \\ \leq & \frac{(t\_2 - t\_1)}{2} D(\mathbf{x}, y) .\end{split} \tag{3.13}$$

From (3.12) and (3.13), we obtain easily

$$D(T\_1\mathbf{x}, T\_2\mathbf{y}) \le \left(\gamma \frac{\left(t\_2 - t\_1\right)^2}{8} + \delta \frac{\left(t\_2 - t\_1\right)}{2}\right) D(\mathbf{x}, \mathbf{y}) \dots$$

Consequently, in view of hypothesis (d), the contractive condition (5) is satisfied with

$$\eta = \gamma \frac{\left(t\_2 - t\_1\right)^2}{8} + \delta \frac{\left(t\_2 - t\_1\right)}{2} < 1.1$$

Therefore, Corollary 2.16 applied to T<sup>1</sup> and T2, which have common fixed point x<sup>∗</sup> ∈ C, that is, x<sup>∗</sup> is a common solution of (3.9). □

#### 3.3. Second-order differential equation

Now, we consider the boundary value problem for second order differential equation

$$\begin{cases} \mathbf{x}''(t) = -f(t, \mathbf{x}(t)), & t \in I, \\ \mathbf{x}(0) = \mathbf{x}(1) = \mathbf{0}, \end{cases} \tag{3.14}$$

Proof. Define the mapping T : X ! X by

∣Tx tð Þ� Ty tð Þ∣ ¼ ∣

<sup>8</sup> : Therefore,

Let x, y∈ X, we obtain

sup<sup>t</sup>∈<sup>I</sup> Ð 1

Hence,

<sup>0</sup> G tð Þ ;<sup>s</sup> ds <sup>¼</sup> <sup>1</sup>

Similar method, we obtain

Moreover, we have

Tx tðÞ¼ <sup>ð</sup><sup>1</sup>

ð1 0

≤ 8αλ1∥x � y∥<sup>∞</sup>

¼ αλ1∥x � y∥∞:

≤ ð1 0

≤ 8 ð1 0

In the above equality, we used that for each t ∈I, we have Ð <sup>1</sup>

Tx tðÞ¼ ∣

ð1 0

≤ 8αλ2∥x∥∞:

<sup>∞</sup> þ jTxj

<sup>∞</sup> þ αλ2jxj

≤ ð Þ λ<sup>1</sup> þ 2λ<sup>2</sup> ð Þj α Tx � Tyj<sup>∞</sup> þ jTxj

≤ 8 ð1 0

Let <sup>e</sup>�<sup>τ</sup> <sup>¼</sup> <sup>λ</sup><sup>1</sup> <sup>þ</sup> <sup>2</sup>λ<sup>2</sup> <sup>&</sup>lt; 1 where <sup>τ</sup> <sup>&</sup>gt; <sup>0</sup>: Form (3.16), (3.17) and (3.18), we obtain

≤ αλ1jx � yj

<sup>¼</sup> <sup>e</sup>�<sup>τ</sup> ð Þασð Þ <sup>x</sup>; <sup>y</sup> :

σð Þ¼j Tx; Ty Tx � Tyj

0

for all x ∈ X and t∈ T: Then the problem (3.14) is equivalent to finding a fixed point u of T in X.

G tð Þ ;s f sð Þ ; x sð Þ ds �

G tð Þ ;s ∣f sð Þ� ; x sð Þ fðs, y sð Þ∣ds

G tð Þ ;s p sð Þ∣x sð Þ� y sð Þ∣ds

Ð 1 <sup>0</sup> G tð Þ ;s ds

G tð Þ ;s f sð Þ ; x sð Þ ds∣

G tð Þ ;s q sð Þ∣x sð Þ∣ds

<sup>∞</sup> þ jTyj<sup>∞</sup>

<sup>∞</sup> þ αλ2jyj

<sup>∞</sup> ½ � ð Þ

∞

G tð Þ ;s f sð Þ ; x sð Þ ds,

ð1 0

Fixed Point Theory Approach to Existence of Solutions with Differential Equations

G tð Þ ;s f sð Þ ; y sð Þ ds∣

http://dx.doi.org/10.5772/intechopen.74560

19

<sup>0</sup> G tð Þ ;<sup>s</sup> ds <sup>¼</sup> <sup>t</sup>

∥Tx � Ty∥<sup>∞</sup> ≤ αλ1∥x � y∥∞: (3.16)

∥Tx∥<sup>∞</sup> ≤ αλ2∥x∥∞: (3.17)

∥Ty∥<sup>∞</sup> ≤ αλ2∥y∥∞: (3.18)

<sup>∞</sup> þ jTyj

<sup>2</sup> ð Þ 1 � t and so

(3.19)

where I ¼ ½ � 0; 1 and f : I � R ! R: is a continuous function.

It is known, and easy to check, that the problem (3.14) is equivalent to the integral equation

$$\mathbf{x}(t) = \int\_{0}^{1} \mathbf{G}(t, \mathbf{s}) f(\mathbf{s}, \mathbf{x}(\mathbf{s})) d\mathbf{s}, \quad \text{for} \quad t \in I,\tag{3.15}$$

where G is the Green's function define by

$$G(t,s) = \begin{cases} t(1-s) & \text{if} \quad 0 \le t \le s \le 1\\ s(1-t) & \text{if} \quad 0 \le s \le t \le 1. \end{cases}$$

That is, if x ∈C<sup>2</sup> ð Þ I; R , then x is a solution of problem (3.14) iff x is a solution of the integral Eq. (3.15).

Let X ¼ C Ið Þ be the space of all continuous functions defined on I. Consider the metric-like σ on X define by

$$\sigma(\mathbf{x}, \mathbf{y}) = \|\mathbf{x} - \mathbf{y}\|\_{\circ} + \|\mathbf{x}\|\_{\circ} + \|\mathbf{y}\|\_{\circ} \text{ for all } \mathbf{x}, \mathbf{y} \in \mathbf{X}\_{\prime}$$

where ∥u∥<sup>∞</sup> ¼ max<sup>t</sup> <sup>∈</sup>½ � <sup>0</sup>;<sup>1</sup> ∣u tð Þ∣ for all u∈ X.

Note that σ is also a partial metric on X and since

$$d\_{\sigma}(\mathbf{x}, \mathbf{y}) \coloneqq 2\sigma(\mathbf{x}, \mathbf{y}) - \sigma(\mathbf{x}, \mathbf{x}) - \sigma(\mathbf{y}, \mathbf{y}) = 2\|\mathbf{x} - \mathbf{y}\|\_{\infty}.$$

By Lemma 2.20, hence ð Þ X; σ is complete since the metric space ð Þ X; ∥ � ∥ is complete.

Theorem 3.6. ref. [12] Suppose the following conditions:

i. there exist continuous functions p : I ! R<sup>þ</sup> such that

$$|f(\mathbf{s}, a) - f(\mathbf{s}, b)| \le 8p(\mathbf{s})|a - b|.$$

for all s ∈I and a, b∈ R;

ii. there exist continuous functions q : I ! R<sup>þ</sup> such that

$$|f(s, a)| \le 8q(s)|a|.$$

for all s ∈I and a ∈ R;

$$\text{iiii.} \quad \max\_{\mathbf{s}\in I} p(\mathbf{s}) = \alpha \lambda\_1 < \frac{1}{49}, \text{ which is } 0 \le \alpha < \frac{1}{7}?$$

iv. max<sup>s</sup>∈Iq sð Þ¼ αλ<sup>2</sup> <sup>&</sup>lt; <sup>1</sup> <sup>49</sup> which is 0 ≤ α < <sup>1</sup> 7 :

Then problem (3.14) has a unique solution u∈ X ¼ C Ið Þ ; R .

Proof. Define the mapping T : X ! X by

x00ðÞ¼� t f tð Þ ; x tð Þ , t ∈I,

(3.14)

xð Þ¼ 0 xð Þ¼ 1 0,

It is known, and easy to check, that the problem (3.14) is equivalent to the integral equation

tð Þ 1 � s if 0 ≤ t ≤ s ≤ 1 sð Þ 1 � t if 0 ≤ s ≤ t ≤ 1:

Let X ¼ C Ið Þ be the space of all continuous functions defined on I. Consider the metric-like σ

σð Þ¼ x; y ∥x � y∥<sup>∞</sup> þ ∥x∥<sup>∞</sup> þ ∥y∥<sup>∞</sup> for all x, y ∈ X,

dσð Þ x; y ≔2σð Þ� x; y σð Þ� x; x σð Þ¼ y; y 2∥x � y∥∞:

∣f sð Þ� ; a f sð Þ ; b ∣ ≤ 8p sð Þ∣a � b∣

∣f sð Þ ; a ∣ ≤ 8q sð Þ∣a∣

7;

7 :

By Lemma 2.20, hence ð Þ X; σ is complete since the metric space ð Þ X; ∥ � ∥ is complete.

ð Þ I; R , then x is a solution of problem (3.14) iff x is a solution of the integral

G tð Þ ;s f sð Þ ; x sð Þ ds, for t∈ I, (3.15)

�

where I ¼ ½ � 0; 1 and f : I � R ! R: is a continuous function.

x tðÞ¼

G tð Þ¼ ;s

where G is the Green's function define by

18 Differential Equations - Theory and Current Research

where ∥u∥<sup>∞</sup> ¼ max<sup>t</sup> <sup>∈</sup>½ � <sup>0</sup>;<sup>1</sup> ∣u tð Þ∣ for all u∈ X.

for all s ∈I and a, b∈ R;

for all s ∈I and a ∈ R;

iii. max<sup>s</sup>∈Ip sð Þ¼ αλ<sup>1</sup> <sup>&</sup>lt; <sup>1</sup>

iv. max<sup>s</sup>∈Iq sð Þ¼ αλ<sup>2</sup> <sup>&</sup>lt; <sup>1</sup>

Note that σ is also a partial metric on X and since

Theorem 3.6. ref. [12] Suppose the following conditions:

i. there exist continuous functions p : I ! R<sup>þ</sup> such that

ii. there exist continuous functions q : I ! R<sup>þ</sup> such that

Then problem (3.14) has a unique solution u∈ X ¼ C Ið Þ ; R .

<sup>49</sup> , which is 0 ≤ α < <sup>1</sup>

<sup>49</sup> which is 0 ≤ α < <sup>1</sup>

That is, if x ∈C<sup>2</sup>

on X define by

Eq. (3.15).

ð1 0

�

$$T\mathfrak{x}(t) = \int\_0^1 G(t, \mathfrak{s}) f(\mathfrak{s}, \mathfrak{x}(\mathfrak{s})) d\mathfrak{s} \,\,\!$$

for all x ∈ X and t∈ T: Then the problem (3.14) is equivalent to finding a fixed point u of T in X. Let x, y∈ X, we obtain

$$\begin{aligned} |T\mathbf{x}(t) - T\mathbf{y}(t)| &= \ |\int\_0^1 G(t,s)f(s,\mathbf{x}(s))ds - \int\_0^1 G(t,s)f(s,\mathbf{y}(s))ds|\\ &\leq \int\_0^1 G(t,s)|f(s,\mathbf{x}(s)) - f(s,\mathbf{y}(s))|ds\\ &\leq \ 8\int\_0^1 G(t,s)p(s)|\mathbf{x}(s) - \mathbf{y}(s)|ds\\ &\leq \ 8\alpha\lambda\_1\|\mathbf{x} - \mathbf{y}\|\_\infty\int\_0^1 G(t,s)ds\\ &= \ \alpha\lambda\_1\|\mathbf{x} - \mathbf{y}\|\_\infty. \end{aligned}$$

In the above equality, we used that for each t ∈I, we have Ð <sup>1</sup> <sup>0</sup> G tð Þ ;<sup>s</sup> ds <sup>¼</sup> <sup>t</sup> <sup>2</sup> ð Þ 1 � t and so sup<sup>t</sup>∈<sup>I</sup> Ð 1 <sup>0</sup> G tð Þ ;<sup>s</sup> ds <sup>¼</sup> <sup>1</sup> <sup>8</sup> : Therefore,

$$\|\|Tx - Ty\|\|\_{\circ} \le \alpha \lambda\_1 \|\|x - y\|\|\_{\circ}.\tag{3.16}$$

Moreover, we have

$$\begin{aligned} Tx(t) &= |\int\_0^1 G(t,s)f(s,x(s))ds| \\ &\le 8 \int\_0^1 G(t,s)q(s)|x(s)|ds \\ &\le 8\alpha\lambda\_2 \|x\|\_\infty. \end{aligned}$$

Hence,

$$\|\|Tx\|\|\_{\circ} \le \alpha \lambda\_2 \|\|x\|\|\_{\circ}.\tag{3.17}$$

Similar method, we obtain

$$\|\|Ty\|\|\_{\circ} \le \alpha \lambda\_2 \|\|y\|\|\_{\circ}.\tag{3.18}$$

Let <sup>e</sup>�<sup>τ</sup> <sup>¼</sup> <sup>λ</sup><sup>1</sup> <sup>þ</sup> <sup>2</sup>λ<sup>2</sup> <sup>&</sup>lt; 1 where <sup>τ</sup> <sup>&</sup>gt; <sup>0</sup>: Form (3.16), (3.17) and (3.18), we obtain

$$\begin{split} \sigma(\mathbf{Tx}, \mathbf{Ty}) &= |\mathbf{Tx} - \mathbf{Ty}|\_{\boldsymbol{\alpha}} + |\mathbf{Tx}|\_{\boldsymbol{\alpha}} + |\mathbf{Ty}|\_{\boldsymbol{\alpha}} \\ &\leq a\lambda\_1 |\mathbf{x} - \mathbf{y}|\_{\boldsymbol{\alpha}} + a\lambda\_2 |\mathbf{x}|\_{\boldsymbol{\alpha}} + a\lambda\_2 |\mathbf{y}|\_{\boldsymbol{\alpha}} \\ &\leq (\lambda\_1 + 2\lambda\_2) [(\alpha)(|\mathbf{Tx} - \mathbf{Ty}|\_{\boldsymbol{\alpha}} + |\mathbf{Tx}|\_{\boldsymbol{\alpha}} + |\mathbf{Ty}|\_{\boldsymbol{\alpha}})] \\ &= (e^{-\tau})a\sigma(\mathbf{x}, \mathbf{y}). \end{split} \tag{3.19}$$

Let β, γ, η, δ > 0 where β < <sup>1</sup> <sup>7</sup> , γ < <sup>1</sup> <sup>7</sup> , η < <sup>1</sup> <sup>7</sup> , δ < <sup>1</sup> <sup>7</sup> : It following (3.19), we get

$$\sigma(\mathbf{Tx}, \mathbf{Ty}) \le (e^{-\mathbf{r}}) \left[ a\sigma(\mathbf{x}, \mathbf{y}) + \beta\sigma(\mathbf{x}, \mathbf{Tx}) + \gamma\sigma(\mathbf{y}, \mathbf{Ty}) + \eta\sigma(\mathbf{x}, \mathbf{Ty}) + \delta\sigma(\mathbf{y}, \mathbf{Tx}) \right],\tag{3.20}$$

where α þ β þ γ þ 2η þ 2δ < 1. Taking the function F : R<sup>þ</sup> ! R in (3.20), where F tðÞ¼ ln ð Þt , which is F∈ F, we get

$$
\tau + F(\sigma(T\mathbf{x}, T\mathbf{y})) \le F(a\sigma(\mathbf{x}, \mathbf{y}) + \beta \sigma(\mathbf{x}, T\mathbf{x}) + \gamma \sigma(\mathbf{y}, T\mathbf{y}) + \eta \sigma(\mathbf{x}, T\mathbf{y}) + \delta \sigma(\mathbf{y}, T\mathbf{x})).
$$

Therefore all hypothesis of Theorem (2.25) are satisfied, and so T has a unique fixed point u∈ X, that is, the problem (3.14) has a unique solution u∈ C<sup>2</sup> ð Þ<sup>I</sup> : □

### 3.4. Partial differential equation

Consider the Laplace operator is a second order differential operator in the n-dimensional Euclidean space, defined as the divergence ð Þ ∇� of the gradient ð Þ ∇f . Thus if f is a twicedifferentiable real-valued function, then the Laplacian of f is defined by

$$
\Delta f = \nabla^2 f = \nabla \cdot \nabla f \tag{3.21}
$$

Proof. Our strategy is to apply Schauder's Fixed Point Theorem to the map

T : L<sup>2</sup>

∇Tu � ∇Tudx ¼

Corollary (2.10) shows that ð Þ �<sup>Δ</sup> �<sup>1</sup> is continuous form <sup>L</sup><sup>2</sup>

Cauchy-Schwarz. T here fore, using Ponincare's inequality

∥Tu∥<sup>2</sup>

Thus if we set <sup>R</sup> <sup>¼</sup> <sup>a</sup>∣Ω∣Cð Þ <sup>Ω</sup> and choose <sup>M</sup> <sup>¼</sup> <sup>u</sup> : <sup>∥</sup>u∥<sup>2</sup>

We consider the following initial value problem

I≔½ � 0; T , satisfying the following conditions:

ii. u and ux are bounded in R � I,

iv. u xð Þ¼ ; 0 φð Þx for all x∈ R.

ing integral equation problem

ously embedded in L<sup>2</sup>

ð Þ Ω , Tu satisfies

ð Ω

Given u∈L<sup>2</sup>

obtain. ∥∇Tu∥<sup>2</sup>

ð Þ <sup>Ω</sup> into <sup>L</sup><sup>2</sup>

�

of H<sup>1</sup>

bounded.

tions},

where <sup>T</sup> is continuous. Lemma (2.9) show that <sup>u</sup> ! f uð Þ is continuous form <sup>L</sup><sup>2</sup>

<sup>L</sup>2ð Þ <sup>Ω</sup> <sup>≤</sup> <sup>C</sup>ð Þ <sup>Ω</sup> <sup>∥</sup>Tu∥<sup>2</sup>

<sup>L</sup>2ð Þ <sup>Ω</sup> <sup>≤</sup>R∥∇Tu∥L2ð Þ <sup>Ω</sup> . Thus T Mð Þ <sup>⊂</sup> <sup>u</sup> : <sup>∥</sup>u∥H1ð Þ <sup>Ω</sup> <sup>≤</sup> <sup>R</sup>

3.5. A non-homogeneous linear parabolic partial differential equation

iii. utð Þ¼ x; t uxxð Þþ x; t H xð Þ ; t; u xð Þ ; t ; uxð Þ x; t for all ð Þ x; t ∈ R � I,

ð Þ! <sup>Ω</sup> <sup>L</sup><sup>2</sup>

ð Ω

T : M ! M, T is compact. Using Poincare's inequality on the right-hand-side in (3.25), we

utð Þ¼ x; t uxxð Þþ x; t H xð Þ ; t; u xð Þ ; t ; uxð Þ x; t , �∞ < x < ∞, 0 < t ≤ T,

where H is continuous and φ assume to be continuously differentiable such that φ and φ<sup>0</sup> are

By a solution of the problem (3.26), we mean a function u � u xð Þ ; t defined on R � I, where

i. u, ut, ux, uxx ∈Cð Þ R � I : {Cð Þ R � I denote the space of all continuous real valued func-

It is important to note that the differential equation problem (3.26) is equivalent to the follow-

u xð Þ¼ ; 0 φð Þx ≥ 0, �∞ < x < ∞,

ð Þ Ω <sup>u</sup> <sup>↦</sup> ð Þ �<sup>Δ</sup> �<sup>1</sup>

ð Þ f uð Þ ,

Fixed Point Theory Approach to Existence of Solutions with Differential Equations

ð Þ Ω . Find a closed, non-empty bounded convex set such that T : M ! M.

<sup>L</sup>2ð Þ <sup>Ω</sup> <sup>≤</sup> <sup>a</sup>∣Ω∣∥Tu∥<sup>2</sup>

ð Þ <sup>Ω</sup> is compact, <sup>T</sup> is compact. □

ð Þ <sup>Ω</sup> into <sup>H</sup><sup>1</sup>

<sup>L</sup>2ð Þ <sup>Ω</sup> :

<sup>L</sup>2ð Þ <sup>Ω</sup> <sup>≤</sup><sup>R</sup>

f uð ÞTudx <sup>≤</sup> <sup>a</sup>∣Ω∣∥Tu∥L2ð Þ <sup>Ω</sup> (3.25)

n o. We have established that

n o, and since the embedding

ð Þ Ω into itself.

21

(3.26)

<sup>0</sup>ð Þ Ω , which is continu-

http://dx.doi.org/10.5772/intechopen.74560

where the latter notations derive from formally writing <sup>∇</sup> <sup>¼</sup> <sup>∂</sup> <sup>∂</sup>x<sup>1</sup> ; <sup>∂</sup> <sup>∂</sup>x<sup>2</sup> ; <sup>⋯</sup>; <sup>∂</sup> ∂xn � �. Equivalently, the Laplacian of f the sum of all the unmixed

$$
\Delta f = \sum\_{i=0}^{n} \frac{\partial^2 f}{\partial x\_i^2}. \tag{3.22}
$$

As a second-order differential operator, the Laplace operator maps Ck functions to C<sup>k</sup>�<sup>2</sup> functions for <sup>k</sup> <sup>≥</sup> 2. the expression (3.21)(or equivalently(3.22)) defines an operator <sup>Δ</sup> : <sup>C</sup>ð Þ<sup>k</sup> <sup>R</sup><sup>n</sup> ð Þ! <sup>C</sup>ð Þ <sup>k</sup>�<sup>2</sup> <sup>R</sup><sup>n</sup> ð Þ or more generator <sup>Δ</sup> : <sup>C</sup>ð Þ<sup>k</sup> ð Þ! <sup>Ω</sup> <sup>C</sup>ð Þ <sup>k</sup>�<sup>2</sup> ð Þ <sup>Ω</sup> for any open set <sup>Ω</sup> Consider semilinear elliptic equation. Look for a function <sup>u</sup> : <sup>Ω</sup> <sup>⊂</sup> <sup>R</sup><sup>n</sup> ! <sup>R</sup><sup>m</sup> that solves

$$-\Delta u = f(u) \quad \text{in} \quad \Omega \tag{3.23}$$

$$
\mu = \mu\_0 \quad \text{on} \quad \partial \Omega \tag{3.24}
$$

where <sup>f</sup> : <sup>R</sup><sup>n</sup> ! <sup>R</sup><sup>m</sup> is a typically nonlinear function. Equivalently look for a fixed point of Tu≔ð Þ �Δu<sup>0</sup> �<sup>1</sup> ð Þ f uð Þ .

Theorem 3.7. ref. [5] Let f ∈Cð Þ R and sup<sup>x</sup><sup>∈</sup> <sup>R</sup>∣f xð Þ∣ ¼ a < ∞. then (3.23) has a weak solution u∈ H<sup>1</sup> <sup>0</sup>ð Þ Ω , i.e.

$$\int\_{\Omega} \nabla \boldsymbol{\mu} \cdot \nabla \boldsymbol{\Phi} d\mathbf{x} = \int\_{\Omega} f(\boldsymbol{\mu}) \boldsymbol{\Phi} d\mathbf{x}, \qquad \forall \boldsymbol{\Phi} \in \mathsf{C}\_{0}^{\boldsymbol{\sigma}}(\Omega).$$

Proof. Our strategy is to apply Schauder's Fixed Point Theorem to the map

$$\begin{aligned} T: L^2(\Omega) &\to L^2(\Omega) \\ u &\mapsto \left(-\Delta\right)^{-1}(f(u))\_{\mathcal{H}} \end{aligned}$$

where <sup>T</sup> is continuous. Lemma (2.9) show that <sup>u</sup> ! f uð Þ is continuous form <sup>L</sup><sup>2</sup> ð Þ Ω into itself. Corollary (2.10) shows that ð Þ �<sup>Δ</sup> �<sup>1</sup> is continuous form <sup>L</sup><sup>2</sup> ð Þ <sup>Ω</sup> into <sup>H</sup><sup>1</sup> <sup>0</sup>ð Þ Ω , which is continuously embedded in L<sup>2</sup> ð Þ Ω . Find a closed, non-empty bounded convex set such that T : M ! M. Given u∈L<sup>2</sup> ð Þ Ω , Tu satisfies

$$\int\_{\Omega} \nabla Tu \cdot \nabla T u dx = \int\_{\Omega} f(u) T u dx \le a \|\Omega\| \|Tu\|\_{L^2(\Omega)}\tag{3.25}$$

Cauchy-Schwarz. T here fore, using Ponincare's inequality

Let β, γ, η, δ > 0 where β < <sup>1</sup>

20 Differential Equations - Theory and Current Research

σð Þ Tx; Ty ≤ e

3.4. Partial differential equation

Laplacian of f the sum of all the unmixed

Tu≔ð Þ �Δu<sup>0</sup> �<sup>1</sup>

<sup>0</sup>ð Þ Ω , i.e.

u∈ H<sup>1</sup>

ð Þ f uð Þ .

ð Ω

∇u � ∇Φdx ¼

which is F∈ F, we get

<sup>7</sup> , γ < <sup>1</sup>

u∈ X, that is, the problem (3.14) has a unique solution u∈ C<sup>2</sup>

differentiable real-valued function, then the Laplacian of f is defined by

where the latter notations derive from formally writing <sup>∇</sup> <sup>¼</sup> <sup>∂</sup>

elliptic equation. Look for a function <sup>u</sup> : <sup>Ω</sup> <sup>⊂</sup> <sup>R</sup><sup>n</sup> ! <sup>R</sup><sup>m</sup> that solves

<sup>Δ</sup><sup>f</sup> <sup>¼</sup> <sup>∇</sup><sup>2</sup>

<sup>Δ</sup><sup>f</sup> <sup>¼</sup> <sup>X</sup><sup>n</sup> i¼0

As a second-order differential operator, the Laplace operator maps Ck functions to C<sup>k</sup>�<sup>2</sup> functions for <sup>k</sup> <sup>≥</sup> 2. the expression (3.21)(or equivalently(3.22)) defines an operator <sup>Δ</sup> : <sup>C</sup>ð Þ<sup>k</sup> <sup>R</sup><sup>n</sup> ð Þ! <sup>C</sup>ð Þ <sup>k</sup>�<sup>2</sup> <sup>R</sup><sup>n</sup> ð Þ or more generator <sup>Δ</sup> : <sup>C</sup>ð Þ<sup>k</sup> ð Þ! <sup>Ω</sup> <sup>C</sup>ð Þ <sup>k</sup>�<sup>2</sup> ð Þ <sup>Ω</sup> for any open set <sup>Ω</sup> Consider semilinear

where <sup>f</sup> : <sup>R</sup><sup>n</sup> ! <sup>R</sup><sup>m</sup> is a typically nonlinear function. Equivalently look for a fixed point of

Theorem 3.7. ref. [5] Let f ∈Cð Þ R and sup<sup>x</sup><sup>∈</sup> <sup>R</sup>∣f xð Þ∣ ¼ a < ∞. then (3.23) has a weak solution

f uð ÞΦdx, <sup>∀</sup><sup>Φ</sup> <sup>∈</sup> <sup>C</sup><sup>∞</sup>

ð Ω ∂2 f ∂x<sup>2</sup> i

<sup>7</sup> , η < <sup>1</sup>

<sup>7</sup> , δ < <sup>1</sup>

where α þ β þ γ þ 2η þ 2δ < 1. Taking the function F : R<sup>þ</sup> ! R in (3.20), where F tðÞ¼ ln ð Þt ,

<sup>τ</sup> <sup>þ</sup> <sup>F</sup>ð Þ <sup>σ</sup>ð Þ Tx; Ty <sup>≤</sup> <sup>F</sup> ασð Þþ <sup>x</sup>; <sup>y</sup> βσð Þþ <sup>x</sup>; Tx γσð Þþ <sup>y</sup>; Ty ησð Þþ <sup>x</sup>; Ty δσð Þ <sup>y</sup>; Tx � �:

Therefore all hypothesis of Theorem (2.25) are satisfied, and so T has a unique fixed point

Consider the Laplace operator is a second order differential operator in the n-dimensional Euclidean space, defined as the divergence ð Þ ∇� of the gradient ð Þ ∇f . Thus if f is a twice-

<sup>7</sup> : It following (3.19), we get

ð Þ<sup>I</sup> : □

. Equivalently, the

f ¼ ∇ � ∇f (3.21)

<sup>∂</sup>x<sup>2</sup> ; <sup>⋯</sup>; <sup>∂</sup> ∂xn

: (3.22)

� �

<sup>∂</sup>x<sup>1</sup> ; <sup>∂</sup>

�Δu ¼ f uð Þ in Ω (3.23)

u ¼ u<sup>0</sup> on ∂Ω (3.24)

<sup>0</sup> ð Þ Ω :

�<sup>τ</sup> ð Þ ασð Þþ <sup>x</sup>; <sup>y</sup> βσð Þþ <sup>x</sup>; Tx γσð Þþ <sup>y</sup>; Ty ησð Þþ <sup>x</sup>; Ty δσð Þ <sup>y</sup>; Tx � �, (3.20)

$$\|\|Tu\|\|\_{L^2(\Omega)}^2 \le \mathsf{C}(\Omega) \|\|Tu\|\|\_{L^2(\Omega)}^2 \le a|\Omega| \|\|Tu\|\|\_{L^2(\Omega)}^2.$$

Thus if we set <sup>R</sup> <sup>¼</sup> <sup>a</sup>∣Ω∣Cð Þ <sup>Ω</sup> and choose <sup>M</sup> <sup>¼</sup> <sup>u</sup> : <sup>∥</sup>u∥<sup>2</sup> <sup>L</sup>2ð Þ <sup>Ω</sup> <sup>≤</sup><sup>R</sup> n o. We have established that T : M ! M, T is compact. Using Poincare's inequality on the right-hand-side in (3.25), we obtain. ∥∇Tu∥<sup>2</sup> <sup>L</sup>2ð Þ <sup>Ω</sup> <sup>≤</sup>R∥∇Tu∥L2ð Þ <sup>Ω</sup> . Thus T Mð Þ <sup>⊂</sup> <sup>u</sup> : <sup>∥</sup>u∥H1ð Þ <sup>Ω</sup> <sup>≤</sup> <sup>R</sup> n o, and since the embedding of H<sup>1</sup> ð Þ <sup>Ω</sup> into <sup>L</sup><sup>2</sup> ð Þ <sup>Ω</sup> is compact, <sup>T</sup> is compact. □

#### 3.5. A non-homogeneous linear parabolic partial differential equation

We consider the following initial value problem

$$\begin{cases} \boldsymbol{u}\_{l}(\mathbf{x},t) = \boldsymbol{u}\_{\text{xx}}(\mathbf{x},t) + H(\mathbf{x},t,\mathbf{u}(\mathbf{x},t),\boldsymbol{u}\_{\mathbf{x}}(\mathbf{x},t)), & -\boldsymbol{\alpha} < \mathbf{x} < \boldsymbol{\alpha}, 0 < t \le T, \\\boldsymbol{u}(\mathbf{x},0) = \boldsymbol{\varphi}(\mathbf{x}) \ge 0, & -\boldsymbol{\alpha} < \mathbf{x} < \boldsymbol{\alpha}, \end{cases} \tag{3.26}$$

where H is continuous and φ assume to be continuously differentiable such that φ and φ<sup>0</sup> are bounded.

By a solution of the problem (3.26), we mean a function u � u xð Þ ; t defined on R � I, where I≔½ � 0; T , satisfying the following conditions:


$$\mathbf{iii}. \quad u\_t(\mathbf{x}, t) = u\_{\mathbf{x}t}(\mathbf{x}, t) + H(\mathbf{x}, t, u(\mathbf{x}, t), u\_{\mathbf{x}}(\mathbf{x}, t)) \text{ for all } (\mathbf{x}, t) \in \mathbb{R} \times I\_{\mathbf{x}}$$

iv. u xð Þ¼ ; 0 φð Þx for all x∈ R.

It is important to note that the differential equation problem (3.26) is equivalent to the following integral equation problem

$$u(\mathbf{x},t) = \int\_{-\infty}^{\infty} k(\mathbf{x}-\xi,t)\varphi(\xi)d\xi + \int\_{0}^{t} \int\_{-\infty}^{\infty} k(\mathbf{x}-\xi,t-\tau)H(\xi,\tau,u(\xi,\tau),u\_{\mathbf{x}}(\xi,\tau))d\xi d\tau \tag{3.27}$$

Consider the graph G with V Gð Þ¼ D ¼ Ω<sup>ω</sup> and E Gð Þ¼ fð Þ u; v ∈ D � D : u xð Þ ; t ≤ v xð Þ ; t and uxð Þ x; t ≤ vxð Þ x; t at each ð Þ x; t ∈ R � Ig. Clearly E Gð Þ is partial ordered and ð Þ D; E Gð Þ satisfy

for all ð Þ x; t ∈ R � I. Then, finding solution of problem (3.27) is equivalent to the ensuring the

k xð Þ � ξ; t � τ ∣Hðξ; τ; vð Þ ξ; τ ; vxð Þ ξ; τ Þ � Hð Þ ξ; τ; uð Þ ξ; τ ; uxð Þ ξ; τ ∣dξdτ

≤ 2π�<sup>1</sup> 2T 1

ωλð Þ Λu; Λv ≤ qωλð Þ u; v , q∈ð Þ 0; 1

�τ

τ þ ln ð Þ dð Þ Λu; Λv ≤ ln ð Þ d uð Þ ; v :

ωλð Þ <sup>Λ</sup>u; <sup>Λ</sup><sup>v</sup> <sup>≤</sup> <sup>T</sup> <sup>þ</sup> <sup>2</sup>π�<sup>1</sup>

dð Þ Λu; Λv ≤ e

ln ð Þ dð Þ Λu; Λv ≤ ln e

<sup>λ</sup> jðvð Þ� <sup>ξ</sup>; <sup>τ</sup> <sup>u</sup>ðξ; <sup>τ</sup>Þ þ vxðξ; <sup>τ</sup>Þ � uxðξ; <sup>τ</sup>ÞÞj � �dξd<sup>τ</sup>

ðt 0 ð∞ �∞

<sup>2</sup>cHωλð Þ u; v :

2T 1 2 � �cHωλð Þ <sup>u</sup>; <sup>v</sup>

d uð Þ ; v , τ > 0:

�<sup>τ</sup> ð Þ d uð Þ ; <sup>v</sup>

k xð Þ � ξ; t � τ Hð Þ ξ; τ; uð Þ ξ; τ ; uxð Þ ξ; τ dξdτ

http://dx.doi.org/10.5772/intechopen.74560

Fixed Point Theory Approach to Existence of Solutions with Differential Equations

∣kxð Þ x � ξ; t � τ ∣dξdτ

(3.28)

23

(3.29)

ðt 0 ð∞ �∞

Since ð Þ u; v ∈E Gð Þ, uð Þ <sup>x</sup>; vx ∈ E Gð Þ and hence ð Þ Λu; Λv ∈ E Gð Þ,ð Þ Λux; Λvx ∈ E Gð Þ:

property (A).

1

≤ 1 λ ðt 0 ð∞ �∞

≤ ðt 0 ð∞ �∞

i.e.

i.e.

≤ cHωλð Þ u; v T:

1

Similarly, we have

ð Þ Λu ð Þ x; t ≔

existence of fixed point of Λ.

<sup>λ</sup> <sup>∣</sup>ð Þ <sup>Λ</sup><sup>v</sup> ð Þ� <sup>x</sup>; <sup>t</sup> ð Þ <sup>Λ</sup><sup>u</sup> ð Þ <sup>x</sup>; <sup>t</sup> <sup>∣</sup>

k xð Þ � ξ; t � τ cH

Therefore, from (3.28) and (3.29) we have

Now, by passing to logarithms, we can write this as

Also, define a mapping Λ : Ω<sup>ω</sup> ! Ω<sup>ω</sup> by

k xð Þ � ξ; t φ ξð Þdξ þ

1

<sup>λ</sup> <sup>∣</sup>ð Þ <sup>Λ</sup><sup>v</sup> <sup>x</sup>ð Þ� <sup>x</sup>; <sup>t</sup> ð Þ <sup>Λ</sup><sup>u</sup> <sup>x</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>∣</sup> <sup>≤</sup> cHωλð Þ <sup>u</sup>; <sup>v</sup>

Thus, from the definition of Λ and by (ii) we have

ð∞ �∞

for all x∈ R and 0 < t ≤ T, where

$$k(\mathfrak{x}, t) \coloneqq \frac{1}{\sqrt{4\pi t}} e^{-\frac{x^2}{4t}}.$$

The problem (3.26) admits a solution if and only if the corresponding problem (3.27) has a solution.

Let

$$\Omega \coloneqq \{ \mu(\mathbf{x}, t); \boldsymbol{\mu}, \boldsymbol{\mu}\_x \in \mathbb{C}(\mathbb{R} \times I) \quad \text{and} \quad \|\boldsymbol{\mu}\| <^{\infty} \}\_{\prime \prime}$$

where

$$\|u\|\| := \sup\_{\mathbf{x} \in \mathbb{R}\_{\succ}} |u(\mathbf{x}, t)| + \sup\_{\mathbf{x} \in \mathbb{R}\_{\succ}} |u\_{\mathbf{x}}(\mathbf{x}, t)|.$$

Obviously, the function ω : R<sup>þ</sup> � Ω � Ω ! R<sup>þ</sup> given by

$$\omega\_{\lambda}(\mu, \upsilon) \coloneqq \frac{1}{\lambda} \|\mu - \upsilon\|\quad = \frac{1}{\lambda}d(\mu, \upsilon).$$

is a metric modular on Ω. Clearly, the set Ω<sup>ω</sup> is a complete modular metric space independent of generators.

Theorem 3.8. ref. [17] Consider the problem (3.26) and assume the following:


$$\begin{aligned} 0 \le \frac{1}{\lambda} \left[ H(\mathbf{x}, t, s\_2, p\_2) - H(\mathbf{x}, t, s\_1, p\_1) \right] \\ \le c\_H \left[ \frac{s\_2 - s\_1 + p\_2 - p\_1}{\lambda} \right] \end{aligned}$$

for all s1; p<sup>1</sup> � �, s2; <sup>p</sup><sup>2</sup> � �<sup>∈</sup> <sup>R</sup> � <sup>R</sup> with <sup>s</sup><sup>1</sup> <sup>≤</sup> <sup>s</sup><sup>2</sup> and <sup>p</sup><sup>1</sup> <sup>≤</sup> <sup>p</sup>2,

iii. H is bounded for bounded s and p:

Then the problem (3.26) admits a solution.

Proof. It is well known that u∈ Ω<sup>ω</sup> is a solution (3.26) iff u∈ Ω<sup>ω</sup> is a solution integral Eq. (3.27).

Consider the graph G with V Gð Þ¼ D ¼ Ω<sup>ω</sup> and E Gð Þ¼ fð Þ u; v ∈ D � D : u xð Þ ; t ≤ v xð Þ ; t and uxð Þ x; t ≤ vxð Þ x; t at each ð Þ x; t ∈ R � Ig. Clearly E Gð Þ is partial ordered and ð Þ D; E Gð Þ satisfy property (A).

Also, define a mapping Λ : Ω<sup>ω</sup> ! Ω<sup>ω</sup> by

$$(\Lambda u)(\mathbf{x}, t) \coloneqq \int\_{-\infty}^{\infty} k(\mathbf{x} - \boldsymbol{\xi}, t) \boldsymbol{\varrho}(\boldsymbol{\xi}) d\boldsymbol{\xi} + \int\_{0}^{t} \int\_{-\infty}^{\infty} k(\mathbf{x} - \boldsymbol{\xi}, t - \boldsymbol{\tau}) H(\boldsymbol{\xi}, \boldsymbol{\tau}, u(\boldsymbol{\xi}, \boldsymbol{\tau}), u\_{\mathbf{x}}(\boldsymbol{\xi}, \boldsymbol{\tau})) d\boldsymbol{\xi} d\boldsymbol{\tau}$$

for all ð Þ x; t ∈ R � I. Then, finding solution of problem (3.27) is equivalent to the ensuring the existence of fixed point of Λ.

Since ð Þ u; v ∈E Gð Þ, uð Þ <sup>x</sup>; vx ∈ E Gð Þ and hence ð Þ Λu; Λv ∈ E Gð Þ,ð Þ Λux; Λvx ∈ E Gð Þ:

Thus, from the definition of Λ and by (ii) we have

$$\begin{split} &\frac{1}{\lambda}|(\Lambda v)(\mathbf{x},t)-(\Lambda u)(\mathbf{x},t)| \\ &\leq \frac{1}{\lambda} \Big[ \int\_{0}^{\tau} k(\mathbf{x}-\xi,t-\tau)|H(\xi,\tau,v(\xi,\tau),v\_{\mathbf{x}}(\xi,\tau))-H(\xi,\tau,u(\xi,\tau),u\_{\mathbf{x}}(\xi,\tau))|d\xi d\tau \\ &\leq \int\_{0}^{t} \int\_{-\infty}^{\infty} k(\mathbf{x}-\xi,t-\tau)c\_{H} \Big[ \frac{1}{\lambda}|(v(\xi,\tau)-u(\xi,\tau)+v\_{\mathbf{x}}(\xi,\tau)-u\_{\mathbf{x}}(\xi,\tau))| \Big] d\xi d\tau \\ &\leq c\_{H}u\omega\_{\lambda}(\mathbf{u},\mathbf{v})T. \end{split} \tag{3.28}$$

Similarly, we have

$$\begin{split} \frac{1}{\lambda} |(\Lambda \boldsymbol{\sigma})\_{\boldsymbol{x}}(\boldsymbol{x}, t) - (\Lambda \boldsymbol{u})\_{\boldsymbol{x}}(\boldsymbol{x}, t)| &\leq c\_{\mathcal{H}} \omega\_{\lambda}(\boldsymbol{u}, \boldsymbol{v}) \int\_{0}^{t} \int\_{-\infty}^{\infty} |k\_{\boldsymbol{x}}(\boldsymbol{x} - \boldsymbol{\xi}, t - \boldsymbol{\tau})| d\boldsymbol{\xi} d\boldsymbol{\tau} \\ &\leq 2\pi^{-\frac{t}{2}} T^{\frac{1}{2}} c\_{\mathcal{H}} \omega\_{\lambda}(\boldsymbol{u}, \boldsymbol{v}). \end{split} \tag{3.29}$$

Therefore, from (3.28) and (3.29) we have

$$
\omega\_{\lambda}(\Lambda \mu, \Lambda \upsilon) \le \left( T + 2\pi^{-\frac{1}{2}} T^{\frac{1}{2}} \right) c\_{H} \omega\_{\lambda}(\mu, \upsilon),
$$

i.e.

u xð Þ¼ ; t

solution.

Let

where

of generators.

for all s1; p<sup>1</sup>

� �, s2; <sup>p</sup><sup>2</sup>

iii. H is bounded for bounded s and p:

Then the problem (3.26) admits a solution.

ð∞ �∞

22 Differential Equations - Theory and Current Research

for all x∈ R and 0 < t ≤ T, where

k xð Þ � ξ; t φ ξð Þdξ þ

ðt 0 ð∞ �∞

∥u∥ ≔ sup

Obviously, the function ω : R<sup>þ</sup> � Ω � Ω ! R<sup>þ</sup> given by

in x and t for each compact subset of R � I,

0 ≤ 1

≤ cH

ii. there exists a constant cH <sup>≤</sup> <sup>T</sup> <sup>þ</sup> <sup>2</sup>π�<sup>1</sup>

<sup>x</sup><sup>∈</sup> <sup>R</sup>, <sup>t</sup><sup>∈</sup> <sup>I</sup>

ωλð Þ <sup>u</sup>; <sup>v</sup> <sup>≔</sup> <sup>1</sup>

Theorem 3.8. ref. [17] Consider the problem (3.26) and assume the following:

k xð Þ ; <sup>t</sup> <sup>≔</sup> <sup>1</sup>

ffiffiffiffiffiffiffi <sup>4</sup>π<sup>t</sup> <sup>p</sup> <sup>e</sup> �x2 4t :

The problem (3.26) admits a solution if and only if the corresponding problem (3.27) has a

Ω≔f g u xð Þ ; t ; u; ux ∈Cð Þ R � I and ∥u∥ < ∞ ,

∣u xð Þ ; t ∣ þ sup

<sup>λ</sup> <sup>∥</sup><sup>u</sup> � <sup>v</sup><sup>∥</sup> <sup>¼</sup> <sup>1</sup>

is a metric modular on Ω. Clearly, the set Ω<sup>ω</sup> is a complete modular metric space independent

i. for c > 0 with ∣s∣ < c and ∣p∣ < c, the function F xð Þ ; t;s; p is uniformly Hölder continuous

� � � H x; <sup>t</sup>;s1; <sup>p</sup><sup>1</sup> � � � �

2T 1 2 � ��<sup>1</sup>

> s<sup>2</sup> � s<sup>1</sup> þ p<sup>2</sup> � p<sup>1</sup> λ � �

Proof. It is well known that u∈ Ω<sup>ω</sup> is a solution (3.26) iff u∈ Ω<sup>ω</sup> is a solution integral Eq. (3.27).

<sup>λ</sup> H x; <sup>t</sup>;s2; <sup>p</sup><sup>2</sup>

� �<sup>∈</sup> <sup>R</sup> � <sup>R</sup> with <sup>s</sup><sup>1</sup> <sup>≤</sup> <sup>s</sup><sup>2</sup> and <sup>p</sup><sup>1</sup> <sup>≤</sup> <sup>p</sup>2,

<sup>x</sup><sup>∈</sup> <sup>R</sup>, <sup>t</sup><sup>∈</sup> <sup>I</sup>

<sup>λ</sup> d uð Þ ; <sup>v</sup>

∣uxð Þ x; t ∣:

≤ q, where q∈ ð Þ 0; 1 such that

k xð Þ � ξ; t � τ Hð Þ ξ; τ; uð Þ ξ; τ ; uxð Þ ξ; τ dξdτ (3.27)

$$
\omega\_\lambda(\Lambda u, \Lambda v) \le q \omega\_\lambda(u, v), \quad q \in (0, 1).
$$

i.e.

$$d(\Lambda u, \Lambda v) \le e^{-\tau} d(u, v), \tau > 0.$$

Now, by passing to logarithms, we can write this as

$$
\ln \left( d(\Lambda \mu, \Lambda \upsilon) \right) \le \ln \left( e^{-\tau} d(\mu, \upsilon) \right)
$$

$$
\tau + \ln \left( d(\Lambda \mu, \Lambda \upsilon) \right) \le \ln \left( d(\mu, \upsilon) \right).
$$

Now, from example 2.22 (i) and taking T ¼ Λ and R ¼ I (Identity map), we deduce that the operator T satisfies all the hypothesis of theorem 2.32.

Integrating a fractional number of time with this formula is straightforward, one can use fractional n by interpreting ð Þ n � 1 ! as Γð Þ n , that is the Riemann-Liouville integral which is

> ðx a

This also makes sense if a ¼ �∞, with suitable restriction on f . The fundamental relation hold

f xð Þ¼ I

the latter of which is semigroup properties. These properties make possible not only the

dx½ � <sup>α</sup> <sup>I</sup>

where ½ �� denote the ceilling function. One also obtains a differintegral interpolation between

An alternative fractional derivative was introduced by Caputo in 1967, and produce a derivative that has different properties it produces zero from constant function and more importantly the initial value terms of the Laplace Transform are expressed by means of the value of that function and of its derivative of integer order rather than the derivative of fractional order as in the Riemann-Liouville derivative. The Caputo fractional derivative with base point x is then

½ ��<sup>α</sup> <sup>α</sup> <sup>d</sup>½ � <sup>α</sup>

Lemma 3.9. ref. [24] Let u : ½ �! 0; ∞ X be continuous function such that u∈ Cð Þ ½ � 0; τ ; X for all

dx½ � <sup>α</sup> f xð Þ:

<sup>t</sup> u tðÞ¼ Bu tð Þ; t > 0 (3.30)

uð Þ¼ 0 u<sup>0</sup> ∈ X (3.31)

d½ � <sup>α</sup> dx½ � <sup>α</sup> <sup>I</sup>

I

½ ��<sup>α</sup> <sup>α</sup>f

½ ��<sup>α</sup> <sup>α</sup>f xð Þ if <sup>α</sup> <sup>&</sup>gt; <sup>0</sup>

f xð Þ if α ¼ 0

�<sup>α</sup>f xð Þ if <sup>α</sup> <sup>&</sup>lt; <sup>0</sup>:

f tð Þð Þ <sup>x</sup> � <sup>t</sup> <sup>α</sup>�<sup>1</sup>

<sup>α</sup>f xð Þ

αþβ f xð Þ dt:

Fixed Point Theory Approach to Existence of Solutions with Differential Equations

http://dx.doi.org/10.5772/intechopen.74560

25

<sup>α</sup>f . One can define

I

<sup>α</sup>f xð Þ¼ <sup>1</sup>

d dx I αþ1

definition of fractional differentiation by taking enough derivative of I

dα dx<sup>α</sup> <sup>f</sup> <sup>¼</sup> <sup>d</sup>½ � <sup>α</sup>

8 >>>><

>>>>:

c Dα <sup>x</sup> f xð Þ¼ I

> c Dα

fractional-order derivative of as well by

differential and integration by defining

τ > 0. Then u is a global solution of

if and only if u the integral equation

Dα <sup>x</sup> f xð Þ¼ I <sup>α</sup> I β <sup>f</sup> � � <sup>¼</sup> <sup>I</sup>

Γð Þ α

defined by

Therefore, as an application of theorem 2.32 we conclude the existence of u<sup>∗</sup> ∈ Ω<sup>ω</sup> such that <sup>u</sup><sup>∗</sup> <sup>¼</sup> <sup>Λ</sup>u<sup>∗</sup> and so <sup>u</sup><sup>∗</sup> is a solution of the problem 3.26.

#### 3.6. Fractional differential equation

Before we will discuss the source of fractional differential equation.

Cauchy's formula for repeated integration. Let f be a continuous function on the real line. Then the nth repeated integral of f based at a,

$$f^{(-\eta)}(\mathbf{x}) = \int\_{a}^{\mathbf{x}} \int\_{a}^{\sigma\_{1}} \int\_{a}^{\sigma\_{2}} \dots \int\_{a}^{\sigma\_{n-1}} f(\sigma\_{n}) d\sigma\_{n} \dots d\sigma\_{3} d\sigma\_{2} d\sigma\_{1}$$

is given by single integration

$$f^{(-n)}(\mathbf{x}) = \frac{1}{(n-1)!} \int\_{a}^{\mathbf{x}} \left(\mathbf{x} - t\right)^{n-1} f(t) dt.$$

A proof is given by mathematical induction. Since f is continuous, the base case follows from the fundamental theorem of calculus.

$$\frac{d}{d\mathfrak{x}}f^{-1}(\mathfrak{x}) = \frac{d}{d\mathfrak{x}}\int\_{a}^{\mathfrak{x}}f(t)dt = f(\mathfrak{x})$$

where

$$f^{-1}(a) = \int\_a^a f(t)dt = 0.1$$

Now, suppose this is true for n, and let us prove it for n þ 1.

Firstly, using the Leibniz integral rule. Then applying the induction hypothesis

$$\begin{split} f^{(-n+1)}(\mathbf{x}) &= \int\_{a}^{\mathbf{x}} \int\_{a}^{\sigma\_{1}} \int\_{a}^{\sigma\_{2}} \dots \int\_{a}^{\sigma\_{n}} f(\sigma\_{n+1}) d\sigma\_{n} \dots d\sigma\_{3} d\sigma\_{2} d\sigma\_{1} \\ &= \int\_{a}^{\mathbf{x}} \frac{1}{(n-1)!} \int\_{a}^{\sigma\_{1}} (\sigma\_{1}-t)^{n-1} f(t) dt d\sigma\_{1} \\ &= \int\_{a}^{\mathbf{x}} \frac{d}{d\sigma\_{1}} \left[ \frac{1}{n!} \int\_{a}^{\sigma\_{1}} (\sigma\_{1}-t)^{n} f(t) dt \right] d\sigma\_{1} \\ &= \frac{1}{n!} \int\_{a}^{\mathbf{x}} (\mathbf{x}-t)^{n} f(t) dt. \end{split}$$

This completes the proof. In fractional calculus, this formula can be used to construct a notion of differintegral, allowing one to differentiate or integrate a fractional number of time. Integrating a fractional number of time with this formula is straightforward, one can use fractional n by interpreting ð Þ n � 1 ! as Γð Þ n , that is the Riemann-Liouville integral which is defined by

$$I^a f(\mathbf{x}) = \frac{1}{\Gamma(\alpha)} \int\_a^\mathbf{x} f(t) (\mathbf{x} - t)^{\alpha - 1} dt.$$

This also makes sense if a ¼ �∞, with suitable restriction on f . The fundamental relation hold

$$\frac{d}{d\mathbf{x}} I^{\alpha+1} f(\mathbf{x}) = I^{\alpha} f(\mathbf{x})$$

$$I^{\alpha} \left( I^{\beta} f \right) = I^{\alpha + \beta} f(\mathbf{x})$$

the latter of which is semigroup properties. These properties make possible not only the definition of fractional differentiation by taking enough derivative of I <sup>α</sup>f . One can define fractional-order derivative of as well by

$$\frac{d^{\alpha}}{d\mathfrak{x}^{a}}f = \frac{d^{[\alpha]}}{d\mathfrak{x}^{[\alpha]}}I^{[\alpha]-\alpha}f$$

where ½ �� denote the ceilling function. One also obtains a differintegral interpolation between differential and integration by defining

$$D\_{\alpha}^{\alpha}f(\mathbf{x}) = \begin{cases} \frac{d^{[\alpha]}}{d\mathbf{x}^{[\alpha]}} I^{[\alpha]-\alpha} f(\mathbf{x}) & \text{if } \alpha > 0 \\ f(\mathbf{x}) & \text{if } \alpha = 0 \\ I^{-\alpha}f(\mathbf{x}) & \text{if } \alpha < 0. \end{cases}$$

An alternative fractional derivative was introduced by Caputo in 1967, and produce a derivative that has different properties it produces zero from constant function and more importantly the initial value terms of the Laplace Transform are expressed by means of the value of that function and of its derivative of integer order rather than the derivative of fractional order as in the Riemann-Liouville derivative. The Caputo fractional derivative with base point x is then

$${}^cD\_x^\alpha f(\mathbf{x}) = I^{[\alpha]-\alpha} \frac{d^{[\alpha]}}{d\mathbf{x}^{[\alpha]}} f(\mathbf{x}) \dots$$

Lemma 3.9. ref. [24] Let u : ½ �! 0; ∞ X be continuous function such that u∈ Cð Þ ½ � 0; τ ; X for all τ > 0. Then u is a global solution of

$$\mathbf{u}^c D\_t^\alpha \mu(t) = B\mu(t); t > 0 \tag{3.30}$$

$$
\mu(0) = \mu\_0 \in X \tag{3.31}
$$

if and only if u the integral equation

Now, from example 2.22 (i) and taking T ¼ Λ and R ¼ I (Identity map), we deduce that the

Therefore, as an application of theorem 2.32 we conclude the existence of u<sup>∗</sup> ∈ Ω<sup>ω</sup> such that

Cauchy's formula for repeated integration. Let f be a continuous function on the real line. Then the

fð Þ σ<sup>n</sup> dσn…dσ3dσ2dσ<sup>1</sup>

f tð Þdt:

fð Þ σ<sup>n</sup>þ<sup>1</sup> dσn…dσ3dσ2dσ<sup>1</sup>

f tð Þdt

f tð Þdtdσ<sup>1</sup>

dσ<sup>1</sup>

operator T satisfies all the hypothesis of theorem 2.32.

<sup>u</sup><sup>∗</sup> <sup>¼</sup> <sup>Λ</sup>u<sup>∗</sup> and so <sup>u</sup><sup>∗</sup> is a solution of the problem 3.26.

Before we will discuss the source of fractional differential equation.

ðx a

ð Þ �<sup>n</sup> ð Þ¼ <sup>x</sup>

d dx f �1 ð Þ¼ x

> f �1 ð Þ¼ a ða a

ðx a

¼ ðx a

¼ ðx a d dσ<sup>1</sup>

¼ 1 n! ðx a

Firstly, using the Leibniz integral rule. Then applying the induction hypothesis

ð<sup>σ</sup><sup>2</sup> a … ð<sup>σ</sup><sup>n</sup> a

1 ð Þ n � 1 !

1 n! ð<sup>σ</sup><sup>1</sup> a

ð Þ <sup>x</sup> � <sup>t</sup> <sup>n</sup>

ð<sup>σ</sup><sup>1</sup> a

ð<sup>σ</sup><sup>1</sup> a

Now, suppose this is true for n, and let us prove it for n þ 1.

ð Þ �nþ<sup>1</sup> ð Þ¼ <sup>x</sup>

f

ð<sup>σ</sup><sup>1</sup> a

ð<sup>σ</sup><sup>2</sup> a … ð<sup>σ</sup>n�<sup>1</sup> a

1 ð Þ n � 1 ! ðx a

A proof is given by mathematical induction. Since f is continuous, the base case follows from

d dx ðx a

ð Þ <sup>x</sup> � <sup>t</sup> <sup>n</sup>�<sup>1</sup>

f tð Þdt ¼ f xð Þ

f tð Þdt ¼ 0:

ð Þ <sup>σ</sup><sup>1</sup> � <sup>t</sup> <sup>n</sup>�<sup>1</sup>

ð Þ <sup>σ</sup><sup>1</sup> � <sup>t</sup> <sup>n</sup>

f tð Þdt:

This completes the proof. In fractional calculus, this formula can be used to construct a notion of differintegral, allowing one to differentiate or integrate a fractional number of time.

� �

3.6. Fractional differential equation

24 Differential Equations - Theory and Current Research

nth repeated integral of f based at a,

the fundamental theorem of calculus.

is given by single integration

where

f

ð Þ �<sup>n</sup> ð Þ¼ <sup>x</sup>

f

$$
\mu(t) = \mu\_0 + \frac{1}{\Gamma(\alpha)} \int\_0^t (t - s)^{\alpha - 1} B \mu(s) ds, \; t \ge 0.
$$

<sup>k</sup>Tut ð Þ� ð Þ Tvt ð Þk ¼ k ð Þ <sup>1</sup>

By iterating this relation, we find that

ð Þ� u tð Þ <sup>T</sup><sup>2</sup>

ð Þ� u tð Þ <sup>T</sup><sup>3</sup>

<sup>k</sup>T<sup>n</sup>ð Þ� u tð Þ <sup>T</sup><sup>n</sup>ð Þk v tð Þ <sup>≤</sup>

kT2

kT3

i. Let f gj Unð Þ<sup>t</sup> <sup>∞</sup>

ii. It holds that

<sup>U</sup>0ðÞ¼ <sup>t</sup> <sup>u</sup>0, Un <sup>¼</sup> <sup>u</sup><sup>0</sup> <sup>þ</sup> <sup>1</sup>

Γð Þ α

<sup>≤</sup> <sup>k</sup> <sup>1</sup> Γð Þ α

≤ t <sup>α</sup>k k<sup>B</sup> L Xð Þ

≤

≤

ð Þk v tð Þ ≤

ð Þk v tð Þ ≤

≤

≤

Corollary 3.11. ref. [24] Consider the same hypothesis of theorem (3.10).

<sup>0</sup> ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

U tðÞ¼ <sup>X</sup><sup>∞</sup>

k¼0

Γð Þ α Ðt <sup>≤</sup> k k<sup>B</sup> L Xð Þ

t α

t α

> t α

t 2α

t α

t 3α

t nα <sup>Γ</sup><sup>n</sup>ð Þ <sup>α</sup> <sup>þ</sup> <sup>1</sup> k k<sup>B</sup> <sup>n</sup>

Γ2

Γ3

≤ ⋯

ðt 0

> ðt 0

<sup>Γ</sup>ð Þ <sup>α</sup> <sup>k</sup>u sð Þ� v sð ÞÞk <sup>ð</sup><sup>t</sup>

<sup>α</sup>Γð Þ <sup>α</sup> <sup>k</sup>u sð Þ� v sð ÞÞk

<sup>Γ</sup>ð Þ <sup>α</sup> <sup>þ</sup> <sup>1</sup> k k<sup>B</sup> L Xð Þ sup

<sup>Γ</sup>ð Þ <sup>α</sup> <sup>þ</sup> <sup>1</sup> k k<sup>B</sup> L Xð Þ sup

<sup>Γ</sup>ð Þ <sup>α</sup> <sup>þ</sup> <sup>1</sup> k k<sup>B</sup> L Xð Þ sup

ð Þ <sup>α</sup> <sup>þ</sup> <sup>1</sup> k k<sup>B</sup> <sup>2</sup>

ð Þ <sup>α</sup> <sup>þ</sup> <sup>1</sup> k k<sup>B</sup> <sup>3</sup>

and for an sufficiently large n,the constant in question is less than 1, i.e., there exists a fixed point u∈Kτ. Observe now that τ > 0 was an arbitrary choice, so we conclude that the fixed point u∈ Cð Þ ½ � 0; τ ; X for all τ > 0 and Lemma (3.9), we obtain the existence and uniqueness of a global solution to the problem (3.30). □

<sup>Γ</sup>ð Þ <sup>α</sup> <sup>þ</sup> <sup>1</sup> k k<sup>B</sup> L Xð Þku sð Þ� v sð ÞÞk

0⩽ s⩽τ

0⩽s⩽ τ

0⩽s⩽ τ

L Xð Þ sup 0⩽ s⩽τ

L Xð Þ sup 0⩽ s⩽τ

<sup>n</sup>¼<sup>0</sup> be a sequence of continuous functions Un : ½ Þ! <sup>0</sup>; <sup>∞</sup> <sup>X</sup> given by

BUn�<sup>1</sup>ð Þs ds, n∈ f g 1; 2; … .

Then there exists a continuous function U : ½ Þ! 0; ∞ X, such that for any τ > 0, we conclude that Un ! U in Cð Þ ½ � 0; τ ; X . Moreover, U tð Þ is the unique global solution of (3.30).

> t <sup>α</sup> ð Þ <sup>B</sup> <sup>k</sup> u0 <sup>Γ</sup>ð Þ <sup>α</sup><sup>k</sup> <sup>þ</sup> <sup>1</sup> :

kT2

L Xð Þ sup 0⩽ s⩽τ

ð Þ <sup>t</sup> � <sup>s</sup> ð Þ <sup>α</sup>�<sup>1</sup> Bus ð Þ ð Þ� v sð Þ ds<sup>k</sup>

Fixed Point Theory Approach to Existence of Solutions with Differential Equations

ð Þ <sup>t</sup> � <sup>s</sup> ð Þ <sup>α</sup>�<sup>1</sup> k k<sup>B</sup> L Xð Þku sð Þ� v sð ÞÞkds

ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

ku sð Þ� v sð ÞÞk:

kTu sð Þ� Tv sð ÞÞk

ku sð Þ� v sð ÞÞk

u sð Þ� <sup>T</sup><sup>2</sup>

ku sð Þ� v sð ÞÞk

ku sð Þ� v sð ÞÞk

v sð ÞÞk

ds

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27

0

Proof. ð Þ ) Let <sup>τ</sup> <sup>&</sup>gt; 0. Since <sup>u</sup> is a global solution of (3.30), then <sup>u</sup>∈Cð Þ ½ � <sup>0</sup>; <sup>τ</sup> ; <sup>X</sup> , <sup>c</sup> Dα <sup>t</sup> u∈Cð Þ ½ � 0; τ ; X andt

$$^cD\_t^\alpha u(t) = Bu(t), \quad t \in (0, \pi].$$

Thus, by applying I α <sup>t</sup> in both sides of the equality (since <sup>c</sup> Dα <sup>t</sup> u∈L<sup>1</sup> ð Þ 0; τ; X ) we obtain

$$u(t) = u(0) + I\_t^\alpha Bu(t) = u\_0 + \frac{1}{\Gamma(\alpha)} \int\_0^t (t - s)^{\alpha - 1} Bu(s)ds, t \ge 0.$$

Since τ > 0 was an arbitrary choice, u satisfies the integral equation for all t ≥ 0, as we wish.

ð Þ ( On the other hand, choose τ > 0 (but arbitrary). By hypothesis, u∈Cð Þ ½ � 0; τ ; X , and satisfies the integral equation,

$$
\mu(t) = \mu\_0 + \frac{1}{\Gamma(\alpha)} \int\_0^t (t - s)^{\alpha - 1} B \mu(s) ds, \quad t \in [0, \tau].
$$

Observing also uð Þ¼ 0 u<sup>0</sup> and rewriting the equality above, we obtain

$$
\mu(t) = \mu(0) + I\_t^\alpha B \mu(\mathbf{s}), \quad t \in [0, \pi].
$$

Since Bu sð Þ <sup>∈</sup>Cð Þ ½ � <sup>0</sup>; <sup>τ</sup> ; <sup>X</sup> , we conclude, by <sup>c</sup> Dα t I α <sup>t</sup> f tðÞ¼ f tð Þ of the fractional integral and derivative property that we can apply <sup>c</sup> Dα <sup>t</sup> in both sides of the integral equation, obtaining

$$^cD\_t^\alpha u(t) = Bu(t), \quad t \in [0, \pi].$$

what lead us to verify that <sup>c</sup> Dα <sup>t</sup> u∈ Cð Þ ½ � 0; τ ; X . Since τ > 0 was an arbitrary choice, we conclude that the function u is a global solution of (3.30). □

Theorem 3.10. ref. [24] Let α∈ ð Þ 0; 1 , B∈L Xð Þ and u<sup>0</sup> ∈ X then the problem (3.30).

have a unique global solution.

Proof. Choose τ > 0. then consider K<sup>τ</sup> ¼ u∈Cð Þ ½ � 0; τ ; X ; uð Þ¼ 0 u<sup>0</sup> and operator.

T : K<sup>τ</sup> ! K<sup>τ</sup> given by

$$T(\mu(t)) = \mu\_0 + \frac{1}{\Gamma(\alpha)} \int\_0^t B \mu(t) dt.$$

We will show that a power (with respect to be composition) of this operator is a contraction and therefore by Banach's Fixed Point Theorem, T have a unique fixed point in K<sup>τ</sup> to this end, observe that for any u, v∈ K<sup>τ</sup>

Fixed Point Theory Approach to Existence of Solutions with Differential Equations http://dx.doi.org/10.5772/intechopen.74560 27

$$\begin{split} \|T(u(t)) - T(v(t))\| &= \quad \|\frac{1}{\Gamma(\alpha)}\int\_{0}^{t} (t-s)^{(\alpha-1)} B(u(s) - v(s)) ds\| \\ &\leq \quad \|\frac{1}{\Gamma(\alpha)}\int\_{0}^{t} (t-s)^{(\alpha-1)} \|B\|\_{L(X)} \|u(s) - v(s)\| \|ds \\ &\leq \quad \frac{\|\|B\|\_{L(X)}}{\Gamma(\alpha)} \|u(s) - v(s)\| \|\int\_{0}^{t} (t-s)^{\alpha-1} ds \\ &\leq \quad \frac{t^{\alpha} \|B\|\_{L(X)}}{\alpha \Gamma(\alpha)} \|u(s) - v(s)\| \| \\ &\leq \quad \frac{t^{\alpha}}{\Gamma(\alpha+1)} \|\|B\|\_{L(X)} \|u(s) - v(s)\| \| \\ &\leq \quad \frac{t^{\alpha}}{\Gamma(\alpha+1)} \|\|B\|\_{L(X)} \sup\_{0 \leq s \leq \tau} \|u(s) - v(s)\| \|. \end{split}$$

By iterating this relation, we find that

u tðÞ¼ u<sup>0</sup> þ

c Dα

α

u tðÞ¼ u<sup>0</sup> þ

andt

Thus, by applying I

satisfies the integral equation,

α

26 Differential Equations - Theory and Current Research

Since Bu sð Þ <sup>∈</sup>Cð Þ ½ � <sup>0</sup>; <sup>τ</sup> ; <sup>X</sup> , we conclude, by <sup>c</sup>

ative property that we can apply <sup>c</sup>

what lead us to verify that <sup>c</sup>

have a unique global solution.

observe that for any u, v∈ K<sup>τ</sup>

T : K<sup>τ</sup> ! K<sup>τ</sup> given by

u tðÞ¼ uð Þþ 0 I

1 Γð Þ α

Proof. ð Þ ) Let <sup>τ</sup> <sup>&</sup>gt; 0. Since <sup>u</sup> is a global solution of (3.30), then <sup>u</sup>∈Cð Þ ½ � <sup>0</sup>; <sup>τ</sup> ; <sup>X</sup> , <sup>c</sup>

<sup>t</sup> in both sides of the equality (since <sup>c</sup>

<sup>t</sup> Bu tðÞ¼ u<sup>0</sup> þ

1 Γð Þ α

u tðÞ¼ uð Þþ 0 I

Observing also uð Þ¼ 0 u<sup>0</sup> and rewriting the equality above, we obtain

Dα

c Dα

Dα

ðt 0 ðt 0

ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

<sup>t</sup> u tðÞ¼ Bu tð Þ, t∈ ð � 0; τ :

1 Γð Þ α

Since τ > 0 was an arbitrary choice, u satisfies the integral equation for all t ≥ 0, as we wish.

ð Þ ( On the other hand, choose τ > 0 (but arbitrary). By hypothesis, u∈Cð Þ ½ � 0; τ ; X , and

ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

α

<sup>t</sup> u tðÞ¼ Bu tð Þ, t∈½ � 0; τ

that the function u is a global solution of (3.30). □

1 Γð Þ α

We will show that a power (with respect to be composition) of this operator is a contraction and therefore by Banach's Fixed Point Theorem, T have a unique fixed point in K<sup>τ</sup> to this end,

ðt 0

Bu tð Þdt:

Theorem 3.10. ref. [24] Let α∈ ð Þ 0; 1 , B∈L Xð Þ and u<sup>0</sup> ∈ X then the problem (3.30).

Proof. Choose τ > 0. then consider K<sup>τ</sup> ¼ u∈Cð Þ ½ � 0; τ ; X ; uð Þ¼ 0 u<sup>0</sup> and operator.

Tut ð Þ¼ ð Þ u<sup>0</sup> þ

Dα t I α

<sup>t</sup> Bu sð Þ, t ∈½ � 0; τ :

ðt 0 Bu sð Þds, t ≥ 0:

Dα <sup>t</sup> u∈L<sup>1</sup>

ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

Bu sð Þds, t ∈½ � 0; τ :

<sup>t</sup> in both sides of the integral equation, obtaining

<sup>t</sup> u∈ Cð Þ ½ � 0; τ ; X . Since τ > 0 was an arbitrary choice, we conclude

Dα

ð Þ 0; τ; X ) we obtain

Bu sð Þds, t ≥ 0:

<sup>t</sup> f tðÞ¼ f tð Þ of the fractional integral and deriv-

<sup>t</sup> u∈Cð Þ ½ � 0; τ ; X

$$\begin{split} \| |T^2(\boldsymbol{u}(t)) - T^2(\boldsymbol{v}(t)) | &\leq \frac{t^\alpha}{\Gamma(\alpha+1)} \| \boldsymbol{B} \|\_{L(X)} \sup\_{0 \leq s \leq \pi} \| |T\boldsymbol{u}(s) - T\boldsymbol{v}(s)| \| \\ &\leq \frac{t^2 \alpha}{\Gamma^2(\alpha+1)} \| \boldsymbol{B} \|\_{L(X)}^2 \sup\_{0 \leq s \leq \pi} \| |\boldsymbol{u}(s) - \boldsymbol{v}(s)| \| \\ \| |T^3(\boldsymbol{u}(t)) - T^3(\boldsymbol{v}(t))| &\leq \frac{t^\alpha}{\Gamma(\alpha+1)} \| \boldsymbol{B} \|\_{L(X)} \sup\_{0 \leq s \leq \pi} \| |T^2\boldsymbol{u}(s) - T^2\boldsymbol{v}(s)| \| \\ &\leq \frac{t^3 \alpha}{\Gamma^3(\alpha+1)} \| \boldsymbol{B} \|\_{L(X)}^3 \sup\_{0 \leq s \leq \pi} \| \boldsymbol{u}(s) - \boldsymbol{v}(s) \| \| \\ &\leq \dots \end{split}$$
 
$$\| |T^n(\boldsymbol{u}(t)) - T^n(\boldsymbol{v}(t)) | \| \leq \frac{t^n \alpha}{\Gamma^n(\alpha+1)} \| \boldsymbol{B} \|\_{L(X)}^n \sup\_{0 \leq s \leq \pi} \| \boldsymbol{u}(s) - \boldsymbol{v}(s) \| \| $$

and for an sufficiently large n,the constant in question is less than 1, i.e., there exists a fixed point u∈Kτ. Observe now that τ > 0 was an arbitrary choice, so we conclude that the fixed point u∈ Cð Þ ½ � 0; τ ; X for all τ > 0 and Lemma (3.9), we obtain the existence and uniqueness of a global solution to the problem (3.30). □

Corollary 3.11. ref. [24] Consider the same hypothesis of theorem (3.10).

i. Let f gj Unð Þ<sup>t</sup> <sup>∞</sup> <sup>n</sup>¼<sup>0</sup> be a sequence of continuous functions Un : ½ Þ! <sup>0</sup>; <sup>∞</sup> <sup>X</sup> given by <sup>U</sup>0ðÞ¼ <sup>t</sup> <sup>u</sup>0, Un <sup>¼</sup> <sup>u</sup><sup>0</sup> <sup>þ</sup> <sup>1</sup> Γð Þ α Ðt <sup>0</sup> ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup> BUn�<sup>1</sup>ð Þs ds, n∈ f g 1; 2; … .

Then there exists a continuous function U : ½ Þ! 0; ∞ X, such that for any τ > 0, we conclude that Un ! U in Cð Þ ½ � 0; τ ; X . Moreover, U tð Þ is the unique global solution of (3.30).

ii. It holds that

$$U(t) = \sum\_{k=0}^{\infty} \frac{(t^{\alpha}B)^k u\_0}{\Gamma(\alpha k + 1)}.$$

Proof. (i) It follows directly from proof of Theorem (3.10).

(ii) It is trivial that U0ðÞ¼ t u0. So we compute, using the gamma function properties, that

$$\mathcal{U}\_1(t) = \mu\_0 + \frac{1}{\Gamma(\alpha)} \int\_0^t (t - s)^{\alpha - 1} B \mu\_0(s) ds = \mu\_0 + \frac{t^{\alpha} B \mu\_0}{\alpha \Gamma(\alpha)} = \mu\_0 + \frac{t^{\alpha} B \mu\_0}{\Gamma(\alpha + 1)}.$$

By a simple induction process, we conclude that

$$\mathcal{U}\_n(t) = \sum\_{k=0}^n \frac{(t^\alpha B)^k u\_0}{\Gamma(\alpha k + 1)}$$

and therefore

$$MI(t) = \lim\_{n \to \infty} \sum\_{k=0}^{n} \frac{\left(t^{\alpha}B\right)^{k} \mu\_{0}}{\Gamma(\alpha k + 1)} = \sum\_{k=0}^{\infty} \frac{\left(t^{\alpha}B\right)^{k} \mu\_{0}}{\Gamma(\alpha k + 1)} \coloneqq E\_{\alpha}(t^{\alpha}B)\mu\_{0}.$$

□

I α <sup>0</sup>þD<sup>α</sup>

for some Ci ∈ R, i ¼ 1, 2, ⋯, N, N ¼ ½ �þ α 1.

boundary value condition.

has a unique solution

G tð Þ¼ ;s

Take (3.40) into (3.39), we have

u tðÞ¼� <sup>ð</sup><sup>1</sup>

0

8 >>><

>>>:

where

<sup>0</sup>þu tðÞ¼ u tðÞ� <sup>C</sup>1<sup>t</sup>

Dα

½ � <sup>t</sup>ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

½ � <sup>t</sup>ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

for some C1, C<sup>2</sup> ∈ R. Therefore, the general solution of (3.36) is

u tðÞ¼� <sup>ð</sup><sup>1</sup>

C<sup>1</sup> ¼ ð1 0

<sup>Γ</sup>ð Þ <sup>α</sup> y sð Þds <sup>þ</sup> <sup>t</sup>

ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

By uð Þ¼ 0 0, we can get C<sup>2</sup> ¼ 0. In addition, uð Þ¼� 1

ð Þ α � 1 Γð Þ α

u tðÞ¼�I

0

<sup>α</sup>�<sup>1</sup> � <sup>C</sup>2<sup>t</sup>

ð1 0

In the following, we present the Green function of fractional differential equation with integral

Theorem 3.14. ref. [26] Let 1 < α < 2, Assume y tð Þ∈ C½ � 0; 1 , then the following equation

uð Þ¼ 0 0, uð Þ¼ 1

u tðÞ¼ <sup>ð</sup><sup>1</sup>

0

ð Þ� <sup>α</sup> � <sup>1</sup> <sup>þ</sup> <sup>s</sup> ½ � <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

ð Þ α � 1 Γð Þ α

Proof. We may apply Lemma (3.13) to reduce Eq. (3.36) to an equivalent integral equation

<sup>0</sup>þy tð Þþ <sup>C</sup>1<sup>t</sup>

<sup>Γ</sup>ð Þ <sup>α</sup> y sð Þds <sup>þ</sup> <sup>C</sup>1<sup>t</sup>

<sup>Γ</sup>ð Þ <sup>α</sup> y sð Þds <sup>þ</sup>

α�1 ð1 0

ð Þ α � 1 þ s

α

ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

<sup>α</sup>�<sup>2</sup> � <sup>⋯</sup> � CNt

Fixed Point Theory Approach to Existence of Solutions with Differential Equations

<sup>0</sup>þu tðÞþ y tðÞ¼ <sup>0</sup>, <sup>0</sup> <sup>&</sup>lt; <sup>t</sup> <sup>&</sup>lt; <sup>1</sup> (3.36)

ð Þ α � 1

α�2

<sup>α</sup>�<sup>1</sup> <sup>þ</sup> <sup>C</sup>2<sup>t</sup>

ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

<sup>Γ</sup>ð Þ <sup>α</sup> y sð Þds <sup>þ</sup> <sup>t</sup>

α�2

<sup>Γ</sup>ð Þ <sup>α</sup> y sð Þds <sup>þ</sup> <sup>C</sup>1, it follows

α�1 ð1 0

u sð Þds: (3.40)

<sup>α</sup>�<sup>1</sup> <sup>þ</sup> <sup>C</sup>2<sup>t</sup>

ðt 0

ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

ð1 0

u sð Þds, (3.37)

if 0 ≤ s ≤ t ≤ 1

if 0 ≤ t ≤ s ≤ 1:

: (3.39)

u sð Þds: (3.41)

G tð Þ ;s y sð Þds (3.38)

<sup>α</sup>�<sup>N</sup> (3.35)

29

http://dx.doi.org/10.5772/intechopen.74560

From the above works, we can see a fact, although the fractional boundary value problems have been studied, to the best of our knowledge, there have been a few works using the lower and upper solution method. However, only positive solution are useful for many application, motivated by the above works, we study the existence and uniqueness of positive solution of the following integral boundary value problem.

$$D\_{0+}^{a}u(t) + f(t, u(t)) = 0, \ 0 < t < 1, \ 1 < a \le 2 \tag{3.32}$$

$$
\mu(0) = 0, \ \mu(1) = \int\_0^1 \mu(s) ds,\tag{3.33}
$$

where <sup>f</sup> : ½ �� <sup>0</sup>; <sup>1</sup> ½ Þ! <sup>0</sup>; <sup>∞</sup> ½ Þ <sup>0</sup>; <sup>∞</sup> is a continuous function and <sup>D</sup><sup>α</sup> <sup>0</sup><sup>þ</sup> is the standard Riemann-Liouville fractional derivative.

We need the following lemmas that will be used to prove our main results.

Lemma 3.12. ref. [25] Let α > 0 and u∈ Cð Þ 0; 1 ∩ Lð Þ 0; 1 . Then fractional differential equation

$$D\_{0+}^{\\\alpha}u(t) = 0$$

has

$$\mu(t) = \mathbb{C}\_1 t^{\alpha - 1} + \mathbb{C}\_2 t^{\alpha - 2} + \dots + \mathbb{C}\_N t^{\alpha - N},\tag{3.34}$$

Ci ∈ R, i ¼ 1, 2, ⋯, N, N ¼ ½ �þ α 1 as unique solution.

Lemma 3.13. ref. [25] Assume that u∈Cð Þ 0; 1 ∩ Lð Þ 0; 1 with a fractional derivative of order α > 0 that belongs to Cð Þ 0; 1 ∩ Lð Þ 0; 1 . Then

Fixed Point Theory Approach to Existence of Solutions with Differential Equations http://dx.doi.org/10.5772/intechopen.74560 29

$$I\_{0+}^{a}D\_{0+}^{a}\mu(t) = \mu(t) - \mathsf{C}\_{1}t^{a-1} - \mathsf{C}\_{2}t^{a-2} - \dots - \mathsf{C}\_{N}t^{a-N} \tag{3.35}$$

for some Ci ∈ R, i ¼ 1, 2, ⋯, N, N ¼ ½ �þ α 1.

In the following, we present the Green function of fractional differential equation with integral boundary value condition.

Theorem 3.14. ref. [26] Let 1 < α < 2, Assume y tð Þ∈ C½ � 0; 1 , then the following equation

$$D\_{0+}^{a}u(t) + y(t) = 0, \ 0 < t < 1\tag{3.36}$$

$$
\mu(0) = 0, \ \mu(1) = \int\_0^1 \mu(s) ds,\tag{3.37}
$$

has a unique solution

$$u(t) = \int\_0^1 G(t, s) y(s) ds\tag{3.38}$$

where

Proof. (i) It follows directly from proof of Theorem (3.10).

1 Γð Þ α

By a simple induction process, we conclude that

U tðÞ¼ lim<sup>n</sup>!<sup>∞</sup>

the following integral boundary value problem.

Liouville fractional derivative.

has

Dα

where <sup>f</sup> : ½ �� <sup>0</sup>; <sup>1</sup> ½ Þ! <sup>0</sup>; <sup>∞</sup> ½ Þ <sup>0</sup>; <sup>∞</sup> is a continuous function and <sup>D</sup><sup>α</sup>

u tðÞ¼ C1t

Ci ∈ R, i ¼ 1, 2, ⋯, N, N ¼ ½ �þ α 1 as unique solution.

α > 0 that belongs to Cð Þ 0; 1 ∩ Lð Þ 0; 1 . Then

We need the following lemmas that will be used to prove our main results.

Xn k¼0

t <sup>α</sup> ð Þ <sup>B</sup> <sup>k</sup> u0 <sup>Γ</sup>ð Þ <sup>α</sup><sup>k</sup> <sup>þ</sup> <sup>1</sup> <sup>¼</sup> <sup>X</sup><sup>∞</sup>

ðt 0

ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

U1ðÞ¼ t u<sup>0</sup> þ

28 Differential Equations - Theory and Current Research

and therefore

(ii) It is trivial that U0ðÞ¼ t u0. So we compute, using the gamma function properties, that

UnðÞ¼ <sup>t</sup> <sup>X</sup><sup>n</sup>

k¼0

Bu0ð Þs ds ¼ u<sup>0</sup> þ

t <sup>α</sup> ð Þ <sup>B</sup> <sup>k</sup> u0 Γð Þ αk þ 1

k¼0

From the above works, we can see a fact, although the fractional boundary value problems have been studied, to the best of our knowledge, there have been a few works using the lower and upper solution method. However, only positive solution are useful for many application, motivated by the above works, we study the existence and uniqueness of positive solution of

uð Þ¼ 0 0, uð Þ¼ 1

Lemma 3.12. ref. [25] Let α > 0 and u∈ Cð Þ 0; 1 ∩ Lð Þ 0; 1 . Then fractional differential equation

<sup>0</sup>þu tðÞ¼ <sup>0</sup>

Lemma 3.13. ref. [25] Assume that u∈Cð Þ 0; 1 ∩ Lð Þ 0; 1 with a fractional derivative of order

<sup>α</sup>�<sup>2</sup> <sup>þ</sup> <sup>⋯</sup> <sup>þ</sup> CNt

Dα

<sup>α</sup>�<sup>1</sup> <sup>þ</sup> <sup>C</sup>2<sup>t</sup>

t <sup>α</sup> ð Þ <sup>B</sup> <sup>k</sup> u0 <sup>Γ</sup>ð Þ <sup>α</sup><sup>k</sup> <sup>þ</sup> <sup>1</sup> <sup>≔</sup>E<sup>α</sup> <sup>t</sup>

<sup>0</sup>þu tð Þþ f tð Þ¼ ; u tð Þ <sup>0</sup>, <sup>0</sup> <sup>&</sup>lt; <sup>t</sup> <sup>&</sup>lt; <sup>1</sup>, <sup>1</sup> <sup>&</sup>lt; <sup>α</sup> <sup>≤</sup> <sup>2</sup> (3.32)

ð1 0

t <sup>α</sup>Bu<sup>0</sup> <sup>α</sup>Γð Þ <sup>α</sup> <sup>¼</sup> <sup>u</sup><sup>0</sup> <sup>þ</sup>

t <sup>α</sup>Bu<sup>0</sup> <sup>Γ</sup>ð Þ <sup>α</sup> <sup>þ</sup> <sup>1</sup> :

<sup>α</sup> ð Þ <sup>B</sup> <sup>u</sup>0:

u sð Þds, (3.33)

<sup>0</sup><sup>þ</sup> is the standard Riemann-

<sup>α</sup>�N, (3.34)

□

$$G(t,s) = \begin{cases} \frac{[t(1-s)]^{\alpha-1}(\alpha-1+s)-[t-s]^{\alpha-1}(\alpha-1)}{(\alpha-1)\Gamma(\alpha)} & \text{if } 0 \le s \le t \le 1\\\frac{[t(1-s)]^{\alpha-1}(\alpha-1+s)}{(\alpha-1)\Gamma(\alpha)} & \text{if } 0 \le t \le s \le 1. \end{cases}$$

Proof. We may apply Lemma (3.13) to reduce Eq. (3.36) to an equivalent integral equation

$$u(t) = -I\_{0+}^{\alpha}y(t) + \mathcal{C}\_1 t^{\alpha - 1} + \mathcal{C}\_2 t^{\alpha - 2}$$

for some C1, C<sup>2</sup> ∈ R. Therefore, the general solution of (3.36) is

$$u(t) = -\int\_0^1 \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} y(s)ds + \mathcal{C}\_1 t^{\alpha-1} + \mathcal{C}\_2 t^{\alpha-2}.\tag{3.39}$$

By uð Þ¼ 0 0, we can get C<sup>2</sup> ¼ 0. In addition, uð Þ¼� 1 ðt 0 ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup> <sup>Γ</sup>ð Þ <sup>α</sup> y sð Þds <sup>þ</sup> <sup>C</sup>1, it follows

$$\mathcal{C}\_1 = \int\_0^1 \frac{\left(t - s\right)^{\alpha - 1}}{\Gamma(\alpha)} y(s) ds + \int\_0^1 u(s) ds. \tag{3.40}$$

Take (3.40) into (3.39), we have

$$u(t) = -\int\_0^1 \frac{(t-s)^{a-1}}{\Gamma(a)} y(s) ds + t^{a-1} \int\_0^1 \frac{(1-s)^{a-1}}{\Gamma(a)} y(s) ds + t^{a-1} \int\_0^1 u(s) ds. \tag{3.41}$$

Let Ð <sup>1</sup> <sup>0</sup> u sð Þds ¼ A,by (3.41), we can get

$$\begin{split} \int\_{0}^{1} u(t)dt &= -\int\_{0}^{1} \int\_{0}^{t} \frac{(t-s)^{a-1}}{\Gamma(\alpha)} y(s) ds dt + \int\_{0}^{1} t^{a-1} \int\_{0}^{t} \frac{(1-s)^{a-1}}{\Gamma(\alpha)} y(s) ds dt + A \int\_{0}^{1} t^{a-1} dt \\ &= -\int\_{0}^{1} \frac{(1-s)^{a}}{a \Gamma(\alpha)} y(s) ds + \int\_{0}^{1} \frac{(1-s)^{a-1}}{a \Gamma(\alpha)} y(s) ds + \frac{A}{\alpha} \\ &= \int\_{0}^{1} \frac{s(1-s)^{a-1}}{a \Gamma(\alpha)} y(s) ds + \frac{A}{\alpha} .\end{split}$$

In fact, by the definition of Green function G tð Þ ;s , for u, v∈ P, we have the estimate

<sup>½</sup>tð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

α�1

G tð Þ ;s ∣f sð Þ� ; u sð Þ f sð Þ ; v sð Þ ∣ds

ð Þ α � 1 þ s

ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

Kt<sup>α</sup>�<sup>1</sup>

G tð Þ ;s ∣f sð Þ� ; Tu sð Þ f sð Þ ; Tv sð Þ ∣ds

G tð Þ ;s a sð Þ∣Tu sð Þ� Tv sð Þ∣ds

Γ2 ð Þ α

ku � vk

KH<sup>n</sup>�<sup>1</sup> t α�1

ð Þ <sup>α</sup> � <sup>1</sup> <sup>n</sup>

ð Þ α � 1 þ s a sð Þds. By mathematical induction, it follows

<sup>Γ</sup><sup>n</sup>ð Þ <sup>α</sup> <sup>k</sup><sup>u</sup> � <sup>v</sup><sup>k</sup>

H ð Þ α � 1 Γð Þ α � �<sup>n</sup>�<sup>1</sup>

ð1 0 s α�1

G tð Þ ;<sup>s</sup> a sð Þ Ks<sup>α</sup>�<sup>1</sup>

ð Þ <sup>α</sup> � <sup>1</sup> <sup>2</sup>

α�1

K t½ � ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

Γ2 ð Þ α

Γ2 ð Þ α

Kku � vkt

ð Þ <sup>α</sup> � <sup>1</sup> <sup>2</sup>

<sup>¼</sup> KHt<sup>α</sup>�<sup>1</sup> ð Þ <sup>α</sup> � <sup>1</sup> <sup>2</sup>

<sup>∣</sup>Tnu tð Þ� <sup>T</sup>nv tð Þ<sup>∣</sup> <sup>≤</sup>

<sup>Γ</sup><sup>n</sup>ð Þ <sup>α</sup> <sup>¼</sup> <sup>K</sup>

ð Þ α � 1 Γð Þ α

ð Þ <sup>α</sup> � <sup>1</sup> <sup>Γ</sup>ð Þ <sup>α</sup> <sup>k</sup><sup>u</sup> � <sup>v</sup>k:

ð Þ <sup>α</sup> � <sup>1</sup> <sup>Γ</sup>ð Þ <sup>α</sup> <sup>k</sup><sup>u</sup> � <sup>v</sup>kds

ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

a sð Þs α�1

ku � vkds

ð Þ α � 1 þ s a sð Þds

< 1:

ð Þ α � 1 þ s

a sð Þku � vkds

Fixed Point Theory Approach to Existence of Solutions with Differential Equations

http://dx.doi.org/10.5772/intechopen.74560

31

ð Þ α � 1 þ s a sð Þds:

G tð Þ ;s a sð Þ∣u sð Þ� v sð Þ∣ds

ð Þ α � 1 Γð Þ α

ð1 0

ð1 0

≤ ð1 0

≤ ð1 0

<sup>≤</sup> <sup>k</sup><sup>u</sup> � <sup>v</sup>k<sup>t</sup>

ð Þ α � 1 þ s a sð Þds, then

∣Tu tð Þ� Tv tð Þ∣ ≤

ð1 0

ð Þ α � 1 Γð Þ α

∣Tu tð Þ� Tv tð Þ∣ ¼

Denote K ¼

Similarly,

where H ¼

Hence, it holds

ð1 0 s α�1

ð1 0

∣T<sup>2</sup>

u tð Þ� <sup>T</sup><sup>2</sup>

ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

KH<sup>n</sup>�<sup>1</sup> t α�1

ð Þ <sup>α</sup> � <sup>1</sup> <sup>n</sup>

by (3.43), for n large enough, we have

v tð Þ∣ ¼

≤ ð1 0

≤ ð1 0

≤ ð1 0

≤

ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

So,

$$A = \frac{\alpha}{\alpha - 1} \int\_0^1 \frac{s(1 - s)^{\alpha - 1}}{\alpha \Gamma(\alpha)} y(s) ds = \int\_0^1 \frac{s(1 - s)^{\alpha - 1}}{(\alpha - 1)\Gamma(\alpha)} y(s) ds.$$

Combine with (3.41), we have

$$\begin{split} u(t) &= -\int\_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} y(s) ds + t^{\alpha-1} \int\_0^1 \frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)} y(s) ds + t^{\alpha-1} \int\_0^1 \frac{s(1-s)^{\alpha-1}}{(\alpha-1)\Gamma(\alpha)} y(s) ds \\ &= -\int\_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} y(s) ds + \int\_0^1 \frac{[t(1-s)^{\alpha-1}(\alpha-1+s)]}{(\alpha-1)\Gamma(\alpha)} y(s) ds \\ &= \int\_0^1 \frac{[t(1-s)^{\alpha-1}(\alpha-1+s) - (t-s)^{\alpha-1}(\alpha-1)]}{(\alpha-1)\Gamma(\alpha)} y(s) ds + \int\_t^1 \frac{[t(1-s)^{\alpha-1}(\alpha-1+s)]}{(\alpha-1)\Gamma(\alpha)} y(s) ds \\ &= \int\_0^1 G(t,s) y(s) ds. \end{split}$$

This complete the proof.

Remark 3.15. Obviously, the Green function G tð Þ ;s satisfies the following properties:

i. G tð Þ ;s > 0, t, s ∈ð Þ 0; 1 ; ii. G tð Þ ;<sup>s</sup> <sup>≤</sup> <sup>2</sup> ð Þ <sup>α</sup>�<sup>1</sup> <sup>Γ</sup>ð Þ <sup>α</sup> ; <sup>0</sup> <sup>≤</sup> t, s <sup>≤</sup> 1.

Theorem 3.16. ref. [26] Assume that function f satisfies

$$|f(t,u) - f(t,v)| \le a(t)|u - v|\tag{3.42}$$

where t∈ ½ � 0; 1 , u, v∈ ½ Þ 0; ∞ , a : ½ �! 0; 1 ½ Þ 0; ∞ is a continuous function. If

$$\int\_0^1 s^{\alpha - 1} (\alpha - 1 + s) a(s) ds < (\alpha - 1) \Gamma(\alpha) \tag{3.43}$$

then the Eq. (3.32) has a unique positive solution.

Proof. If T<sup>n</sup> is a contraction operator for n sufficiently large, then the Eq. (3.32) has a unique positive solution.

In fact, by the definition of Green function G tð Þ ;s , for u, v∈ P, we have the estimate

$$\begin{aligned} |Tu(t) - Tv(t)| &= \int\_0^1 G(t, s) |f(s, u(s)) - f(s, v(s))| ds \\ &\le \int\_0^1 G(t, s) a(s) |u(s) - v(s)| ds \\ &\le \int\_0^1 \frac{[t(1 - s)^{\alpha - 1}(\alpha - 1 + s)}{(\alpha - 1)\Gamma(\alpha)} a(s) ||u - v|| ds \\ &\le \frac{||u - v||t^{\alpha - 1}}{(\alpha - 1)\Gamma(\alpha)} \int\_0^1 (1 - s)^{\alpha - 1}(\alpha - 1 + s) a(s) ds. \end{aligned}$$

Denote K ¼ ð1 0 ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup> ð Þ α � 1 þ s a sð Þds, then

$$|Tu(t) - Tv(t)| \le \frac{Kt^{\alpha - 1}}{(\alpha - 1)\Gamma(\alpha)}||u - v||.$$

Similarly,

Let Ð <sup>1</sup>

So,

u tðÞ¼�

¼ � ðt 0

¼ ð1 0

¼ ð1 0

ii. G tð Þ ;<sup>s</sup> <sup>≤</sup> <sup>2</sup>

positive solution.

This complete the proof.

i. G tð Þ ;s > 0, t, s ∈ð Þ 0; 1 ;

ð1 0

<sup>0</sup> u sð Þds ¼ A,by (3.41), we can get

30 Differential Equations - Theory and Current Research

ð1 0 ðt 0 ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

<sup>α</sup>Γð Þ <sup>α</sup> y sð Þds <sup>þ</sup>

<sup>α</sup>Γð Þ <sup>α</sup> y sð Þds <sup>þ</sup>

ð1 0

> α�1 ð1 0

ð1 0

h i

ð Þ� <sup>α</sup> � <sup>1</sup> <sup>þ</sup> <sup>s</sup> ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>

<sup>s</sup>ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

<sup>A</sup> <sup>¼</sup> <sup>α</sup> α � 1

<sup>Γ</sup>ð Þ <sup>α</sup> y sð Þds <sup>þ</sup> <sup>t</sup>

<sup>Γ</sup>ð Þ <sup>α</sup> y sð Þds <sup>þ</sup>

ð Þ <sup>α</sup>�<sup>1</sup> <sup>Γ</sup>ð Þ <sup>α</sup> ; <sup>0</sup> <sup>≤</sup> t, s <sup>≤</sup> 1.

Theorem 3.16. ref. [26] Assume that function f satisfies

ð1 0 s α�1

then the Eq. (3.32) has a unique positive solution.

<sup>Γ</sup>ð Þ <sup>α</sup> y sð Þdsdt <sup>þ</sup>

<sup>s</sup>ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

ð1 0

> A α :

<sup>α</sup>Γð Þ <sup>α</sup> y sð Þds <sup>¼</sup>

ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

h i

ð Þ <sup>α</sup> � <sup>1</sup> <sup>Γ</sup>ð Þ <sup>α</sup> y sð Þds <sup>þ</sup>

Remark 3.15. Obviously, the Green function G tð Þ ;s satisfies the following properties:

Proof. If T<sup>n</sup> is a contraction operator for n sufficiently large, then the Eq. (3.32) has a unique

where t∈ ½ � 0; 1 , u, v∈ ½ Þ 0; ∞ , a : ½ �! 0; 1 ½ Þ 0; ∞ is a continuous function. If

<sup>t</sup>ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

<sup>Γ</sup>ð Þ <sup>α</sup> y sð Þds <sup>þ</sup> <sup>t</sup>

ð Þ α � 1

ð Þ α � 1 þ s

ð Þ <sup>α</sup> � <sup>1</sup> <sup>Γ</sup>ð Þ <sup>α</sup> y sð Þds

ð1 0 t α�1 ðt 0

ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

<sup>α</sup>Γð Þ <sup>α</sup> y sð Þds <sup>þ</sup>

ð1 0 ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

A α

<sup>s</sup>ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup> ð Þ <sup>α</sup> � <sup>1</sup> <sup>Γ</sup>ð Þ <sup>α</sup> y sð Þds:

> α�1 ð1 0

> > ð1 t

∣f tð Þ� ; u f tð Þ ; v ∣ ≤ a tð Þ∣u � v∣ (3.42)

ð Þ α � 1 þ s a sð Þds < ð Þ α � 1 Γð Þ α (3.43)

<sup>Γ</sup>ð Þ <sup>α</sup> y sð Þdsdt <sup>þ</sup> <sup>A</sup>

<sup>s</sup>ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup> ð Þ <sup>α</sup> � <sup>1</sup> <sup>Γ</sup>ðÞ<sup>α</sup> y sð Þds

<sup>½</sup>tð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

ð Þ α � 1 þ s ð Þ <sup>α</sup> � <sup>1</sup> <sup>Γ</sup>ð Þ <sup>α</sup> y sð Þds

ð1 0 t α�1 dt

u tð Þdt ¼ �

¼ � ð1 0

¼ ð1 0

Combine with (3.41), we have

ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

ð Þ <sup>t</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

<sup>t</sup>ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup>

G tð Þ ;s y sð Þds:

ðt 0

$$\begin{split} |T^2u(t) - T^2v(t)| &= \int\_0^1 G(t,s)[f(s,Tu(s)) - f(s,Tv(s))]ds \\ &\le \int\_0^1 G(t,s)a(s)|Tu(s) - Tv(s)|ds \\ &\le \int\_0^1 G(t,s)a(s)\frac{Ks^{\alpha-1}}{(\alpha-1)\Gamma(\alpha)}||u-v||ds \\ &\le \int\_0^1 \frac{K|t(1-s)|^{\alpha-1}(\alpha-1+s)}{(\alpha-1)^2\Gamma^2(\alpha)}a(s)s^{\alpha-1}||u-v||ds \\ &\le \frac{K||u-v||t^{\alpha-1}}{(\alpha-1)^2\Gamma^2(\alpha)}\int\_0^1 s^{\alpha-1}(1-s)^{\alpha-1}(\alpha-1+s)a(s)ds \\ &= \frac{KHt^{\alpha-1}}{(\alpha-1)^2\Gamma^2(\alpha)}||u-v|| \end{split}$$

where H ¼ ð1 0 s α�1 ð Þ <sup>1</sup> � <sup>s</sup> <sup>α</sup>�<sup>1</sup> ð Þ α � 1 þ s a sð Þds. By mathematical induction, it follows <sup>∣</sup>Tnu tð Þ� <sup>T</sup>nv tð Þ<sup>∣</sup> <sup>≤</sup> KH<sup>n</sup>�<sup>1</sup> t α�1 ð Þ <sup>α</sup> � <sup>1</sup> <sup>n</sup> <sup>Γ</sup><sup>n</sup>ð Þ <sup>α</sup> <sup>k</sup><sup>u</sup> � <sup>v</sup><sup>k</sup>

by (3.43), for n large enough, we have

$$\frac{K H^{n-1} t^{a-1}}{(\alpha - 1)^{\
u} \Gamma^{\boldsymbol{u}}(\alpha)} = \frac{K}{(\alpha - 1)\Gamma(\alpha)} \left( \frac{H}{(\alpha - 1)\Gamma(\alpha)} \right)^{n-1} < 1.$$

Hence, it holds

$$\|\|T^\nu u - T^\nu v\|\| < \|u - v\|\|\_\nu$$

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which implies T<sup>n</sup> is a contraction operator for n sufficiently large, then the Eq. (3.32) has a unique positive solution. □
