Sturm-Liouville problem involving $dy/dx$

Question

$$ \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + \lambda y=0 \ , y(-1)=y(1) = 0 $$

I have 2 ideas

1- $ (\frac{dy}{dx} +2y )' + \lambda y = 0 \\ \text{which means weighted function is } w=1 $

2- $ \frac{dy}{dx} (p \frac{dy}{dx}) + \lambda w(x)y = 0 \\ \text{again weighted function is w=1.} $

I ignored dy/dx term .. I saw an example which just did that.

Please guide me. I need to construct Strum-Liouville system and find its orthogonality using weighted function.

Thank you

Answer 1

You given ODE: $$y''+2y'+\lambda y=0 ~~~(1) $$ Can be writte in S_L form as $$e^{-2x}\left(\frac{d}{dx} e^{2x} \frac{dy}{dx}\right)+\lambda y=0.~~~~(2)$$ ODE (1) can be solved by letting $y=e^{mx}$, we get $$m^2+2m+\lambda=0 \implies m_{1,2}=-1\pm \sqrt{1-\lambda}.$$ In order to have periodic solutions, let $q^2=\lambda-1$ So $$Y=e^{-x}[C_1 e^{iqx} +C_2 e^{-iqx}]$$ Applying $y(-1)=0=y(1)$, we get $$y=A e^{-x} \cos qx, x \in[-1,1], q=(n+1/2)\pi, n=0,2,3,..$$ So the eigenvalues are $\lambda_n=(n+1/2)^2\pi^2+1$, with eigenfunctions as $$y_n(x)=A e^{-x} \cos [(n+1/2) \pi x].$$ The given ODE can be cast in a simple S-L problem as (2), then $$-\frac{1}{w(x)}\frac{d}{dx} \left(p(x)\frac{dy_n}{dx}\right) =\mu_n y_n, x \in[a,b]$$ implies that the two solutions $y_m, y_n$ are orthogonal as $$\int_{a}^{b} w(x) y_m(x) y_n (x) dx=C \delta_{m,n}.$$ Here, in this case (1,2), $w(x)=p(x)=e^{2x}.$ We see the orthogonality here as $$A^2\int_{-1}^{1} e^{2x} e^{-2x} \cos[(m+1/2)\pi x] \cos[(n+1/2)\pi x] dx= A^2 \int_{0}^{1}[\cos[(m+n)\pi x]+ \cos[(m-n)\pi x]] dx=A^2\frac{\sin (m-n) \pi}{(m-n) \pi}=A^2 \delta_{m,n} $$

Answer 2

Let $y=e^{-x}f$. Then \begin{align} y' &= \;\;\;\;\;\;\;\;\;\;\;\;\;e^{-x}f'-e^{-x}f \\ y'' &= e^{-x}f''-2e^{-x}f'+e^{-x}f\\ y''+2y'+\lambda y &= e^{-x}f''+(\lambda -1)e^{-x}f.\\ \end{align} The equation in $f$ becomes $$ f''+(\lambda-1)f=0 \\ f(-1)=f(1)=0. $$ The corresponding solutions $f$ are scalar multiples of the following for $n=0,1,2,3,\cdots$: $$ s_n(x)= \sin(n\pi x),\;\; \lambda_n=1+n^2\pi^2 \\ c_n(x) = \cos((n+1/2)\pi x),\;\;\; \lambda_n=1+(n+1/2)^2\pi^2 $$ The original solutions $y_n$ are scalar multiples of $e^{-x}s_n(x)$ and $e^{-x}c_n(x)$.