How to solve this Sturm-Liouville problem $y''+2y'+\lambda y = 0$?

Question

I have the following Sturm-Liouville problem:

$$\left\{ \begin{aligned} & y''+2y'+\lambda y = 0, 0<x<a \\ & y(0) = 0, y'(a)=0 \end{aligned} \right.$$

I have tried to reduce it to Sturm-Liouville form, got this: $$\left\{ \begin{aligned} & (e^{2x}y')'+e^{2x}\lambda y = 0, 0<x<a \\ & y(0) = 0, y'(a)=0 \end{aligned} \right.$$

Then, I checked whether there exist negative lambdas via: $$ p(x)y(x)y'(x)\Big|_0^a $$ where $p(x)=e^{2x}$

So it evaluated 0, so we know that for $\lambda < 0$ there is no non-trivial solutions.

But reducing didn't help much, since I anyway had to find the general solution of the equation.

Solving, I got $$ \alpha_{1,2}=-1\pm\sqrt{1-\lambda}\\ y(x)=C_1e^{\alpha_1}+C_2e^{\alpha_2} $$

So I then checked for $\lambda=0$ and got: $$ y(x) = C_1 + C_2e^{-x}\\ y'(x) = -C_2e^{-x} $$ And substituting the initial conditions: $$\left\{ \begin{aligned} & C_1 + C_2 = 0\\ & -C_2e^{-a} = 0 \end{aligned}\right. \iff\left\{ \begin{aligned} & C_1 = 0\\ & C_2 = 0 \end{aligned} \right.$$ So also no nontrivial solutions for this case.

Then I checked $\lambda=1$: $$ y(x) = C_1e^{-x} + C_2xe^{-x}\\ y'(x) = -C_1e^{-x} + C_2e^{-x}-C_2xe^{-x} $$ And with initial conditions this evaluates to: $$ \left\{ \begin{aligned} & C_1=0\\ & C_2-C_2a=0 \end{aligned} \right. $$ So if $a=1$ we have non-trivial solutions, where $C_1=0$ and $C_2$ is any non-zero number. So we can pick 1.

So if a = 1: $$ y(x)=xe^{-x} $$

Now all that's left is to check $\lambda>1$: $$ \alpha_{1,2}=-1\pm i \sqrt{\lambda-1}\\ y(x)=e^{-x}C_1\cos(\sqrt{\lambda-1}x)+e^{-x}C_2\sin(\sqrt{\lambda-1}x)\\ y'(x)=-C_1\sqrt{\lambda-1}e^{-x}\sin(\sqrt{\lambda-1}x)-C_2e^{-x}\sin(\sqrt{\lambda-1}x) +C_2\sqrt{\lambda-1}e^{-x}\cos(\sqrt{\lambda-1}x)-C_1e^{-x}\sin(\sqrt{\lambda-1}x) $$ With initial conditions we got: $$ \left\{ \begin{aligned} & C_1=0\\ & C_2e^{-a}(\sqrt{\lambda-1}\cos(\sqrt{\lambda-1}a)-\sin(\sqrt{\lambda-1}a))=0 \end{aligned} \right. $$ And here I stuck, because i don't know how to solve: $$ \sqrt{\lambda-1}\cos(\sqrt{\lambda-1}a)-\sin(\sqrt{\lambda-1}a) = 0 $$

Where did I go wrong or what could I do to solve this?

Answer 1

The equation $$ y''+2y'+\lambda y = 0 $$ can be transformed by multiplying by $e^x$: $$ e^x y''+2e^x y'+\lambda e^x y =0 \\ (e^x y)''+(\lambda-1) e^x y = 0 $$ The solutions are $$ e^x y = A\cos(\sqrt{\lambda-1}x)+B\frac{\sin(\sqrt{\lambda-1}x)}{\sqrt{\lambda-1}} \\ y = Ae^{-x}\cos(\sqrt{\lambda-1}x)+Be^{-x}\frac{\sin(\sqrt{\lambda-1}x)}{\sqrt{\lambda-1}}. $$ The limiting form as $\lambda\rightarrow 1$ is $$ y=Ae^{-x}+Bxe^{-x}. $$ For all $\lambda$, these solutions satisfy $$ y(0)=A,\;\; y'(0)=(B-A) $$ You want the solutions to satisfy $y(0)=0$, which gives $$ y=Be^{-x}\frac{\sin(\sqrt{\lambda-1}x)}{\sqrt{\lambda-1}}. $$ In order for $y'(a)=0$ to hold, it is necessary and sufficient that $\lambda$ satisfy the eigenvalue equation: $$ 0=y'(a)=-Be^{-a}\frac{\sin(\sqrt{\lambda-1}a)}{\sqrt{\lambda-1}}+Be^{-a}\cos(\sqrt{\lambda-1}a) \\ \frac{\sin(\sqrt{\lambda-1}a)}{\sqrt{\lambda-1}}=\cos(\sqrt{\lambda-1}a) $$ $\lambda=1$ is a solution iff $a=1$. Otherwise, the eigenvalue equation is transcendental: $$ \tan(\sqrt{\lambda-1}a)=\sqrt{\lambda-1}. $$