I have the eigenvalue problem, $\frac{d}{dx}\big((1-x^2)\frac{du}{dx}\big)+\lambda u=0$, on $[-1,1]$ subject to single boundary condiction $u(-1) = u(1)$. Assume that there is an eigenfunction of the form $u(x)=a_0+a_1x+a_2x^2$. Find the possible eigenvalues for such an eigenfunction. To solve this problem, I plug the $u(x)$ and the 2nd derivative $u(x)$ into the $\frac{d}{dx}\big((1-x^2)\frac{du}{dx}\big)+\lambda u=0$, solve for $x$. But I couldn't get any eigenvalues. Any help will be appreciated.
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Please use Latex/MathJax to format your post. – Stockfish Nov 18 '18 at 23:11
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Include your calculations, then someone might help. And you shouldn't solve for $x$, but rather choose $a_0,a_1,a_2$ so that the equality $\frac{d}{dx}\big((1-x^2)\frac{du}{dx}\big)+\lambda u=0$ is true for all $x$. – Michał Miśkiewicz Nov 18 '18 at 23:32
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@MichałMiśkiewicz Hi,sir. Could you please tell me – Ray Nov 18 '18 at 23:44
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@MichałMiśkiewicz why do I need to solve for a0,a2. Cause on my textbook, they used to solve for x, and get 3 different cases. like lambda greater than zero or less zero or equal zero. – Ray Nov 18 '18 at 23:46
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Well, the thing you're looking for is the function $u$, which is of the form $u(x)=a_0+a_1 x+a_2 x^2$. Thus, finding $u$ is equivalent to finding $a_0,a_1,a_2$. On the other hand, solving for $x$ makes no sense. – Michał Miśkiewicz Nov 19 '18 at 09:24
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Well, I assume it is sufficiently late I am not doing your homework for you. The boundary condition tells you that $a_1=0$, so it's missing from your Ansatz. Plugging the surviving piece into the equation yields $$ \partial ((1-x^2)2a_2)=-\lambda(a_0+a_2x^2), $$ so, comparing powers, $\lambda=6$ and $a_2=-3a_0$.
In fact, do you now see how your particular case (n=2) falls into the general order-n polynomial solution $$ P_n\propto \partial^n (x^2-1)^n $$ of $$ P_n(x)= \partial ((1-x^2)\partial P_n) +n(n+1)P_n=0 ~~? $$
Cosmas Zachos
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