How do you solve the problem with these boundary conditions. The equation $u_{xx}-(4-\lambda)u=0$ is clearly in Sturm-Liouville form, and the characteristic equation is $z^2-(4+\lambda)=0$ which has roots $\pm \sqrt{4+\lambda}$.
If $\sqrt{4+\lambda} > 0$, then the general solution is $u = A\exp{(\sqrt{4+\lambda}x)}+B\exp{(-\sqrt{4+\lambda}x)}$. I thought that normally, this can be shown to only have the trivial solution $A=B=0$ from the boundary conditions, but in this case, they imply
$$\sqrt{4+x}(A-B) = A\exp{(\sqrt{4+\lambda}\pi)}+B\exp{(-\sqrt{4+\lambda}\pi)}$$
Already, this looks like $B$ can be written in the form $B=f(\lambda)A$ by solving the equation, which implies an infinite number of solutions for A and B. How can you show that the only solution to this is $A=B=0$? And for these type of problems, I saw that only the case $\sqrt{4+\lambda} < 0$ generates non-trivial solutions, and the eigenfunctions are usually trig functions. But in that case, how do you solve for the boundary conditions for something like
$$\sqrt{4+\lambda}B=A\sin(\sqrt{4+\lambda}\pi)+B\cos(\sqrt{4+\lambda}\pi)$$?
Would an extra equality condition like $u_x(0)=u(\pi)=0$ be needed?
Any help is appreciated. Thanks!
Best Answer
The ambiguity in your problem comes from the fact that scalar multiples of solutions are also solutions. You can eliminate this by adding a second non-homogeneous normalization condition. For example, $$ u_x(0)-u(\pi)=0 \\ u(0) = 1. $$ The solution of $u_{xx}=(4+\lambda)u$ subject to $u(0)=1$ is $$ \cosh(\sqrt{4+\lambda}x). $$ Then the condition $u_{x}(0)-u(\pi)=0$ gives $$ \cosh(\sqrt{4+\lambda}\pi)=0 \\ \implies \sqrt{4+\lambda}\pi=i(n+1/2)\pi \\ \implies (4+\lambda)\pi^2=-(n+1/2)^2\pi^2 \\ \implies \lambda=-4-(n+1/2)^2 $$ The eigenfunctions are $$ \cosh(i(n+1/2)x)=\cos((n+1/2)x),\;\; n=0,1,2,3,\cdots. $$ The orthogonality conditions are not difficult to verify.