# find the solution of this PDE !

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

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

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

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

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

Now treat $u_y+ u= f(y)$ 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 $\mu(y)$ such that $\frac{d\mu u}{dy}= \mu\frac{du}{dy}+ \mu u$.

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

That is, $e^yu_y+ e^yu= (e^yu)_y= f(y)$ so that $e^y u= F(y)+ C$ 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-

$e^yu(x,y)= F(y)+ G(x)$ so $u(x,y)= e^{-y}(F(y)+ G(x))$ 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):

$u(x,y)= e^{-y}(F(y)+ G(x))$ so $u(x, 0)= G(x)= x^2$. Now you have $u(x,y)= e^{-y}(F(y)+ x^2)$ but you still need another condition, perhaps something of the form u(0,y)= f(y), to determine F(y).
• April 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 $u=\frac{1}{2}e^{y}\left(y-\frac{1}{2}\right)+e^{-y}\left(x^2+\frac{1}{4}\right)$
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 $u_{y}(0,y) + u(0,y)$ gives the equation $G\prime(y)=ye^{2y}$ it is easy to find the value of G from here, then the last condition makes it easy to find the function in x.
• April 18th 2010, 06:13 AM
Jester
Your answer is correct! Another way would do is something a little different. I'd let $v = u_x$ so the PDE is

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

The PDE is separable so the solution is

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

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

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

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

$
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.
• April 19th 2010, 10:06 AM
punkstart
Much simpler ! Thank you!
Will remember that aproach!