When can grad(u)=(b-u)N be solved, and how?

[syra]

b(x) is a scalar-valued function, and N(x) a vector valued-function, taking on arguments in R^n. Not actually for schoolwork, but academic nonetheless.

 

[fzero]

If we can ignore singular behavior (\vec{N}\neq 0, b(x)-u(x) \neq 0, etc), one way to try to solve

\nabla u = (b-u)\vec{N}

is the following. Define a new orthonormal basis

\vec{v} = \frac{1}{\sum_i N_i^2} (N_1, \ldots, N_n),

\vec{v}^{(1)}_\perp = \frac{1}{N_1^2+N_2^2} (N_2, -N_1,0,\ldots,0).

The other \vec{v}^{(i)}_\perp can be constructed by Gram-Schmidt. So these are all an orthonormal set.

We then note that

\vec{v} \cdot \nabla u = (b-u)

\vec{v}^{(i)}_\perp \cdot \nabla u = 0.

If we can define new coordinates (X,Y_i) by inverting

\frac{\partial}{\partial X} = \vec{v} \cdot \nabla_x,~~\frac{\partial}{\partial Y_i} = \vec{v}^{(i)}_\perp \cdot \nabla_x,

then we obtain a first-order ordinary differential equation for u(X):

\frac{\partial u}{\partial X} = b- u, ~~ \frac{\partial u}{\partial Y^i} =0.

I suspect that one of the requirements for these new coordinates to exist is that N = \nabla\phi for some scalar \phi so that 2nd derivatives match.

 

[syra]

Excuse my laziness for not using latex.

The coordinate transformation doesn't require an orthonormal set of v's, it just requires that ||v||=1/||N|| and orthogonality. The new coordinates require grad(X) = v = N/||N||^2, but we can lax this condition up by not requiring ||v||=1, in which case the final equations will have du/dX = (b-u) ||v||/||N|| instead, and there need only exist some scalar function phi such that grad(X) = phi N for some X.

In any case, the resulting differential equation is not in fact ordinary, and I still don't know how to solve it...

 

[fzero]

If \partial u/\partial Y^i =0, then u = u(X), so the remaining equation is an ODE. The equations that determine the coordinate X are still PDEs, but are likely to be simpler than the original equation for u.

 

[syra]

Oh woops, I was being silly.