Answer to Question #128950 in Differential Equations for Sehrish

Question

Find the eigenvalues and associated eigenfunction of the strum-Liouville problem xy"+y'+lamda/2 y=0,y'(1)=0,y'(e^2π)=0

Answer 1

$$Solution$$

$$xy''+y'+ \frac{\lambda}{2 }y=0$$

$$y'(1)=0,y'(e^2π)=0$$

We begin by making some assumptions which will simplify the problem. This will turn our differential equation into a constant-coefficient equation.

Let

$t=lnx$

And

$y(x)=\phi (ln x)=\phi(t)$

$y'(x)=\frac{1}{x}(\frac{d \phi (t)}{dt})$

$y''(x)=\frac{1}{x^2}(\frac{d^2 \phi (t)}{dt^2}-\frac{d \phi (t)}{dt})$

Next, we substitute our new y's to the original equation and simplify until we get a second order, constant coefficient equation.

$$x(\frac{1}{x^2}(\frac{d^2 \phi (t)}{dt^2}-\frac{d \phi (t)}{dt}))+\frac{1}{x}(\frac{d \phi (t)}{dt})+\frac{\lambda}{2 }\phi(t)=0$$

$$(\frac{d^2 \phi (t)}{xdt^2}-\frac{d \phi (t)}{xdt})+(\frac{d \phi (t)}{xdt})+\frac{\lambda}{2 }\phi(t)=0$$

$$\frac{d^2 \phi (t)}{xdt^2}+\frac{\lambda}{2 }\phi(t)=0$$

This is easy to solve with a characteristic equation.

$$0=r^2+\frac{\lambda}{2}x$$

Solving for r we get

$\sqrt{r^2}=\sqrt{-\frac{\lambda}{2}x} \implies r=\sqrt{-\frac{\lambda}{2}x}$

We now write out the solution in exponential form.

$$\phi(t)=C_1 \epsilon^{(\sqrt{\frac{-\lambda x}{2}})t}+C_2 \epsilon^{(\sqrt{\frac{-\lambda x}{2}})t}$$

Converting back to $y(x)$ we get

$$y(x)=C_1 \epsilon^{(\sqrt{\frac{-\lambda x}{2}})ln x}+C_2 \epsilon^{(\sqrt{\frac{-\lambda x}{2}})ln x}$$

Applying Euler's identities and picking two linear independent solutions

$$y(x)=C_3 \epsilon^{0}sin(\sqrt{\frac{-\lambda x}{2}})ln x+C_4 \epsilon^{0} cos (\sqrt{\frac{-\lambda x}{2}})ln x$$

$\epsilon ^{0} = 1$

$$y(x)=C_3 sin(\sqrt{\frac{-\lambda x}{2}})ln x+C_4 cos (\sqrt{\frac{-\lambda x}{2}})ln x$$

Substituting our boundary conditions.

$$y=0,y'(1)=0,y'(e^2π)=0$$

Starting with y'(1) we get

$$y(1)=C_3(1)sin(\sqrt{\frac{-\lambda x}{2}ln(1)}+C_4(1)cos(\sqrt{\frac{-\lambda x}{2}ln(1)}$$

$$y(1)=0=C_3sin0+C_4cos0=C_4 \implies C_4=0$$

$$y(x)=C_3(1)sin(\sqrt{\frac{-\lambda x}{2}ln(1)}$$

$y'(\epsilon ^{2\pi})=0$

$y'(\epsilon^{2\pi})=0=C_3(\epsilon^{2\pi})sin(\sqrt{\frac{-\lambda x}{2}ln(\epsilon^{2\pi})}$

$n \pi=\sqrt{\frac{-\lambda x}{2}ln(\epsilon^{2\pi})} \implies \lambda=(\frac{2n\pi}{x \epsilon^{2\pi}})^2--->Eigenvalue$

Simplifying to get the eigenfunction, we get

$y(x)=Cx\sin \sqrt{x(\frac{2n\pi}{2x \epsilon^{2\pi}})^2}lnx \implies y(x)=Cx\sin \frac{n\pi}{x \epsilon^{2\pi}}\sqrt{lnx}--->Eigenfunction$