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I have a problem with solving PDEs with the fourier trasform method when the function not depends only on x and t but also on the y variable.

In particular, when I have to solve this equation $u_t=u_x+u_y$ with the initial condition $$u(x,y,0)=f(x,y)$$.
If I have only the $x$ variable I have no problem to solve it but in this case what I have to do? I mean.. Do I take the fourier transform on x and the on y? Thank you for all attention.. I hope that someone would like to help me.

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  • $\begingroup$ You need 2D fourier transform -- It's basically the same as doing FT over x, and then FT the result over y (rows & columns), but I suggest you read about it first. $\endgroup$
    – orion
    Commented Apr 3, 2014 at 18:02

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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{dx}{ds}=-1$ , letting $x(0)=x_0$ , we have $x=x_0-s=x_0-t$

$\dfrac{dy}{ds}=-1$ , letting $y(0)=y_0$ , we have $y=y_0-s=y_0-t$

$\dfrac{du}{ds}=0$ , letting $u(0)=F(x_0,y_0)$ , we have $u(x,y,t)=F(x_0,y_0)=F(x+t,y+t)$

$u(x,y,0)=f(x,y)$ :

$F(x,y)=f(x,y)$

$\therefore u(x,y,t)=f(x+t,y+t)$

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  • $\begingroup$ Thank you very much for your answer, but the request is that unfortunately I have to solve it with the Fourier transform.. $\endgroup$ Commented Apr 4, 2014 at 7:04

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