# Normal form of p.d.e?

1. Jun 10, 2006

### asdf1

how do you change the following p.d.e to normal form and solve it?
uxx -4uxy+3uyy=0?

2. Jun 10, 2006

### HallsofIvy

Staff Emeritus
How would you eliminate the "xy" term in x2- 4xy+ 3y2= 0 in order to determine what type of conic it was?

One method would be to write it as a matrix formula:
$$\left( \begin{array} {cc} x & y \end{array} \right) \left( \begin{array} {cc} 1 & -2 \\-2 & 3 \end{array} \right) \left( \begin{array} {c} x \\ y \end{array}\right)= 0$$
and find the eigenvalues and eigenvectors. The eigenvector point in the "characteristic directions" and taking your axes in those directions reduces to "normal form".

With a PDE, the same thing happens: using the characteristic directions as the new variables reduces the equation to normal form.

This is the second problem in a row in which you have essentially said "how do I solve partial differential equations". What have you learned in class so far?

3. Jun 10, 2006

### asdf1

why would you think of writing it as a matrix formula?

4. Jun 10, 2006

### HallsofIvy

Staff Emeritus
I said that was one method.

I'll ask again: "How would you eliminate the "xy" term in $x^2- 4xy+ 3y^2= 0$ in order to determine what type of conic it was?"

5. Jun 11, 2006

### asdf1

shift the x and y values?

6. Jun 11, 2006

### HallsofIvy

Staff Emeritus
I have no idea what you mean by "shift the x and y values". My point was that eliminating the uxy in a pde is essentially the same as eliminating the xy term in a conic (by rotating the coordinate system) and wanted to know if you knew how to do that.

7. Jun 16, 2006

### asdf1

i've looked that part up~ thank you very much!