0
$\begingroup$

Let the boundary problem be: $x^2y''+\lambda y=0$ with $1<x<2$ and boundary conditions $y(1) = 0 = y(2)$. Assume $\lambda = \mu^2>0$I am trying to solve for $r$ by first dividing $x^2y''+\lambda y=0$ by $x^2$ to get $y''+ \frac{\lambda}{x^2} y =0$. This will give me $r = \pm i \frac{\mu}{x}$.

However, this will give me $y(x) = C_1\cos(\mu)+C_2\sin(\mu)$. I'm not quite sure how to proceed from here because it seems like the $x$ terms just cancel out unless I am doing something wrong.

$\endgroup$
1
$\begingroup$

This is a Cauchy-Euler equation, so we assume $y=x^r.$ Then we have $y'=rx^{r-1}$ and $y''=r(r-1)x^{r-2},$ which we insert into the original DE as \begin{align*} r(r-1)x^r+\lambda x^r&=0\\ r^2-r+\lambda&=0\\ r&=\frac{1\pm\sqrt{1-4\lambda}}{2}. \end{align*} Let \begin{align*} r_+&=\frac{1+\sqrt{1-4\lambda}}{2}\\ r_-&=\frac{1-\sqrt{1-4\lambda}}{2}. \end{align*} Then $$y=c_+ x^{r_+}+c_- x^{r_-}. $$ The boundary conditions yield \begin{align*} 0&=c_+ +c_- \\ 0&=c_+ 2^{r_+}+c_- 2^{r_-}. \end{align*} Unfortunately, the only solution is the trivial solution $c_-=c_+=0,$ implying that there are no solutions to this eigenvalue problem (since eigenvectors are nonzero by definition).

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Since $y$ is, in general, a function of $x$, we will first need to convert the equation into an equation with constant coefficients so that your method can be applied. Notice that this can be done by replacing $x$ by $e^{z}$, where since $1 < x < 2$, we have $0 < z < \ln 2$.

Also, if we denote $D \equiv \dfrac{d}{dx}$ and $\theta \equiv \dfrac{d}{dz}$, then

$$x^2 y'' = \theta \left( \theta - 1 \right) y$$

Hence, we get the boundary value problem

$$\theta^2 y - \theta y + \lambda y = 0, \ 0 < z < \ln 2$$

and the boundary conditions $y \left( 0 \right) = y \left( \ln 2 \right) = 1$.

Now, you can proceed by your method to find the eigenvalues and eigenfunctions for the BVP. However, in the end, do not forget to replace $z$ by suitable function of $x$, so that the solution is in terms of what was given in the problem statement.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.