# Finding eigenfunctions and eigenvalues from a differential equation

Consider the differential equation $$X''(x)+\lambda X=0$$ on $$0 \leq x \leq 1$$with boundary conditions $$X'(0)+X(0)=0 \ \ \ \ \text{and} \ \ \ \ X(1)=0.$$ I have a few problems here that I think I figured out, but I would appreciate another look or some hints as to what I can fix. Or, maybe I am totally wrong!

$$\textbf{My first goal:}$$

Find an eigenfunction associated with eigenvalue $$\lambda=0.$$

An eigenvalue $$\lambda =0$$ would mean that $$X''(x)=0$$. This means that the solution takes the form $$X(x)=Ax+B.$$ Since $$X'(0)=A$$ and $$X(0)=B$$, $$X'(0)+X(0)=0 \iff A=-B.$$ Therefore an eigenfunction that works would be $$\boxed{X_{0}(x)=-2x+2}.$$

$$\textbf{My second goal:}$$

Find an expression for all eigenvalues $$\lambda = \beta ^2>0.$$

This one requires a little more work. The solution to $$X'(x)+\beta ^2 X=0$$ takes the form $$X(x)=A\cos\beta x +B\sin \beta x.$$ Taking derivatives, one easily finds that $$X'(0)=B\beta$$ and $$X(0)=A.$$ Thus we obtain $$B\beta + A=0 \implies \beta= \frac{-A}{B}.$$ Finally, this gives $$\boxed{\beta=\frac{A^2}{B^2}}$$

Edit: I just realized that my last problem did not necessarily satisfy $$X(1)=0.$$
• First goal, a better eigenfunction would be $$X(x) = Ax - A, \quad \forall A \in \mathbb{R}$$ Always try to keep it as general as possible. Second goal, applying both boundary conditions yields \begin{align} \beta &= -\frac{A}{B} \\ &= \tan \beta \end{align} and using the oddness of $\tan( \cdot)$, we get $$\lambda = \tan^{2}(\sqrt{\lambda})$$ – Mattos Nov 15 '18 at 6:10
Since you want $$X(1)=0$$, you cat set $$X'(1)$$ to any non-zero constant ($$0$$ leads to $$X\equiv 0$$.) So I'd pick a value of $$1$$. Then $$X_{\lambda}(x) = \frac{\sin(\sqrt{\lambda}(x-1))}{\sqrt{\lambda}}$$ Choosing a constant value for $$X'(1)$$ forces the limiting case as $$\lambda\rightarrow 0$$ to be the correct solution for $$\lambda=0$$ as well, which is $$X_{0} = x-1.$$ You can check that this is an eigenfunction. So $$\lambda=0$$ is an eigenvalue, with corresponding eigenfucntion $$x-1$$.
The general eigenvalue equation becomes $$X_{\lambda}(0)+X_{\lambda}'(0)=0 \\ -\frac{\sin(\sqrt{\lambda})}{\sqrt{\lambda}}+\cos(\sqrt{\lambda})=0$$ The limit at $$\lambda\rightarrow 0$$ is $$0$$. So $$\lambda_0=0$$ is an eigenvalue with $$X_{0}=x-1$$. For $$\lambda\ne 0$$, the solutions are zeros of $$\tan(\sqrt{\lambda})=\sqrt{\lambda}.$$ This is a transcendental equation. You can plot the graphs of $$y=\tan(x)$$ and $$y=x$$, and check the intersections of the graphs for $$x \ge 0$$. The negative values of $$\sqrt{\lambda}$$ can be ignored because they lead to duplicate values of $$\lambda$$. The non-negative solutions are ordered as follows: $$\sqrt{\lambda_0} = 0 < \sqrt{\lambda_1} < \frac{\pi}{2} < \sqrt{\lambda_2} < \frac{3\pi}{2} < \sqrt{\lambda_3} < \frac{5\pi}{2} < \cdots$$ Asymptotically, $$\sqrt{\lambda_n} \approx \frac{(2n-1)\pi}{2}$$ or $$\lambda_n \approx \frac{(2n-1)^2\pi^2}{4}$$ as $$n\rightarrow\infty$$.