Integrating derivatives in matrix elements

  • Thread starter jostpuur
  • Start date
2,100
16
First a little warm up problem. Suppose [itex]g:\mathbb{R}^N\to\mathbb{C}[/itex] is some fixed function, and we want to find [itex]f:\mathbb{R}^N\to\mathbb{C}[/itex] such that

[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.
 
2,100
16
One possible way to start is to first consider the special case [itex]g=0[/itex]. The equation [itex]0=u\cdot\nabla_x f(x)[/itex] with my [itex]u[/itex] is the same as

[tex]
\left(\begin{array}{c}
\partial_3 f_1 \\ \partial_3 f_2 \\
\end{array}\right)
= -\frac{i}{\sqrt{2}}
\left(\begin{array}{cc}
\partial_1 - \partial_2 & -\partial_1 - \partial_2 \\
-\partial_1 - \partial_2 & -\partial_1 + \partial_2 \\
\end{array}\right)
\left(\begin{array}{c}
f_1 \\ f_2 \\
\end{array}\right)
[/tex]

Unfortunately, I was unable to learn anything about this equation either. A formal solution can be written as

[tex]
\left(\begin{array}{c}
f_1(x_1,x_2,x_3) \\ f_2(x_1,x_2,x_3) \\
\end{array}\right)
= \exp\Big(-\frac{ix_3}{\sqrt{2}}
\left(\begin{array}{cc}
\partial_1 - \partial_2 & -\partial_1 - \partial_2 \\
-\partial_1 - \partial_2 & -\partial_1 + \partial_2 \\
\end{array}\right)\Big)
\left(\begin{array}{c}
f_1(x_1,x_2,0) \\ f_2(x_1,x_2,0) \\
\end{array}\right)
[/tex]

Unfortunately, the matrices

[tex]
\left(\begin{array}{cc}
1 & -1 \\ -1 & -1 \\
\end{array}\right),\quad\quad
\left(\begin{array}{cc}
-1 & -1 \\ -1 & 1 \\
\end{array}\right)
[/tex]

cannot be diagonalized simultaneously, and therefore the exponential series cannot be written in terms of Taylor series after a linear transform. I cannot see what kind of transformation the series represents.
 
2,100
16
If it is assumed that [itex]f,g[/itex] are differentiable, the formula can be checked to work.
I think we'll have to assume that the partial derivatives of [itex]g[/itex] are bounded too. This is probably not relevant for the discussion, since I'm interested in finding formulas mostly, but anyway, I wanted to fix the original claim...
 

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