# Second order Sturm-Liouville ODE with difficult boundary conditions

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!

• I would think that the typing of the BC was interrupted. You need two BC, but in the current form it is only one. So quite probably homogeneous BC $u_x(0)=u(\pi)=0$ were intended. This is the same as looking for the even solutions with $u(-\pi)=u(\pi)=0$. Commented Mar 14, 2022 at 7:13

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.