# Computing eigenvalues for specific initial conditions

Take the following equation $$-x'' = \gamma x$$ which has the known solution $$x = A\cos(\sqrt{\gamma}t ) + B \sin(\sqrt{\gamma}t).$$Now assume that $$x(0) = x(\alpha)$$ and $$x'(0) = -x'(\alpha)$$. I am looking for the eigenvalues $$\gamma$$.

Approach: I tried constructing the system that results from the two equations: $$\begin{cases}A = A\cos(\sqrt{\gamma}\alpha) + B\sin(\sqrt{\gamma}\alpha),\\ B\sqrt{\gamma} = A\sqrt{\gamma}\sin(\sqrt{\gamma}\alpha) - B\sqrt{\gamma}\cos(\sqrt{\gamma}\alpha),\end{cases}$$which can then be transformed into a matrix when solving for $$A$$ and $$B$$, $$M = \begin{bmatrix}1- \cos(\sqrt{\gamma}\alpha) & - \sin(\sqrt{\gamma}\alpha)\\ - \sqrt{\gamma} \sin(\sqrt{\gamma}\alpha) & \sqrt{\gamma}+ \sqrt{\gamma}\cos(\sqrt{\gamma}\alpha) \end{bmatrix}.$$ But this matrix has determinant zero. I am stuck now.

Question: How can I get the eigenvalue $$\gamma$$ from these conditions?

• Is that not the matrix $M$ used to solve $M[A,B]^t =[0,0]^t$? If so, it better have determinant $0$. Commented Aug 1 at 15:08
• @Steen82. Well.. it is zero indeed. I thought this was the method to find $\gamma$? For more info math.stackexchange.com/questions/676293/… Commented Aug 1 at 15:13

The fact that $$\det M(\gamma)=0$$ for any $$\gamma$$ implies that any $$\gamma\in\mathbb{R}$$ is an eigenvalue of $$-\frac{d^2}{dt^2}$$ with the given boundary conditions. Indeed, one can easily check that the functions below are nontrivial solutions to $$-x''=\gamma x$$ satisfying $$x(0)=x(\alpha)$$ and $$x'(0)=-x'(\alpha)$$: $$x_{\gamma}(t)= \begin{cases} \sin (\sqrt{\gamma}\,x)+\sin (\sqrt{\gamma}\,(\alpha-x))&\text{if \gamma>0,\,\sqrt{\gamma}\neq \frac{2n\pi}{\alpha}\,(n\in\mathbb{N}_{>0}),} \\ \cos\!\left(\frac{2n\pi x}{\alpha}\right)&\text{if \sqrt{\gamma}=\frac{2n\pi}{\alpha}\,(n\in\mathbb{N}_{>0}),} \\ 1&\text{if \gamma=0}, \\ \sinh (\sqrt{-\gamma}\, x)+\sinh (\sqrt{-\gamma}\,(\alpha- x))&\text{if \gamma<0}. \end{cases}$$

• This is exactly the answer that I was looking for. Thank you very much! Commented Aug 2 at 9:41

Expanding my comment here, since I think this is too long for a comment.

For any $$\gamma$$ and any $$A,B$$ the function $$x=A\cos(\sqrt{\gamma}t)+B\sin(\sqrt{\gamma}t)$$ is a solution to $$-x''=\gamma x.$$ Simply setting $$A=B=0$$ always solves your problem including the initial values.

Now, you can in a manner similar to my original answer, convert your initial values into matrix form and ask under which conditions that problem has non-zero solutions for $$A,B.$$ I think this is what you did, and you found the matrix has determinant zero, so non-zero solutions are possible.

My original point was that the determinant being zero, was a good thing. I interpreted your statement "But this matrix has determinant zero." to imply it was not desirable.

My previous expansion. $$M\begin{bmatrix} A \\ B \end{bmatrix} = \begin{bmatrix}1- \cos(\sqrt{\gamma}\beta) & - \sin(\sqrt{\gamma}\beta)\\ - \sqrt{\gamma} \sin(\sqrt{\gamma}\beta) & \sqrt{\gamma}+ \sqrt{\gamma}\cos(\sqrt{\gamma}\beta) \end{bmatrix} \begin{bmatrix} A \\ B \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$ Is the matrix form of the system \begin{align*} A - A\cos(\sqrt{\gamma}\beta) - B\sin(\sqrt{\gamma}\beta) &= 0 \\ -A\sqrt{\gamma} \sin(\sqrt{\gamma}\beta)+B\sqrt{\gamma}+ B\sqrt{\gamma}\cos(\sqrt{\gamma}\beta) &=0 \end{align*} Which rearranges to your system.

Let me solve this system. If $$\gamma=0$$ the system is $$0=0,$$ so any $$A,B$$ gives a solution. Assume $$\gamma\neq 0$$ and divide the second equation by $$\sqrt{\gamma}.$$ Let $$c:=\cos\sqrt{\gamma}\beta)$$, and $$s:=\sin(\sqrt{\gamma}\beta).$$ Then the matrix form is $$\begin{bmatrix} 1- c & - s \\ - s & 1+c \end{bmatrix} \begin{bmatrix} A \\ B \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ If $$c=1$$, then $$s=0$$ and the equations are $$0\cdot A = 0$$ and $$2B=0$$, so $$A$$ can be anything and $$B=0.$$ If $$c=-1$$ then the equations are $$2A=0$$ and $$0\cdot B=0$$ so $$A=0$$ and $$B$$ can be anything. If $$c\neq \pm1$$ then $$s\neq 0.$$ Multiplying the first row by $$1+c$$ and the second row by $$-s$$ we get the system $$\begin{bmatrix} s^2 & - s(1+c) \\ s^2 & - s(1+c) \end{bmatrix} \begin{bmatrix} A \\ B \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ Hence, if $$A$$ is anything and $$B=\tfrac{1+c}{c}A,$$ then we have a solution for $$A,B.$$ The relationship between $$A,B$$ depends on $$\gamma$$ and on $$\beta.$$

• Thank you for the attention but that has nothing to do with my question. I want to know how to find $\gamma$. Commented Aug 1 at 15:37
• Unfortunately, it is not possible to find $\gamma$ based on the info in your problem. Commented Aug 1 at 15:43
• If I may, can you elaborate on why is that? Commented Aug 1 at 15:46
• Exactly. But lets forget about the $A = B = 0$ since they are not interesting. You mentioned that non-zero solutions are possible. Can I find them? I am essentially looking for an expression $\gamma$ equals to something so that $x = A\cos(\sqrt{\gamma}t) + B\sin(\sqrt{\gamma}t)$ is a non-trivial solution. Commented Aug 1 at 16:31
• That is your choice, I just proved you can't have what you want. Commented Aug 1 at 18:33