2
$\begingroup$

Consider the eigenvalue problem, $$x^2u_{xx}-2xu_x+2u=\lambda x^2u$$ for $0<x<1$, with boundary conditions $u_x(0)=0$ and $u(1)=u_x(1)$. Determine a function $M(x)$ so that, under the change of variable $u(x)=M(x)w(x)$ , it can be transformed to the form $$w_{xx}=-\lambda w$$ for $0<x<1$.

My attempt:

If $u(x)=M(x)w(x)$, then $u'=Mw'+wM'$ and $u''=Mw''+wM''+2w'M'$. Substituting this into the intial equation gives, \begin{align} x^2(Mw''+wM''+2w'M')-2x(Mw'+wM')+2wM&=\lambda x^2wM \\ x^2Mw''+x^2wM''+2x^2w'M'-2xMw'-2xwM'+2wm&=\lambda x^2wM. \end{align} Observing the coefficient of $w''$ indicates that $$M=\frac{1}{x^2}.$$ However, applying this value of $M$ will make the RHS positive, not negative. Substituting this value of $M$ in to the equation does not yield the required form.

I do not understand what to do. Any advice/hints are appreciated.

$\endgroup$
  • $\begingroup$ Where is your $M''$? $\endgroup$ – xpaul Mar 13 at 1:27
  • $\begingroup$ Isn't it there? With coefficient $x^2w$. $\endgroup$ – user557493 Mar 13 at 1:28
  • $\begingroup$ Sorry, I didn't see before. I see now. $\endgroup$ – xpaul Mar 13 at 1:32
2
$\begingroup$

Let the coefficient of $w'(x)$ be zero, namely $$ 2x^2M'(x)-2xM(x)=0. $$ This gives $M(x)=x$ and hence $$ w''(x)=\lambda w(x). $$

$\endgroup$
1
+100
$\begingroup$

I try to elaborate on the accepted answer:

We have \begin{equation} x^2\left[M''(x)w(x)+2M'(x)w'(x)+M(x)w''(x)\right] -2x\left[M'(x)w(x)+M(x)w'(x)\right]+2m(x)w(x)=\lambda x^2M(x)w(x). \hspace{3cm} (1) \end{equation} On rearranging terms, we get \begin{equation} x^2M(x)w''(x) + \left[2x^2M'(x)-2xM(x)\right]w'(x)+\left[x^2M''(x)-2xM'(x) +2M(x)-\lambda x^2M(x)\right]w(x)=0. \end{equation} Comparing the coefficients of the above equation with $w''(x)-\lambda w(x) = 0$, we get that \begin{align} x^2M(x) &= g(x) \\ 2x^2M'(x)-2xM(x)&= 0 \\ x^2M''(x)-2xM'(x) +2M(x)-\lambda x^2M(x)&=-\lambda g(x) \end{align} From the seond equation, we get that \begin{multline} 2x^2M'(x)-2xM(x)= 0 \implies xM'(x)-M(x)=0\implies \frac{1}{M(x)}M'(x)=\frac{1}{x}\\\implies \int \frac{1}{M(x)} d(M(x))=\int \frac{1}{x} dx\implies \ln M(x) = \ln x\implies M(x)=x. \end{multline} Substituting this relation back to our set of three equations, we that $g(x)=x^3$. We can crosscheck that the solution is correct using (1), where we see that the factor of $g(x)=x^3$ beautifully cancels out to give the desired result.

$\endgroup$

Your Answer

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