$\displaystyle u_{xy}+u_{x} = 0$

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 ?!

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- Apr 17th 2010, 07:34 AMpunkstartfind 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 AMHallsofIvy
$\displaystyle u_{xy}+ u_x= (u_y+ u)_x= 0$

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 AMpunkstartSolved !
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 AMJester
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 AMpunkstartMuch simpler ! Thank you!
Will remember that aproach!