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?


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 $k=\frac32\pi-\frac2{3\pi}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.