 # How to solve this linear pde for y ( x ) (other functions are known and &#x03BB;< herbariak1 2022-05-21 Answered
How to solve this linear pde for $y\left(x\right)$ (other functions are known and $\lambda$ is a constant):
$\frac{d}{dx}\left(g\left(x\right){y}^{\prime }\left(x\right)\right)={\lambda }^{2}g\left(x\right)y\left(x\right)$
Everything I know about this equation is that it is called the Sturmian equation. I did some research, but the theory, which is for a more general form of the equation, is too hard for me to understand.
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Consider the operator
$Lf=\frac{1}{g}\frac{d}{dx}\left(g\frac{df}{dx}\right).$
You want to solve $Lf={\lambda }^{2}f$. Start by solving for ${f}_{0}$ such that
$L{f}_{0}=0,\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{f}_{0}\left(0\right)=A,\phantom{\rule{thickmathspace}{0ex}}{f}_{0}^{\prime }\left(0\right)=B.$
This is done by integrating the following in $x$:
$\frac{d}{dx}\left(g\frac{d{f}_{0}}{dx}\right)=0\phantom{\rule{0ex}{0ex}}g\frac{d{f}_{0}}{dx}-g\left(0\right)B=0\phantom{\rule{0ex}{0ex}}\frac{d{f}_{0}}{dx}=\frac{Bg\left(0\right)}{g\left(x\right)}\phantom{\rule{0ex}{0ex}}{f}_{0}\left(x\right)=A+B{\int }_{0}^{x}\frac{g\left(0\right)}{g\left(x\right)}dx.$
Then recursively solve
$\frac{1}{g}\frac{d}{dx}\left(g\frac{d{f}_{n}}{dx}\right)={f}_{n-1},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{f}_{n}\left(0\right)=0,\phantom{\rule{thickmathspace}{0ex}}{f}_{n}^{\prime }\left(0\right)=0\phantom{\rule{0ex}{0ex}}g\frac{d{f}_{n}}{dx}={\int }_{0}^{x}g\left(y\right){f}_{n-1}\left(y\right)dy\phantom{\rule{0ex}{0ex}}{f}_{n}\left(x\right)={\int }_{0}^{x}\frac{1}{g\left(z\right)}{\int }_{0}^{z}g\left(y\right){f}_{n-1}\left(y\right)dydz$
Now form the sum
$f\left(x,\lambda \right)=\sum _{n=0}^{\mathrm{\infty }}{\lambda }^{2n}{f}_{n}\left(x\right).$
You can verify that $f\left(0\right)={f}_{0}\left(0\right)=A$ and ${f}^{\prime }\left(0\right)={f}_{0}^{\prime }\left(0\right)=B$. And,
$Lf=\sum _{n=1}^{\mathrm{\infty }}{\lambda }^{2n}{f}_{n-1}={\lambda }^{2}\sum _{n=0}^{\mathrm{\infty }}{\lambda }^{2n}{f}_{n}={\lambda }^{2}f.$