Show that lamda is greater than or equal to zero for a sturm liouville problem

Question

To show that this problem can be put into S-L form for an eigenvalue problem,

Observe that

The S-L form is of $$\text{p'(x)}\phi _x\text{+p(x)}\phi _{\text{xx}}\text{+q(x)$\phi $+$\lambda \phi $w(x)=0}$$

And multiplying the equation by $$1/x$$, we have

$$\text{x$\phi $''+$\phi $'+}\text{$\lambda $x}^{-1}\text{$\phi $=0}$$

we have that $$p'(x)=1,p(x)=x,q(x)=0 and w(x)=1/x$$

How do I show that $$\lambda \geq 0$$? Am I to invoke Rayleight's quotient?

Answer 1

You shoud subsitute in formla then show that every term is equal or grater than zero

Answer 2

First note that $$ x\phi''(x)+\phi'(x)+\frac{\lambda}{x}\phi(x) = 0 $$ can be written as $$ -(x\phi'(x))' =\frac{\lambda}{x}\phi(x) $$ Now suppose that $\phi$ is a solution of the above for which $\phi(1)=\phi(2)=0$. Then $$ \lambda\int_{1}^{2}\phi(x)^{2}\frac{1}{x}dx = -\int_{1}^{2}(x\phi'(x))'\phi(x)dx \\ = \left\{ -(x\phi'(x))\phi(x)|_{x=1}^{2}+\int_{1}^{2}x\phi'(x)\phi'(x)dx\right\} \\ = \int_{1}^{2}x\phi'(x)^{2}dx. $$ If $\phi$ is not identically $0$, then $$ \lambda = \frac{\int_{1}^{2}x\phi'(x)^{2}dx}{\int_{1}^{2}\phi(x)^{2}\frac{1}{x}dx} \ge 0. $$ In fact, the equality is strict because if $\lambda = 0$, then $\phi'\equiv 0$ and, hence, $\phi\equiv C$ which forces $C=0$ because $\phi(1)=\phi(2)=0$. Therefore $\lambda > 0$.

The Eigenfunctions and Eigenvalues: The original ODE is Euler's equation $$ x^{2}\phi''(x)+x\phi'(x)+\lambda\phi(x)=0. $$ The solutions for $\lambda > 0$ are of the form $x^{\alpha}$ where $\alpha$ satisfies $$ \alpha(\alpha-1)+\alpha+\lambda = 0,\\ \alpha^{2}=-\lambda \\ \alpha = \pm i\sqrt{\lambda}. $$ The eigenfunctions $\phi_{\lambda}=Ax^{i\sqrt{\lambda}}+Bx^{-i\sqrt{\lambda}}$ require $\lambda$ to satisfy $$ \phi_{\lambda}(1)=A+B = 0 \implies A=-B \\ \phi_{\lambda}(2)=A2^{i\sqrt{\lambda}}-A2^{-i\sqrt{\lambda}}=0 \implies 2^{2i\sqrt{\lambda}}=1 $$ So the eigenvalues are the $\lambda$ for which $$ e^{2i\ln(2)\sqrt{\lambda}}=1 \\ \implies 2\ln(2)\sqrt{\lambda} = 2\pi n,\;\;\;n=1,2,3,\cdot \\ \implies \lambda = \frac{\pi^{2}n^{2}}{\ln(2)^{2}} $$ Substituting back into the expression for $\phi_{\lambda}$ above, the eigenfunctions are non-zero constant multiples of $$ \phi_{n}(x) = \sin\left(n\pi\frac{\ln x}{\ln 2}\right),\;\;\; n=1,2,3,\cdots. $$