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$$\left(u(1-u^2)\frac{d^2}{du^2}-(u^2+1)\frac{d}{du}-\frac{au}{(1-u^2)}-\frac{bu^3}{(1-u^2)}+c\right)G(u,u')=mu^2\delta(u-u'),$$ where $a$, $b$, $c$, $m$ are constants, $u\in(-1,1)$ and the boundary conditions are $G(u,u')\rightarrow 0$ for $u=\pm 1$. Thank you very much!

The constants are proper input parameters. Can mathematica read Dirac delta functions directly? I particularly want to know how to reformulate this question so that it is available to numerically solve this equation.

Wein Eld
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  • @AccidentalFourierTransform, I am very sorry, I corrected my question. Though, it is still not a standard Green's function. – Wein Eld Apr 12 '16 at 14:10
  • @MichaelSeifert, Thank you for your remind. I added the boundary conditions. – Wein Eld Apr 12 '16 at 14:19
  • You can use DiracDelta[] in the expression. This however, is only handled correctly by DSolve, which is limited to analytically solvable differential equations. Possibly related question with good hints by Leonid Shifrin how to tackle this kind of problem: http://mathematica.stackexchange.com/questions/61176/using-ndsolve-to-solve-a-pde-with-a-dirac-delta-function – Thies Heidecke Apr 12 '16 at 15:13
  • Here's one idea (don't know if it will work nicely or not): for each $u'$, solve the differential equation to the right and left of $u'$ (so that the delta function is actually zero) using NDSolve, then match the two functions at $u'$ with a Wronskian. – march Apr 12 '16 at 15:35
  • @ThiesHeidecke That's very useful information. Thanks! – Wein Eld Apr 12 '16 at 18:11
  • @march Yes, I know the Wronskian in the analytical method. But in numerical method, there are no two different explicit solutions, hence the Wronskian can not be defined properly. – Wein Eld Apr 12 '16 at 18:14
  • Perhaps I'm totally misremembering how this works (it's been a while), but what about if you used ParametricNDSolve? That way, you can solve the problem for $u_{<}$ that matches the left boundary condition with the right boundary condition as a parameter, and the corresponding problem for $u_{>}$ with the left boundary condition as a parameter, and then vary that parameter until the Wronskian matches? – march Apr 12 '16 at 18:47
  • @march: I tried that approach myself yesterday. Unfortunately, the ODE in question has regular singular points at the boundary $u = \pm 1$, which Mathematica really doesn't like. I suspect you could do a power-series expansion about these points, which would allow you to move the boundary to $\pm (1 - \epsilon)$; but I haven't tried it out myself. – Michael Seifert Apr 13 '16 at 13:29
  • @MichaelSeifert, Thank you very much. I think we can replace the boundary condition appeared in the question by $G= 0$ at, say $u=\pm 0.999999$. But I really want to know how did you handle the Dirac delta function? – Wein Eld Apr 15 '16 at 10:29
  • By integrating the ODE twice in a small region about $u'$, we find that $G$ is continuous at $u'$ and that its derivative has a discontinuous change of $m u' / (1 - {u'}^2)$ at $u'$. This can be viewed as a boundary condition between the solutions in $[-1,u')$ and $(u',+1]$. You can use ParametricNDSolve to solve for $G$ in each of these regions in terms of some parameters (either the value of $G'$ at the boundaries, or the value of $G$ at $u'$). Finally, you use FindRoot to try to vary your parameter(s) so that your two solutions satisfy the correct boundary conditions at $u'$. – Michael Seifert Apr 15 '16 at 13:44
  • When I tried this on a similar equation, I ran into problems with the values at $u'$ being too sensitive to the values at $\pm 1$. If this happens to you, you could try playing around with the Method settings in ParametricNDSolve. – Michael Seifert Apr 15 '16 at 13:45
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    @ThiesHeidecke Perhaps things have advanced since you last tried them: ListLinePlot@NDSolveValue[{y'[x] == y[x] + DiracDelta[x - 1], y[0] == 0}, y, {x, 0, 2}] --> http://i.stack.imgur.com/QdR3j.png (V10.4.1, but it works in V9.0.1, too) – Michael E2 Jul 11 '16 at 18:47
  • @MichaelE2 Cool! Nice to see it working in NDSolve, too now! – Thies Heidecke Jul 11 '16 at 21:48

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