On Wed, Feb 15, 2012 at 1:13 AM, Daniel Wheeler <[email protected]>wrote:
> > > On Mon, Feb 13, 2012 at 7:13 PM, Michael Brown < > [email protected]> wrote: > >> >> Thanks for your help. Even if I don't end up using FiPy for my current >> project, it looks like a great package and I'll definitely keep it in mind >> for the future/recommend it. >> > > Cool. Thanks. > No problem. > I don't think you need to split between the x and u parts of f as long as > you can construct a g such that g_x = f. g can be arbitrary. > > Okay, but the problem is f depends on u, so to solve g_x = f(u(x, t), x) I need to know u, the unknown! That makes g a functional of u, not a local function of u(x). That's why I tried to split off the u dependence, but it doesn't look like it's going to help. Either way I get an integro-differential equation to solve. Either what I had before or: u_t - v_x = 0 v_t - u_x = g = \int f(u(x, t), x) dx It's starting to look like a nonlocal problem. >> Also, how would this method extend to two or three dimensions? >> > > You could use > > u_t - v_x - w_y = 0 > > v_t - u_x = g > w_t - u_y = h > > where g and h are arbitrary as long as > > g_x + h_y = f > > Either one could be zero for instance. It might matter in some numerical > way, but not for the maths. > I get it now. That's a neat trick. > > Good luck! > -- > Daniel Wheeler > Thanks!
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