0
$\begingroup$

Given the linear homogeneous PDE

$$ y'' + p(x)y' + q(x)y = 0 $$

and the transformation

$$ u = y\exp(\frac{1}{2}\int_{a}^{x} p(s) ds) $$

I need to show

$$ u'' + r(x)u = 0 $$

where

$$ r = q - \frac{p'}{2} - \frac{p^2}{4} $$

I am trying to show this by calculating $y'$ and $y''$ via $u$ and plugging them into the first equation to get the second equation.

I use the fact that $y' = \frac{dy}{du} \frac{du}{dx}$ and $y''$ = $\frac{d^{2}u}{dx^2}(\frac{dx}{du})^2 + \frac{dy}{du} \frac{d^{2}u}{dx^2}$

And this leads me to

$$ y' = \frac{pu}{2\exp(\frac{1}{2}\int_{a}^{x} p(s) ds)} $$

and

$$ y'' = \frac{p^{2}u}{4\exp(\frac{1}{2}\int_{a}^{x} p(s) ds)} $$

Transforming my equation into this (by plugging in $\frac{p^{2}u}{4}$ for $u''$):

$$ u'' + \frac{p^{2}u}{2} + qu = 0 $$

However, this is not the correct equation I am trying to show.

Can someone tell me where I made a mistake in my calculations? I suspect it involves my chain rule application.

$\endgroup$
1
  • $\begingroup$ I think the use of the chain rule is wrong. You seem to assume that $$y(x) = y(u(x))$$ or something similar. But actually this is not the case, since $y$ depends also explicitly on the $x$ variable through $p$. $\endgroup$ – Kore-N Dec 5 '17 at 10:10
1
$\begingroup$

$$ y'' + p(x)y' + q(x)y = 0.$$ $$ u = y\exp(\frac{1}{2}\int_{a}^{x} p(s) ds)\Rightarrow y = \displaystyle ue^{-\frac{1}{2}\int_{a}^{x} p(s) ds}=uv, $$ where $v=-\frac{1}{2}\int_{a}^{x} p(s) ds$. Observe that $v'=-\frac{1}{2}pv$ and $v^{''}=-\frac{1}{2}p'v+\frac{1}{4}p^2v=(-\frac{1}{2}p'+\frac{1}{4}p^2)v$.

Since $y=uv$, differentiating we get $y'=u'v+uv'=(u'-\frac{1}{2}pu)v$ and $y^{''}=u{''}v+2u'v'+uv^{''}=(u{''}-\frac{1}{2}p'u-\frac{1}{2}pu')v-\frac{1}{2}p(u'-\frac{1}{2}pu)v=[u^{''}-\frac{1}{2}p'u-\frac{1}{2}pu'+\frac{1}{4}p^2u]v$.
Now substituting $y,y'$ and $y^{''}$ into the equation $$y'' + p(x)y' + q(x)y = 0$$ we get the desired transformed equation $$ u'' + \big(q-\frac{1}{2}p'-\frac{1}{4}p^2\bigg)u = 0.$$

$\endgroup$
1
  • $\begingroup$ Perfect, it makes sense now. Thank you! $\endgroup$ – Tetramputechture Dec 5 '17 at 16:23
0
$\begingroup$

Let us write $$P(x) = \frac{1}{2}\int_0^x p(y) dy.$$ We evaluate the derivatives in the following order:

$$ \partial_x u = \partial_x y e^P + y e^P\frac{1}{2}p, $$ as well as $$ \partial_x^2 u = \underbrace{\partial_x^2 y e^P + 2\partial_xy e^P\frac{1}{2}p}_{-ye^P q} + y e^P \Big( \frac{1}{4}p^2 + \frac{1}{2}p' \Big). $$ So that substituting $u = y e^P$on the right hand-side you find the required identity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.