Spatially Dispersionless, Unconditionally Stable FC–AD Solvers for Variable-Coefficient PDEs

Abstract

We present fast, spatially dispersionless and unconditionally stable high-order solvers for partial differential equations (PDEs) with variable coefficients in general smooth domains. Our solvers, which are based on (i) A certain “Fourier continuation” (FC) method for the resolution of the Gibbs phenomenon on equi-spaced Cartesian grids, together with (ii) A new, preconditioned, FC-based solver for two-point boundary value problems for variable-coefficient Ordinary Differential Equations, and (iii) An Alternating Direction strategy, generalize significantly a class of FC-based solvers introduced recently for constant-coefficient PDEs. The present algorithms, which are applicable, with high-order accuracy, to variable-coefficient elliptic, parabolic and hyperbolic PDEs in general domains with smooth boundaries, are unconditionally stable, do not suffer from spatial numerical dispersion, and they run at Fast Fourier Transform speeds. The accuracy, efficiency and overall capabilities of our methods are demonstrated by means of applications to challenging problems of diffusion and wave propagation in heterogeneous media.

Appendix: An Auxiliary Lemma

Lemma 1

Let $\tilde{q}^{\ell }, g_{a}$ and $g_{b}$ be smooth functions defined in the interval $[b,c]$, and let $\tilde{q}^{\ell }$ be strictly positive in that interval. If $g_{a}$ and $g_{b}$ satisfy the conditions (25), then the overdetermined ODE problem

$$\begin{aligned}&v-\tilde{p}\frac{dv}{dx} -\tilde{q}^{\ell }\frac{d^{2}v}{dx^2}=g_{a}+\mu g_{b} \quad \mathrm{in}\ (b,c),\end{aligned}$$

(42)

$$\begin{aligned}&v(b)=\frac{dv}{dx}(b)=0, \quad v(c)=\frac{dv}{dx}(c)=0, \end{aligned}$$

(43)

is not solvable: Eqs. (42)–(43) do not admit solutions $v$ for any real value of the constant $\mu$.

Proof

Assume a solution $v$ of the problem (42)–(43) exists. Denoting by $G(x,\xi )$ the Green function of the problem,

$$\begin{aligned}&G(x,\xi )-\tilde{p}\frac{\partial }{\partial x} G(x,\xi ) -\tilde{q}^{\ell }\frac{\partial ^{2}}{\partial x^2}G(x,\xi )=\delta _{(x=\xi )} \quad \text{ in } (b,c),\\&G(b,\xi )=G(c,\xi )=0, \end{aligned}$$

and letting

$$\begin{aligned} h=g_{a}+\mu g_{b}, \end{aligned}$$

(44)

the solution $v$ can be expressed in the form

$$\begin{aligned} v(x)=\int \limits _{b}^{c}G(x,\xi )h(\xi )\,\mathrm{d}\xi . \end{aligned}$$

Taking into account the Neumann boundary conditions (43) we then obtain

$$\begin{aligned} \frac{dv}{dx}(b)&= \int \limits _{b}^{c}\frac{\partial G}{\partial x}(b,\xi )h(\xi )\,\mathrm{d}\xi =0,\end{aligned}$$

(45)

$$\begin{aligned} \frac{dv}{dx}(c)&= \int \limits _{b}^{c}\frac{\partial G}{\partial x}(c,\xi )h(\xi )\,\mathrm{d}\xi =0. \end{aligned}$$

(46)

Now, as is known (see e.g. in [18, Ch. V.28]), the function $\partial G/\partial \xi$ satisfies the ODE problems

$$\begin{aligned}&\frac{\partial G}{\partial \xi }(x,b)- \tilde{p}\frac{\partial }{\partial x}\left( \frac{\partial G}{\partial \xi }(x,b)\right) -\tilde{q}^{\ell } \frac{\partial ^{2}}{\partial x^2} \left( \frac{\partial G}{\partial \xi }(x,b)\right) =0 \quad \text{ in } (b,c),\\&\frac{\partial G}{\partial \xi }(b,b)=\frac{1}{\tilde{q}^{\ell }(b)},\quad \frac{\partial G}{\partial \xi }(c,b)=0 \end{aligned}$$

and

$$\begin{aligned}&\frac{\partial G}{\partial \xi }(x,c)- \tilde{p}\frac{\partial }{\partial x}\left( \frac{\partial G}{\partial \xi }(x,c)\right) -\tilde{q}^{\ell }\frac{\partial ^{2}}{\partial x^2} \left( \frac{\partial G}{\partial \xi }(x,c)\right) =0 \quad \text{ in } (b,c),\\&\frac{\partial G}{\partial \xi }(b,c)=0,\quad \frac{\partial G}{\partial \xi }(c,c)=-\frac{1}{\tilde{q}^{\ell }(c)}. \end{aligned}$$

In view of the identity $\frac{\partial G}{\partial x}(x,\xi )=\frac{\partial G}{\partial \xi }(\xi ,x)$ (which follows from the symmetry $G(x,\xi )=G(\xi ,x)$ of the Green function) it follows that the function

$$\begin{aligned} H (x) = \frac{\partial G}{\partial x}(b,x)- \frac{\partial G}{\partial x}(c,x) \end{aligned}$$

satisfies the two-point boundary-value problem

$$\begin{aligned}&H- \tilde{p}\frac{dH}{dx} -\tilde{q}^{\ell }\frac{d^{2}H}{d x^2}=0 \quad \text{ in } (b,c),\\&H(b)=\frac{1}{\tilde{q}^{\ell }(b)},\quad H(c)=\frac{1}{\tilde{q}^{\ell }(c)}. \end{aligned}$$

Applying the strong maximum principle [9] to this elliptic equation we obtain the estimate

$$\begin{aligned} H (x) = \frac{\partial G}{\partial x}(b,x)-\frac{\partial G}{\partial x}(c,x) \ge C > 0\quad \text{ for } x\in [b,c], \end{aligned}$$

where $C$ is the strictly positive constant $C=\min \{1/\tilde{q}^{\ell }(b),1/\tilde{q}^{\ell }(c)\}$. From (25), (44), (45) and (46) we thus obtain

$$\begin{aligned} 0= \int \limits _{b}^{c}\left( \frac{\partial G}{\partial x}(b,x)- \frac{\partial G}{\partial x}(c,x)\right) h(x)\,\mathrm{d}x \!\ge \! C\int \limits _{b}^{c} (g_{a}(x)+\mu g_{b}(x))\,\mathrm{d}x \!=\! C\int \limits _{b}^{c}g_{a}(x)\,\mathrm{d}x\!>\!0, \end{aligned}$$

which is a contradiction, and the lemma follows. $\square$