Why do all solutions to this equation have the same form?

Question

In this paper on page 45, the authors state that

Let's assume we know that $$ w \times dw = d\varphi + \sum_{j=1}^3\alpha_jdx_j, \tag{1}$$ where $\varphi \in H^1(\mathbb{R}^3, \mathbb{R})$, $\alpha_j$ are real numbers and $\vert w \vert=1$. One then checks that $$ w(x)= \exp i \left(\varphi(x)+\sum_{j=1}^3 \alpha_j x_j + \theta\right). \tag{2}$$ for some $\theta \in \mathbb{R}$.

Here $$w \times dw= \sum_{i=1}^3 (w_1 \partial_i w_2 - w_2 \partial_i w_1) dx_i$$ for $$w=w_1 +i w_2 : \mathbb{R}^3 \to \mathbb{C}.$$

I do not understand this statement. In fact, one can easily check that $w$ from (2) satisfies equation (1). But why does every $w$ that satisfies equation (1) have the form (2)? Any hint would be much appreciated!

Answer 1

I too, like Anthony Carapetis, find the notation $w \times dw$ somewhat obscure and confusing; I'm used to using the wedge symbol, "$\wedge$", in the context of differential forms, and I can't yet see how the "$\times$" symbol is used an any manner with which I am already familiar, e.g. as it occurs in ordinary vector calculus on $\Bbb R^3$.

I checked over (very briefly) the paper cited by mjb in his post, and I couldn't find, in the short amount of time I spent, any definition or justification for the $w \times dw$ notation. I suspect it's probably in there, and perhaps a careful reading of the paper will allow me to find it.

Nevertheless, we can simply take the notation $w \times dw$ at face value, based upon the OP's inclusion of the definition

$w \times dw = \sum_{i = 1}^3 (w_1 \partial_i w_2 - w_2 \partial_i w_2)dx_i$,

and work with this seemingly somewhat ad hoc notation as it is given to us; the above equation defines the operator $w \times dw$, for the present purposes, regardless of the niceties of traditional mathematical symbolism. Note that I have inserted parentheses to group the expression $w_1 \partial_i w_2 - w_2 \partial_i w_2$, so the whole thing multiplies $dx_i$; this is the easiest way I can make consistent sense of this equation. Accepting this (hopefully) minor modification, we observe that

$w \times dw = \sum_{i = 1}^3 (w_1 \partial_i w_2 - w_2 \partial_i w_2)dx_i = w_1dw_2 - w_2dw_1$,

the $dw_i$ being ordinary differential forms; and I guess the $\times$-product notation makes a little more sense from this point of view: taking $w = (w_1, w_2, 0)^T$ and $dw = (dw_1, dw_2, 0)^T$, the ordinary $\times$ operation may be morphed into a sensible expression similar to the usual determinant-based definition:

$w \times dw = \det \begin {bmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\w_1 & w_2 & 0 \\ dw_1 & dw_2 & 0 \end{bmatrix} = (w_1dw_2 - w_2dw_1) \mathbf{k}$;

that's as far as I wish to take this discussion of notation, simply accepting

$w \times dw = w_1dw_2 - w_2dw_1$

for the present purposes.

These things being said, it is certainly easy to check that (1) follows from (2) based on this ad hoc formula for $w \times dw$; indeed we have,

$w = w_1 + i w_2 = \exp i(\varphi(x) + \sum_{j = 1}^3 \alpha_j x_j + \theta)$

$= \cos(\varphi(x) + \sum_{j = 1}^3 \alpha_j x_j + \theta) + i \sin(\varphi(x) + \sum_{j = 1}^3 \alpha_j x_j + \theta)$,

whence

$dw_1 = -\sin(\varphi(x) + \sum_{j = 1}^3 \alpha_j x_j + \theta)(d\varphi + \sum_{j = 1}^3 \alpha_j dx_j)$,

and

$dw_2 = \cos(\varphi(x) + \sum_{j = 1}^3 \alpha_j x_j + \theta)(d\varphi + \sum_{j = 1}^3 \alpha_j dx_j)$,

so that

$w_1 dw_2 = \cos^2(\varphi(x) + \sum_{j = 1}^3 \alpha_j x_j + \theta)(d\varphi + \sum_{j = 1}^3 \alpha_j dx_j)$,

and

$w_2 dw_1 = -\sin^2(\varphi(x) + \sum_{j = 1}^3 \alpha_j x_j + \theta)(d\varphi + \sum_{j = 1}^3 \alpha_j dx_j)$,

whence

$w \times dw = w_1dw_2 - w_2dw_1$

$= (\cos^2(\varphi(x) + \sum_{j = 1}^3 \alpha_j x_j + \theta) + \sin^2(\varphi(x) + \sum_{j = 1}^3 \alpha_j x_j + \theta))(d\varphi + \sum_{j = 1}^3 \alpha_j dx_j)$

$= d\varphi + \sum_{j = 1}^3 \alpha_j dx_j$,

using the standard, elementary trigonometric identity $\cos^2 \beta + \sin^2 \beta = 1$.

The key to making it go the other way is, I think, to fully exploit the hypotheses given. That is, in addition to assuming

$w \times dw = d\varphi + \sum_{j = 1}^3 \alpha_j dx_j$,

one needs to realize "we know that . . . $\vert w \vert = 1$". For then we may write

$w = \cos \psi + i \sin \psi = \exp(i\psi)$,

whence

$w_1 = \cos \psi$

and

$w_2 = \sin \psi$,

and voila! the key turns in the lock, since now

$dw_1 = -\sin \psi d\psi$,

$dw_2 = \cos \psi d\psi$,

yielding

$w \times dw = w_1dw_2 - w_2dw_1 = (\cos^2 \psi + \sin^2 \psi) d\psi = d\psi$,

so that setting

$d\psi = w \times dw = d\varphi + \sum_{j = 1}^3 \alpha_j dx_j$,

we have

$\psi = \varphi + \sum_{j = 1}^3 \alpha_j x_j + \theta$

for constant $\theta$; this last feasible, locally at least, since

$d\psi - (d\varphi + \sum_{j = 1}^3 \alpha_j dx_j) = d(\psi - (\varphi + \sum_{j = 1}^3 \alpha_j x_j)) = 0$.

Setting

$\psi = \varphi + \sum_{j = 1}^3 \alpha_j x_j + \theta$

in the formula $w = \exp (i\psi)$ yields the desired result.

It should be observed that the assumption $\vert w \vert = 1$ appears to be essential here; indeed, if we allow $\vert w \vert \ne 1$, we may have $w = r \exp(i\psi)$, $r$ a non-constant function, and

$dw_1 = \cos \psi dr - r \sin \psi d \psi$,

$dw_2 = \sin \psi dr + r \cos \psi d \psi$,

and an easy calculation reveals that

$w \times dw$ = $w_1 dw_2 - w_2 dw_1 = r^2 d \psi$,

whence

$d(w \times dw) = 2 r dr \wedge d \psi \ne 0$

in general, so that $w \times dw$ won't be integrable, i.e. an exact differential form. If $r \ne 1$ is constant then we apparently have a solution with

$\psi = r^{-2}(\varphi + \sum_{j = 1}^3 \alpha_j x_j) + \theta$,

viz.

$w = r \exp (i( r^{-2}(\varphi + \sum_{j = 1}^3 \alpha_j x_j) + \theta))$,

but here $\vert w \vert = r \ne 1$.

Finally, I should add that in carrying out the above discussion I have passed over certain analytic niceties, such as the exact interpretation of the meaning of $d \varphi$ for $\varphi \in H^1(\Bbb{R}^3, \Bbb{R})$, whether the derivatives should be taken as weak or strong, and how these concerns might affect the result; nor have I addressed questions which might arise regarding the global existence and/or smoothness of $\psi$. But I think, based upon what I have been able to read, that the arguments will fly. If someone knows or believes otherwise, I'd appreciate hearing about it. Anyway, I hope I have covered the essential points.

Cheers!

Answer 2

mjb! and any other readers as well!

Since this post is too long to be a comment, and since long posts with lots of Latex take an excruciatingly long time to render on my computer, I have chosen to add a separate answer to the question; it's really a continuation of the first one.

Hope I'm not spoiling the fun, but here's how I show existence and smoothness of $\psi$: since $\vert w \vert = 1$, we have $w_1^2 + w_2^2 = 1$, which means there always exists a $\psi$ with $w_1 = \cos \psi$ and $w_2 = \sin \psi$; just think of the unit circle in the $w_1$-$w_2$ plane, and take $\psi$ to be, say, the angle that the segment joining the point $(0, 0)$ with $(w_1, w_2)$ makes with the $w_1$-axis. What I have just typed is of course a verbal description of the usual picture from elementary geometry which relates $\sin$ and $\cos$ to the coordinates of points on the unit circle; I'd include a graphic if I could but I don't have the SW tools to so on my 'droid, from which I write this response. But I'm pretty sure you are familiar with the picture I am describing; anyway, it's all over the web in one form or another. So such a $\psi$ exists. It is of course not unique as the transformation $\psi \to \psi + 2\pi$ yields another solution to the equations $\cos \psi = w_1$, $\sin \psi = w_2$. However, $\psi$ is easily seen to be a smooth function of $w_1$, $w_2$: in the vicinity of a point on the unit circle $(w_1, w_2)$ such that $w_1 \ne 0$, observe that we have

$\tan \psi = \frac{\sin \psi}{\cos \psi} = \frac{w_2}{w_1}$,

which in itself is sufficient to establish that $\psi$ is a smooth function of $w_1$ and $w_2$ since, writing

$g(\psi, w_1, w_2) = \tan \psi - w_1^{-1}w_2$

we have, by the implicit function theorem, that the $\psi(w_1, w_2)$ such that

$g(\psi(w_1, w_2), w_1, w_2) = \tan \psi(w_1, w_2) - w_1^{-1}w_2 = 0$

is in fact smooth, since

$\frac{\partial g}{\partial \psi} = \frac{\partial \tan \psi}{\partial \psi} = \sec^2 \psi = \cos^{-2} \psi = w_1^{-2} \ne 0$.

Near points on the circle $w_1^2 + w_2^2 = 1$ where $w_1 = 0$, we have $w_2 \ne 0$, so we can work with $\cot \psi$ instead of $\tan \phi$; one needs simply reverse the roles of $w_1$ and $w_2$ in the preceding argument to show $\psi$ smooth in this case.

That's my take establishing that $\psi(w_1, w_2)$ with $\exp(i \psi(w_1, w_2)) = w_1 + iw_2$ is smooth.

Next, let me refine my remarks concerning

$d\psi = w \times dw = d\varphi + \sum_{j = 1}^3 \alpha_j dx_j$,

implying

$\psi = \varphi + \sum_{j = 1}^3 \alpha_j x_j + \theta$

for constant $\theta$; when I said that this last step is feasible, locally at least, since

$d\psi - (d\varphi + \sum_{j = 1}^3 \alpha_j dx_j) = d(\psi - (\varphi + \sum_{j = 1}^3 \alpha_j x_j)) = 0$,

I was referring to the fact that, for a function $f$, $df = 0$ is only possible if, and hence implies, $f$ is a constant. I introduced the notion of locality to cover the possibility that the domain of the functions might not be connected, in which case we would have something like $\theta$ is constant on each (topological) component of their domain. This of course is a case of verbal and mathematical overkill, since we know from the problem statement that the domain of everything is $\Bbb{R}^3$. When I wrote those words, I was in fact moving very fast (I had to leave for work momentarily); different differential forms were flying through my mind, and, truth to tell, I kept thinking of the related fact that a form $\rho$ such that $d \rho = 0$ satisfies $\rho = d \sigma$ for some form $\sigma$, which of course is true locally. That idea inadvertantly crept into, and influenced, my write-up. In summary, I think what I said is essentially correct, certainly we can take $\theta$ to be contstant "locally at least", since in fact the conclusion holds in a global sense on $\Bbb{R}^3$. I hope these remarks clarify my previous response.

Finally, in thinking about this problem, I realized something interesting about our friend $w \times dw$ which I think is worth passing on. We have observed that

$w \times dw = w_1 dw_2 - w_2 dw_1$;

note now that we also have

$w_1 d w_2 - w_2 dw_1 = w_1^2 d(w_1^{-1}w_2) = w_1^2d(\tan \psi)$,

which indicates that the "operator" $w \times dw$ is somehow deeply connected to the phasic or angular representation $w = \exp(i\psi)$. Of course, some of this notion has been broken out, to a certain extent, in our discussion; but I suspect there may be a lot more to be seen, though I will have to leave that topic for a future discussion. Phase relationships for complex-valued functions on manifolds form a fascinating and far-reaching subject in their own right; I'm always interested in hearing more about them.

Hope these remarks clarified my previous post.

Cheers.