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 $\boxed{\beta=\frac{A^2}{B^2}}$

What are your thoughts? Thanks in advance!

Edit: I just realized that my last problem did not necessarily satisfy $X(1)=0.$

  • 1
    $\begingroup$ 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 $$\lambda = \tan^{2}(\sqrt{\lambda})$$ $\endgroup$ – Mattos Nov 15 '18 at 6:10
  • $\begingroup$ Why do you have deleted your question without leaving a comment? It takes some time to post an appropriate answer. $\endgroup$ – callculus Nov 26 '18 at 20:30

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 $n\rightarrow\infty$.


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