# Thread: Change of coordinates question for a pde

1. ## Change of coordinates question for a pde

Hello,

I am struggling with the following question:

Find all the solutions of the equation

(d/dt (u(x,t))) -(d/dx (u(x,t))) +2u(x,t)=0

Hint: a change on coordinates to simplify (d/dt (u(x,t))) -(d/dx (u(x,t)))

Any help woyld be appreciated.

2. ## Re: Change of coordinates question for a pde

Originally Posted by lordslytherin
Hello,

I am struggling with the following question:

Find all the solutions of the equation

(d/dt (u(x,t))) -(d/dx (u(x,t))) +2u(x,t)=0

Hint: a change on coordinates to simplify (d/dt (u(x,t))) -(d/dx (u(x,t)))

Any help woyld be appreciated.
Maybe this will help you. See my last post here:

http://www.mathhelpforum.com/math-he...-182858-2.html

3. ## Re: Change of coordinates question for a pde

Originally Posted by Also sprach Zarathustra
Maybe this will help you. See my last post here:

http://www.mathhelpforum.com/math-he...-182858-2.html
I think you and the OP are talking about different problems.

To the OP. If you introduce a change of variable from $(t,x)$ to $(r,s)$ then from the chain rule

$u_t = u_r r_t + u_s s_t$
$u_x = u_r r_x + u_s s_x$

so your original problem becomes

$u_r r_t + u_s s_t - \left( u_r r_x + u_s s_x\right) + 2u = 0$
or

$\left(r_t- r_x\right)u_r +\left(s_t - s_x\right)u_s + 2 u = 0$.

If you can chose r and s such that

$r_t - r_x = 0$ and $s_t-s_x = 1$, the your PDE becomes $u_s + 2u = 0$ - and ODE!

Here's one choice
Spoiler:

$r = t+x, s = t$

See how that goes.