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I am trying to find the Green's function for the operator $Au=((1-x^2)u')'$ with boundary conditions $|u(\pm 1)|<\infty$. The general solution of $Au=0$ gives $u=c_1+c_2\log{\frac{1-x}{1+x}}$. To satisfy either left or right boundary condition, the logarithmic term needs to vanish and both $u_L$ and $u_R$ are constant. By continuity we also have $u_L=u_R=const.$ which gives $A_xG(x,\xi)=0$ not $A_xG(x,\xi)=\delta(x-\xi)$.

Am I making a mistake somewhere? or is there some other approach that will work?

an instance
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    Start with $\langle Au,\phi \rangle = 0$, get $\int_{-1}^1 ((1-x^2)u')' \phi dx = 0$, formally integrate by parts to get $\left. (1-x^2)u' \phi \right |{-1}^1 - \int{-1}^1 (1-x^2) u' \phi' dx = 0$. Assuming $u'$ (not $u$ itself) is not too singular at $\pm 1$ (e.g. if it is bounded), then you get $\int_{-1}^1 (1-x^2) u' \phi' dx = 0$ and the weak problem makes sense. – Ian Mar 08 '22 at 19:13
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    You are correct that the heuristic of assuming a strong solution on both sides of the excitation does not make sense. – Ian Mar 08 '22 at 19:14
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    @JeanMarie I saw that question too, it's very close but the boundaries are different there (0 and 1 instead of $\pm 1$) and that eliminates the problem that I am running into. – an instance Mar 08 '22 at 20:42

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