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This is for $ x \in [0,1]$ with initial conditions $ y(0) = 0$ and $2y(1) = y'(1)$.

My attempt:

for $\lambda = 0$ we get the eigenvalues $0$.

for $\lambda > 0$ I get the solution as $y(x) = A\sin \sqrt{\lambda}x + B \cos \sqrt{\lambda}x$ and after using the initial condition I get that the eigenvalues satisfy $$\tan \sqrt{\lambda} = \sqrt{\lambda} /2$$ which has an infinite number of solutions. I also get that $A = 0$.

for $\lambda < 0$ I get the solution $y(x) = A \sinh \sqrt{-\lambda}x + B \cosh \sqrt{-\lambda} x$. Again, using the initial conditions I get the eigenvalues satisfy $$\tanh \sqrt{-\lambda} = \sqrt{-\lambda}/2$$.

Now, for this question I have to "find an expression that defines the eigenvalues" and the corresponding eigenfunctions. I have found TWO expressions, which express two sets of eigenvalues for $\lambda > 0$ and $\lambda < 0$ - how do I combine these for a single expression?

I am also asked to use this to find the coefficients $B_k$ so that $$x^2 = B_0 \sinh \sqrt{-\lambda_0}x + \sum_{k=1}^\infty B_k \sin\sqrt{\lambda_k} x $$ for $x \in (0,1)$.

I would appreciate any help please.

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Any eigenfunction must be a constant multiple of the solution of $$ -y''+\lambda y = 0,\;\;\; y(0)=0,\; y'(0)=1. $$ This is because $y'(0)\ne 0$ for any such eigenfunction solution, because the only solution of $y''+\lambda y=0$ with $y(0)=y'(0)=0$ is the trivial $0$ solution.

The advantage of using fixed values of $y(0)$ and $y'(0)$ is that the resulting solution will have an everywhere-convergent power series expansion in $\lambda$: $$ y(x,\lambda) = \frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}\lambda^{n}}{(2n+1)!}. $$ The branch cut for $\sqrt{\lambda}$ is chosen along the positive real axis. It's not an accident that the square roots disappear in this expansion. This always happens for fixed endpoint conditions as a consequence of general theorems of ODEs. One advantage of this approach is that special case $\lambda=0$ becomes a limiting case as $\lambda\rightarrow 0$: $$ y(x,0) = \lim_{\lambda\rightarrow 0}(x,\lambda) = x. $$ Taking this limit is justified because you know the solution $y(x,\lambda)$ for the fixed left endpoint conditions must have an everywhere convergent power series expansion in $\lambda$. Of course you can check that the unique solution of $y''=0$, $y(0)=0$, $y'(0)=1$ is $y(x,0)=x$.

The only way for $\lambda$ to be an eigenvalue is if $$ 0=2y(1)-y'(1)=2\frac{\sin(\sqrt{\lambda})}{\sqrt{\lambda}}-\cos(\sqrt{\lambda}). $$ The eigenvalues of such a selfadjoint problem must always be real. The eigenvalues for regular problems of this type are always real zeros of a power series in $\lambda$; note that the expression on the right of the above equation is an everywhere convergent power series in $\lambda$. The bulk of the eigenvalues will be positive, but it is possible for some to be negative. If $\lambda < 0$, then \begin{align} \sin(\sqrt{\lambda}) &=\sin(i\sqrt{-\lambda}) =\frac{e^{-\sqrt{-\lambda}}-e^{\sqrt{-\lambda}}}{2i}=i\sinh(\sqrt{-\lambda}) \\ \cos(\sqrt{\lambda})&=\cosh(\sqrt{-\lambda}) \\ \sqrt{\lambda} & = i\sqrt{-\lambda} \end{align} So the eigenvalue equation for $\lambda < 0$ can be written as $$ 0=2\frac{\sinh(\sqrt{-\lambda})}{\sqrt{-\lambda}}-\cosh(\sqrt{-\lambda}). $$ $\lambda=0$ is not an eigenvalue because $y(x)=x$ does not satisfy $2y(1)=y'(1)$, which can be directly verified, or seen from $$ \lim_{\lambda\rightarrow 0}\left[2\frac{\sin(\sqrt{\lambda})}{\sqrt{\lambda}}-\cos(\sqrt{\lambda})\right]= 2-1 = 1 \ne 0. $$ The positive eigenvalues must satisfy $$ \tan(\sqrt{\lambda}) = \frac{\sqrt{\lambda}}{2} $$ The positive solutions $\lambda$ are the squares of the positive solutions of $\tan\mu = \frac{\mu}{2}$. This is a transcendental equation whose solutions can be visualized by plotting $\frac{\mu}{2}$ on top of $\tan \mu$ for $\mu > 0$. The positive eigenvalues can be seen from the graph to have a simple asymptotic behavior for large $n$: $$ 0 < \lambda_0 < \lambda_1 < \lambda_2 < \lambda_3 < \cdots, \\ \mu_n \approx (2n+1)\frac{\pi}{2} \implies \lambda_n =\mu_n^2\approx (2n+1)^2\frac{\pi^2}{4}. $$ The eigenfunctions for the positive eigenvalues are $y_n(x)=\sin(\sqrt{\lambda_n}x)/\sqrt{\lambda_n}$, or just $y_n(x)=\sin(\sqrt{\lambda_n}x)$. For large $n$, the eigenfunctions are approximately $\sin((n+\frac{1}{2})\pi x)$, up to normalization constants.

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Your two statements are equivalent.

This is because: $i\tan(x) \equiv \tanh(ix)$

Let's start from the statement $\tanh(\sqrt{-\lambda})=\frac {\sqrt{-\lambda}} 2$

This is equivalent to $\tanh(i\sqrt{\lambda})=\frac {i\sqrt{\lambda}} 2$

The identity above changes the RHS, giving $i\tan(\sqrt{\lambda})=\frac {i\sqrt{\lambda}} 2$

and finally $\tan(\sqrt{\lambda})=\frac {\sqrt{\lambda}} 2$

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  • $\begingroup$ Yes I tried that too, $\tanh (\sqrt{-\lambda}) = \tanh i \sqrt{\lambda} = i \tan \sqrt{\lambda} $ but I don't follow how they are equivalent as one holds for $\lambda < 0$ and the other for positive lambda.. $\endgroup$ – wire2 Feb 17 '16 at 0:02
  • $\begingroup$ Plotting these on wolframalpha I don't get the same soltuions, tanh should give 3 solutions, whereas tan one gives an infinite number. $\endgroup$ – wire2 Feb 17 '16 at 0:11
  • $\begingroup$ alright I think I get what you mean now - I am still stuck on finding the corresponding eigenfunctions if you can help with that. I get two sets of eigenfunctions now, again one for $\lambda > 0$ and one with less than. $\endgroup$ – wire2 Feb 17 '16 at 0:29

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