4. BEM numerical implementation for displacement field

Based on Eqs. (2) and (3), Eq. (1) can be rewritten as

$$L\_{\rm fl}u\_k = \rho \ddot{u}\_p \cdot \left(D\_{\rm fl}u\_k + D\_{\rm pj}T\_a\right) = \rho \ddot{u}\_p \cdot \rho b\_p = \rho b\_p^\circ \tag{39}$$

where

$$L\_{fl} = D\_l \frac{\partial}{\partial x\_l}, D\_{fl} = \mu H\_0^2 D\_l \frac{\partial}{\partial x\_l}, D\_{pf} = \cdot\\\beta\_{pf} D\_{fl}, D\_l = C\_{pfkl} \frac{\partial}{\partial x\_l}, D\_f = \left(\frac{\partial}{\partial x\_l} + \Lambda\right), \Lambda = \frac{m}{x+1} \tag{40}$$

when the temperatures are known, the displacement can be computed by solving (39) using BEM. By choosing u<sup>∗</sup> <sup>p</sup> as the weight function and applying the weighted residual method, Eq. (39) can be reexpressed as

Mechanics of Functionally Graded Materials and Structures

$$\int\_{\mathbf{R}} \left< L\_{fl} u\_{k^\*} \cdot \rho b\_p^{\ast} \right> u\_p^{\ast} d\mathbf{R} = \mathbf{0} \tag{41}$$

From (39), (50) and (51), the representation formula may be written as

Boundary Element Model for Nonlinear Fractional-Order Heat Transfer in Magneto…

The displacement particular solution may be defined as

Now, we obtain the traction particular solution t

t q

The use of (57) together with the dual reciprocity

mp � Ljlu<sup>∗</sup>

mkuq pn � �d<sup>R</sup> <sup>¼</sup>

ð R

Leads to

37

Ljlu<sup>q</sup> knu<sup>∗</sup>

The domain integral may be approximated as follows

pn <sup>¼</sup> Cpjklu<sup>q</sup>

kn,l

nj, Ljlu<sup>q</sup>

ð C u∗ mpt q pn � <sup>t</sup> <sup>∗</sup> mpuq pn � �dC (58)

q

kn ¼ f q

pn and source function f

pn (56)

Differentiation of (54) leads to.

DOI: http://dx.doi.org/10.5772/intechopen.88255

Let

ð52Þ

ð53Þ

ð54Þ

ð55Þ

ð57Þ

ð59Þ

q pn as

The first term in (41) can be integrated partially using Gau β theory yields

$$\int\_{\mathcal{R}} \mathbf{C}\_{pjkl} u\_{k,lj} u\_p^\* \, d\mathcal{R} = \int\_{\mathcal{C}} \mathbf{C}\_{pjkl} u\_{k,l} u\_p^\* \, n\_j d\mathcal{C} - \int\_{\mathcal{R}} \mathbf{C}\_{pjkl} u\_{k,l} u\_{p,j}^\* d\mathcal{R} \tag{42}$$

The last term in (42) can be integrated partially twice using Gau β theory yields

$$\int\_{\mathcal{R}} \mathbf{C}\_{pjkl} u\_{k,l} u\_{p,j}^\* d\mathbf{R} = \int\_{\mathcal{C}} \mathbf{C}\_{pjkl} u\_k u\_{p,j}^\* n\_l d\mathbf{C} - \int\_{\mathcal{R}} \mathbf{C}\_{pjkl} u\_k u\_{p,j}^\* d\mathbf{R} \tag{43}$$

Based on Eq. (43), Eq. (42) can be rewritten as

$$\int\_{\mathcal{R}} \mathbf{C}\_{\mathrm{pjk}l} \boldsymbol{\mu}\_{k,l} \boldsymbol{\mu}\_{p}^{\*} \, d\mathcal{R} - \int\_{\mathcal{R}} \mathbf{C}\_{\mathrm{pjk}l} \boldsymbol{\mu}\_{k} \boldsymbol{u}\_{p,j}^{\*} \, d\mathcal{R} = \int\_{\mathcal{C}} \mathbf{C}\_{\mathrm{pjk}l} \boldsymbol{\mu}\_{k,l} \boldsymbol{u}\_{p}^{\*} \boldsymbol{n}\_{j} \mathrm{d}\mathcal{C} - \int\_{\mathcal{C}} \mathbf{C}\_{\mathrm{pjk}l} \boldsymbol{\mu}\_{k} \boldsymbol{u}\_{p,j}^{\*} \boldsymbol{n}\_{l} \mathrm{d}\mathcal{C} \tag{44}$$

which can be written as

$$\int\_{\mathbb{R}} \left\{ L\_{jl} u\_k \cdot u\_p^\* \cdot L\_{jl}^\* u\_k^\* \cdot u\_p \right\} d\mathbb{R} = \int\_{\mathbb{C}} \left\{ G\_{jl} u\_k \cdot u\_p^\* \cdot G\_{jl}^\* u\_k^\* \cdot u\_p \right\} d\mathbb{C} \tag{45}$$

The boundary tractions are

$$t\_p = \mathcal{C}\_{\text{pjk}} u\_{k,l} n\_j = \mathcal{G}\_{\text{jl}} u\_k \text{ and } t\_p^\* = \mathcal{C}\_{\text{pjk}} u\_{k,j}^\* n\_l = \mathcal{G}\_{\text{jl}}^\* u\_k^\* \tag{46}$$

By using the symmetry relation of elasticity tensor, we obtain

$$L\_{fl} = C\_{pfkl} \Re \frac{\partial^2}{\partial \mathbf{x}\_l \partial \mathbf{x}\_l} = C\_{kfpl} \Re \frac{\partial^2}{\partial \mathbf{x}\_l \partial \mathbf{x}\_l} = L\_{fl}^\* \tag{47}$$

$$\mathcal{L}\_{fl} = \mathcal{L}\_{pflkl} \mathfrak{N} n\_f \frac{\partial}{\partial \mathfrak{X}\_l} = \mathcal{L}\_{kfpl} \mathfrak{N} n\_l \frac{\partial}{\partial \mathfrak{X}\_l} = \mathcal{G}\_{fl}^\* \tag{48}$$

Using Eqs. (46)–(48), the Eq. (45) can be reexpressed as

$$\int\_{\mathbb{R}} \left\{ \mathfrak{C}\_{pjkl} u\_{k,lj} \cdot u\_p^\* \cdot \mathcal{C}\_{pjkl} u\_{k,lj}^\* \cdot u\_p \right\} d\mathbb{R} = \int\_{\mathbb{C}} \left\{ \mathfrak{t}\_p u\_p^\* \cdot \mathfrak{t}\_p^\* u\_p \right\} d\mathbb{C} \tag{49}$$

We define the fundamental solution u<sup>∗</sup> mk by the relation

$$L\_{jl}u\_{mk}^{"\prime} = \cdot \delta(\mathfrak{x}, \xi)\delta\_{pm} \tag{50}$$

By modifying the weighting functions, Eq. (49) can be written as

$$\int\_{\mathbb{R}} \left\{ \mathcal{L}\_{pflkl} u\_{k,lf} \cdot u\_{mp}^\* \cdot \mathcal{L}\_{pflkl} u\_{mk,lf}^\* \cdot u\_p \right\} d\mathbb{R} = \int\_{\mathbb{C}} \left\{ \mathfrak{t}\_p u\_{mp}^\* \cdot \mathfrak{t}\_{mp}^\* u\_p \right\} d\mathbb{C} \qquad (51)$$

Boundary Element Model for Nonlinear Fractional-Order Heat Transfer in Magneto… DOI: http://dx.doi.org/10.5772/intechopen.88255

From (39), (50) and (51), the representation formula may be written as

$$\begin{aligned} &u\_m(\xi) \\ &= \int\_{\mathbf{C}} \left\{ u\_{mp}^\*(\mathbf{x}, \xi) t\_p(\mathbf{x}) \cdot t\_{mp}^\*(\mathbf{x}, \xi) u\_p(\mathbf{x}) \right\} d\mathbf{C} \cdot \int\_{\mathbf{R}} u\_{mp}^\*(\mathbf{x}, \xi) \rho b\_p^"(\mathbf{x}) d\mathbf{R} \end{aligned} \tag{52}$$

Let

ð41Þ

dR (42)

p,jldR (43)

nldC (44)

<sup>k</sup> (46)

ð45Þ

ð47Þ

ð48Þ

ð49Þ

ð50Þ

ð51Þ

The first term in (41) can be integrated partially using Gau β theory yields

Cpjkluk,lu<sup>∗</sup>

Cpjkluku<sup>∗</sup> p,j nldC �

> ð C

The last term in (42) can be integrated partially twice using Gau β theory yields

Cpjkluk,lu<sup>∗</sup>

<sup>p</sup> njdC �

ð R

ð R

<sup>p</sup> njdC �

<sup>p</sup> <sup>¼</sup> Cpjklu<sup>∗</sup>

mk by the relation

k,j

nl <sup>¼</sup> <sup>G</sup><sup>∗</sup>

jl u<sup>∗</sup>

Cpjkluk,lu<sup>∗</sup>

Cpjkluku<sup>∗</sup>

ð C p,j

Cpjkluku<sup>∗</sup> p,j

ð R

> ð R

<sup>p</sup> dR �

which can be written as

The boundary tractions are

ð R

36

Cpjkluk,lju<sup>∗</sup>

Cpjkluk,lju<sup>∗</sup>

Cpjkluk,lu<sup>∗</sup>

ð R <sup>p</sup> dR ¼

Mechanics of Functionally Graded Materials and Structures

p,j dR ¼ ð C

Based on Eq. (43), Eq. (42) can be rewritten as

Cpjkluku<sup>∗</sup>

ð C

p,jldR ¼

tp <sup>¼</sup> Cpjkluk,lnj <sup>¼</sup> Gjluk and <sup>t</sup> <sup>∗</sup>

Using Eqs. (46)–(48), the Eq. (45) can be reexpressed as

By modifying the weighting functions, Eq. (49) can be written as

We define the fundamental solution u<sup>∗</sup>

By using the symmetry relation of elasticity tensor, we obtain

$$
\rho \boldsymbol{b}\_{\mathcal{p}}^{\cdot} = \cdot \left( \left( \boldsymbol{D}\_{jl} + \boldsymbol{D}\_{pk} + \boldsymbol{\Lambda} \boldsymbol{D}\_{l} \right) \boldsymbol{u}\_{k} + \boldsymbol{D}\_{pj} \boldsymbol{\tau}\_{\boldsymbol{\sigma}} \right),
$$

$$\rho \approx \sum\_{a=1}^{N} f\_{pn}^{q} \propto\_{n}^{q} = \sum\_{a=1}^{N} \left( L\_{ll} u\_{kn}^{q} \right) \propto\_{n}^{q}, D\_{pk} = \cdot \rho \delta\_{pk} \frac{\partial^{2}}{\partial \tau^{2}} \tag{53}$$

The displacement particular solution may be defined as

$$
\mu\_{kn}^q = \delta\_{kn} (r^2 + r^3) \tag{54}
$$

Differentiation of (54) leads to.

$$u\_{kn.l}^q = \delta\_{kn} (2r + 3r^2) r\_{\mu}, \qquad u\_{kn.l i}^q = \delta\_{kn} ((2 + 3r)\delta\_{lj} + 3rr\_{jl}r\_{\mu}) \tag{55}$$

Now, we obtain the traction particular solution t q pn and source function f q pn as

$$\mathbf{t}\_{\rm pn}^{\rm q} = \mathbf{C}\_{\rm pjkl} \mathbf{u}\_{\rm kn, l}^{\rm q} \mathbf{n}\_{\rm j}, \mathbf{L}\_{\rm jl} \mathbf{u}\_{\rm kn}^{\rm q} = \mathbf{f}\_{\rm pn}^{\rm q} \tag{56}$$

The domain integral may be approximated as follows

$$\int\_{\mathbb{R}} u\_{mp}^\* \rho \, b\_p^\dagger d\mathbb{R} \approx \sum\_{q=1}^{\mathbb{N}} \left( \int\_{\mathbb{R}} \left( L\_{fl} u\_{kn}^q u\_{mp}^\* \right) d\mathbb{R} \right) \ll\_n^q \tag{57}$$

The use of (57) together with the dual reciprocity

$$\int\_{\mathcal{R}} \left( L\_{jl} u\_{kn}^{q} u\_{mp}^{\*} - L\_{jl} u\_{mk}^{\*} u\_{pn}^{q} \right) d\mathcal{R} = \int\_{\mathcal{C}} \left( u\_{mp}^{\*} t\_{pn}^{q} - t\_{mp}^{\*} u\_{pn}^{q} \right) d\mathcal{C} \tag{58}$$

Leads to

$$\begin{aligned} \int\_{\mathbb{R}} \
u\_{mp}^{\*} \rho \mathbf{b}\_{p}^{\*} d\mathbf{R} &= \sum\_{\mathbf{q}=1}^{N} \int\_{\mathcal{R}} \left( \mathbb{L}\_{fl} \boldsymbol{u}\_{mk}^{\*} \boldsymbol{u}\_{pn}^{q} \right) d\mathbf{R} \\ &+ \int\_{\mathcal{C}} \left( \boldsymbol{u}\_{mp}^{\*} \boldsymbol{t}\_{pn}^{q} \cdot \boldsymbol{t}\_{mp}^{\*} \boldsymbol{u}\_{pn}^{q} \right) d\mathbf{C} \right) \boldsymbol{\alpha}\_{n}^{q} \end{aligned} (59)$$

From (50), we can write

$$\int\_{\mathbb{R}} L\_{fl} u\_{mk}^{\ast} u\_{pn}^{q} \, d\mathbb{R} = \int\_{\mathbb{R}} \cdot \delta\left(\mathfrak{x}, \xi\right) \delta\_{pm} u\_{pn}^{q} \, d\mathbb{R} = \cdot u\_{mn}^{q}(\xi) \tag{60}$$

By using (64) and (65), the dual reciprocity boundary integral equation becomes

Boundary Element Model for Nonlinear Fractional-Order Heat Transfer in Magneto…

By implementing the point collocation procedure and using (67) and (68),

By using the point collocation procedure, can be calculated from (53) as

On the basis of isoparametric concept, we can write

DOI: http://dx.doi.org/10.5772/intechopen.88255

Eq. (66) may be reexpressed as

Let us suppose that

We can write (69) as follows

Now, from (74), we may derive

From (73) using (75) we have

39

ð66Þ

ð67Þ

ð68Þ

ð69Þ

ð70Þ

ð71Þ

ð72Þ

ð73Þ

ð74Þ

ð75Þ

ð76Þ

By using (52), (59) and (60), we obtain

$$\begin{split} \mathfrak{u}u\_m(\xi) &= \int\_{\mathbb{C}} \left< u\_{mp}^\* t\_p \cdot t\_{mp}^\* u\_p \right> d\mathbb{C} \\ &+ \sum\_{q=1}^N \left( u\_{mn}^q(\xi) \cdot \int\_{\mathbb{C}} \left< u\_{mp}^\* t\_{pn}^q \cdot t\_{mp}^\* u\_{pn}^q \right> d\mathbb{C} \right) \mathfrak{a}\_n^{\mathbf{q}} \end{split} \tag{61}$$

According to Fahmy [9–11], the right-hand side integrals of (61) can be reexpressed as

$$\begin{split} \int\_{\mathbb{C}} \left( u\_{mp}^{\*} (\mathbf{x}, \xi) t\_{p} (\mathbf{x}) \cdot t\_{mp}^{\*} (\mathbf{x}, \xi) u\_{p} (\xi) \right) d\mathbb{C} \\ = & \lim\_{\varepsilon \to 0} \int\_{\mathbb{C} \cdot \mathbb{C}\_{\varepsilon}} \left( u\_{mp}^{\*} (\mathbf{x}, \xi) t\_{p} (\mathbf{x}) \cdot t\_{mp}^{\*} (\mathbf{x}, \xi) u\_{p} (\xi) \right) d\mathbb{C} \\ \quad + & \lim\_{\varepsilon \to 0} \int\_{\mathbb{C}\_{\varepsilon}} \left( u\_{mp}^{\*} (\mathbf{x}, \xi) t\_{p} (\mathbf{x}) \cdot t\_{mp}^{\*} (\mathbf{x}, \xi) u\_{p} (\xi) \right) d\mathbb{C}\_{\varepsilon} \end{split} (62)$$

and

$$\int\_{\mathbb{C}} \left( u\_{mp}^{\*} (\boldsymbol{\chi}, \boldsymbol{\xi}) t\_{pn}^{q} (\boldsymbol{\chi}) \cdot t\_{mp}^{\*} (\boldsymbol{\chi}, \boldsymbol{\xi}) u\_{pn}^{q} (\boldsymbol{\xi}) \right) d\mathbb{C} = $$
 
$$\lim\_{\boldsymbol{\xi} \to \boldsymbol{0}} \int\_{\mathbb{C} \cdot \mathbb{C}\_{\boldsymbol{\sigma}}} \left( u\_{mp}^{\*} (\boldsymbol{\chi}, \boldsymbol{\xi}) t\_{pn}^{q} (\boldsymbol{\chi}) \cdot t\_{mp}^{\*} (\boldsymbol{\chi}, \boldsymbol{\xi}) u\_{pn}^{q} (\boldsymbol{\xi}) \right) d\mathbb{C} \tag{63} $$
 
$$\quad + \lim\_{\boldsymbol{\xi} \to \boldsymbol{0}} \int\_{\mathbb{C}\_{\boldsymbol{\sigma}}} \left( u\_{mp}^{\*} (\boldsymbol{\chi}, \boldsymbol{\xi}) t\_{pn}^{q} (\boldsymbol{\chi}) \cdot t\_{mp}^{\*} (\boldsymbol{\chi}, \boldsymbol{\xi}) u\_{pn}^{q} (\boldsymbol{\xi}) \right) d\mathbb{C}\_{\boldsymbol{\mathfrak{s}}} $$

According to Fahmy [12], Guiggiani and Gigante [77] and Mantič [78] Eqs. (62) and (63) can respectively be expressed as

$$\begin{aligned} \int\_{\mathbb{C}} \left( u\_{mp}^{\*} (\mathbf{x}, \xi) t\_{p} (\mathbf{x}) \cdot t\_{mp}^{\*} (\mathbf{x}, \xi) u\_{p} (\xi) \right) d\mathbb{C} \\ &= \int\_{\mathbb{C}} u\_{mp}^{\*} t\_{p} d\Gamma \cdot \oint\_{\mathbb{C}} u\_{p} \, t\_{mp}^{\*} d\Gamma \cdot c\_{p} \mu\_{p} (\xi) \end{aligned} \quad (64)$$
 
$$\begin{aligned} \int\_{\mathbb{C}} \left( u\_{mp}^{\*} (\mathbf{x}, \xi) t\_{pm}^{q} (\mathbf{x}) \cdot t\_{mp}^{\*} (\mathbf{x}, \xi) u\_{pm}^{q} (\xi) \right) d\mathbb{C} \\ &= \int\_{\mathbb{C}} u\_{mp}^{\*} t\_{pm}^{q} d\Gamma \cdot \oint\_{\mathbb{C}} u\_{pn}^{q} t\_{mp}^{\*} d\Gamma \cdot c\_{p} \mu\_{p}^{q} (\xi) \end{aligned} \quad (65)$$

Boundary Element Model for Nonlinear Fractional-Order Heat Transfer in Magneto… DOI: http://dx.doi.org/10.5772/intechopen.88255

By using (64) and (65), the dual reciprocity boundary integral equation becomes

$$\begin{split} c\_{p\_l} u\_p(\varepsilon) &+ \oint\_{\mathbb{C}} u\_p \, t\_{mp}^\* d\Gamma - \int\_{\mathbb{C}} u\_{mp}^\* \, t\_p d\Gamma \\ &= \sum\_{q=1}^N \left( c\_{p\_l} u\_{pn}^q(\varepsilon) + \oint\_{\mathbb{C}} u\_{pn}^q \, t\_{mp}^\* d\Gamma - \int\_{\mathbb{C}} u\_{mp}^\* \, t\_{pn}^q d\Gamma \right) a\_n^{\mathbf{q}} \end{split} (66)$$

On the basis of isoparametric concept, we can write

$$\{\!\!\!\!\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/)^{\times}\tag{67}$$

$$\{u^q, t^q\} \approx \sum\_{k=1}^N \wp\_k \left\{\vec{u}^q\_k, \vec{t}^q\_k\right\} = \Phi^\tau \{\vec{u}^q, \vec{t}^q\} \tag{68}$$

By implementing the point collocation procedure and using (67) and (68), Eq. (66) may be reexpressed as

$$\zeta \uplus \eta \pounds = \sum\_{q=1}^{N} (\zeta \uplus^q \cdot \eta \pounds^q) \alpha^q(\mathfrak{r}) \tag{69}$$

Let us suppose that

From (50), we can write

reexpressed as

and

38

By using (52), (59) and (60), we obtain

Mechanics of Functionally Graded Materials and Structures

and (63) can respectively be expressed as

According to Fahmy [9–11], the right-hand side integrals of (61) can be

According to Fahmy [12], Guiggiani and Gigante [77] and Mantič [78] Eqs. (62)

ð60Þ

ð61Þ

ð62Þ

ð63Þ

ð64Þ

ð65Þ

$$\breve{U} = [\breve{u}^1 \ \breve{u}^2 \ \dots \ \breve{u}^N] \tag{70}$$

$$\ngg = [\overleftarrow{t}^1 \overleftarrow{t}^2 \dots \overleftarrow{t}^N] \tag{71}$$

$$\alpha = [\alpha^1 \ \alpha^2 \ \dots \ \alpha^N]^T \tag{72}$$

We can write (69) as follows

$$
\zeta \sharp \mathfrak{i}(\mathfrak{r}) \cdot \mathfrak{r} \acute{\mathfrak{i}}(\mathfrak{r}) = \left( \zeta \breve{U} \cdot \eta \breve{\wparrow} \right) \mathfrak{a} \left( \mathfrak{r} \right) \tag{73}
$$

By using the point collocation procedure, can be calculated from (53) as

$$
\rho \circ \tilde{\mathfrak{u}}(\mathfrak{r}) \cdot \rho \check{b}(\mathfrak{r}) = F a(\mathfrak{r}) \tag{74}
$$

Now, from (74), we may derive

$$a(\mathbf{r}) = F^{-1}\{\rho \check{\tilde{u}}(\mathbf{r}) \cdot \rho \check{b}(\mathbf{r})\}\tag{75}$$

From (73) using (75) we have

$$
\mathcal{M}\check{\tilde{u}} + \zeta \check{u} = \eta \check{\mathfrak{z}}(\tau) + \check{\mathcal{B}}(\tau) \tag{76}
$$

where

$$\mathcal{U} = \{\eta \check{\rho} \text{-} \zeta \check{U}\} F^{-1}, \qquad \mathcal{M} = \rho \mathfrak{U}, \qquad \check{\mathcal{B}}(\mathfrak{r}) = \rho \mathfrak{U} \check{\mathcal{B}}(\mathfrak{r}). \tag{77}$$

By considering the following known k and unknown u superscripts nodal vectors

$$\{\check{u}^{k},\check{t}^{u}\} \in \mathbb{C}\_{\simeq} \{\check{u}^{u},\check{t}^{k}\} \in \mathbb{C}\_{4} \tag{78}$$

ð85Þ

ð86Þ

ð87Þ

ð88Þ

ð89Þ

By using (85) and (86), we have from (83)

DOI: http://dx.doi.org/10.5772/intechopen.88255

Boundary Element Model for Nonlinear Fractional-Order Heat Transfer in Magneto…

u

5. Numerical results and discussion

<sup>n</sup>þ<sup>1</sup> from (73).

We implement the successive over-relaxation (SOR) of Golub and Van Loan [79] for solving (87) to obtain . Then, the unknown and can be obtained from (76) and (77), respectively. By using the procedure of Bathe [80], we

The BEM that has been used in the current paper can be applicable to a wide variety of FGA structures problems associated with the proposed theory of three temperatures nonlinear uncoupled magneto-thermoelasticity. In order to evaluate the influence of graded parameter on the three temperatures and displacements, the

numerical results are carried out and depicted graphically for homogeneous

(m ¼ 0) and functionally graded (m ¼ 0:5 and 1:0) structures.

Variation of the electron temperature Te through the thickness coordinate x.

In which.

obtain the traction vector t

Figure 2.

41

Hence (76) may be written as

$$
\begin{bmatrix} \mathcal{M}^{11} & \mathcal{M}^{12} \\ \mathcal{M}^{21} & \mathcal{M}^{22} \end{bmatrix} \begin{bmatrix} \widetilde{\mathfrak{u}}^{k}(\mathfrak{r}) \\ \widetilde{\mathfrak{u}}^{u}(\mathfrak{r}) \end{bmatrix} + \begin{bmatrix} \zeta^{11} & \zeta^{12} \\ \zeta^{21} & \zeta^{22} \end{bmatrix} \begin{bmatrix} \widetilde{\mathfrak{u}}^{k}(\mathfrak{r}) \\ \widetilde{\mathfrak{u}}^{u}(\mathfrak{r}) \end{bmatrix} = \begin{bmatrix} \eta^{11} & \eta^{12} \\ \eta^{21} & \eta^{22} \end{bmatrix} \begin{bmatrix} \widetilde{\mathfrak{r}}^{k}(\mathfrak{r}) \\ \widetilde{\mathfrak{r}}^{u}(\mathfrak{r}) \end{bmatrix} + \begin{bmatrix} \widetilde{\mathfrak{B}}^{1}(\mathfrak{r}) \\ \widetilde{\mathfrak{B}}^{2}(\mathfrak{r}) \end{bmatrix} \begin{pmatrix} \mathcal{M}^{1}(\mathfrak{r}) \\ \mathcal{M}^{2}(\mathfrak{r}) \end{pmatrix}
$$

From the first row of (79), we can calculate the unknown fluxes as follows

$$\begin{cases} \dot{t}^{u}(\tau) \\ = (\eta^{12})^{\cdot 1} [\mathcal{M}^{11}\dot{\tilde{u}}^{k}(\tau) + \mathcal{M}^{12}\dot{\tilde{u}}^{u}(\tau) + \zeta^{11}\tilde{u}^{k}(\tau) \\ + \zeta^{12}\tilde{u}^{u}(\tau)\cdot\eta^{11}\dot{t}^{k}(\tau)\cdot\tilde{\mathcal{B}}^{1}(\tau)] \end{cases} \tag{80}$$

From the second row of (79) and using (80) we get

$$\mathcal{M}^u \check{\mathfrak{U}}^u(\mathfrak{r}) + \zeta^u \check{\mathfrak{u}}^u(\mathfrak{r}) = Q^k(\mathfrak{r}) \tag{81}$$

where

$$Q^k(\tau) = \breve{\mathcal{B}}^k(\tau) + \eta^k \breve{\xi}^k(\tau) \cdot \mathcal{M}^k \breve{\mathcal{U}}^k(\tau) \cdot \zeta^k \breve{\mathcal{U}}^k(\tau)$$

$$\mathcal{M}^u = \mathcal{M}^{22} \cdot \eta^{22} (\eta^{12})^{-1} \mathcal{M}^{12}$$

$$\zeta^u = \zeta^{22} \cdot \eta^{22} (\eta^{12})^{-1} \zeta^{12} \tag{82}$$

$$\mathcal{M}^k = \mathcal{M}^{21} \cdot \eta^{22} (\eta^{12})^{-1} \mathcal{M}^{11}$$

$$\zeta^k = \zeta^{21} \cdot \eta^{22} (\eta^{12})^{-1} \zeta^{11}$$

$$\breve{\mathcal{B}}^k(\tau) = \mathcal{B}^2(\tau) \cdot \eta^{22} (\eta^{12})^{-1} \mathcal{B}^1(\tau)$$

Eq. (81) can be written at ð Þ n þ 1 time step as

$$\mathcal{M}^u \check{u}\_{n+1}^u(\tau) + \zeta^u \check{u}\_{n+1}^u(\tau) = Q\_{n+1}^k(\tau) \tag{83}$$

where

$$\mathcal{Q}\_{n+1}^{k}(\tau) = \check{\mathcal{B}}\_{n+1}^{k}(\tau) + \eta^{k} \check{t}\_{n+1}^{k}(\tau) \cdot \mathcal{M}^{k} \check{\mathcal{U}}\_{n+1}^{k}(\tau) \cdot \zeta^{k} \check{u}\_{n+1}^{k}(\tau) \tag{84}$$

In order to solve (83), The implicit backward finite difference scheme has been applied based on the Houbolt's algorithm and the following approximations

Boundary Element Model for Nonlinear Fractional-Order Heat Transfer in Magneto… DOI: http://dx.doi.org/10.5772/intechopen.88255

$$\check{\mathfrak{u}}\_{n+1} \approx \frac{1}{6\Delta t} (1\,\mathrm{1\,\upmu\_{n+1}} \cdot 18\,\upmu\_n + 9\,\upmu\_{n\cdot 1} \cdot 2\,\upmu\_{n\cdot 2})\tag{85}$$

$$\check{\mathfrak{u}}\_{n+1} \approx \frac{1}{\Delta t^2} (2\check{\mathfrak{u}}\_{n+1} \cdot 5\check{\mathfrak{u}}\_n + 4\check{\mathfrak{u}}\_{n\cdot 1} \cdot \check{\mathfrak{u}}\_{n\cdot 2}) \tag{86}$$

By using (85) and (86), we have from (83)

$$
\zeta^u \sharp\_{n+1}^u(\tau) = \mathbb{Q}\_{n+1}^k(\tau) \tag{87}
$$

In which.

ð77Þ

ð78Þ

ð79Þ

ð80Þ

ð81Þ

ð82Þ

ð83Þ

ð84Þ

By considering the following known k and unknown u superscripts nodal vectors

From the first row of (79), we can calculate the unknown fluxes as follows

In order to solve (83), The implicit backward finite difference scheme has been

applied based on the Houbolt's algorithm and the following approximations

where

where

where

40

Hence (76) may be written as

Mechanics of Functionally Graded Materials and Structures

From the second row of (79) and using (80) we get

Eq. (81) can be written at ð Þ n þ 1 time step as

$$
\zeta^u = \frac{2\mathcal{M}^u}{\star^z} + \zeta^u \tag{88}
$$

$$\mathbb{Q}\_{n+1}^k = Q\_{n+1}^k + \frac{\mathcal{M}^u}{\Delta t^2} (\mathbb{S}\mathbb{I}\_n \cdot \mathbb{A}\mathbb{I}\_{n\cdot 1} + \mathbb{I}\_{n\cdot 2}) \tag{89}$$

We implement the successive over-relaxation (SOR) of Golub and Van Loan [79] for solving (87) to obtain . Then, the unknown and can be obtained from (76) and (77), respectively. By using the procedure of Bathe [80], we obtain the traction vector t u <sup>n</sup>þ<sup>1</sup> from (73).
