Solving a P.D.E in Normal Form: Tips and Tricks

[asdf1]

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

 

[HallsofIvy]

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?

 

[asdf1]

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

 

[HallsofIvy]

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?"

 

[asdf1]

shift the x and y values?

 

[HallsofIvy]

I have no idea what you mean by "shift the x and y values". My point was that eliminating the u_xy 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.

 

[asdf1]

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