- 2,100

- 16

[tex]

g(x) = u\cdot\nabla_x f(x)

[/tex]

holds, where [itex]u\in\mathbb{R}^N[/itex] is some constant. The problem is not extremely difficult, and after some work the following formula can be found:

[tex]

f(x) = f\Big(x - \frac{u\cdot x}{\|u\|^2}u\Big) + \int\limits_0^{(x\cdot u)/\|u\|^2} g\Big(x - \frac{u\cdot x}{\|u\|^2}u+ \nu u\Big) d\nu

[/tex]

If it is assumed that [itex]f,g[/itex] are differentiable, the formula can be checked to work. The idea is that [itex]f[/itex] can be chosen arbitrarily as initial condition in the orthogonal complement of [itex]\langle u\rangle[/itex].

Now, my real problem. Assume that [itex]u[/itex] is [itex]NM^2[/itex] component object, whose elements have been arranged so that first [itex]u=(u^1,\ldots, u^N)[/itex], and then each [itex]u^n[/itex] is a [itex]M\times M[/itex] matrix. Assume that [itex]g:\mathbb{R}^N\to\mathbb{C}^M[/itex] is some fixed function. Then we want to find [itex]f:\mathbb{R}^N\to\mathbb{C}^M[/itex] such that symbolically the same formula

[tex]

g(x) = u\cdot\nabla_x f(x)

[/tex]

holds. Now the previous solution formula doesn't work anymore, and I don't know how to fix the situation.

I am in particular interested in the case [itex]N=3,M=2[/itex] and

[tex]

u=\Big(

\left(\begin{array}{cc}

\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\

\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\

\end{array}\right),\;

\left(\begin{array}{cc}

\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\

-\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\

\end{array}\right),\;

\left(\begin{array}{cc}

0 & i \\

-i & 0 \\

\end{array}\right)\Big)

[/tex]

So this case is the same as

[tex]

g_1(x) = \frac{1}{\sqrt{2}}\big(\partial_1 f_1(x) + \partial_2 f_1(x) + \partial_1 f_2(x) - \partial_2 f_2(x) \big) + i \partial_3 f_2(x)

[/tex]

[tex]

g_2(x) = \frac{1}{\sqrt{2}}\big(\partial_1 f_1(x) - \partial_2 f_1(x) - \partial_1 f_2(x) - \partial_2 f_2(x)\big) - i \partial_3 f_1(x)

[/tex]

I don't know if this special case is any easier than the general case, though.