# Limits of the wave equation with piecewise constant propagation speed

Consider a wave equation $$\frac{\partial^2 u}{\partial t^2} = c(x)^2 \frac{\partial^2 u}{\partial x^2} \tag{1}$$ In frequency domain this becomes an ODE: $$-\omega^2 u = c(x)^2 \frac{\partial^2 u}{\partial x^2} \tag{2}$$ We can solve (2) analytically when $$c(x)$$ is constant, then the solution is $$\exp\left(\pm\frac{i \omega}{c} x\right)$$. We can also solve it analytically if $$c(x)$$ is piecewise constant: write in every region-of-constant-$$c$$ a linear combination of leftward and rightward propagating waves, impose the appropriate continuity conditions on the interfaces between regions with different values of $$c$$, solve for the coefficients of the linear combinations.

Suppose now that $$c(x)$$ is some arbitrary function. We can still write it as a limit of piecewise constant functions: $$c(x)=\lim_{\Delta\rightarrow 0^+} \sum_{i=-\infty}^{\infty} \left\{\begin{matrix}c(i\Delta) & x\in [i\Delta,(i+1)\Delta[ \\ 0 & \text{otherwise}\end{matrix}\right.$$

Let us call these piecewise constant approximations to $$c$$, $$c_{\Delta}$$ $$c_{\Delta}(x)=\sum_{i=-\infty}^{\infty} \left\{\begin{matrix}c(i\Delta) & x\in [i\Delta,(i+1)\Delta[ \\ 0 & \text{otherwise}\end{matrix}\right.$$ We can analytically solve, for any strictly positive $$\Delta$$, $$-\omega^2 u_{\Delta}= c_{\Delta}(x)^2 \frac{\partial^2 u_{\Delta}}{\partial x^2}$$

Question

Is $$\lim_{\Delta\rightarrow 0^+} u_{\Delta} = u$$?

In other words, can we approximate general solutions of (2) by substituting in a piecewise constant approximation of $$c(x)$$, and then solving analytically?

I expect the answer to be "no".

• because $$\lim_{\Delta\rightarrow 0^+} c_{\Delta} = c$$, but $$\lim_{\Delta\rightarrow 0^+} \frac{\partial}{\partial x}c_{\Delta} \neq \frac{\partial}{\partial x} c$$, in fact the limit doesn't even exist
• In literature on numerical solutions of PDEs, I have never seen this suggested as a viable way of approximating solutions of wave equations.

New:

For the initial conditions of $$u_\Delta$$ choose $$u_\Delta(0) = u(0)$$ and $$u'_\Delta(0) = u'(0)$$. Define $$v = u - u_\Delta$$ and $$\delta c = c - c_\Delta$$. So $$v(0) = v'(0) = 0$$. Suppose $$c$$ is sufficiently well behaved so $$u$$ is also. $$u_\Delta$$ should be well behaved excluding a discontinuous second derivative and if ever $$c_\Delta = 0$$ then $$u_\Delta$$ and $$u''_\Delta$$ jumps to zero.

Define $$v = u - u_\Delta$$ and $$\delta c = c - c_\Delta$$. Subtracting $$-\omega^2 v = c_\Delta^2 v'' - (2c\delta c - \delta c^2) u''$$ Suppose for some $$x_0 \geq 0$$ we want to know $$v(x_0)$$. Note if $$\Delta \rightarrow 0$$ then the sup norm $$||\delta c||_{\infty, [0,x_0]} \rightarrow 0$$. Since $$u''$$ and $$c$$ are well behaved we're arguing $$||(2c\delta c - \delta c^2) u''||_{\infty,[0,x_0]} \rightarrow 0$$. If this term on the RHS were zero, by the uniqueness theorem $$v(x)=0$$ on $$[0,x_0]$$. The fact it approaches zero simply means it perturbs $$v(x)$$ from $$0$$ an arbitrarily small amount.

So $$u_\Delta \rightarrow u$$ as $$\Delta \rightarrow 0$$ if $$c$$ is sufficiently nice. I haven't tried to address what a necessary and sufficient condition may be.

Original:

It's hard for me to picture the limit does not hold for a large class of well behaved functions. On the other hand surely we can contrive functions which will break the limit. By work such as Lebesgue's I feel finding a necessary and sufficient condition on functions for the limit to hold may not be easy.

Thoughts for proof: If we can prove the limit holds when $$c(x)$$ is a polynomial perhaps we can take a limit on the degree to argue the limit holds for many Taylor series.

For convenience scale $$c(x)$$ by $$\omega$$. Suppose $$c(x)$$ is a polynomial and denote $$c(x)^2 = c_0 + c_1 x + \cdots + c_n x^{2n}$$. If we were able to solve for $$u$$ and $$u_\Delta$$ we'd be able to take the limit and see if they match. Applying the Laplace transform which is generally a simplification

\begin{align} \mathcal{L}\{-u\} = -U(s) =& \sum_{0\leq k \leq 2n} \mathcal{L}\{c_k x^k u''\} \\ =& \sum_{0\leq k \leq 2n} (-1)^k c_k \frac{d^k}{ds^k}\bigg[s^2U-su(0)-u'(0)\bigg] \\ =& -c_0 (su(0)+ u'(0)) + c_1 u(0) + \sum_{0\leq k\leq 2n} (-1)^k (c_k s^2-2c_{k+1}s+2c_{k+2}) U^{(k)} \end{align} where $$c_{2n+1},c_{2n+2} = 0$$.

\begin{align} \mathcal{L}\{-u_\Delta\}=-U_\Delta(s)= & \sum_{0\leq k} \mathcal{L}\Big\{\big(H(x-k\Delta)-H(x-(k+1)\Delta)\big)c(k\Delta)^2 u_\Delta''(x)\Big\} \\ =& \sum_{0\leq k} c(k\Delta)^2 e^{-sk\Delta}\Big[\mathcal{L}\{u''_\Delta(x+k\Delta)\} - e^{-s\Delta}\mathcal{L}\{u''_\Delta(x+(k+1)\Delta)\}\Big] \end{align}

using $$\mathcal{L}\{H(x-k\Delta)u''_\Delta(x)\}=\int_0^\infty e^{-sx}H(x-k\Delta)u''_\Delta(x) dx = e^{-sk\Delta}\mathcal{L}\{u''_\Delta(x+k\Delta)\}$$ where $$H$$ is the heaviside function. Unfortunately $$u''_\Delta(x)$$ is not of the right form. Yet going back and directly solving the original equations for any $$\Delta$$ seems recursive without a generating formula. Perhaps there's less explicit theorems which can be used.