**4. Calculation of sensitivity coefficients**

Generally, there have two approaches for determining the gradient; the first is a discretize-then-differentiate approach and the second is a differentiate-thendiscretize approach.

The first approach is to approximate the gradient of the functional by a finite difference quotient approximation, but in general, we cannot determine the sensitivities exactly, so this method may led to larger error.

Here we intend to use differentiate-then-discretize approach which we refer to as the sensitivity equation method. This method can be determined more efficiently with the help of the sensitivities

In the case of a nonlinear inverse problem, the matrix **J** has some functional dependence on the vector **p**. The solution of Eq. (13) requires an iterative procedure, which is obtained by linearizing the vector **U P**ð Þ with a Taylor series expansion around the current solution at iteration *k*. Such a linearization is given by

*<sup>k</sup>* **<sup>P</sup>**‐**P***<sup>k</sup>* � �, (15)

**<sup>k</sup>** � �*<sup>T</sup>* **<sup>G</sup>**‐**U P***<sup>k</sup>* � � � � *:* (16)

*<sup>T</sup>***J** to be nonsingular, or

� 6¼ 0, (17)

�≈0 are denoted ill-conditioned. Inverse heat transfer

**J**

*<sup>T</sup>***J** if necessary. *μ<sup>k</sup>* is made large in the beginning of the

, for *j* ¼ 1, 2, ⋯, m þ 1,

*<sup>k</sup>* � �*<sup>T</sup>* **<sup>G</sup>**‐**U P***<sup>k</sup>* � � � � , (18)

*<sup>T</sup>***J** is not required to be non-

*F* **p***<sup>k</sup>*þ<sup>1</sup> � �<*ε*1, (19)

*<sup>k</sup>* are the estimated temperatures and the sensitivity matrix

**J**

**U P**ð Þ¼ **U P***<sup>k</sup>* � � <sup>þ</sup> **<sup>J</sup>**

evaluated at iteration *k*, respectively. Eq. (15) is substituted into (14) and the resulting expression is rearranged to yield the following iterative procedure to

> *<sup>k</sup>* � �*<sup>T</sup>* **J** *<sup>k</sup>* h i‐<sup>1</sup>

> > **J** *<sup>T</sup>***J** � � �

*<sup>T</sup>***J** is zero, or even very small, the parameters *pj*

cannot be determined by using the iterative procedure of Eq. (16).

*<sup>k</sup>* � �*<sup>T</sup>* **J** *<sup>k</sup>* <sup>þ</sup> *<sup>μ</sup><sup>k</sup> <sup>Ω</sup><sup>k</sup>* h i‐<sup>1</sup>

Formula (17) gives the so called identifiability condition, that is, if the determi-

problems are generally very ill-conditioned, especially near the initial guess used for the unknown parameters, creating difficulties in the application of Eqs. (14) or (16). The Levenberg-Marquardt method alleviates such difficulties by utilizing an itera-

where *μ<sup>k</sup>* is a positive scalar named damping parameter and *Ω<sup>k</sup>* is a diagonal

iterations, since the problem is generally ill-conditioned in the region around the initial guess used for iterative procedure, which can be quite far from the exact

singular in the beginning of iterations and the Levenberg-Marquardt method tends to the steepest descent method, that is, a very small step is taken in the negative gradient direction. The parameter *μ<sup>k</sup>* is then gradually reduced as the iteration procedure advances to the solution of the parameter estimation problem, and then the Levenberg-Marquardt method tends to the Gauss method given by (16). The following criteria were suggested in literature [13] to stop the iterative procedure of

The purpose of the matrix term *μ<sup>k</sup>Ω<sup>k</sup>* is to damp oscillations and instabilities due to the ill-conditioned character of the problem, by making its components large as

The iterative procedure given by Eq. (16) is called the Gauss method. Such method is actually an approximation for the Newton (or Newton-Raphson) method. We note that Eq. (14), as well as the implementation of the iterative

where **U P***<sup>k</sup>* � � and **J**

obtain the vector of unknown parameters **P**:

*Inverse Heat Conduction and Heat Exchangers*

procedure given by Eq. (16), require the matrix **J**

*<sup>T</sup>***J** � � �

**<sup>P</sup>***<sup>k</sup>*þ<sup>1</sup> <sup>¼</sup> **<sup>P</sup>***<sup>k</sup>* <sup>þ</sup> **<sup>J</sup>**

parameters. With such an approach, the matrix **J**

the Levenberg-Marquardt method given by Eq. (18):

where j j� is the determinant.

Problems satisfying **J**

tive procedure in the form:

compared to those of **J**

nant of **J**

matrix.

**108**

**<sup>P</sup>***<sup>k</sup>*þ<sup>1</sup> <sup>¼</sup> **<sup>P</sup>***<sup>k</sup>* <sup>þ</sup> **<sup>J</sup>**

*Inverse Heat Conduction and Heat Exchangers*

$$\mathbf{u}\_k = \frac{\partial u}{\partial p\_k}, k = 1, 2, \cdots m+1. \tag{23}$$

We can change the first derivative in time and the integral. We have

ð*L* 0 **Γ** �

ð Þ¼ *f*, *g*

We also can define the following bilinear form:

Ð *L* 0

8

>>>>>>>>>>>>>><

Ð *L* 0

Ð *L*

⋯ Ð *L*

>>>>>>>>>>>>>>:

ð Þ **P**2

find **U**ð Þ *x*, *t* satisfying

*d dt* ðL 0

*L*2 ð Þ 0, *L*

*a*ð Þ¼ **U**, *v*

ð*L* 0 **Γ** �

problem in time.

**111**

*<sup>∂</sup>v x*ð Þ , *<sup>t</sup> <sup>∂</sup><sup>x</sup> dx* <sup>¼</sup>

for all *v*∈*V*<sup>0</sup> ≔ *H*<sup>1</sup>

**U**ð Þ� *x*, *t v x*ð Þ*dx* þ

*DOI: http://dx.doi.org/10.5772/intechopen.89096*

*v*ð Þ¼ 0 0 ¼ *v L*ð Þ, because *v*∈*V*0. This leads to an equivalent problem to ð Þ **P**1 : ∀*t*>0,

*A Numerical Approach to Solving an Inverse Heat Conduction Problem Using the Levenberg…*

*<sup>∂</sup>v x*ð Þ , *<sup>t</sup> ∂x*

> ð*L* 0

� *<sup>p</sup>*<sup>1</sup> <sup>þ</sup> *<sup>p</sup>*2*<sup>x</sup>* <sup>þ</sup> <sup>⋯</sup> <sup>þ</sup> *pm*þ<sup>1</sup>*x<sup>m</sup>* � � *<sup>∂</sup><sup>u</sup>*

� *<sup>p</sup>*<sup>1</sup> <sup>þ</sup> *<sup>p</sup>*2*<sup>x</sup>* <sup>þ</sup> <sup>⋯</sup> <sup>þ</sup> *pm*þ<sup>1</sup>*x<sup>m</sup>* � � *<sup>∂</sup>u*<sup>1</sup>

<sup>0</sup> � *<sup>p</sup>*<sup>1</sup> <sup>þ</sup> *<sup>p</sup>*2*<sup>x</sup>* <sup>þ</sup> <sup>⋯</sup> <sup>þ</sup> *pm*þ<sup>1</sup>*x<sup>m</sup>* � � *<sup>∂</sup>u*<sup>2</sup>

<sup>0</sup> � *<sup>p</sup>*<sup>1</sup> <sup>þ</sup> *<sup>p</sup>*2*<sup>x</sup>* <sup>þ</sup> <sup>⋯</sup> <sup>þ</sup> *pm*þ<sup>1</sup>*x<sup>m</sup>* � � *<sup>∂</sup>um*þ<sup>1</sup>

*dt*ð Þ **<sup>U</sup>**, *<sup>v</sup> <sup>L</sup>*<sup>2</sup> <sup>þ</sup> *<sup>a</sup>*ð Þ¼ **<sup>U</sup>**, *<sup>v</sup>* ð Þ **<sup>F</sup>**, *<sup>v</sup> <sup>L</sup>*<sup>2</sup>

In this section, we search a semi-discrete approximation of the weak problem ð Þ **P**2 , using the Galerkin finite element method. This leads to a first order Cauchy-

Let *Vh* be a *Nx* þ 1 dimensional subspace of *V* and *V*0,*<sup>h</sup>* ¼ *Vh*∩*V*0. Then the

*dt*ð Þþ **<sup>U</sup>***h*, *vh <sup>a</sup>*ð Þ¼ **<sup>U</sup>***h*, *vh* ð Þ **<sup>F</sup>**, *vh*

*T*.

following problem is an approximation of the weak problem, find

**U***h*ð Þ¼ x, 0 **U**0,*<sup>h</sup>*ð Þ *x* **U***h*ð Þ¼ 0, *t* **G**1ð Þ*t* **U***h*ð Þ¼ *L*, *t* **G**2ð Þ*t*

Finally, we obtain with this notations the weak problem of ð Þ **P**1 :

**U**ð Þ¼ *x*, 0 **U**0ð Þ *x* **U**ð Þ¼ 0, *t* **G**1ð Þ*t* **U**ð Þ¼ *L*, *t* **G**2ð Þ*t*

*d*

8 >>>>>><

>>>>>>:

**4.1 Space-discretization with the Galerkin method**

*d*

for all *vh* ∈*V*0,*<sup>h</sup>*. where **U***<sup>h</sup>* ¼ ½ � *uh*, *u*1,*<sup>h</sup>*, *u*2,*<sup>h</sup>*⋯*um*þ1*:<sup>h</sup>*

8 >>>>>><

>>>>>>:

*uh*, *u*1,*<sup>h</sup>*, *u*2,*<sup>h</sup>*⋯*um*þ1*:<sup>h</sup>* ∈*Vh* that satisfies:

*dx* ¼

ð*L* 0

<sup>0</sup>ð Þ 0, *L* . To simplify the notation we use the scalar product in

*∂x ∂v <sup>∂</sup><sup>x</sup> dx*

� �*dx*

� �*dx*

� �*dx*

*∂x ∂v <sup>∂</sup><sup>x</sup>* � *<sup>∂</sup><sup>u</sup> ∂x ∂v ∂x*

*∂x ∂v <sup>∂</sup><sup>x</sup>* � *<sup>x</sup> <sup>∂</sup><sup>u</sup> ∂x ∂v ∂x*

*∂x ∂v <sup>∂</sup><sup>x</sup>* � *xm*þ<sup>1</sup> *<sup>∂</sup><sup>u</sup> ∂x ∂v ∂x*

**F**ð Þ� *x*, *t v x*ð Þ*dx*, (30)

*:*

*:* (33)

, (34)

(32)

*f x*ð Þ� *g x*ð Þ*dx:* (31)

We first differentiate the flow system (1)–(4) with respect to each of the design parameters *<sup>p</sup>*1, *<sup>p</sup>*2, …, *pm*þ<sup>1</sup> � �, to obtain the *<sup>m</sup>* <sup>þ</sup> 1 continuous sensitivity systems: for *k* ¼ 1, 2, ⋯, *m* þ 1

$$\begin{cases} \frac{\partial u\_k}{\partial t} = \frac{\partial}{\partial \mathbf{x}} \left[ \left( p\_1 + p\_2 \mathbf{x} + \dots + p\_{m+1} \mathbf{x}^m \right) \frac{\partial u\_k}{\partial \mathbf{x}} + \mathbf{x}^{k-1} \frac{\partial u}{\partial \mathbf{x}} \right] \\\ u\_k(\mathbf{x}, \mathbf{0}) = \mathbf{0} \\\ u\_k(\mathbf{0}, t) = \mathbf{0} \\\ u\_k(L, t) = \mathbf{0} \end{cases} . \tag{24}$$

There have (*m* þ 2) equations, we can make them in one system equation and use the finite element methods to solve the system of equation. Here, we give the vector form of the equation as follow:

$$\begin{aligned} \mathbf{(P1)} \begin{cases} \frac{\partial \overrightarrow{\mathbf{U}}}{\partial t} + \nabla \cdot \overrightarrow{\Gamma} = \overrightarrow{\mathbf{F}} \\ \overrightarrow{\mathbf{U}}(\varkappa, \mathbf{0}) = \overrightarrow{\mathbf{U}}\_{0}(\varkappa) \end{cases} \end{aligned} \tag{25}$$

$$\begin{aligned} \overrightarrow{\mathbf{U}}(\mathbf{0}, t) = \overrightarrow{\mathbf{G}}\_{1}(t) \\ \overrightarrow{\mathbf{U}}(L, t) = \overrightarrow{\mathbf{G}}\_{2}(t) \end{aligned}$$

where

$$\begin{aligned} \stackrel{\cdot}{\mathbf{U}} = \begin{bmatrix} u \\ u\_1 \\ u\_2 \\ \dots \\ u\_{m+1} \end{bmatrix}, \stackrel{\cdot}{\mathbf{T}} = \begin{bmatrix} -(p\_1 + p\_2\mathbf{x} + \cdots + p\_{m+1}\mathbf{x}^m)\frac{\partial u}{\partial \mathbf{x}} \\ -(p\_1 + p\_2\mathbf{x} + \cdots + p\_{m+1}\mathbf{x}^m)\frac{\partial u\_1}{\partial \mathbf{x}} - \left(\frac{\partial u}{\partial \mathbf{x}}\right) \\ -(p\_1 + p\_2\mathbf{x} + \cdots + p\_{m+1}\mathbf{x}^m)\frac{\partial u\_2}{\partial \mathbf{x}} - \mathbf{x}\left(\frac{\partial u}{\partial \mathbf{x}}\right) \\ \dots \\ -(p\_1 + p\_2\mathbf{x} + \cdots + p\_{m+1}\mathbf{x}^m)\frac{\partial u\_{m+1}}{\partial \mathbf{x}} - \left(\mathbf{x}^m \frac{\partial u}{\partial \mathbf{x}}\right) \end{bmatrix}, \stackrel{\cdot}{\mathbf{F}} = \begin{bmatrix} f(\mathbf{x}, \mathbf{t}) \\ 0 \\ 0 \\ \dots \\ 0 \end{bmatrix}. \tag{26}$$

$$\stackrel{\cdot}{\mathbf{U}}\_0(\mathbf{x}) = \begin{bmatrix} \mathbf{u}\_0(\mathbf{x}), \mathbf{0}, \mathbf{0}, \cdots, \mathbf{0} \end{bmatrix}^T, \stackrel{\cdot}{\mathbf{G}}\_1(t) = \begin{bmatrix} g\_1(t), \mathbf{0}, \mathbf{0}, \cdots, \mathbf{0} \end{bmatrix}^T, \stackrel{\cdot}{\mathbf{G}}\_2(t) = \begin{bmatrix} g\_2(t), \mathbf{0}, \mathbf{0}, \cdots, \mathbf{0} \end{bmatrix}^T. \tag{27}$$

We use the Galerkin finite element method approximation for discretizing problem (25). For this, we multiply the Eq. (25) by a test function *v* : ½ �! 0, *L R*, *v*∈*V*<sup>0</sup> ≔ *H*<sup>1</sup> <sup>0</sup>ð Þ 0, *L* and integrate the obtained equation in space form 0 to *L*. We obtain the following equation:

$$\int\_{0}^{L} \frac{\partial \mathbf{U}(\mathbf{x}, t)}{\partial t} \cdot \boldsymbol{\nu}(\mathbf{x}) d\mathbf{x} \cdot \int\_{0}^{L} \nabla \Gamma \cdot \boldsymbol{\nu}(\mathbf{x}) d\mathbf{x} = \int\_{0}^{L} \mathbf{F}(\mathbf{x}, t) \cdot \boldsymbol{\nu}(\mathbf{x}) d\mathbf{x},\tag{28}$$

integrating by parts gives

$$\int\_{0}^{L} \nabla \Gamma \cdot \boldsymbol{\nu}(\mathbf{x}) d\mathbf{x} = (\Gamma \cdot \boldsymbol{\nu}(\mathbf{x})) \Big|\_{0}^{L} - \int\_{0}^{L} \Gamma \cdot \frac{\partial \boldsymbol{\nu}(\mathbf{x}, t)}{\partial \mathbf{x}} d\mathbf{x}. \tag{29}$$

*A Numerical Approach to Solving an Inverse Heat Conduction Problem Using the Levenberg… DOI: http://dx.doi.org/10.5772/intechopen.89096*

We can change the first derivative in time and the integral. We have *v*ð Þ¼ 0 0 ¼ *v L*ð Þ, because *v*∈*V*0. This leads to an equivalent problem to ð Þ **P**1 : ∀*t*>0, find **U**ð Þ *x*, *t* satisfying

$$\frac{d}{dt}\int\_{0}^{L}\mathbf{U}(\mathbf{x},t)\cdot\boldsymbol{v}(\mathbf{x})d\mathbf{x}+\int\_{0}^{L}\boldsymbol{\Gamma}\cdot\frac{\partial\boldsymbol{v}(\mathbf{x},t)}{\partial\mathbf{x}}d\mathbf{x}=\int\_{0}^{L}\mathbf{F}(\mathbf{x},t)\cdot\boldsymbol{v}(\mathbf{x})d\mathbf{x},\tag{30}$$

for all *v*∈*V*<sup>0</sup> ≔ *H*<sup>1</sup> <sup>0</sup>ð Þ 0, *L* . To simplify the notation we use the scalar product in *L*2 ð Þ 0, *L*

$$f(f, \mathbf{g}) = \int\_0^L f(\mathbf{x}) \cdot \mathbf{g}(\mathbf{x}) d\mathbf{x}.\tag{31}$$

We also can define the following bilinear form:

$$a(\mathbf{U}, \nu) = \int\_0^L \Gamma \cdot \frac{\partial \nu(\mathbf{x}, t)}{\partial \mathbf{x}} d\mathbf{x} = \begin{cases} \int\_0^L \left( -\left( p\_1 + p\_2 \mathbf{x} + \dots + p\_{m+1} \mathbf{x}^m \right) \frac{\partial u}{\partial \mathbf{x}} \frac{\partial \nu}{\partial \mathbf{x}} d\mathbf{x} \\\\ \int\_0^L \left( -\left( p\_1 + p\_2 \mathbf{x} + \dots + p\_{m+1} \mathbf{x}^m \right) \frac{\partial u\_1}{\partial \mathbf{x}} \frac{\partial \nu}{\partial \mathbf{x}} - \frac{\partial u}{\partial \mathbf{x}} \frac{\partial \nu}{\partial \mathbf{x}} \right) d\mathbf{x} \\\\ \int\_0^L \left( -\left( p\_1 + p\_2 \mathbf{x} + \dots + p\_{m+1} \mathbf{x}^m \right) \frac{\partial u\_2}{\partial \mathbf{x}} \frac{\partial \nu}{\partial \mathbf{x}} - \mathbf{x} \frac{\partial u}{\partial \mathbf{x}} \frac{\partial \nu}{\partial \mathbf{x}} \right) d\mathbf{x} \\\\ \int\_0^L \left( -\left( p\_1 + p\_2 \mathbf{x} + \dots + p\_{m+1} \mathbf{x}^m \right) \frac{\partial u\_{m+1}}{\partial \mathbf{x}} \frac{\partial \nu}{\partial \mathbf{x}} - \mathbf{x}^{m+1} \frac{\partial u}{\partial \mathbf{x}} \frac{\partial \nu}{\partial \mathbf{x}} \right) d\mathbf{x} \end{cases} \tag{32}$$

Finally, we obtain with this notations the weak problem of ð Þ **P**1 :

$$\mathbf{U}(\mathbf{P}2)\begin{cases} \frac{d}{dt}(\mathbf{U},\nu)\_{L^{2}} + a(\mathbf{U},\nu) = (\mathbf{F},\nu)\_{L^{2}}\\ \mathbf{U}(\mathbf{x},\mathbf{0}) = \mathbf{U}\_{0}(\mathbf{x}) \\ \mathbf{U}(0,t) = \mathbf{G}\_{1}(t) \\ \mathbf{U}(L,t) = \mathbf{G}\_{2}(t) \end{cases} \tag{33}$$

#### **4.1 Space-discretization with the Galerkin method**

In this section, we search a semi-discrete approximation of the weak problem ð Þ **P**2 , using the Galerkin finite element method. This leads to a first order Cauchyproblem in time.

Let *Vh* be a *Nx* þ 1 dimensional subspace of *V* and *V*0,*<sup>h</sup>* ¼ *Vh*∩*V*0. Then the following problem is an approximation of the weak problem, find *uh*, *u*1,*<sup>h</sup>*, *u*2,*<sup>h</sup>*⋯*um*þ1*:<sup>h</sup>* ∈*Vh* that satisfies:

$$\begin{cases} \frac{d}{dt}(\mathbf{U}\_h, v\_h) + a(\mathbf{U}\_h, v\_h) = (\mathbf{F}, v\_h) \\ \mathbf{U}\_h(\mathbf{x}, 0) = \mathbf{U}\_{0,h}(\mathbf{x}) \\ \mathbf{U}\_h(0, t) = \mathbf{G}\_1(t) \\ \mathbf{U}\_h(L, t) = \mathbf{G}\_2(t) \end{cases}, \tag{34}$$

for all *vh* ∈*V*0,*<sup>h</sup>*. where **U***<sup>h</sup>* ¼ ½ � *uh*, *u*1,*<sup>h</sup>*, *u*2,*<sup>h</sup>*⋯*um*þ1*:<sup>h</sup> T*.

<sup>u</sup>*<sup>k</sup>* <sup>¼</sup> *<sup>∂</sup><sup>u</sup> ∂pk*

ð Þ **P**1

parameters *<sup>p</sup>*1, *<sup>p</sup>*2, …, *pm*þ<sup>1</sup>

*∂uk <sup>∂</sup><sup>t</sup>* <sup>¼</sup> *<sup>∂</sup>*

*Inverse Heat Conduction and Heat Exchangers*

vector form of the equation as follow:

8 >>>>><

>>>>>:

*uk*ð Þ¼ *x*, 0 0 *uk*ð Þ¼ 0, *t* 0 *uk*ð Þ¼ *L*, *t* 0

*k* ¼ 1, 2, ⋯, *m* þ 1

where

*u u*1 *u*2 ⋯ *um*þ<sup>1</sup>

⋯

<sup>0</sup>ð Þ¼ *<sup>x</sup>* ½ � u0ð Þ *<sup>x</sup>* , 0, 0, <sup>⋯</sup><sup>0</sup> *<sup>T</sup>*, **<sup>G</sup>**

obtain the following equation:

ðL 0

integrating by parts gives

*<sup>∂</sup>***U**ð Þ *<sup>x</sup>*, *<sup>t</sup>*

ð*L* 0

**U** ! ¼

**U** !

**110**

*v*∈*V*<sup>0</sup> ≔ *H*<sup>1</sup>

We first differentiate the flow system (1)–(4) with respect to each of the design

*<sup>∂</sup><sup>x</sup> <sup>p</sup>*<sup>1</sup> <sup>þ</sup> *<sup>p</sup>*2*<sup>x</sup>* <sup>þ</sup> <sup>⋯</sup> <sup>þ</sup> *pm*þ<sup>1</sup>*xm* � � *<sup>∂</sup>uk*

There have (*m* þ 2) equations, we can make them in one system equation and use the finite element methods to solve the system of equation. Here, we give the

> þ ∇ � **Γ** ! ¼ **F** !

ð Þ¼ *x*, 0 **U** ! <sup>0</sup>ð Þ *x*

ð Þ¼ 0, *t* **G** ! <sup>1</sup>ð Þ*t*

ð Þ¼ *L*, *t* **G** ! <sup>2</sup>ð Þ*t*

*∂x*

*<sup>∂</sup><sup>x</sup>* � *<sup>∂</sup><sup>u</sup> ∂x* � �

*<sup>∂</sup><sup>x</sup>* � *<sup>x</sup> <sup>∂</sup><sup>u</sup> ∂x* � �

*<sup>∂</sup><sup>x</sup>* � *xm <sup>∂</sup><sup>u</sup>*

, **G** !

> ð*L* 0

*∂x* � �

> *<sup>∂</sup>v x*ð Þ , *<sup>t</sup> ∂x*

, **F** ! ¼

<sup>2</sup>ðÞ¼ *<sup>t</sup> <sup>g</sup>*2ð Þ*<sup>t</sup>* , 0, 0, <sup>⋯</sup><sup>0</sup> � �*<sup>T</sup>*

**F**ð Þ� *x*, *t v x*ð Þ*dx*, (28)

*dx:* (29)

*∂***U** ! *∂t*

8 >>>>>>><

**U** !

**U** !

>>>>>>>:

� *<sup>p</sup>*<sup>1</sup> <sup>þ</sup> *<sup>p</sup>*2*<sup>x</sup>* <sup>þ</sup> <sup>⋯</sup> <sup>þ</sup> *pm*þ<sup>1</sup>*xm* � � *<sup>∂</sup><sup>u</sup>*

� *<sup>p</sup>*<sup>1</sup> <sup>þ</sup> *<sup>p</sup>*2*<sup>x</sup>* <sup>þ</sup> <sup>⋯</sup> <sup>þ</sup> *pm*þ<sup>1</sup>*xm* � � *<sup>∂</sup>u*<sup>1</sup>

� *<sup>p</sup>*<sup>1</sup> <sup>þ</sup> *<sup>p</sup>*2*<sup>x</sup>* <sup>þ</sup> <sup>⋯</sup> <sup>þ</sup> *pm*þ<sup>1</sup>*xm* � � *<sup>∂</sup>u*<sup>2</sup>

� *<sup>p</sup>*<sup>1</sup> <sup>þ</sup> *<sup>p</sup>*2*<sup>x</sup>* <sup>þ</sup> <sup>⋯</sup> <sup>þ</sup> *pm*þ<sup>1</sup>*xm* � � *<sup>∂</sup>um*þ<sup>1</sup>

ð*L* 0

<sup>∇</sup>**<sup>Γ</sup>** � *v x*ð Þ*dx* <sup>¼</sup> ð Þ **<sup>Γ</sup>** � *v x*ð Þ *<sup>L</sup>*

<sup>1</sup>ðÞ¼ *<sup>t</sup> <sup>g</sup>*1ð Þ*<sup>t</sup>* , 0, 0, <sup>⋯</sup><sup>0</sup> � �*<sup>T</sup>*

We use the Galerkin finite element method approximation for discretizing problem (25). For this, we multiply the Eq. (25) by a test function *v* : ½ �! 0, *L R*,

<sup>0</sup>ð Þ 0, *L* and integrate the obtained equation in space form 0 to *L*. We

∇**Γ** � *v x*ð Þ*dx* ¼

0 � � � ð*L* 0 **Γ** �

!

*<sup>∂</sup><sup>t</sup>* � *v x*ð Þ*dx*‐

**U** !

� �, to obtain the *<sup>m</sup>* <sup>þ</sup> 1 continuous sensitivity systems: for

� �

, *k* ¼ 1, 2, ⋯*m* þ 1*:* (23)

<sup>þ</sup> *<sup>x</sup>k*�<sup>1</sup> *<sup>∂</sup><sup>u</sup> ∂x*

, (25)

*f*ð Þ x, t 0 0 ⋯ 0

*:* (26)

*:* (27)

*:* (24)

*∂x*

The choice of *Vh* is completely arbitrary. So we can choose it the way that for later treatment, it will be as easy as possible. For example, we subdivide the interval ½ � 0, *L* into partitions of equal distances *h*:

$$0 = a\_1 < a\_2 < \cdots < a\_{N\_x} < a\_{N\_x + 1} = L \,, \, a\_i = (i - 1) \cdot h \tag{35}$$

$$\mathcal{V}\_h = \left\{ v\_h \in \mathbb{C}^0[0, L] : v\_h \big|\_{\left[a\_{i, \stackrel{\scriptstyle \alpha}{i \cdot \cdot} \right]}} \in \mathbb{P}\_1, \forall i = 1 \cdots N\_x \right\}, \tag{36}$$

$$V\_{0,h} = \{ v\_h \in V\_h : v\_h(\mathbf{0}) = v\_h(L) = \mathbf{0} \}. \tag{37}$$

Note that **M**, **A** ∈ **R***Nx*�1�*Nx*þ<sup>1</sup>

*DOI: http://dx.doi.org/10.5772/intechopen.89096*

*Δt* !

> !ð Þ *tk* , **F** ! **<sup>k</sup>** ¼ **F** !

considered and the solution is obtained. **Example 1.** Consider (1)–(4) with

> <sup>4</sup> <sup>þ</sup> *<sup>x</sup>* <sup>2</sup> <sup>þ</sup> <sup>1</sup> � � � � *<sup>e</sup>*

We take the observed data *g* as

where *p*1, *p*2, *p*<sup>3</sup> are unknown coefficients.

We obtain the unique exact solution

� **u** !

the Cauchy problem

**<sup>M</sup>** � **<sup>u</sup>** ! *<sup>k</sup>*þ<sup>1</sup> � **u** ! *k*

**M** þ *Δt* 2 **A** � �

**5. Numerical experiment**

*f x*ð Þ¼ , *<sup>t</sup>* sin *<sup>x</sup> <sup>x</sup>*<sup>2</sup>

And

**113**

where **u** ! **<sup>k</sup>** ¼ **u** , **u** ! ∈*RNx*þ<sup>1</sup>

**<sup>M</sup>** *<sup>d</sup> dx* **<sup>u</sup>**

**u** !ð Þ¼ *<sup>t</sup>*<sup>0</sup> **<sup>u</sup>** ! **0**

8 <

þ 1 2 **A** � **u** ! **<sup>k</sup>** þ **<sup>1</sup>** þ

The Eq. (47) can be written in simple form as

**<sup>k</sup>** <sup>þ</sup> **<sup>1</sup>** <sup>¼</sup> **<sup>M</sup>** � *<sup>Δ</sup><sup>t</sup>*

, *and* **F** !

> !ðÞ¼ *<sup>t</sup>* **<sup>F</sup>** ! ð Þ*t*

!ðÞþ*<sup>t</sup>* **<sup>A</sup>** � **<sup>u</sup>**

*A Numerical Approach to Solving an Inverse Heat Conduction Problem Using the Levenberg…*

the Crank-Nicolson method can be applied to (46) at time *tk*, resulting in

ð Þ *tk* , *k* ¼ 0, 1, ⋯*:*.

2 **A** � �

In this section, we are going to demonstrate some numerical results for ð Þ u x, t ð Þ,q xð Þ in the inverse problem (1)–(5). Therefore the following examples are

the algebraic system (48) is solved by Gauss elimination method.

*<sup>u</sup>*ð Þ¼ 1, t sin 1ð Þe‐*<sup>t</sup>*

�*<sup>t</sup>* � ð Þ *<sup>x</sup>* <sup>þ</sup> <sup>1</sup>

<sup>2</sup> cosð Þ *<sup>x</sup> <sup>e</sup>*

*<sup>u</sup>*ð Þ¼ x, t sin xð Þe‐*<sup>t</sup>*

^ *<sup>q</sup>*ð Þ¼ <sup>x</sup> *<sup>p</sup>*<sup>1</sup> <sup>þ</sup> *<sup>p</sup>*2*<sup>x</sup>* <sup>þ</sup> *<sup>p</sup>*3*x*<sup>2</sup>

*g t*ðÞ¼ *<sup>u</sup>*ð Þ¼ <sup>0</sup>*:*5, t sin 0ð Þ *:*<sup>5</sup> <sup>e</sup>‐*<sup>t</sup>*

The unknown function *q*ð Þ x defined as the following form

� **u** ! **<sup>k</sup>** þ *Δt* 2 **F** ! **<sup>k</sup>** þ **F** ! **k** þ **1** � �, (48)

1 2 **A** � **u** ! **<sup>k</sup>** <sup>¼</sup> <sup>1</sup> 2 **F** ! **<sup>k</sup>** þ **F** ! **k** þ **1** � �, (47)

∈ **R***Nx*�<sup>1</sup>

: , (46)

*u*ð Þ¼ x, 0 sinx, 0≤ *x*≤ 1, (49) *u*ð Þ¼ 0, t 0, 0 ≤*t* ≤1 (50)

�*<sup>t</sup>* � sin *xe*�*<sup>t</sup>*

*<sup>q</sup>*ð Þ¼ <sup>x</sup> <sup>1</sup> <sup>þ</sup> <sup>0</sup>*:*5x <sup>þ</sup> <sup>0</sup>*:*25x<sup>2</sup> (53)

, 0≤*t*≤1, (51)

, 0≤*x* ≤10≤*t*≤1 (52)

*:* (54)

0≤*t*≤ 1 (55)

, (56)

. So that (43) is equal to

Note, that the finite dimension allows us to build a finite base for the corresponding space. In the case of *V*0,*h*, we have: f g *φ<sup>i</sup> Nx <sup>i</sup>*¼<sup>2</sup> where ∀*i* ¼ 2⋯*Nx*.

$$\rho\_i(\mathbf{x}) = \begin{cases} 0 & \mathbf{x} \in [a\_0, a\_{i-1}] \\ \frac{\mathbf{x}}{h} - (i - 2) & \mathbf{x} \in [a\_{i-1}, a\_i] \\ i - \frac{\mathbf{x}}{h} & \mathbf{x} \in [a\_i, a\_{i+1}] \\ 0 & \mathbf{x} \in [a\_{i+1}, a\_{N\_x + 1}] \end{cases},\tag{38}$$

while we add for *Vh* the two functions *φ*<sup>1</sup> and *φNx*þ<sup>1</sup> defined as:

$$\rho\_1(\mathbf{x}) = \begin{cases} 1 - \frac{\mathbf{x}}{h}, & \text{if } \mathbf{x} \in [a\_1, a\_2] \\ 0, & \text{if } \mathbf{x} \in [a\_2, a\_{N\_x + 1}] \end{cases},\tag{39}$$

$$\rho\_{N\_x+1}(\mathbf{x}) = \begin{cases} \mathbf{0}, & \text{if } \mathbf{x} \in [a\_1, a\_{N\_x}] \\ \mathbf{x} - N\_x + \mathbf{1}, & \text{if } \mathbf{x} \in [a\_{N\_x}, a\_{N\_x+1}] \end{cases}, \tag{40}$$

so that we can write **U***<sup>h</sup>* as a linear combination of the basic elements:

$$\mathbf{U}\_{h}(\mathbf{x},t) = \sum\_{j=1}^{N\_{x}+1} \tilde{\mathbf{U}}\_{j}(t) \cdot \boldsymbol{\wp}\_{j}(\mathbf{x}),\tag{41}$$

$$\mathbf{U}\_{0,h}(\mathbf{x},t) = \sum\_{j=1}^{N\_x+1} \mathbf{U}\_0(\mathbf{x}\_j) \cdot \boldsymbol{\rho}\_j(\mathbf{x}),\tag{42}$$

where **<sup>U</sup>**<sup>~</sup> <sup>1</sup>ðÞ¼ *<sup>t</sup>* **<sup>G</sup>**0ð Þ*<sup>t</sup>* and **<sup>U</sup>**<sup>~</sup> *Nx*þ<sup>1</sup>ðÞ¼ *<sup>t</sup>* **<sup>G</sup>**1ð Þ*<sup>t</sup>* . Using that *<sup>a</sup>*ð Þ �, � is bilinear form and that Eq. (34) is valid for each element of the base f g *φ<sup>i</sup> Nx <sup>i</sup>*¼2, we obtain

$$\sum\_{j=1}^{N\_x+1} \frac{d}{dt} \tilde{\mathbf{U}}\_j(t) \cdot \left(\boldsymbol{\rho}\_j, \boldsymbol{\rho}\_i\right) + \sum\_{j=1}^{N\_x+1} \tilde{\mathbf{U}}\_j(t) \cdot \boldsymbol{a}\left(\boldsymbol{\rho}\_j, \boldsymbol{\rho}\_i\right) = (\mathbf{F}, \boldsymbol{\rho}\_i), \tag{43}$$

∀*i* ¼ 2⋯*Nx*. This equation can be written in a vector form. For this we define the vectors **u** ! , **u** ! **<sup>0</sup>** and **F** ! with components

$$\mathbf{F}\_i(t) = (\mathbf{F}, \rho\_i)\_{L^2}, \mathbf{u}\_j(t) \coloneqq \bar{\mathbf{u}}\_j(t), \mathbf{u}\_{\mathbf{0}}, \mathbf{j} = \mathbf{u}\_0(\mathbf{x}\_{\mathbf{j}}), \tag{44}$$

and matrices **M** and **A** as

$$m\_{i\bar{j}} \coloneqq \left(\rho\_i, \rho\_{\bar{j}}\right)\_{L^2}, a\_{i\bar{j}} \coloneqq a\left(\rho\_i, \rho\_{\bar{j}}\right),\tag{45}$$

*A Numerical Approach to Solving an Inverse Heat Conduction Problem Using the Levenberg… DOI: http://dx.doi.org/10.5772/intechopen.89096*

Note that **M**, **A** ∈ **R***Nx*�1�*Nx*þ<sup>1</sup> , **u** ! ∈*RNx*þ<sup>1</sup> , *and* **F** ! ∈ **R***Nx*�<sup>1</sup> . So that (43) is equal to the Cauchy problem

$$\begin{cases} \mathbf{M} \frac{d}{d\mathbf{x}} \overrightarrow{\mathbf{u}}(t) + \mathbf{A} \cdot \overrightarrow{\mathbf{u}}(t) = \overrightarrow{\mathbf{F}}(t) \\\\ \overrightarrow{\mathbf{u}}(t\_0) = \overrightarrow{\mathbf{u}}\_0 \end{cases},\tag{46}$$

the Crank-Nicolson method can be applied to (46) at time *tk*, resulting in

$$\mathbf{M} \cdot \left(\frac{\overrightarrow{\mathbf{u}}\_{k+1} - \overrightarrow{\mathbf{u}}\_{k}}{\Delta t}\right) + \frac{1}{2}\mathbf{A} \cdot \overrightarrow{\mathbf{u}}\_{k+1} + \frac{1}{2}\mathbf{A} \cdot \overrightarrow{\mathbf{u}}\_{k} = \frac{1}{2}\left(\overrightarrow{\mathbf{F}}\_{k} + \overrightarrow{\mathbf{F}}\_{k+1}\right),\tag{47}$$

where **u** ! **<sup>k</sup>** ¼ **u** !ð Þ *tk* , **F** ! **<sup>k</sup>** ¼ **F** ! ð Þ *tk* , *k* ¼ 0, 1, ⋯*:*. The Eq. (47) can be written in simple form as

$$\left(\mathbf{M} + \frac{\Delta t}{2}\mathbf{A}\right) \cdot \overrightarrow{\mathbf{u}}\_{\mathbf{k}+\mathbf{1}} = \left(\mathbf{M} - \frac{\Delta t}{2}\mathbf{A}\right) \cdot \overrightarrow{\mathbf{u}}\_{\mathbf{k}} + \frac{\Delta t}{2} \left(\overrightarrow{\mathbf{F}}\_{\mathbf{k}} + \overrightarrow{\mathbf{F}}\_{\mathbf{k}+\mathbf{1}}\right),\tag{48}$$

the algebraic system (48) is solved by Gauss elimination method.

#### **5. Numerical experiment**

The choice of *Vh* is completely arbitrary. So we can choose it the way that for later treatment, it will be as easy as possible. For example, we subdivide the interval

�

0 *x*∈½ � *a*0, *ai*�<sup>1</sup>

*<sup>h</sup>* � ð Þ *<sup>i</sup>* � <sup>2</sup> *<sup>x</sup>*<sup>∈</sup> *ai*�1, *ai* ½ �

*<sup>h</sup> <sup>x</sup>*<sup>∈</sup> *ai* ½ � , *ai*þ<sup>1</sup> 0 *x*∈½ � *ai*þ1, *aNx*þ<sup>1</sup>

*<sup>h</sup>* , *if x*∈½ � *<sup>a</sup>*1, *<sup>a</sup>*<sup>2</sup> 0, *if x*∈½ � *a*2, *aNx*þ<sup>1</sup>

0, *if x*∈ *a*1, *aNx* ½ �

*<sup>h</sup>* � *Nx* <sup>þ</sup> 1, *if x*<sup>∈</sup> *aNx* ½ � , *aNx*þ<sup>1</sup>

**<sup>U</sup>**<sup>~</sup> *<sup>j</sup>*ðÞ� *<sup>t</sup> <sup>φ</sup><sup>j</sup>*

**U**<sup>0</sup> *xj* � � � *<sup>φ</sup><sup>j</sup>*

where **<sup>U</sup>**<sup>~</sup> <sup>1</sup>ðÞ¼ *<sup>t</sup>* **<sup>G</sup>**0ð Þ*<sup>t</sup>* and **<sup>U</sup>**<sup>~</sup> *Nx*þ<sup>1</sup>ðÞ¼ *<sup>t</sup>* **<sup>G</sup>**1ð Þ*<sup>t</sup>* . Using that *<sup>a</sup>*ð Þ �, � is bilinear form and

∀*i* ¼ 2⋯*Nx*. This equation can be written in a vector form. For this we define the

**F***i*ðÞ¼ *t* **F**, *φ<sup>i</sup>* ð Þ*L*<sup>2</sup> , **u***j*ð Þ*t* ≔ **u**~ **<sup>j</sup>**ð Þ*t* , **u0** , **<sup>j</sup>** ¼ **u0 xj**

*L*2

Note, that the finite dimension allows us to build a finite base for the

*Vh* <sup>¼</sup> *vh* <sup>∈</sup>*C*0½ � 0, *<sup>L</sup>* : *vh* ½ � *ai*,*ai*þ<sup>1</sup>

*x*

8 >>>>>><

>>>>>>:

*<sup>i</sup>* � *<sup>x</sup>*

while we add for *Vh* the two functions *φ*<sup>1</sup> and *φNx*þ<sup>1</sup> defined as:

8 <

*x*

**U***h*ð Þ¼ *x*, *t*

**U**0,*<sup>h</sup>*ð Þ¼ *x*, *t*

� �

with components

*mij* ≔ *φi*, *φ<sup>j</sup>* � �

þ *N* X*<sup>x</sup>*þ<sup>1</sup> *j*¼1

that Eq. (34) is valid for each element of the base f g *φ<sup>i</sup>*

*dt* **<sup>U</sup>**<sup>~</sup> *<sup>j</sup>*ðÞ� *<sup>t</sup> <sup>φ</sup>j*, *<sup>φ</sup><sup>i</sup>*

*N* X*<sup>x</sup>*þ<sup>1</sup> *j*¼1

and matrices **M** and **A** as

vectors **u**

**112**

! , **u** ! **<sup>0</sup>** and **F** !

*d*

(

<sup>1</sup> � *<sup>x</sup>*

so that we can write **U***<sup>h</sup>* as a linear combination of the basic elements:

*N* X*<sup>x</sup>*þ<sup>1</sup> *j*¼1

*N* X*<sup>x</sup>*þ<sup>1</sup> *j*¼1

corresponding space. In the case of *V*0,*h*, we have: f g *φ<sup>i</sup>*

*φi*ð Þ¼ *x*

*φ*1ð Þ¼ *x*

*φNx*þ<sup>1</sup>ð Þ¼ *x*

0 ¼ *a*<sup>1</sup> <*a*<sup>2</sup> < ⋯ < *aNx* <*aNx*þ<sup>1</sup> ¼ *L*, , *ai* ¼ ð Þ� *i* � 1 *h* (35)

� ∈P1, ∀*i* ¼ 1⋯*Nx* � �, (36)

*V*0,*<sup>h</sup>* ¼ f g *vh* ∈*Vh* : *vh*ð Þ¼ 0 *vh*ð Þ¼ *L* 0 *:* (37)

*Nx*

: , (39)

*Nx*

� �

**<sup>U</sup>**<sup>~</sup> *<sup>j</sup>*ðÞ� *<sup>t</sup> <sup>a</sup> <sup>φ</sup>j*, *<sup>φ</sup><sup>i</sup>*

, *aij* ≔ *a φi*, *φ<sup>j</sup>* � �

*<sup>i</sup>*¼2, we obtain

*<sup>i</sup>*¼<sup>2</sup> where ∀*i* ¼ 2⋯*Nx*.

, (38)

, (40)

ð Þ *x* , (41)

ð Þ *x* , (42)

¼ **F**, *φ<sup>i</sup>* ð Þ, (43)

� �, (44)

, (45)

½ � 0, *L* into partitions of equal distances *h*:

*Inverse Heat Conduction and Heat Exchangers*

In this section, we are going to demonstrate some numerical results for ð Þ u x, t ð Þ,q xð Þ in the inverse problem (1)–(5). Therefore the following examples are considered and the solution is obtained.

**Example 1.** Consider (1)–(4) with

$$u(\mathbf{x}, \mathbf{0}) = \sin \mathbf{x}, \quad \mathbf{0} \le \mathbf{x} \le \mathbf{1},\tag{49}$$

$$u(\mathbf{0}, \mathbf{t}) = \mathbf{0}, \quad \mathbf{0} \le t \le \mathbf{1} \tag{50}$$

$$u(\mathbf{1}, \mathbf{t}) = \sin(\mathbf{1})\mathbf{e}^{\cdot t}, \quad \mathbf{0} \le t \le \mathbf{1},\tag{51}$$

$$f(\mathbf{x}, t) = \left(\sin \mathbf{x} \left(\frac{\mathbf{x}^2}{4} + \frac{\mathbf{x}}{2} + \mathbf{1}\right)\right) e^{-t} - \frac{(\mathbf{x} + \mathbf{1})}{2} \cos \left(\mathbf{x}\right) e^{-t} - \sin \mathbf{x} e^{-t}, \quad \mathbf{0} \le \mathbf{x} \le \mathbf{1} \\ \mathbf{0} \le t \le \mathbf{1} \quad \{\mathbf{52}\}$$

We obtain the unique exact solution

$$q(\mathbf{x}) = \mathbf{1} + \mathbf{0}.5\mathbf{x} + \mathbf{0}.25\mathbf{x}^2 \tag{53}$$

And

$$u(\mathbf{x}, \mathbf{t}) = \sin\left(\mathbf{x}\right) \mathbf{e}^{\mathbf{t}}.\tag{54}$$

We take the observed data *g* as

$$\mathbf{g}(t) = \boldsymbol{\mu}(\mathbf{0}.5, \mathbf{t}) = \sin\left(\mathbf{0}.5\right) \mathbf{e}^{\mathbf{t}} \mathbf{0} \le t \le \mathbf{1} \tag{55}$$

The unknown function *q*ð Þ x defined as the following form

$$q(\hat{\mathbf{x}}) = p\_1 + p\_2 \mathbf{x} + p\_3 \mathbf{x}^2,\tag{56}$$

where *p*1, *p*2, *p*<sup>3</sup> are unknown coefficients.

**Table 1** shows how the Levenberg-Marquardt algorithm can find the best parameters after 12 iterations when it is initialized in four different points.

**Figures 1**–**4** show the fitness of the estimated parameters and the rate of convergence.

**Figures 5**–**8** show the comparison between the inversion results *q x* ^ð Þ and the exact value *q x*ð Þ:

**Table 2** shows the values of *q*ð Þ j*Δ*x and *u*ð Þ j*Δ*x, 0*:*5 in *x* ¼ *jΔx* with the all the initial values are set 1.

**Example 2.** Consider (1)–(4) with

$$u(\mathbf{x}, \mathbf{0}) = \mathbf{x}e^{\mathbf{x}}, \quad \mathbf{0} \le \mathbf{x} \le \mathbf{1},\tag{57}$$

$$u(\mathbf{0}, \mathbf{t}) = \mathbf{t} \mathbf{e}^t, \quad \mathbf{0} \le t \le \mathbf{1} \tag{58}$$

$$\mu(\mathbf{1}, \mathbf{t}) = (\mathbf{1} + \mathbf{t})\mathbf{e}^{\mathbf{1} \cdot \mathbf{t}}, \quad \mathbf{0} \le t \le \mathbf{1} \tag{59}$$

$$f(\mathbf{x}, \mathbf{t}) = \mathbf{e}^{\cdot(\mathbf{t}+\mathbf{x})} \cdot (\mathbf{t}+\mathbf{x}) \mathbf{e}^{\cdot(\mathbf{t}+\mathbf{x})} \cdot \mathbf{e}^{\mathbf{x}} \left(\mathbf{e}^{\cdot(\mathbf{t}+\mathbf{x})} \cdot (\mathbf{t}+\mathbf{x}) \mathbf{e}^{\cdot(\mathbf{t}+\mathbf{x})}\right)$$

$$+ \mathbf{e}^{\mathbf{x}} \Big(2\mathbf{e}^{\cdot(\mathbf{t}+\mathbf{x})} \cdot (\mathbf{t}+\mathbf{x}) \mathbf{e}^{\cdot(\mathbf{t}+\mathbf{x})}\Big), \quad 0 \le \mathbf{x} \le \mathbf{1} \\ 0 \le t \le 1 \tag{60}$$


**Table 1.**

*Performance of the algorithm when it is run to solve the model using three different parameters guesses.*

We obtain the unique exact solution

*All the initial values for the parameters are set 10.*

We take the observed data *g* as

And

**115**

**Figure 3.**

**Figure 2.**

*All the initial values for the parameters are set 1.*

*DOI: http://dx.doi.org/10.5772/intechopen.89096*

*q*ð Þ¼ x *e*

*A Numerical Approach to Solving an Inverse Heat Conduction Problem Using the Levenberg…*

*<sup>u</sup>*ð Þ¼ x, t ð Þ <sup>x</sup> <sup>þ</sup> <sup>t</sup> <sup>e</sup>‐*x*‐*<sup>t</sup>*

*<sup>x</sup>*, (61)

*:* (62)

**Figure 1.** *All the initial values for the parameters are set 0.5.*

*A Numerical Approach to Solving an Inverse Heat Conduction Problem Using the Levenberg… DOI: http://dx.doi.org/10.5772/intechopen.89096*

**Figure 2.** *All the initial values for the parameters are set 1.*

**Table 1** shows how the Levenberg-Marquardt algorithm can find the best parameters after 12 iterations when it is initialized in four different points. **Figures 1**–**4** show the fitness of the estimated parameters and the rate of

**Figures 5**–**8** show the comparison between the inversion results *q x* ^ð Þ and the

**Table 2** shows the values of *q*ð Þ j*Δ*x and *u*ð Þ j*Δ*x, 0*:*5 in *x* ¼ *jΔx* with the all the

*<sup>u</sup>*ð Þ¼ 0, t te‐*<sup>t</sup>*

*<sup>u</sup>*ð Þ¼ 1, t ð Þ <sup>1</sup> <sup>þ</sup> <sup>t</sup> <sup>e</sup>‐1‐*<sup>t</sup>*

<sup>þ</sup> <sup>e</sup>*<sup>x</sup>* 2e‐ð Þ <sup>t</sup>þ<sup>x</sup> ‐ð Þ <sup>t</sup> <sup>þ</sup> <sup>x</sup> <sup>e</sup>‐ð Þ <sup>t</sup>þ<sup>x</sup>

0.999729028233183 0.499885876453056 0.252009862457315

Error F 8.7564944405 � <sup>10</sup>�<sup>14</sup> 8.7564944427 � <sup>10</sup>�<sup>14</sup> 8.7564944420 � <sup>10</sup>�<sup>14</sup> 8.7564944420 � <sup>10</sup>�<sup>14</sup>

*Performance of the algorithm when it is run to solve the model using three different parameters guesses.*

*<sup>f</sup>*ð Þ¼ x, t <sup>e</sup>‐ð Þ <sup>t</sup>þ<sup>x</sup> ‐ð Þ <sup>t</sup> <sup>þ</sup> <sup>x</sup> <sup>e</sup>‐ð Þ <sup>t</sup>þ<sup>x</sup> ‐*<sup>e</sup>*

*<sup>u</sup>*ð Þ¼ x, 0 xe‐*x*, 0≤*x*≤1, (57)

*<sup>x</sup>* <sup>e</sup>‐ð Þ <sup>t</sup>þ<sup>x</sup> ‐ð Þ <sup>t</sup> <sup>þ</sup> <sup>x</sup> <sup>e</sup>‐ð Þ <sup>t</sup>þ<sup>x</sup>

0.999729028233194 0.499885876453057 0.252009862457325

**0.5 0.5 0.5 1 1 1 10 10 10 50 50 50**

, 0≤*t*≤ 1 (58)

, 0≤*t*≤1 (59)

, 0≤ *x*≤ 10≤*t*≤1 (60)

0.999729028307261 0.499885876454169 0.25200986249336

convergence.

**Starting point**

**Table 1.**

**Figure 1.**

**114**

*All the initial values for the parameters are set 0.5.*

Iteration 12 0.999729028233135

0.499885876453067 0.252009862457275

exact value *q x*ð Þ:

initial values are set 1.

**Example 2.** Consider (1)–(4) with

*Inverse Heat Conduction and Heat Exchangers*

**Figure 3.** *All the initial values for the parameters are set 10.*

We obtain the unique exact solution

$$q(\mathbf{x}) = e^{\mathbf{x}},\tag{61}$$

And

$$u(\mathbf{x}, \mathbf{t}) = (\mathbf{x} + \mathbf{t})\mathbf{e}^{\mathbf{x} \cdot \mathbf{t}}.\tag{62}$$

We take the observed data *g* as

**Figure 4.** *All the initial values for the parameters are set 50.*

**Figure 5.** *The comparison chart with all the initial values for the parameters is set 0.5.*

*g t*ðÞ¼ *<sup>u</sup>*ð Þ¼ <sup>0</sup>*:*5, t ð Þ <sup>0</sup>*:*<sup>5</sup> <sup>þ</sup> <sup>t</sup> <sup>e</sup>ð Þ ‐0*:*5‐<sup>t</sup> <sup>0</sup><sup>≤</sup> *<sup>t</sup>*≤1*:* (63)

The unknown function *q*ð Þ x defined as the following form

*The comparison chart with all the initial values for the parameters is set 50.*

*The comparison chart with all the initial values for the parameters is set 10.*

*DOI: http://dx.doi.org/10.5772/intechopen.89096*

*A Numerical Approach to Solving an Inverse Heat Conduction Problem Using the Levenberg…*

*p*1, *p*2, ⋯, *p*7, *p*<sup>8</sup> are unknown coefficients.

convergence.

**117**

**Figure 8.**

**Figure 7.**

exact value *q x*ð Þ:

^ *<sup>q</sup>*ð Þ¼ <sup>x</sup> *<sup>p</sup>*<sup>1</sup> <sup>þ</sup> *<sup>p</sup>*2*<sup>x</sup>* <sup>þ</sup> *<sup>p</sup>*3*x*<sup>2</sup> <sup>þ</sup> *<sup>p</sup>*4*x*<sup>3</sup> <sup>þ</sup> *<sup>p</sup>*5*x*<sup>4</sup> <sup>þ</sup> *<sup>p</sup>*6*x*<sup>5</sup> <sup>þ</sup> *<sup>p</sup>*7*x*<sup>6</sup> <sup>þ</sup> *<sup>p</sup>*8*x*7, where

**Table 3** shows how the Levenberg-Marquardt algorithm can find the best parameters after 20 iterations when it is initialized in four different points. **Figures 9**–**12** show the fitness of the estimated parameters and the rate of

**Figures 13**–**16** show the comparison between the inversion results *q x* ^ð Þ and the

**Figure 6.** *The comparison chart with all the initial values for the parameters is set 1.*

*A Numerical Approach to Solving an Inverse Heat Conduction Problem Using the Levenberg… DOI: http://dx.doi.org/10.5772/intechopen.89096*

**Figure 7.** *The comparison chart with all the initial values for the parameters is set 10.*

**Figure 8.** *The comparison chart with all the initial values for the parameters is set 50.*

$$\mathbf{g}(\mathbf{t}) = \boldsymbol{\mu}(\mathbf{0}.5, \mathbf{t}) = (\mathbf{0}.5 + \mathbf{t})\mathbf{e}^{(\cdot 0.5 \cdot \mathbf{t})}\mathbf{0} \le t \le 1. \tag{63}$$

The unknown function *q*ð Þ x defined as the following form

^ *<sup>q</sup>*ð Þ¼ <sup>x</sup> *<sup>p</sup>*<sup>1</sup> <sup>þ</sup> *<sup>p</sup>*2*<sup>x</sup>* <sup>þ</sup> *<sup>p</sup>*3*x*<sup>2</sup> <sup>þ</sup> *<sup>p</sup>*4*x*<sup>3</sup> <sup>þ</sup> *<sup>p</sup>*5*x*<sup>4</sup> <sup>þ</sup> *<sup>p</sup>*6*x*<sup>5</sup> <sup>þ</sup> *<sup>p</sup>*7*x*<sup>6</sup> <sup>þ</sup> *<sup>p</sup>*8*x*7, where *p*1, *p*2, ⋯, *p*7, *p*<sup>8</sup> are unknown coefficients.

**Table 3** shows how the Levenberg-Marquardt algorithm can find the best parameters after 20 iterations when it is initialized in four different points.

**Figures 9**–**12** show the fitness of the estimated parameters and the rate of convergence.

**Figures 13**–**16** show the comparison between the inversion results *q x* ^ð Þ and the exact value *q x*ð Þ:

**Figure 4.**

**Figure 5.**

**Figure 6.**

**116**

*All the initial values for the parameters are set 50.*

*Inverse Heat Conduction and Heat Exchangers*

*The comparison chart with all the initial values for the parameters is set 0.5.*

*The comparison chart with all the initial values for the parameters is set 1.*


**Figure 10.**

**Figure 11.**

**Figure 12.**

**119**

*All the initial values for the parameters are set 0.5.*

*DOI: http://dx.doi.org/10.5772/intechopen.89096*

*A Numerical Approach to Solving an Inverse Heat Conduction Problem Using the Levenberg…*

*All the initial values for the parameters are set 1.*

*All the initial values for the parameters are set 2.*

#### **Table 2.**

*The values of q*ð Þ j*Δ*x *and u*ð Þ j*Δ*x, 0*:*5 *in x* ¼ *jΔx with the all the initial values being set to 1.*


#### **Table 3.**

*Performance of the algorithm when it is run to solve the model using four different parameters guesses.*

**Figure 9.** *All the initial values for the parameters are set 0.1.*

*A Numerical Approach to Solving an Inverse Heat Conduction Problem Using the Levenberg… DOI: http://dx.doi.org/10.5772/intechopen.89096*

**Figure 10.** *All the initial values for the parameters are set 0.5.*

**Numerical Exact Numerical Exact**

*j q*ð Þ j*Δ*x *q*ð Þ j*Δ*x *u*ð Þ j*Δ*x, 0*:*5 *u*ð Þ j*Δ*x, 0*:*5 0 0.999729028233183 1 0 0

*The values of q*ð Þ j*Δ*x *and u*ð Þ j*Δ*x, 0*:*5 *in x* ¼ *jΔx with the all the initial values being set to 1.*

**0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5**

1.01536263500695 0.896348850692318 0.954285278486704 –0.890298849338373 1.40131909032315 –0.871276197318301 0.183785492186491 0.0359104061952515

Error *<sup>F</sup>* <sup>7</sup>*:*<sup>89749200363512</sup> � <sup>10</sup>‐<sup>11</sup> <sup>7</sup>*:*<sup>89749200363504</sup> � <sup>10</sup>‐<sup>11</sup> <sup>7</sup>*:*<sup>89749200353888</sup> � <sup>10</sup>‐<sup>11</sup> <sup>7</sup>*:*<sup>8974920035389</sup> � <sup>10</sup>‐<sup>11</sup>

*Performance of the algorithm when it is run to solve the model using four different parameters guesses.*

**1111 1111**

1.01536263525763 0.896348847022403 0.954285302637587 –0.890298935334171 1.40131926588117 –0.871276402487896 0.18378561965322 0.0359103735933848

**2222 2222**

1.01536263525905 0.896348846999736 0.954285302790922 –0.890298935876618 1.40131926696099 –0.87127640370648 0.183785620380814 0.0359103734148875

**Table 2.**

**Table 3.**

**Figure 9.**

**118**

*All the initial values for the parameters are set 0.1.*

**Starting point 0.1 0.1 0.1 0.1**

Iteration 20 1.01536263526644

**0.1 0.1 0.1 0.1**

*Inverse Heat Conduction and Heat Exchangers*

0.896348846894057 0.954285303464511 –0.890298938193057 1.40131927153131 –0.871276408882294 0.183785623507722 0.0359103726343979

1 1.05223771450306 1.0525 0.0605593190239173 0.0605520280601669 2 1.10978659802209 1.11 0.120511797611786 0.120499040271796 3 1.17237567879026 1.1725 0.179257059078521 0.179242065904716 4 1.24000495680758 1.24 0.236207449080344 0.236194164064666 5 1.31267443207404 1.3125 0.290793943250869 0.290786288212692 6 1.39038410458965 1.39 0.342471828361625 0.342472971890064 7 1.47313397435441 1.4725 0.390726114897089 0.390737778838824 8 1.56092404136831 1.56 0.435076630410587 0.435098463062163 9 1.65375430563136 1.6525 0.475082717530532 0.475111787267016 10 1.75162476714355 1.75 0.510347406713368 0.510377951544573

**Figure 11.** *All the initial values for the parameters are set 1.*

**Figure 12.** *All the initial values for the parameters are set 2.*

**Figure 13.** *The comparison chart with all the initial values for the parameters is set 0.1.*

**Figure 14.** *The comparison chart with all the initial values for the parameters is set 0.5.*

**Table 4** shows the values of *q*ð Þ j*Δ*x and *u*ð Þ j*Δ*x, 0*:*5 in *x* ¼ *jΔx* with the all the

**Numerical Exact Numerical Exact**

*j q*ð Þ j*Δ*x *q*ð Þ j*Δ*x *u*ð Þ j*Δ*x, 0*:*5 *u*ð Þ j*Δ*x, 0*:*5 0 1.01536263525763 1 0.303342962644088 0.303265329856317 1 1.11378168059013 1.10517091807565 0.329201964677126 0.329286981656416 2 1.22765694959399 1.22140275816017 0.347492882224886 0.347609712653987 3 1.35549021305874 1.349858807576 0.359347568702678 0.359463171293777 4 1.4973722149265 1.49182469764127 0.365808711321159 0.365912693766539 5 1.65432828415206 1.64872127070013 0.367792378208857 0.367879441171442 6 1.82785056869393 1.82211880039051 0.366091218454182 0.366158192067887 7 2.01963499046464 2.01375270747048 0.361387761473796 0.361433054294643 8 2.23154102006879 2.22554092849247 0.354267869273063 0.354291330944216 9 2.46579237015687 2.45960311115695 0.345233023618059 0.345235749518249 10 2.72543670622333 2.71828182845905 0.334712604803175 0.334695240222645

*A Numerical Approach to Solving an Inverse Heat Conduction Problem Using the Levenberg…*

A numerical method to estimate the temperature u x, t ð Þ and the coefficient *q*ð Þ x

1.The present study, successfully applies the numerical method involving the Levenberg-Marquardt algorithm in conjunction with the Galerkin finite

is proposed for an IHCP and the following results are obtained.

*The values of q*ð Þ j*Δ*x *and u*ð Þ j*Δ*x, 0*:*5 *in x* ¼ *jΔx with the all the initial values are set 1.*

*The comparison chart with all the initial values for the parameters is set 2.*

*DOI: http://dx.doi.org/10.5772/intechopen.89096*

initial values are set 1.

element method to an IHCP.

**6. Conclusions**

**Table 4.**

**121**

**Figure 16.**

**Figure 15.** *The comparison chart with all the initial values for the parameters is set 1.*

*A Numerical Approach to Solving an Inverse Heat Conduction Problem Using the Levenberg… DOI: http://dx.doi.org/10.5772/intechopen.89096*

**Figure 16.** *The comparison chart with all the initial values for the parameters is set 2.*


#### **Table 4.**

**Figure 13.**

**Figure 14.**

**Figure 15.**

**120**

*The comparison chart with all the initial values for the parameters is set 0.1.*

*Inverse Heat Conduction and Heat Exchangers*

*The comparison chart with all the initial values for the parameters is set 0.5.*

*The comparison chart with all the initial values for the parameters is set 1.*

*The values of q*ð Þ j*Δ*x *and u*ð Þ j*Δ*x, 0*:*5 *in x* ¼ *jΔx with the all the initial values are set 1.*

**Table 4** shows the values of *q*ð Þ j*Δ*x and *u*ð Þ j*Δ*x, 0*:*5 in *x* ¼ *jΔx* with the all the initial values are set 1.
