# Partial differential equations: general solution

1. Nov 25, 2005

### sachi

I have a 2nd order homogenous P.D.E:

(d^2)V/((dx)^2) + (d^2)V/((dy)^2) + 6(d^2)V/(dx dy) = 0

where all derivatives are partial derivatives. I need to transform this to form

a*(d^2)f/(dX^2) + b*(d^2)f/(dY^2) = 0 where a, b are constants and the derivatives are again partial, and f(X,Y) = V(x,y)

I I need to make a change of variables e.g x=cX +dY etc. but I get very messy algebraic expressions which I can't simplify. Thanks very much.

2. Nov 25, 2005

### Fermat

You have V = V(x,y).

Then make the change of variables,

X = X(x,y)
Y = Y(x,y)

where,

X = y - m1x
Y = y - m2x

(It doesn't need to be any more complicated)

Now use the chain rule for partial differentials involving change of variables.

Last edited: Nov 25, 2005
3. Nov 25, 2005

### saltydog

You need to rotate axes to eliminate the mixed partial. This is normally done by defining a new function H(w,z) such that:

$$w=xCos(a)+ySin(a)\quad\text{and}\quad z=-xSin(a)+yCos(a)$$

with a to be determined so that the coefficient of the mixed partial in the final results is zero.

So, start calculating all the partials and substitute them into your equation and end up with a PDE in H(w,z), then determine a so that the coefficient of the mixed partial is zero. Solve the PDE in H(w,z), then substitute back x and y:

$$V(x,y)=H(xCos(a)+ySin(a),-xSin(a)+yCos(a))$$

I'll start off the partials:

$$\frac{\partial V}{\partial x}=\frac{\partial H}{\partial w}\frac{\partial w}{\partial x}+\frac{\partial H}{\partial z}\frac{\partial z}{\partial w}$$

so then:

$$\frac{\partial^2 V}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial H}{\partial w}\frac{\partial w}{\partial x}+\frac{\partial H}{\partial z}\frac{\partial z}{\partial w}\right)$$

So do that one, the mixed one, and then for y and then substitute them back into your PDE. It goes fast once you get going. Then solve for a to make the mixed one go away.

Edit: Hum . . . well maybe that too Fermat.

Last edited: Nov 25, 2005
4. Nov 26, 2005

### Fermat

Yeah That's the method we were shown for solving 2nd order pde's.

If the pde was of the form,

$$A\cdot \frac{\partial ^2V}{\partial x^2} + B\cdot \frac{\partial ^2V}{\partial x \partial y} + C\cdot \frac{\partial ^2V}{\partial y^2} = 0$$

we would transform it to,

$$P(m_1)\cdot \frac{\partial ^2V}{\partial X^2} + Q(m_1,m_2)\cdot \frac{\partial ^2V}{\partial X \partial Y} + P(m_2)\cdot \frac{\partial ^2V}{\partial Y^2} = 0$$

$$\mbox{However, rather than making, } Q(m_1,m_2) = 0 \mbox{ we would make } P(m_1) = 0 \mbox{ and } P(m_2) = 0$$

This would give,

$$\frac{\partial ^2V}{\partial X \partial Y} = 0$$

which is then easily integrated to give,

V = f(X) + g(Y)
============

where you would substitute for,

X = y - m1x
Y = y - m2x

You would get m1 and m2 when setting P(m1) = 0 and P(m2) = 0.

So, you could use any functions f() and g() that you liked, as long as their arguments were y - m1x and y - m2x.

Last edited: Nov 26, 2005
5. Nov 27, 2005

### saltydog

Thanks for point that out Fermat. That's a much better approach considering, after I looked into it a bit, that an arbitrary solution of the wave equation:

$$\frac{\partial^2 u}{\partial t^2}=a^2\frac{\partial^2}{\partial x^2}$$

can be written in the form:

$$u(x,t)=F(x+at)+G(x-at)$$

and thus I would suspect your m's above are related to the final coefficient on the x-partial term above. I'll work with it a bit.