- #1
Spinny
- 20
- 0
This problem may be very easy or very difficult (probably the first), but I can't seem to make sense of it, and that annoys me. It's not all that important (at least not yet), but I just can't seem to let it go. Anyways, here it is.
Consider the following PDE:
[tex]\frac{\partial^2 f}{\partial r^2}+\frac{\partial^2 f}{\partial z^2}-\frac{1}{r}\frac{\partial f}{\partial r} = 0[/tex]
What I would like to do is to be able to show explicitly how we can go from the above PDE to this one:
[tex]\frac{\partial^2 g}{\partial r^2}+\frac{\partial^2 g}{\partial z^2}+\frac{1}{r}\frac{\partial g}{\partial r} = 0[/tex]
by using the transformation
[tex]g = \int \frac{f}{r}dr \Rightarrow f = r\frac{\partial g}{\partial r}[/tex]
The functions are [tex]f = f(r,z)[/tex] and [tex]g = g(r,z)[/tex]. And yes, the second PDE is Laplace's equation in cylindrical coordinates, [tex]\nabla^2 g(r,z) = 0[/tex].
Any help would be greatly appreciated!
Consider the following PDE:
[tex]\frac{\partial^2 f}{\partial r^2}+\frac{\partial^2 f}{\partial z^2}-\frac{1}{r}\frac{\partial f}{\partial r} = 0[/tex]
What I would like to do is to be able to show explicitly how we can go from the above PDE to this one:
[tex]\frac{\partial^2 g}{\partial r^2}+\frac{\partial^2 g}{\partial z^2}+\frac{1}{r}\frac{\partial g}{\partial r} = 0[/tex]
by using the transformation
[tex]g = \int \frac{f}{r}dr \Rightarrow f = r\frac{\partial g}{\partial r}[/tex]
The functions are [tex]f = f(r,z)[/tex] and [tex]g = g(r,z)[/tex]. And yes, the second PDE is Laplace's equation in cylindrical coordinates, [tex]\nabla^2 g(r,z) = 0[/tex].
Any help would be greatly appreciated!