What makes classifying PDEs so challenging and rewarding?

pivoxa15
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Is it the case that if you have a t depdence, and one of x or y then replace one of x or y with t and use the master 2 variable 2nd order PDE form. However if you had x, y and t in a PDE than that is a 3 variable PDE and would be different. Why do they only consider PDEs with only 2 variables? Are there classifications for 3 or more variables? Or is it the case that if you want to extend the 2 variables case to more you could use vector calculus and replace the x by (x,y,z) and have the t there so 4 variables.
 
Surely you understand that the classification of P.D.E.s does not depend on what you happen to CALL a variable. If one of the independent variables occurs in a first but not second derivative (and there is at least one other independent variable with a second derivative) that is a parabolic equation.

In particular, the "diffusion" or "heat" equation
\frac{\partial^2 u}{\partial x^2}= \kappa \frac{\partial u}{\partial t} is parabolic.
 
I'm new to PDEs and I think they are exponentially harder than ODEs. There is so much behind an innocent looking PDE like the over you describe above. Is that why there have been fields medals awarded for people who have produced results in PDE theory.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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