# Thread: Partial derivatives (from D'Alembert)

1. ## Partial derivatives (from D'Alembert)

Hello, I'm wondering, how would you answer the following question?

I've put in the conditions, but really can't see how to find the functions. I'd really appreciate some help. Thank you.

2. Originally Posted by jonmondalson
Hello, I'm wondering, how would you answer the following question?

I've put in the conditions, but really can't see how to find the functions. I'd really appreciate some help. Thank you.
So we have that

$\displaystyle u_t(x,t)=-cF'(x-ct)+cG'(x+ct)$

If you plug into the equations you get that

$\displaystyle \frac{1}{x^2+1}=F(x)+G(x)$

$\displaystyle 0=-cF'(x)+cG'(x) \iff F'(x)=G'(x)$

The last line gives

$\displaystyle F(x)=G(x)+k$

This gives

$\displaystyle G(x)=\frac{1}{2}\frac{1}{x^2+1}-\frac{k}{2}$

and

$\displaystyle F(x)=\frac{1}{2}\frac{1}{x^2+1}+\frac{k}{2}$

$\displaystyle F(x-ct)=\frac{1}{2}\frac{1}{(x-ct)^2+1}+\frac{k}{2}$

$\displaystyle G(x+ct)=\frac{1}{2}\frac{1}{(x+ct)^2+1}-\frac{k}{2}$

So the final solution is

$\displaystyle u(x,t)=\frac{1}{2}\frac{1}{(x-ct)^2+1}+\frac{1}{2}\frac{1}{(x+ct)^2+1}$