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.