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

[miniradman]

Homework Statement

find the solution to:

[itex]\frac{\partial^{2}u}{\partial x \partial y} = 0[/itex]

[itex]\frac{\partial^{2}u}{\partial x^{2}} = 0[/itex]

[itex]\frac{\partial^{2}u}{\partial y^{2}} = 0[/itex]

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:

[itex]u(x,y) = xf(y) + g(y)[/itex]

[itex]u(x,y) = yf(x) + g(x)[/itex]

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)

[itex]u(x,y) = F(y) + g(x)[/itex] where [itex]F(x) [/itex] is the integral of [itex]f(x)[/itex]

However the final solution is:

[itex]u(x,y) = Ax + By + C[/itex]

...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 [itex]F(x) [/itex] (what to do with it).

 

[strangerep]

[itex]\frac{\partial^{2}u}{\partial x \partial y} = 0[/itex]
[itex]\frac{\partial^{2}u}{\partial x^{2}} = 0[/itex]
[itex]\frac{\partial^{2}u}{\partial y^{2}} = 0[/itex]

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:

[itex]u(x,y) = xf(y) + g(y)[/itex]
[itex]u(x,y) = yf(x) + g(x)[/itex]

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)

[itex]u(x,y) = F(y) + g(x)[/itex] where [itex]F(x) [/itex] is the integral of [itex]f(x)[/itex]

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,

[itex]\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[/itex]
[itex]\Rightarrow u(x,y)=H(x)+g(y)[/itex], where [itex]H(x)=\int h(x)\, dx[/itex]

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