# Mixed derivative in DE

1. Apr 11, 2005

### Gonzolo

Hi, has anyone here ever seen anything like this?

f = f(x,y,t)
g = g(x,y,t)

$$\frac{ \partial^{2}{f} }{ \partial {y}\partial {t} } } + \frac{ \partial^{2}{f} }{ \partial {x}\partial {t} } }+ \frac{\partial{f}}{\partial{t}}+ \frac{\partial{f}}{\partial{y}}+ \frac{\partial{f}}{\partial{x}}+f = g$$

Personnally, my blood pressure has begun to drop dangerously. You can drop in a real constant by each term. Any info or sites on DE's with mixed derivatives would be helpful. Can any DE with mixed derivatives at all be solved analytically? A good direction for numerical solving would be helpful too.

If it's as satanic as it first seems, I'll probably do approximations to simplify my model.

Thanks.

2. Apr 12, 2005

### Gonzolo

Actually, the units don't work without the constants, so it should be :

$$C_1\frac{ \partial^{2}{f} }{ \partial {y}\partial {t} } } + C_2\frac{ \partial^{2}{f} }{ \partial {x}\partial {t} } }+C_3\frac{\partial{f}}{\partial{t}}+C_4\frac{\partial{f}}{\partial{y}}+C_5\frac{\partial{f}}{\partial{x}}+C_6f = g$$

g(x,y,t) is known
Finding f(x,y,t) is the problem.

3. Apr 12, 2005

### HallsofIvy

Staff Emeritus
It is always possible, by rotating the coordinate system, to convert mixed derivatives to 'non-mixed' derivatives. One way is to set up the second derivative coefficients as a matrix and find the eigen-vectors. Those should be the new axes.

4. Apr 13, 2005

### saltydog

Yea, Gonzolo, I'd like to see that one solved too. However, I think a good approach is to first analyze it in just 2-D. You know, look first at f(x,y) and g(x,y):

$$\frac{\partial^2f}{\partial x\partial y}+\frac{\partial f}{\partial x}+f=g$$

I mean, just any solution, any initial condition, any boundary conditions. Can anyone here propose a method for solving this one?