I'm currently having a class at university that discusses Sturm Liouville Problems. We have to solve a few problems one of which I can't seem to solve. We use Advanced Engineering Mathematics by Erwin Kreyszig.
The problem is a Sturm-Liouville problem:
$$y''-2y'+(\lambda+1)y=0\ ,$$ with boundary conditions $y(0)=0$ and $y(1)=0$.
The problem says the following:
Find the eigenvalues and eigenfunctions. Verify orthogonality. Start by writing the ODE in the form of $$[p(x) y']' + [q(x) + \lambda r(x)]y=0$$ using problem 6.
Problem 6 states the following:
Show that $$y''+f y'+(g+\lambda h)y=0$$ takes the form
$$[p(x) y']' + [q(x) + \lambda r(x)]y=0$$
if you set
$$\begin{align}
p&=e^{\int f\ dx}\ ,\cr
q&=p g\ ,\cr
r&=p h\ .
\end{align}$$
Why would you do such a transformation?
Now the problems are that I can't seem to understand why such a transformation is useful, and I can't seem to find the solutions of the differential equation.
According to problem 6 I find the following:
$$\begin{align}
f&=-2\cr
g&=1\cr
h&=1
\end{align}$$
so:
$$
\begin{align}
p(x) &= e^{-2x}\cr
q(x) &= e^{-2x}\cr
r(x) &= e^{-2x}\ .
\end{align}
$$
Thus I get
$$[e^{x}y']' + [1+\lambda]e^{-2x}y = 0$$ where $y(0)=0$ and $y(1)=0$.
But I can't seem to get any further with this.
Best Answer
To put the equation $$ ay''+by'+cy=\lambda dy $$ into the form $$ (py')'+ry=\lambda wy,\\ py''+p'y'+ry=\lambda wy, $$ it is natural to try to multiplying the first equation by some function $h$ in order to arrange things so that $hb$ is the derivative of $ha$. Then the second form is achieved with $p=ah$. This gives you an equation for $h$ in terms of $a$ and $b$: $$ (ah)'=bh \\ ah'+a'h=bh \\ \frac{h'}{h}=-\frac{a'}{a}+\frac{b}{a} \\ \ln |h| = -\ln |a|+\int\frac{b}{a}\,dx \\ h = \frac{1}{a}\mbox{exp}\left(\int\frac{b}{a}\,dx\right). $$ (I've ignored a multiplicative constant--you can choose it to be $1$.) Multiplying the first equation by this $h$ leads to $$ \left( \mbox{exp}\left\{\int\frac{b}{a}\,dx\right\}y' \right)'+chy=\lambda dhy. $$ In your case, $a=1$, $b=-2$, which gives $h=\frac{1}{a}\mbox{exp}(\int\frac{-2}{1}dx)=e^{-2x}$, and leads to the equation $$ (e^{-2x}y')'+(\lambda+1)e^{-2x}y=0. $$ This is a weighted Sturm-Liouville equation with weight $w(x)=e^{-2x}$. Eigenfunctions $\phi_{\lambda}$, $\phi_{\mu}$ corresponding to different eigenvalues $\lambda$, $\mu$ are orthogonal with respect to the weight $w$: $$ \begin{align} (\lambda-\mu)\int_{0}^{1}e^{-2x}\phi_{\mu}\phi_{\lambda}\,dx & = -\int_{0}^{1}(e^{-2x}\phi_{\lambda}')'\phi_{\mu}-\phi_{\lambda}(e^{-2x}\phi_{\mu}')'\phi_{\lambda}\,dx \\ & = -\int_{0}^{1}\left[(e^{-2x}\phi_{\lambda}')\phi_{\mu}-\phi_{\lambda}(e^{-2x}\phi_{\mu}')\right]'\,dx \\ & = -\left.\left[(e^{-2x}\phi_{\lambda}')\phi_{\mu}-\phi_{\lambda}(e^{-2x}\phi_{\mu}')\right]\right|_{x=0}^{1} = 0. \end{align} $$ Eigenfunctions/Eigenvalues: The Sturm-Liouville normal form does not help you solve the equation; the normal form gives you orthogonality relations and other interesting information, but doesn't usually help you solve. Your original equation is $$ y''-2y'+(1+\lambda)y = 0,\\ y(0)=y(1)=0. $$ Without regard to the endpoint conditions, the general solution is $$ y=Ae^{\lambda_{1}x}+Be^{\lambda_{2}x} $$ where $\lambda_{1},\lambda_{2}$ are the roots of $p(z)=0$, where $$ p(z)=z^{2}-2z+(1+\lambda) = 0,\\ (z-1)^{2}=-\lambda,\\ z = 1 \pm \sqrt{-\lambda} $$ The case of a double root where $\lambda=0$ has to be handled separately. If $\lambda < 0$, then the general solution can be written as $$ y=(A\sinh(\sqrt{-\lambda}x)+B\cosh(\sqrt{-\lambda}x))e^{x}. $$ If $\lambda > 0$, then the general solution is better written as $$ y = (A\sin(\sqrt{\lambda}x)+B\cos(\sqrt{\lambda}x))e^{x}. $$ Because you need $y(0)=0$, then $B=0$ in either case. The second endpoint condition is satisfied only for a discrete set of $\lambda$ which give $y(1)=0$. Otherwise there is no non-zero solution satisfying the endpoint conditions.
There are no non-zero solutions for $\lambda < 0$ because $\sinh$ is only $0$ at $0$. For $\lambda > 0$, the eigenvalues $\lambda$ are the values $\lambda > 0$ for which $\sin(\sqrt{\lambda})=0$, which gives $$ \sqrt{\lambda} = n\pi \implies \lambda = n^{2}\pi^{2}. $$ So the eigenvalues are $\lambda_{n}=n^{2}\pi^{2}$ for $n=1,2,3,\cdots$ and the corresponding eigenfunctions are $$ \phi_{n}(x)=e^{x}\sin(n\pi x),\;\;\; n=1,2,3,\cdots. $$ The special case where $\lambda=0$ does not give an eigenvalue because the general solution is $(A+Bx)e^{x}$, which cannot vanish at $x=0$ and $x=1$ unless $A=B=0$. Notice that you do have the desired orthogonality for the eigenfunctions: for $n \ne m$, $$ \int_{0}^{1}e^{-2x}\phi_{n}(x)\phi_{m}(x)\,dx=\int_{0}^{1}\sin(n\pi x)\sin(m\pi x)\,dx =0,\;\;\; n \ne m. $$