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I want to solve the pde $u_x+u_y+u=e^{x+2y}$ with ivc $u(x,0)=0$, which is a specific case of the general pde $au_x+bu_y+cu=g(x,y)$, where $a,b,c\in \mathbb{R}$.

Consider the change of coordinates $\xi=ax+by, \ \eta =ax-by$. With this change of variable it can be shown that the pde becomes $(a^2+b^2)u_\xi +cu=g(\xi,\eta)$ and the solution is given by

$ u(\xi, \eta)=e^{-\frac{c}{a^2+b^2}\xi} (f(\eta)+\frac{1}{a^2+b^2} \int g(\xi,\eta)e^{\frac{c}{a^2+b^2}\xi} d\xi )$

From the pde I wrote in the beginning I found the solution $u(\xi,\eta)=-f(\eta) e^{-1/2\xi}+\frac{1}{3}e^{\xi+2\eta}$. It can easily be checked that it is a solution. My problem occurs when I have to use the initial value condition. I know it needs to be $u(x,0)=0$ How do I "transform" this ivc into my new variables? Because if I just use the initial value condition $u(\xi,0)=0$ I get

$ u(\xi,0)=f(0)=\frac{2}{3}e^{\frac{3}{2}\xi} $

and I am not sure this is correct nor how to find $f$ from this.

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Go back from the $\xi,\eta$ variables to $x,y$. The solution in the original variables is $$ u(x,y)=-f(x-y)\,e^{-(x+y)/2}+\frac13\,e^{3x-y}. $$ The initial condition is $$ u(x,0)=-f(-y)\,e^{-y/2}+\frac13\,e^{-y}=0. $$ From here you can easily find $f$.

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