So I have the following Sturm-Liouville problem:
$$
y'' + \lambda y = 0
$$
Such that $ \lambda > 0 $ and the initial conditions are as follows:
$$ y (0) + y'(0) = 0 $$
$$ y(1) + y'(1) = 0 $$
So my attempt at this goes something like this:
I know that the $\lambda$ is positive so the solution must be:
$$ y(t) = A\cos(\sqrt(\lambda)t) + B\sin(\sqrt(\lambda)t)$$
and:
$$ y'(t) = -A\sqrt{\lambda}\sin(\sqrt{\lambda}t) + B\sqrt{\lambda}\cos(\sqrt{\lambda})t)$$
Such that $A$ and $B$ are constants.
So I can evaluate the solution at the first initial condition:
\begin{align}
&=A\cos(\sqrt{\lambda}0) + B\sin(\sqrt{\lambda}0) + -A\sqrt{\lambda})\sin(\sqrt{\lambda}0) + B\sqrt{\lambda}\cos(\sqrt{\lambda}0)\\
&=A + B\sqrt{\lambda}
\end{align}
Thus I know:
$$ A = -B\sqrt{\lambda}$$
Evaluating at the second initial condition:
\begin{align}
&=A\cos(\sqrt{\lambda}1) + B\sin(\sqrt{\lambda}1) -A\sqrt{\lambda}\sin(\sqrt{\lambda}1) + B\sqrt{\lambda}\cos(\sqrt{\lambda}1)\\
&=A\cos(\sqrt{\lambda}) + B\sin(\sqrt{\lambda}) -A\sqrt{\lambda}\sin(\sqrt{\lambda}) + B\sqrt{\lambda}\cos(\sqrt{\lambda})
\end{align}
I'm not sure where to go from here, I factored the second initial condition by cos and sin but that led me to a trivial solution for A and B like this:
$$(A + B\sqrt{\lambda})\cos(\sqrt{\lambda}) + (B – A\sqrt{\lambda})\sin(\sqrt{\lambda}) = 0$$
Plugging in $$ A = -B\sqrt{\lambda}$$.
$$(-B\sqrt{\lambda} + B\sqrt{\lambda})\cos(\sqrt{\lambda}) + (B + B\sqrt{\lambda}\sqrt{\lambda})\sin(\sqrt{\lambda}) = 0$$
$$ (B + B\lambda)\sin(\sqrt{\lambda}) = 0$$
So assuming $B$ isn't $0$, then I know $\lambda = (n\pi)^2$ but how do I find the value of B?
Any guidance would be greatly appreciated!
Thank you.
Best Answer
Start by solving $y''+\lambda y = 0$ subject to $$ y'(0)+y(0)=0 \\ y(0)=1. $$ The second condition is arbitrary; you could take $y(0)-y'(0)=1$ for example. The point is that no solution can satisfy $y'(0)+y(0)=0$ and $y(0)=0$ unless it is identically the $0$ solution, which means you can always scale $y$ so that $y'(0)+y(0)=0$ and $y(0)=1$. The solution of this problem is $$ y_{\lambda}(x)=-\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}+\cos(\sqrt{\lambda}x). $$ Notice that the above is the correct solution at $\lambda=0$ when interpreted as a limit as $\lambda\rightarrow 0$, where it gives $-x+1$. In fact, $y_{\lambda}$ is a power series in $\lambda$. This will always be the case.
The added condition that $y'(1)+y(1)=0$ gives an equation in $\lambda$: $$ -\cos(\sqrt{\lambda})-\sqrt{\lambda}\sin(\sqrt{\lambda})-\frac{\sin(\sqrt{\lambda})}{\sqrt{\lambda}}+\cos(\sqrt{\lambda}) = 0. $$ The solutions $\lambda$ are the zeros of a power series in $\lambda$, and $\lambda=0$ is not a solution. The solutions $\lambda$ must satisfy $$ \sqrt{\lambda}\sin(\sqrt{\lambda})+\frac{\sin(\sqrt{\lambda})}{\sqrt{\lambda}}=0. $$ Looks like $\lambda=0$ is not a solution, but $\sqrt{\lambda}=n\pi$ for $n=1,2,3,\cdots$ are solutions. And $\lambda=-1$ looks like a solution. $y_{-1}$ is an exponential solution $e^{-x}$.