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uri
10-11-2008, 06:01 PM
small question:

The PDE: A*Uxx+B*Uxy+C*Uyy=0

is parabolic given B^2-4AC=0.

How do we see from this that the Heat equation Ut=Uxx+Uyy is parabolic?

Thanks!

HallsofIvy
10-12-2008, 02:21 PM
You can't- not from that formula: it only applies pde in two independent variables and your equation has 3- x, y, and t. The actual definition of "parabolic" pde is that, in some coordinate system, there is no second derivative with respect to one of the variables. Using that definition, the fact the Ut= Uxx+ Uyy is parabolic is obvious: it's already in a coordinate system in which there is no second derivative with respect to t.

If you really must use that formula, you need to reduce to two variables.

For example, since x and y appear in the same way, if you ignore the y dependence you get
Ut= Uxx which corresponds to A= 1, B= C= 0 since there are no other second derivative:B^2- 4AC= 0^2- 4(1)(0)= 0.

Here we could "get away" with dropping the y dependence because the equation is symmetric in x and y so we didn't lose anything. If we dropped the t dependence we would get Uxx+ Uyy= 0 which is elliptic.

(There are some textbooks that restrict "parabolic" to equations with only 2 variables and would refer to Ut= Uxx+ Uyy as "parabolic-elliptic".)

uri
10-13-2008, 09:17 AM
Thanks! That really helped.

february1771
02-20-2009, 05:32 AM
potiskanje, tako lepo temo dobro delovno mesto