**2.3 Third case: Δ** ¼ **0**

If the discriminant vanishes ð Þ Δ ¼ 0 , then *q*<0 and the only solution of problem (4) with

$$\left(\mathbf{x}'(\mathbf{0})\right)^2 = \dot{\mathbf{x}}\_0^2 = \frac{\left(p + qy\_0^2\right)^2}{-2q},\tag{29}$$

reads

$$\mathbf{x}(t) = \sqrt{-\frac{p}{q}} \tanh\left[\sqrt{\frac{p}{2}}t \pm \tanh^{-1}\left(\kappa\_0 \sqrt{-\frac{q}{p}}\right)\right].\tag{30}$$

which may be verified by direct computation.

**Figure 2.** *The profile of solution (27) is plotted against t.*

*Engineering Problems - Uncertainties, Constraints and Optimization Techniques*

**Remark 1**. The solution of the i.v.p.

$$\begin{cases} \ddot{\boldsymbol{\kappa}} + p\boldsymbol{\kappa} + q\boldsymbol{\kappa}^3 = \mathbf{0}, \\ \boldsymbol{\kappa}(\mathbf{0}) = \boldsymbol{\kappa}\_0 \, \mathop{\mathbf{\mathcal{E}}}\boldsymbol{\kappa}'(\mathbf{0}) = \mathbf{0}, \end{cases} \tag{31}$$

is given by

$$\mathbf{x}(t) = \mathbf{x}\_0 \mathbf{cn}\left(\sqrt{p + q\mathbf{x}\_0^2}, \frac{q\mathbf{x}\_0^2}{2(p + q\mathbf{x}\_0^2)}\right). \tag{32}$$

**Remark 2**. For *<sup>p</sup>* <sup>þ</sup> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi *<sup>p</sup>*<sup>2</sup> <sup>þ</sup> <sup>2</sup>*<sup>q</sup>* <sup>p</sup> x\_0 <sup>2</sup> <sup>&</sup>gt; 0, then the solution of the i.v.p.

$$\begin{cases} \ddot{\mathbf{x}} + p\mathbf{x} + q\mathbf{x}^3 = \mathbf{0}, \\ \varkappa(\mathbf{0}) = \mathbf{0} \not\approx \varkappa'(\mathbf{0}) = \dot{\mathbf{x}}\_0. \end{cases} \tag{33}$$

is given by

$$\mathbf{x}(t) = \frac{\sqrt{2}\dot{\mathbf{x}}\_0}{\sqrt{\sqrt{p^2 + 2q\dot{\mathbf{x}}\_0^2} + p}} \text{sn}\left(\sqrt{\frac{p + \sqrt{p^2 + 2q\dot{\mathbf{x}}\_0^2}}{2}}, -\frac{p^2 + q\dot{\mathbf{x}}\_0^2 - \sqrt{p^2 + 2q\dot{\mathbf{x}}\_0^2}p}{q\dot{\mathbf{x}}\_0^2}\right). \tag{34}$$

**Remark 3**. According to the following identity

$$\text{cm}(\sqrt{\alpha}t, m) = 1 - \frac{\text{S}\_0}{1 + \text{S}\_1 \wp(t; \text{g}\_2, \text{g}\_3)},\tag{35}$$

with

$$\begin{aligned} S\_0 &= \frac{6}{(4m+1)}, S\_1 = \frac{12}{(4m+1)\alpha}, \\\ g\_2 &= \frac{1}{12} \left( 16m^2 - 16m + 1 \right) \alpha^2, \\\ g\_3 &= \frac{1}{216} (2m-1) \left( 32m^2 - 32m - 1 \right) \alpha^3, \end{aligned}$$

the solution of the i.v.p. (4)-(5) could be written in terms of the Weierstrass elliptic function ℘ � ℘ *t*; *g*2, *g*<sup>3</sup> � �. More precisely, if Δ >0 then

$$\chi(t) = A - \frac{A\left(\frac{4p}{3A^2q + p} + 2\right)}{1 + \frac{12}{3A^2q + p}\wp\left(t + t\_0; \mathbf{g}\_2, \mathbf{g}\_3\right)},\tag{36}$$

with

$$\begin{aligned} t\_0 &= \mathfrak{G}^{-1} \Big( \frac{3A^3 q + 3A^2 q \varkappa\_0 + 5Ap + p\varkappa\_0}{12(A - \varkappa\_0)}; \mathfrak{g}\_2, \mathfrak{g}\_3 \Big), \\ \mathfrak{g}\_2 &= \frac{1}{12} \Big( -3A^4 q^2 - 6A^2 pq + p^2 \Big), \\ \mathfrak{g}\_3 &= \frac{p}{216} \left( 9A^4 q^2 + 18A^2 pq + p^2 \right), \end{aligned} \tag{37}$$

*Analytical Solutions of Some Strong Nonlinear Oscillators DOI: http://dx.doi.org/10.5772/intechopen.97677*

and

$$A = \sqrt{\frac{-p \pm \sqrt{\left(p + q \mathbf{x}\_0^2\right)^2 + 2q \dot{\mathbf{x}}\_0^2}}{q}} = \pm \sqrt{\frac{-p \pm \sqrt{\Delta}}{q}}.\tag{38}$$

The solution (36) is periodic with period

$$T = 2\int\_{\rho}^{+\infty} \frac{d\mathbf{x}}{\sqrt{4\mathbf{x}^3 - \mathbf{g}\_2\mathbf{x} - \mathbf{g}\_3}},\tag{39}$$

where *<sup>ρ</sup>* is the greatest real root of the cubic 4*x*<sup>3</sup> � *<sup>g</sup>*2*<sup>x</sup>* � *<sup>g</sup>*<sup>3</sup> <sup>¼</sup> 0. **Remark 4**. An approximate analytic solution of the i.v.p. (31) is given by

$$\mathbf{x}(t) = \frac{\mathbf{x}\_0 \sqrt{\mathbf{1} + \lambda} \cos \left( wt \right)}{\sqrt{\mathbf{1} + \lambda} \cos^2 \left( wt \right)},\tag{40}$$

where

$$w = \frac{1}{2} \sqrt{\frac{5(p + qx\_0^2)\lambda^2 + (12p + 11qx\_0^2)\lambda + 8p + 6qx\_0^2}{3\lambda + 2}}\tag{41}$$

and *λ* is a root of the cubic

$$2\mathfrak{S}(p+q\mathbf{x}\_0^2)\lambda^3 + (\mathfrak{S}8p + \mathfrak{S}9q\mathbf{x}\_0^2)\lambda^2 + 2(\mathfrak{L}6p + 2\mathfrak{L}q\mathbf{x}\_0^2)\lambda + \mathfrak{S}q\mathbf{x}\_0^2 = \mathbf{0}.\tag{42}$$

#### **Example 3.**

Let us consider the i.v.p.

$$\begin{cases} \ddot{\mathbf{x}} + \mathbf{x} + \mathbf{10}x^3 = \mathbf{0}, \\ \varkappa(\mathbf{0}) = 4, \varkappa'(\mathbf{0}) = \mathbf{0}. \end{cases} \tag{43}$$

The approximate solution in trigonometric form is given by

$$\propto\_{\rm app}(t) = \frac{3.34603 \cos \left(10.7542t\right)}{\sqrt{1 - 0.300255 \cos^2 \left(10.7542t\right)}} \,\mathrm{.}\tag{44}$$

The exact solution reads

$$\mathbf{x}(t) = 4\mathbf{cn}\left(\sqrt{161}t|\frac{80}{161}\right),\tag{45}$$

with period

$$T = \frac{4K\left(\frac{80}{161}\right)}{\sqrt{161}}.$$

The error on the interval 0 ≤*t*≤ *T* equals 0*:*025*:*

The comparison between the approximate analytic solution (44) and the exact analytic solution (45) is illustrated in **Figure 3**.

**Figure 3.** *A comparison between the approximate solution (44) and exact solution (45).*

**Remark 5**. An approximate analytical solution of the i.v.p. (33) is given by

$$\propto\_{\text{app}}(t) = \frac{\dot{\mathbf{x}}\_0 \cdot \sin\left(\sqrt{\alpha}t\right)}{\sqrt{\alpha}\sqrt{1 + \lambda\sin^2\left(\sqrt{\alpha}t\right)}},\tag{46}$$

where

$$\rho = -\frac{\sqrt{\lambda^2 \left(64p^2 - 160q\dot{x}\_0^2\right) + 25p^2\lambda^4 + 80p^2\lambda^3 - 128q\dot{x}\_0^2\lambda} + p\lambda(5\lambda + 8)}{16\lambda} \tag{47}$$

and *λ* is a solution of the quintic

$$\begin{aligned} 125p^2\lambda^5 + 10\left(79p^2 + 125q\dot{\mathbf{x}}\_0^2\right)\lambda^4 + 40\left(43p^2 + 85q\dot{\mathbf{x}}\_0^2\right)\lambda^3 + \\ 8\left(196p^2 + 389q\dot{\mathbf{x}}\_0^2\right)\lambda^2 + 64\left(8p^2 + 17q\dot{\mathbf{x}}\_0^2\right)\lambda + 128q\dot{\mathbf{x}}\_0^2 = \mathbf{0} \end{aligned} \tag{48}$$

#### **Example 4.**

The approximate trigonometric solution of

$$\begin{cases} \ddot{\varkappa} + \mathbf{3}\varkappa + \mathbf{5}\varkappa^3 = \mathbf{0}, \\ \varkappa(\mathbf{0}) = \mathbf{0}, \varkappa'(\mathbf{0}) = \mathbf{1}. \end{cases} \tag{49}$$

reads

$$\varkappa\_{\rm app}(t) = \frac{0.499502 \sin \left(2.00199t\right)}{\sqrt{1 - 0.0817025 \sin^2 \left(2.00199t\right)}} \,\text{.}\tag{50}$$

The exact solution is

$$\mathbf{x}(t) = \sqrt{\frac{2}{3 + \sqrt{19}}} \mathbf{s} \mathbf{n} \left( \sqrt{\frac{1}{2} \left( 3 + \sqrt{19} \right)} t, \frac{1}{5} \left( -14 + 3\sqrt{19} \right) \right), \tag{51}$$

with period

$$T = 4\sqrt{\frac{1}{5}\left(\sqrt{19} - 3\right)}K\left(\frac{1}{5}\left(-14 + 3\sqrt{19}\right)\right) = 3.1383.\tag{52}$$

The error on the interval 0 ≤*t*≤ *T* equals 0*:*00018291*:*

*Analytical Solutions of Some Strong Nonlinear Oscillators DOI: http://dx.doi.org/10.5772/intechopen.97677*

**Figure 4.** *A comparison between the approximate solution (50) and exact solution (51).*

**Figure 4** demonstrates the comparison between the approximate analytic solution (50) and the exact analytic solution.

## **3. An analytical solution of the undamped Duffing-Helmholtz Equation**

The undamped Duffing-Helmholtz equation reads

$$\begin{cases} \ddot{\boldsymbol{\varkappa}} + p\boldsymbol{\varkappa} + q\boldsymbol{\varkappa}^2 + r\boldsymbol{\varkappa}^3 = \mathbf{0}, \\\\ \boldsymbol{\varkappa}(\mathbf{0}) = \boldsymbol{\varkappa}\_0 \, \text{and} \boldsymbol{\varkappa}'(\mathbf{0}) = \dot{\boldsymbol{\varkappa}}\_0. \end{cases} \tag{53}$$

We will give a solution to the i.v.p. (53) in terms of Weierstrass elliptic functions. For solving this problem the following ansatz is considered

$$\varkappa(t) = A + \frac{B}{1 + C\wp(t + t\_0; \mathbf{g}\_2, \mathbf{g}\_3)},\tag{54}$$

where *BC* 6¼ 0*:*

Substituting the ansatz (54) into the ordinary differential equation (ode) � *<sup>x</sup>*€ <sup>þ</sup> *px* <sup>þ</sup> *qx*<sup>2</sup> <sup>þ</sup> *rx*<sup>3</sup> <sup>¼</sup> 0, gives

$$\frac{1}{2(1+C\otimes^3)}\sum\_{j=0}^{3}K\_j\otimes^j=\mathbf{0},\tag{55}$$

with

$$\begin{aligned} K\_3 &= 2C^2(\mathbf{A}^3Cr + \mathbf{A}^2 \mathbf{C}q + \mathbf{A}\mathbf{C}p + 2\mathbf{B}), \\ K\_2 &= 2C(\mathbf{3A}^3Cr + \mathbf{3A}^2 \mathbf{B}\mathbf{C}r + \mathbf{3A}^2 \mathbf{C}q + 2\mathbf{A}\mathbf{B}\mathbf{C}q + \mathbf{3A}\mathbf{C}p + \mathbf{B}\mathbf{C}p - \mathbf{6B}), \\ K\_1 &= C(\mathbf{6A}^3r + 12\mathbf{A}^2\mathbf{B}r + \mathbf{6A}^2q + \mathbf{6A}\mathbf{B}^2r + \mathbf{8AB}q + \mathbf{6A}p + 2\mathbf{B}^2q - \mathbf{3B}\mathbf{C}q\_2 + \mathbf{4B}p), \\ K\_0 &= A^3r + \mathbf{6A}^2\mathbf{B}r + 2\mathbf{A}^2q + \mathbf{6A}\mathbf{B}^2r + \mathbf{4AB}q + 2\mathbf{A}p + 2\mathbf{B}^3r + 2\mathbf{B}^2q - \mathbf{4BC}^2\mathbf{g}\_3 + \mathbf{B}\mathbf{C}\mathbf{g}\_2 + 2\mathbf{B}p. \end{aligned}$$

Equating the coefficients *K <sup>j</sup>* to zero will give us an algebraic system. Solving this system, we finally get

$$\begin{aligned} B &= -\frac{6A(A^2r + Aq + p)}{3A^2r + 2Aq + p}, C = \frac{12}{3A^2r + 2Aq + p}, \\ \mathbf{g}\_2 &= -\frac{1}{12} \left( 3r^2A^4 + 4qrA^3 + 6prA^2 - p^2 \right), \\ \mathbf{g}\_3 &= \frac{1}{216} \left[ (9pr^2 - 3q^2r)A^4 + (12pqr - 4q^3)A^3 + (18p^2r - 6pq^2)A^2 + p^3 \right], \end{aligned} \tag{56}$$

The values of *t*<sup>0</sup> and *A* could be determined from the initial conditions *x*ð Þ¼ 0 *x*<sup>0</sup> and *x*<sup>0</sup> ð Þ¼ 0 *x*\_ <sup>0</sup> and

$$
\ddot{\mathbf{x}}(\mathbf{0}) + p\mathbf{x}(\mathbf{0}) + q\mathbf{x}^2(\mathbf{0}) + r\mathbf{x}^3(\mathbf{0}) = \mathbf{0}.\tag{57}
$$

We have

$$t\_0 = \pm \wp^{-1}(\frac{\varkappa\_0 - A - B}{C(A - \varkappa\_0)}; \mathbf{g}\_2, \mathbf{g}\_3). \tag{58}$$

The number *A* is a solution to the quartic

$$4\Im r A^4 + 4qA^2 + 6pA - \left(3r\varkappa\_0^4 + 4q\varkappa\_0^3 + 6p\varkappa\_0^2 + 6\dot{\varkappa}\_0^2\right) = 0.\tag{59}$$

#### **Example 5.**

The solution of the i.v.p.

$$\begin{cases} \ddot{\mathbf{x}} + \mathbf{x} + 2\mathbf{x}^2 + 3\mathbf{x}^3 = \mathbf{0}, \\ \varkappa(\mathbf{0}) = \mathbf{1} \text{and} \varkappa'(\mathbf{0}) = \mathbf{1}, \end{cases} \tag{60}$$

according to the relation (54) is given by

$$\infty(t) = 1.07627 - \frac{2.72078}{1 + 0.762858 \wp(t - 0.148317; -7.16667, 0.675926)} \,\text{.}\tag{61}$$

In **Figure 5**, the comparison with the approximate analytic solution (61) and the approximate numerical solution using RK4 is investigated.

The periodicity of solution (61) is given by

*A comparison between solution (61) and the approximate numerical solution using RK4.*
