# Transform $x^2u_{xx}-2xu_x+2u=\lambda x^2u$ into $w_{xx}=-\lambda w$ by choosing $M(x)$ where $u(x)=M(x)w(x)$

Consider the eigenvalue problem, $$x^2u_{xx}-2xu_x+2u=\lambda x^2u$$ for $$0, with boundary conditions $$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.

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.

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

Let the coefficient of $$w'(x)$$ be zero, namely $$2x^2M'(x)-2xM(x)=0.$$ This gives $$M(x)=x$$ and hence $$w''(x)=\lambda w(x).$$
We have $$$$x^2\left[M''(x)w(x)+2M'(x)w'(x)+M(x)w''(x)\right] -2x\left[M'(x)w(x)+M(x)w'(x)\right]+2m(x)w(x)=\lambda x^2M(x)w(x). \hspace{3cm} (1)$$$$ On rearranging terms, we get $$$$x^2M(x)w''(x) + \left[2x^2M'(x)-2xM(x)\right]w'(x)+\left[x^2M''(x)-2xM'(x) +2M(x)-\lambda x^2M(x)\right]w(x)=0.$$$$ Comparing the coefficients of the above equation with $$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.