Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A solution to a PDE of my interest is

$u(x,y)=2\left(F''(x)-G''(y)\right)(x-y)^{2}-12\left(F'(x)+G'(y)\right)(x-y)+24\left(F(x)-G(y)\right)$

where $F(x)$ and $G(y)$ are arbitrary functions to be determined from initial conditions. In this case, I know

$u(x,y_{0})=\frac{u_{0}(x-y_{0})^{2}}{(x_{0}-y_{0})^{2}}$,

$u(x_{0},y)=\frac{u_{0}(x_{0}-y)^{2}}{(x_{0}-y_{0})^{2}}$.

I tried all sorts of things to find $F(x)$ and $G(y)$. I tried all sorts of things to no avail; any suggestions what steps/approaches I should take?

share|improve this question

1 Answer 1

I made some progress on this so I thought I would share it. First let $u(x_{0},y_{0})=u_{0}$ then one can write the following

$ u(x_{0},y_{0})=2\left(F''(x_{0}-G''(y_{0})\right)(x_{0}-y_{0})^{2}-12\left(F'(x_{0})+G'(y_{0})\right)(x_{0}-y_{0})+24\left(F(x_{0}-G(y_{0}\right)=u_{0} $

Now choose five values arbitrarily and the 6th value will be determined. I made the following choices: $F''(x_{0})=1\quad F'(x_{0})=1\quad F(x_{0})=1 \quad G''(y_{0})=?\quad G'(y_{0})=-1 \quad G(y_{0})=1$

Then find $G''(y_{0})$ as follows:

$ u_{0}=2(1-G''(y_{0}))(x_{0}-y_{0})^{2} $ and this gives $G''(y_{0})=1-k$ where

$ k=\frac{1}{2}\frac{u_{0}}{(x_{0}-y_{0})^{2}} $

Now apply the initial conditions. Let's start with:

$ u(x,y_{0})=\frac{u_{0}(x-y_{0})^{2}}{(x_{0}-y_{0})^{2}} $

Plugging this into the solution of the pde, using what we have abve and simplifying we obtain

$ (x-y_{0})^{2}F''(x)-6(x-y_{0})F'(x)+12F(x)=H(x) $

where

$H(x)=(x-y_{0})^{2}+6(x-y_{0})+12 $

Now what we have a second order non-homogeneous ODE and to solve for F first compute the fundamental solutions and proceed via the route of variation of parameters. One will use the other initial condition to find G. I will make an update soon on the full solution.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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