Linear Differential Partial eq.

Posted September 16th, 2007 by Anonymous

Analysis questions

What are the characteristics and characteristic coordinates for the eq:

$u_{xx}+5u_{xy}+4u_{yy}+7u_y=\sin(x)$

Comments

Answer

On September 21st, 2007 Structure says:

Characteristic equation is $y'^2-5y'+4=0$ with solution y'=4 and y'=1
so $y-4x=c_1 and y-x=c_2$
You can see that $u_{part}=-sinx$ transform equation taking
$u(x,y)=u_{part}+u_{hom}$ into

$\xi=y-4x$ $\psi=y-x$

$\frac{\partial^2u_{hom}}{\partial x^2}+5 \frac{\partial^2u_{hom}}{\partial x\partial y}+4 \frac{\partial^2u_{hom}}{\partial y^2}+7 \frac{\partial u_{hom}}{\partial y}=0$
So if you take $u_{hom}(x,y)=v(\xi(x,y),\psi(x,y))=v(y-4x,y-x)$ you get
$\frac{\partial u_{hom}}{\partial x}(x,y)=-4\frac{\partial v}{\partial\xi}(y-4x,y-x)-\frac{\partial v}{\partial\psi}(y-4x,y-x)$
$\frac{\partial u_{hom}}{\partial y}(x,y)=\frac{\partial v}{\partial\xi}(y-4x,y-x)+\frac{\partial v}{\partial\psi}(y-4x,y-x)$
$\frac{\partial ^2u_{hom}}{\partial x^2}(x,y)=16\frac{\partial^2 v}{\partial\xi^2}(y-4x,y-x)+8\frac{\partial ^2v}{\partial \xi \partial\psi}(y-4x,y-x)+\frac{\partial^2 v}{\partial\psi^2}(y-4x,y-x)$

$\frac{\partial ^2u_{hom}}{\partial x \partial y}(x,y)=-4\frac{\partial^2 v}{\partial\xi^2}(y-4x,y-x)-5\frac{\partial ^2v}{\partial \xi \partial\psi}(y-4x,y-x)-\frac{\partial^2 v}{\partial\psi^2}(y-4x,y-x)$
$\frac{\partial ^2u_{hom}}{\partial y^2}(x,y)=\frac{\partial^2 v}{\partial\xi^2}(y-4x,y-x)+2\frac{\partial ^2v}{\partial \xi \partial\psi}(y-4x,y-x)+\frac{\partial^2 v}{\partial\psi^2}(y-4x,y-x)$

Finally you get:
$\frac{\partial^2v}{\partial\xi\partial\psi}+7\frac{\partial v}{\partial \xi}+7\frac{\partial v}{\partial \psi}=0$

So the solution is $u(x,y)=-\sin x+u_{hom}(x,y)$