# F(x,y) and change of variables

1. Nov 28, 2007

### gamma-ray-burst

1. The problem statement, all variables and given/known data
f(x,y) a function of two variables.
x = 2u
y = u-3v

Using a change of variables, transform the equation (d²f/dx²) + (df/dy) = 0
into the coordinates system {u,v}.

2. Relevant equations
We have kind of a replacement teacher for the session and it is his first time giving the course. He almost doesn't give examples and well he's pretty bad :( The problem is that the exams and homeworks are the same for every group, so whenever we get an homework much of what is in it is mostly unknown to us (questions made by the other teachers of the other groups).
So anyways, I'll stop moaning. I quite don't understand what I have to do. If someone could just point me in some direction, I'd be glad.

3. The attempt at a solution
Should I do the partial derivatives? Where do I use the change of variables?

Oyoyoy.

Last edited: Nov 28, 2007
2. Nov 28, 2007

### Dick

You need to use the chain rule for partial derivatives. E.g. df/du=(df/dx)*(dx/du)+(df/dy)*(dy/du) (all partial derivatives). Try looking that up.

3. Nov 28, 2007

### gamma-ray-burst

Oooh yes, that definitely rings a bell.
I'll try that tonight, thanks :D

4. Nov 28, 2007

### gamma-ray-burst

Hey hmm, the partial derivative in x of "e^(xy)", is it zero?

5. Nov 28, 2007

### Dick

No, why would you think that? Is this a different problem?

6. Nov 29, 2007

### gamma-ray-burst

Ok now, sorry I was fooling around with the other numbers of the homework.
So should I do something like, assuming f = f(x,y), :

df/du = (df/dx)(dx/du) + (df/dy)(dy/du)
and
df/dv = (df/dx)(dx/dv) + (df/dy)(dy/dv)

But then what does it means to transform the equation into the coordinates system {u,v}?

7. Nov 29, 2007

### HallsofIvy

Staff Emeritus
It means to get a corresponding equation in the variable u and v rather than x and y!
$$\frac{\partial f}{\partial y}= \frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+ \frac{\partial f}{\partial v}\frac{\partial v}{\partial y}$$
In particular, with your x= 2u, y= u- 3v, you will need u and v in terms of x and y so solve the 2equation: u= x/2, y= (x/2)- 3v: 3v= (x/2)- y: v= x/6- y/3.
From that
$$\frac{\partial u}{\partial y}= 0$$
and
$$\frac{\partial v}{\partial y}= -1/3$$
so
$$\frac{\partial f}{\partial y}= -\frac{1}{3}\frac{\partial f}{\partial v}$$

8. Nov 29, 2007

### gamma-ray-burst

And there was light.
Wow thanks now that's clear :D
So now I just have to find d²f/dx² = d/dx [(df/du)(du/dx) + (df/dv)(dv/dx)] (or i'm not sure, gotta check my notes about composed functions) and rewrite the equation with the new values found? Like "what I'm about to find" + -1/3 (df/dv) = 0 ?
That is wonderful.

9. Nov 29, 2007

### HallsofIvy

Staff Emeritus
Yes. Be careful about the second derivative. Write out the first derivative and then apply the same chain rule when differentiating that.

10. Nov 29, 2007

### gamma-ray-burst

Ok so now I have :

d²f/dx² = d/dx [ 1/2 (df/du) + 1/6 (df/dv) ]
Is this right?
Damn I feel so stupid because I feel that it's not that hard :(
I'm not even sure about the next steps.
I think I need a drink.

11. Nov 29, 2007

### gamma-ray-burst

And wouldn't dy/du = 1 because they are independant or something?
f(2u, u-3v)
Or I'm talking crap.
oyoyoy