# find the solution of this PDE !

• Apr 17th 2010, 07:34 AM
punkstart
find the solution of this PDE !
$\displaystyle u_{xy}+u_{x} = 0$
subject to boundary conditions :
$\displaystyle u_{y}(0,y)+u(0,y) = ye^{y}$
$\displaystyle u(x,0)=x^2$

I have never done direct integration with this " mix " of order, how do i go about it ?!
• Apr 17th 2010, 08:01 AM
HallsofIvy
Quote:

Originally Posted by punkstart
$\displaystyle u_{xy}+u_{x} = 0$
subject to boundary conditions :
$\displaystyle u_{y}(0,y)+u(0,y) = ye^{y}$
$\displaystyle u(x,0)=x^2$

I have never done direct integration with this " mix " of order, how do i go about it ?!

$\displaystyle u_{xy}+ u_x= (u_y+ u)_x= 0$
so that $\displaystyle u_y+ u$ is constant with respect to x- that is, $\displaystyle u_y+u= f(y)$ where f is an arbitrary function of y only.

Now treat $\displaystyle u_y+ u= f(y)$ as an ordinary differential equation in y. It is a first order, linear, differential equation so it can be integrated using an "integrating factor"- that is, a function $\displaystyle \mu(y)$ such that $\displaystyle \frac{d\mu u}{dy}= \mu\frac{du}{dy}+ \mu u$.

since $\displaystyle \frac{d\mu u}{dy}= \mu \frac{du}{dy}+ u \frac{d\mu}{dy}= \mu\frac{du}{dy}+ \mu u$ we must have $\displaystyle \frac{d\mu}{dy}= \mu$ or $\displaystyle \frac{d\mu}{\mu}= 1$, $\displaystyle ln(\mu)= y$ and $\displaystyle \mu(y)= e^y$.

That is, $\displaystyle e^yu_y+ e^yu= (e^yu)_y= f(y)$ so that $\displaystyle e^y u= F(y)+ C$ where F(y) is an anti-derivative of f. Since f was arbitrary, F is any arbitrary (differentiable) function of y. However, since we have been doing this with y only, treating x as a constant, that "constant of integration", C, may be an arbitrary function of x-

$\displaystyle e^yu(x,y)= F(y)+ G(x)$ so $\displaystyle u(x,y)= e^{-y}(F(y)+ G(x))$ where F and G can be any arbitrary differentiable functions.

(Yes, just as the solutions to ordinary differential equations involve unknown constants, so the solutions to partial differential equations involve unknown functions).

Now, use the additional conditions (of which you don't have enough):

$\displaystyle u(x,y)= e^{-y}(F(y)+ G(x))$ so $\displaystyle u(x, 0)= G(x)= x^2$. Now you have $\displaystyle u(x,y)= e^{-y}(F(y)+ x^2)$ but you still need another condition, perhaps something of the form u(0,y)= f(y), to determine F(y).
• Apr 18th 2010, 03:09 AM
punkstart
Solved !
Thanx HallsofIvy,i used the integrating factor( i don't think i would have thought of that,since never tried to solve a PDE by making it first order!) and with a little manipulation got my answer. Apparently i can't pm yet, but you can verify that the solution is $\displaystyle u=\frac{1}{2}e^{y}\left(y-\frac{1}{2}\right)+e^{-y}\left(x^2+\frac{1}{4}\right)$
The key here was that you can rename the product of F(y) and the integrating factor,as G(y) since the product still only depends on y,then the condition on $\displaystyle u_{y}(0,y) + u(0,y)$ gives the equation$\displaystyle G\prime(y)=ye^{2y}$ it is easy to find the value of G from here, then the last condition makes it easy to find the function in x.
• Apr 18th 2010, 06:13 AM
Jester
Your answer is correct! Another way would do is something a little different. I'd let $\displaystyle v = u_x$ so the PDE is

$\displaystyle v_y + v = 0$ and the second BC $\displaystyle u(x,0) = x^2$ is $\displaystyle u_x(x,0) = 2x$ so $\displaystyle v(x,0) = 2x$.

The PDE is separable so the solution is

$\displaystyle v = F(x)e^{-y}$ and with the BC gives $\displaystyle F(x) = 2x$.

Thus far $\displaystyle v = 2xe^{-y}.$

Then $\displaystyle v = u_x = 2xe^{-y}$ and integrating gives

$\displaystyle u = x^2 e^{-y} + G(y).$ Now use the first BC

$\displaystyle u_y + u = G'(y) + G(y) = ye^y \; \text{when}\; x = 0$. This is an ODE for G which can be solved by ODE methods.
• Apr 19th 2010, 10:06 AM
punkstart
Much simpler ! Thank you!
Will remember that aproach!