Second order PDE (w.r.t 2 variables)

[miniradman]

Homework Statement

find the solution to:

$\frac{\partial^{2}u}{\partial x \partial y} = 0$

$\frac{\partial^{2}u}{\partial x^{2}} = 0$

$\frac{\partial^{2}u}{\partial y^{2}} = 0$

Homework Equations

theorem of integration

The Attempt at a Solution

now from a previous question I had earlier, I have found that I can simply do integration as per normal. So in doing that I managed to get:

$u(x,y) = xf(y) + g(y)$

$u(x,y) = yf(x) + g(x)$

However I have a problem that arises when I take the integral of two different variables (in the care of the first expression for u)

$u(x,y) = F(y) + g(x)$ where $F(x)$ is the integral of $f(x)$

However the final solution is:

$u(x,y) = Ax + By + C$

...to which I don't how to get to. I understand that when you put back all the partial differentials together, all those arbitrary functions collapse down to one constant of integration. However, I don't see how to get those constant co-efficients in front of x and y. Also I don't know how to treat the $F(x)$ (what to do with it).

 

[strangerep]

$\frac{\partial^{2}u}{\partial x \partial y} = 0$
$\frac{\partial^{2}u}{\partial x^{2}} = 0$
$\frac{\partial^{2}u}{\partial y^{2}} = 0$

The Attempt at a Solution

now from a previous question I had earlier, I have found that I can simply do integration as per normal. So in doing that I managed to get:

$u(x,y) = xf(y) + g(y)$
$u(x,y) = yf(x) + g(x)$

I'm guessing you applied a previous result to the 2nd and 3rd PDEs? But you chose the same function names(!). They should have different names, else confusion will result.

However I have a problem that arises when I take the integral of two different variables (in the care of the first expression for u)

$u(x,y) = F(y) + g(x)$ where $F(x)$ is the integral of $f(x)$

Huh?? Why are you trying to integrate like this? In any case, it's perhaps better to simply start with your solution to the 2nd PDE, i.e.,
$$u(x,y) = xf(y) + g(y)$$ and then substitute that into the 3rd. That gives you some constraints on what $f(y)$ and $g(y)$ can possibly be. In fact, it determines both of these up to 4 constants. Then substitute the solution for $u$ so far into the 1st PDE. That should give the correct answer with 3 constants.

[...] I understand that when you put back all the partial differentials together, all those arbitrary functions collapse down to one constant of integration.

I suspect you're misunderstanding something here. Try the method I outlined above.

 

[benorin]

A simple PDE

Instead of using the previous result:

From the frist equation,

$\frac{\delta ^2 u}{\delta x\delta y}=0\Rightarrow \int \frac{\delta ^2 u}{\delta x\delta y}\, dy=\int 0 \, dy\Rightarrow \frac{\delta u}{\delta x}=h(x)\Rightarrow \int\frac{\delta u}{\delta x}\, dx=\int h(x)\, dx$
$\Rightarrow u(x,y)=H(x)+g(y)$, where $H(x)=\int h(x)\, dx$

now substitute this into $u(x,y)=H(x)+g(y)$ and plug it into the second and third equations to obtain a system of two ordinary differential equations (simple ones), namely $H^{\prime\prime}(x)=0$ and $g^{\prime\prime}(x)=0$. Solve these and substitute the their solutions into $u(x,y)=H(x)+g(y)$ to arrive at $u(x,y)=Ax+By+C$.