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I am trying to solve the variable-coefficient Laplace equation $$\partial(\epsilon\partial u) = (\partial\epsilon)(\partial u) + \epsilon\partial^2 u = 0$$using a finite difference scheme: $$\left(\frac{\epsilon(x+h) - \epsilon(x-h)}{2h}\right)\left(\frac{u(x+h)-u(x-h)}{2h}\right) + \epsilon(x) \frac{u(x+h) - 2u(x) + u(x-h)}{h^2}=0$$This gives us $$u(x) = \frac{1}{2}\left(\frac{\epsilon(x+h) - \epsilon(x-h)}{2}\right)\left(\frac{u(x+h)-u(x-h)}{2}\right) + \frac{\epsilon(x)}{2}\left[u(x+h) + u(x-h)\right]$$Here is my python script that solves the equation on the domain [0,1], and sets the coefficient to 1 everywhere except on the interval [0.4, 0.6] where it is set to 2. Instead of converging, it produces nans. If I set the coefficient to the same everywhere, it works fine.

import numpy as np
import matplotlib.pyplot as plt

num_points = 100
domain = np.linspace(0, 1, num_points)

# Coefficient is 1 everywhere except in the interval [0.4, 0.6]
eps = np.ones(num_points)
for i, x in enumerate(domain):
    if 0.4 <= x <= 0.6:
        eps[i] = 2.0

# Solution
u = np.zeros(num_points)

#  Boundary conditions: u(0) = 1, u(1) = 0
u[0] = 1.0

while True:
    u_old = u.copy()
    for i in range(1, num_points - 1):
        u[i] = 0.125 * (eps[i + 1] - eps[i - 1]) * (u[i + 1] - u[i - 1])
        u[i] += 0.5 * eps[i] * (u[i + 1] + u[i - 1])

    if np.allclose(u, u_old):
        break

plt.plot(domain, u)
plt.show()
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1 Answer 1

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The iteration could be blowing up.

You can see this by looking at your iteration

$|u_{i+1}|\le \frac{1}{4} \omega(\epsilon, 2h) |u_i|_{\infty} + |\epsilon|_{\infty} |u_i|_{\infty}$

where $\omega(\epsilon, 2h) = \sup |\epsilon(x) - \epsilon(x+2h)|$. In your homogeneous case, $\omega = 0$, and $|\epsilon| = 1$, there is no blowing up.

For discontinuous case, $\omega = 1$, then

$$|u_{i+1}|\le \frac{5}{4} |u_i|$$

which does not have any control over the maximum.

To verify this, if you bring down your $\epsilon$ to something like $\epsilon = 0.25$ overall except $\epsilon = 0.5$ on some parts, then there should be no problem.

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  • $\begingroup$ Thanks. Could this issue also be resolved by evaluating the coefficient on a different grid from the solution? I.e. Is an expression like this valid? mathb.in/77138 where h' > h $\endgroup$
    – DJames
    Commented Dec 11, 2023 at 10:35
  • $\begingroup$ yes. But the magnitude of $\epsilon$ maybe the bottleneck. @DJames, it seems your iteration should divide $\epsilon$ on rhs. $\endgroup$
    – Yimin
    Commented Dec 11, 2023 at 17:06
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    $\begingroup$ It may also help to apply the finite-difference scheme to the conservative form of the equation, $(\epsilon u_x)_x=0$. If $\epsilon$ is not differentiable, then applying the product rule may not be justified and your discretization is having problems due to this. $\endgroup$
    – whpowell96
    Commented Dec 11, 2023 at 17:23
  • $\begingroup$ @whpowell96 unfortunately that is likely my problem: In my use case, the coefficient changes abruptly, like a step function, at various interfaces. Here is my original question that motivated this follow-up physics.stackexchange.com/q/791946/136054 $\endgroup$
    – DJames
    Commented Dec 12, 2023 at 16:10
  • 1
    $\begingroup$ It will definitely help to try the conservative discretization. If you still have problems, there are specific methods for interface problems such as this. C.f., The immersed interface method. $\endgroup$
    – whpowell96
    Commented Dec 12, 2023 at 16:34

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