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I have a problem to resolve with the Finite Difference method in $[a,b]$: $$-\frac{d}{dx}(\alpha(x)\frac{du}{dx})= g(x),$$ with $\alpha(x) \in L^{\infty}$ continuous in $]a,c[$ and $]c,b[$ and discontinuous at $c$, $\alpha_{*}< \alpha(x)<\alpha_{**}$, $g \in L^2$, and $u \in C^4([a,b])$.

I want to resolve it with the FDM, I obtain a scheme on $]a,c[$ and $]c,b[$ but I don't know how to deal with the discontinuity at $c$. I take for my discretization $hc_{-}=\frac{c-a}{M+1}$ and $hc_{+}=\frac{b-c}{N+1}$ with $x_{i}^{-} = a+ihc_{-}$ and $x_{j}^{+}=c+jhc_{+}$. So I have for $1< i < M-1$ :$-\frac{\alpha_{i+1}(u_{i+1}-u_{i})}{hc_{-}^2}$ $\frac{-\alpha_{i-1}(u_{i}-u_{i-1})}{hc_{-}^2}$ for $]a,c[$ and for $M+1<i<N+M$: $-\frac{\alpha_{i+1}(u_{i+1}-u_{i})}{hc_{+}^2}$ $\frac{-\alpha_{i-1}(u_{i}-u_{i-1})}{hc_{+}^2}$ for $]c,b[$.

But i have no idea how to do for $x_M$, the point of discontinuity of the function $a$... Can you help me please ?

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  • $\begingroup$ I don't know enough details to constitute a complete answer, but the immersed interface method is one approach to such problems. C.f. Leveque & Li 1994 doi.org/10.1137/0731054 or Li 2003 doi.org/10.11650/twjm/1500407515 $\endgroup$
    – whpowell96
    Commented Nov 25, 2023 at 17:47
  • $\begingroup$ That's all I have as guidance for this exercise, $\alpha_*$ and $\alpha_{**} \in R_{+}^+$ $\endgroup$
    – Kaneki Ken
    Commented Nov 25, 2023 at 19:38
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    $\begingroup$ I will note that you will have to have a very specific $g$ so that you end up with $u\in C^4$ despite the fact that the coefficient has a discontinuity. $\endgroup$ Commented Nov 26, 2023 at 3:35
  • $\begingroup$ I take it in $C^4$ for Taylor expression with "hc-" and "hc+" $\endgroup$
    – Kaneki Ken
    Commented Nov 26, 2023 at 10:28
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    $\begingroup$ @nicoguaro i pose $\varphi(x) = \alpha(x)\frac{du}{dx}$ and $\varphi' \in L^2$ $\endgroup$
    – Kaneki Ken
    Commented Nov 27, 2023 at 10:20

1 Answer 1

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Since this is apparently a homework problem, let's just illustrate the idea on a simple small example. Let's take the domain [0,1], with the discontinuity at $x=0.5$, and assume $\alpha$=1 to the left of $x=0.5$, and $\alpha=2$ to the right; also assume constant $g$. In addition, assume Dirichlet boundary conditions $u(0)=u_L$ and $u(1)=u_R$

Our ODE becomes $ u_{xx} = g $ in the left half-domain, and $ u_{xx} = g/2 $ in the right half-domain. Next, to derive the matching condition at the discontinuity, integrate the ODE over a narrow segment covering the discontinuity,

$ \int_{0.5-\epsilon}^{0.5+\epsilon} \left[ \frac{d}{dx} ( \alpha u') = g \right] dx, $

which leads to

$ ( \alpha u')_{0.5+\epsilon} - ( \alpha u')_{0.5-\epsilon} = 2 g \epsilon, $

and in the limit $\epsilon \rightarrow 0$ for our choice of $\alpha(x)$ it becomes

$u'_{x-} = 2 u'_{x+}$,

using the one-sided derivative notation; this matching condition has the obvious meaning of flux conservation.

Now, let's do FD discretization of the ODE using just five uniformly distributed grid points and using simplest FD approximations; $u_0$ and $u_4$ correspond to the boundaries, and $u_2$ corresponds to $x=0.5$. This FD formulation amounts to solving the following system of five linear equations:

$u_0 = u_L$

$u_0 - 2 u_1 + u_2 = h^2 g $

$2(u_3-u_2) = u_2-u_1$

$u_2 - 2 u_3 + u_4 = h^2 g/2 $

$u_4 = u_R$

Here $h$ is the grid spacing, the third equation is the matching condition at the discontinuity.

It is easy to verify that this linear system is solvable:

import numpy as np
A = np.array([[1,0,0,0,0], [1,-2,1,0,0], [0,1,-3,2,0], [0,0,1,-2,1],[0,0,0,0,1]])
np.linalg.det(A)
-6.000000000000001
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  • $\begingroup$ But we still don't have the form of the approximation for the discontinuity, what do you have for f(c) ? for example, If $c = x_k$, $h^2g_k = ... ?$, I think for the example : 2u_1 - 3u_2 - 2u_3 = f_k ??? $\endgroup$
    – Kaneki Ken
    Commented Nov 28, 2023 at 21:49
  • $\begingroup$ Well, if the RHS source function g does not contain a delta-function at the discontinuity, we have the flux conservation across the discontinuity, $(\alpha u')_{-} = (\alpha u')_{+}$. That's the equation describing the discontinuity. $\endgroup$ Commented Nov 29, 2023 at 0:12
  • $\begingroup$ so what I wrote for the nodes of the discontinuity is good?, and for the generalization I think I have to use $\alpha^-$ and $\alpha^+$ at each side and do the subtraction no? $\endgroup$
    – Kaneki Ken
    Commented Nov 29, 2023 at 7:16
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    $\begingroup$ @KanekiKen For the gird node corresponding to the discontinuity, you'll need to use that special equation expressing the flux conservation $(\alpha u')_{-} = (\alpha u')_{+}$, that's it. $\endgroup$ Commented Nov 30, 2023 at 17:17

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