Efficient MFS Algorithms for Problems in Thermoelasticity

Abstract

We propose efficient fast Fourier transform (FFT)-based algorithms using the method of fundamental solutions (MFS) for the numerical solution of certain problems in planar thermoelasticity. In particular, we consider problems in domains possessing radial symmetry, namely disks and annuli and it is shown that the MFS matrices arising in such problems possess circulant or block-circulant structures. The solution of the resulting systems is facilitated by appropriately diagonalizing these matrices using FFTs. Numerical experiments demonstrating the applicability of these algorithms are also presented.

Appendix A

The following lemma establishes that the matrices A ₂₁ and A ₂₂ in (5.3) in the case of boundary conditions (2.1c) are circulant.

Lemma 1

The matrix

$$ (A )_{kl}= \frac{\partial}{\partial n} \log\sqrt {(x_{1_{k}}- \xi_{1_l})^2 + (x_{2_{k}}-\xi_{2_l})^2}, \quad k,l=1,\ldots, N $$

is circulant.

Proof

We consider the collocation points

$$ (x_{1_{k}}, x_{2_{k}}) = \varrho_{\ell_1} \biggl(\cos\frac{2 (i-1) \pi}{N}, \sin\frac{2 (i-1) \pi}{N} \biggr), \quad i=1,\ldots,N; \quad\ell_1=1 \mbox{ or } 2, $$

(A.1)

and the singularities

(A.2)

Then clearly,

$$ \frac{\partial}{\partial n} \log\sqrt{(x_{1_{k}}- \xi_{1_l})^2 + (x_{2_{k}}-\xi_{2_l})^2} = \frac{(x_{1_{k}}-\xi_{1_l}) {n_1}_k +(x_{2_{k}}-\xi_{2_l}) {n_2}_k}{(x_{1_{k}}-\xi_{1_l})^2 + (x_{2_{k}}-\xi_{2_l})^2}, $$

(A.3)

where ${\boldsymbol{n} }_{k}=(n_{1_{k}},n_{2_{k}})$ is the outward normal vector at the boundary point $(x_{1_{k}}, x_{2_{k}})$, We shall show that both the numerator and the denominator on the right hand side of (A.3) are circulant which yields the required result. Using expressions (A.2) and (A.3), the numerator $(x_{1_{k}}-\xi_{1_{l}}) {n_{1}}_{k} +(x_{2_{k}}-\xi_{2_{l}}) {n_{2}}_{k}$ becomes

which only depends only on l−k, hence the quantity is circulant.

Similarly, the denominator $(x_{1_{k}}-\xi_{1_{l}})^{2} + (x_{2_{k}}-\xi_{2_{l}})^{2}$ becomes

which again depends only on l−k and is therefore circulant. □

The following lemma establishes that the matrices $\tilde{B}_{21}$ and $\tilde{B}_{22}$ in (5.10) are block circulant.

Lemma 2

Let

$$ B_{kl} = \left ( \begin{array}{c@{\quad}c} T_{11}({\boldsymbol{x} }_k^{\ell_1},{\boldsymbol{\xi}}_l^{\ell_2}) & T_{12}({\boldsymbol{x} }_k^{\ell_1},{\boldsymbol{\xi}}_l^{\ell_2}) \\[4pt]T_{21}({\boldsymbol{x} }_k^{\ell _1},{\boldsymbol{\xi}}_l^{\ell_2}) & T_{22}({\boldsymbol{x} }_k^{\ell_1},{\boldsymbol{\xi}}_l^{\ell_2}) \end{array} \right ), \quad \ell_1, \ell_2=1,2. $$

Then

If k+j>N (respectively l+j>N) then k+j is replaced by k+j−N (respectively by l+j−N).

Proof

We shall first prove the results for ℓ ₁=ℓ ₂=ℓ. Using the notation $\alpha_{\ell_{1}}=\alpha_{\ell_{2}}= \alpha_{\ell }=\alpha$,

$$ \Delta x_{k,l} = {x_1}_k^\ell- {\xi_1}_l^\ell, \quad\quad\Delta y_{k,l} = {x_2}_k^\ell- {\xi_2}_l^\ell, \quad\quad r_{k,l} = \sqrt{(\Delta x_{k,l})^2 + ( \Delta y_{k,l})^2}, $$

(A.4)

from (3.4), by differentiation, we get

and

$$U_{{22}_y}(\Delta x_{k,l},\Delta y_{k,l})= \frac{(3-4 \bar{\nu}) (\Delta y_{k,l})^3+(1-4 \bar{\nu}) (\Delta x_{k,l})^2\Delta y_{k,l}}{r_{k,l}^4}. $$

Now, from (3.8) we obtain

(A.5)

(A.6)

(A.7)

and

(A.8)

Following [12, Lemma 3.1], it is sufficient to show that

Now, since

$$ (x_{k},y_{k} ) = \biggl(\varrho\cos{ \frac{2 \pi (k-1)}{N},\varrho\sin\frac{2 \pi(k-1)}{N}} \biggr) $$

(A.9)

and the coordinates of the singularity ξ _l are

$$ (x_{l},y_{l} ) = \biggl({R \cos \frac{2 \pi (l+\alpha-1)}{N},R \sin\frac{2 \pi(l+\alpha-1)}{N}} \biggr), $$

(A.10)

it follows that

(A.11)

From the expressions for the traction fundamental solutions, we obtain that

(A.12)

(A.13)

and

(A.14)

(A.15)

First we want to show, from Eqs. (A.5) and (A.12), that

(A.16)

From (A.9), (A.10) and (A.11)

and

(A.17)

In addition,

(A.18)

Finally,

and

(A.19)

From (A.17), (A.18) and (A.19), it easily follows that (A.16) holds.

Next we want to show, from Eqs. (A.8) and (A.15), that

(A.20)

We have that,

(A.21)

From (A.17), (A.19) and (A.21), it easily follows that (A.20) holds.

Next we want to show, from Eqs. (A.6) and (A.13), that

(A.22)

We have that

and

(A.23)

Clearly, from (A.19), (A.23), (A.17), (A.18) and (A.21), it follows that (A.22) holds.

Finally, we want to show, from Eqs. (A.7) and (A.14), that

(A.24)

Again, from (A.19), (A.23), (A.17), (A.18) and (A.21), it follows that (A.24) holds.

The proof for the case ℓ ₁≠ℓ ₂ is identical if instead of (A.4) we take

$$\Delta x_{k,l} = {x_1}_k^{\ell_1} - { \xi_1}_l^{\ell_2}, \quad\quad\Delta y_{k,l} = {x_2}_k^{\ell_1} - {\xi_2}_l^{\ell_2}, \quad\quad r_{k,l} = \sqrt{(\Delta x_{k,l})^2 + ( \Delta y_{k,l})^2}. $$

 □