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Linear partial differential eqn.

Posted September 8th, 2007 by vic777
in

How can I find a solution for the eqn. with Cauchy's conditions ?
$ xu_x+yu_y=2xy $, with $ u=2 $ on $ y=x^2 $

‹ linear partial differential eq. Linear partial equations ›

Comments

Extremly explicit and

On September 8th, 2007 vic777 says:

Extremly explicit and complete...very gratefull

Don't mention it.

On September 8th, 2007 Structure says:

Hi Vic777

Glad to be able to help you.
Marius

Solution

On September 8th, 2007 Structure says:

This is a quasilinear first order differential equation .
First we look for a solution using a implicit function.
let F(x,y,u(x,y))=0 a function where u(x,y) is the desired function. If we can apply the implicit function theorem then $ \frac{\partial u}{\partial x}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}} $ and $ \frac{\partial u}{\partial y}=-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}} $
Replacing in the given equation we have after nominator elimination
$ x\frac{\partial F}{\partial x}+y\frac{\partial F}{\partial x}+2xy\frac{\partial F}{\partial x}=0 $
From this we have to find two prime integrals for the symmetric system
$ \frac{dx}{x}=\frac{dy}{y}=\frac{dz}{2xy} $
From $ \frac{dx}{x}=\frac{dy}{y} $ we have $ \frac{y}{x}=c_1 $
and we can write $ \frac{ydx}{xy}=\frac{xdy}{xy}=\frac{ydx+xdy}{2xy}=\frac{dz}{2xy} $ so ydx+xdy=d(xy)=dz
and so $ z-xy=c_2 $
Now $ F(x,y,z)=G(\frac{y}{x},z-xy) $ is a solution for the second partial differential equation where G is an arbitrary function $ C^1 $so we can find the solution u(x,y) from the equation $ F(x,y,u(x,y))=G(\frac{y}{x},u(x,y)-xy)=0 $ .

If we can apply implicit function theorem with respect to the second variable we can find
$ u(x,y)-xy=g(\frac{y}{x}) $
Now $ u(x,x^2)=x^3+g(\frac{x^2}{x})=x^3+g(x)=2 $ so $ g(x)=2-x^3 $

Finally $ u(x,y)=2+xy-\frac{y^3}{x^3} $

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