Integrating derivatives in matrix elements

[jostpuur]

First a little warm up problem. Suppose g:\mathbb{R}^N\to\mathbb{C} is some fixed function, and we want to find f:\mathbb{R}^N\to\mathbb{C} such that

<br /> g(x) = u\cdot\nabla_x f(x)<br />

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

<br /> 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<br />

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

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

<br /> g(x) = u\cdot\nabla_x f(x)<br />

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 N=3,M=2 and

<br /> u=\Big(<br /> \left(\begin{array}{cc}<br /> \frac{1}{\sqrt{2}} &amp; \frac{1}{\sqrt{2}} \\<br /> \frac{1}{\sqrt{2}} &amp; -\frac{1}{\sqrt{2}} \\<br /> \end{array}\right),\;<br /> \left(\begin{array}{cc}<br /> \frac{1}{\sqrt{2}} &amp; -\frac{1}{\sqrt{2}} \\<br /> -\frac{1}{\sqrt{2}} &amp; -\frac{1}{\sqrt{2}} \\<br /> \end{array}\right),\;<br /> \left(\begin{array}{cc}<br /> 0 &amp; i \\<br /> -i &amp; 0 \\<br /> \end{array}\right)\Big)<br />

So this case is the same as

<br /> 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) <br />
<br /> 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)<br />

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

 

[jostpuur]

One possible way to start is to first consider the special case g=0. The equation 0=u\cdot\nabla_x f(x) with my u is the same as

<br /> \left(\begin{array}{c}<br /> \partial_3 f_1 \\ \partial_3 f_2 \\<br /> \end{array}\right)<br /> = -\frac{i}{\sqrt{2}}<br /> \left(\begin{array}{cc}<br /> \partial_1 - \partial_2 &amp; -\partial_1 - \partial_2 \\<br /> -\partial_1 - \partial_2 &amp; -\partial_1 + \partial_2 \\<br /> \end{array}\right)<br /> \left(\begin{array}{c}<br /> f_1 \\ f_2 \\<br /> \end{array}\right)<br />

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

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

Unfortunately, the matrices

<br /> \left(\begin{array}{cc}<br /> 1 &amp; -1 \\ -1 &amp; -1 \\<br /> \end{array}\right),\quad\quad<br /> \left(\begin{array}{cc}<br /> -1 &amp; -1 \\ -1 &amp; 1 \\<br /> \end{array}\right)<br />

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.

 

[jostpuur]

If it is assumed that f,g are differentiable, the formula can be checked to work.

I think we'll have to assume that the partial derivatives of g 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...