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I'm looking at the Sturm-Liouville BVP $$\begin{cases} y'' + \lambda y = 0\\ y(0) + y'(0) = 0, y(1) = (0) \end{cases}.$$

I can do the problem but I can't finish it off at the very end (it's probably some easy algebra or trigonometric manipulation that I just can't see).

For the case $\lambda > 0$ my working is as follows:

Set $k^2 = \lambda$. After some working we find that our solution is $y = C_1 \cos kx + C_2 \sin kx$.

Using $y(1) = 0$ we find that $C_1 \cos k + C_2 \sin k = 0$. Using $y(0) + y'(0) = 0$ we find that $0 = C_1 \cos k + C_2 \sin k$. Both of these give $0 = C_2(\sin k - k \cos k)$. So we have a non-trivial solution if $\sin k = k\cos k$, i.e. $\tan k = k$.

My lecture notes are happy to take $k_n \approx (2n-1)\frac{\pi}{2}$, i.e. $\lambda_n \approx \frac{(2n-1)^2\pi^2}{4}$.

I'm happy up to here, but I don't understand how we end up with eigenfunctions $y_n = \sin(k_n(1-x)$.

Also, it seems that we only take $n$ to be non-negative, why is this the case?

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The solution is easier if you start at the right endpoint with $$y=C_1\cos k(1-x)+C_2\sin k(1-x)$$ From this form of the solution we have right away $$y(1)=C_1=0$$ So all we need do is search for solutions of the form $y=\sin k(1-x)$. Applying the boundary condition at the left endpoint $$y(0)+y^{\prime}(0)=\sin k-k\cos k=0$$ Since $\cos k=0$ does not lead to a solution, we divide to get $$\tan k=k$$ If $k=0$, it contradicts our assumption that $\lambda=k^2>0$ and the special case of $\lambda=0$ was already taken into account with the solution $y_0=1-x$, $\lambda_0=0$. $k<0$ leads to the additive inverse of the corresponding $k>0$ solution, so that doesn't yield any new eigenfunctions. If you look at the graph of $y=\tan x$, we can see that it crosses $y=x$ at $x=0$ which we have already handled, and to the left of the asyptotes at $x=\left(n+\frac12\right)\pi$, roughly at $$k=(n+\frac12)\pi-\frac1{(n+\frac12)\pi}$$ for $n$ a positive integer only. The first root is at about $k=\frac32\pi-\frac2{3\pi}$.

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