hypersurface evolving with tangential velocity

Question

If a hypersurface $S_t$ evolves with velocity only in the tangential direction, is $S_t \equiv S_0$ for all $S$? This is what I have read is true (or something very similar). Can someone give me an example so I can see this? I have no idea what tangential velocity means in 1D for example.

Answer 1

Before I say anything else I've got to hand it to Will Jagy for presenting such an elegant and concise answer to this question. I'm afraid mine is anything but concise, and if my readers can find a little elegance in what I write, I will be most deeply gratified. Lauds, Will.

I think that, to get a meaningful answer to this question, you need to come up with some reasonably precise and rigorous definition of what it means for a hypersurface $S$ to evolve in the first place. So I'll try to provide one, and then use it to discuss, and hopefully answer, the OP's question.

I'm going to assume all the action takes place in some differentiable manifold $M$, and that $S \subset M$ is the given hypersurface. I'm not going to go into the details of what constitutes a differentiable manifold here; there are plenty of references for that, both on- and off-line. Now let $\dim M = N$; I shall then assume the term hypersurface refers to an $N - 1$ dimensional submanifold of $M$. As I recall, this is the standard terminology in the event that $M = R^N$, and if I am not mistaken this language is also used for other $M$ as well. Again, suitable rigorous definitions may be had in a multitude of places, both on- and off-line.

Actually, I think most of what I say here applies to general submanifolds of $M$, not only hypersurfaces; nevertheless, I'll stick to hypersurfaces for the bulk of this answer.

Having (hopefully) settled these terminological questions, we next must ask just what it means for a hypersurface to evolve. Though there are many ways of looking at this, some of which are quite deep, complex and sophisticated (I am thinking here of things like mean curvature evolution in a normal direction to $S_t$), in this answer we will stick with a more basic concept; that is: evolution under the action of a one-parameter family of diffeomorphisms $\phi_t$ of the manifold $M$. In this well-known view of things, for each $t \in R$, $\phi_t$ is a diffeomorphism from $M$ to itself: $\phi_t:M \to M$; thus each $\phi_t$, if expressed in local coordinates $x_1, x_2, . . . , x_N$ about a point $x \in M$, would consist of a family of $N$ differentiable functions $\phi_{tj}(x_1, x_2, . . . , x_N)$, $1 \le j \le N$ such that, if $y = \phi_t(x) \in M$, and $y_1, y_2, . . . , y_N$ are coordinates in the vicinity of $y$, we have $y_j(t) = \phi_{tj}(x_1, x_2, . . . , x_N)$ for $1 \le j \le N$; of course, different choices of coordinates would give rise to different functions $\phi_{tj}$, but they are all related by differentiable transformations in a manner which is again well-known. Furthermore, everything is differentiable with respect to $t$ as well, so that for any point $x \in M$, the $y_j(t) = \phi_{tj}(x_1, x_2, . . . , x_N)$ give the coordinates of a differentiable curve in the $y-$coordinate system. Lastly, but certainly not leastly, we assume/arrange things so that $\phi_0$ is the identity map on $M$: $\phi_0(x) = x$ for $x \in M$.

Given such a one-parameter family $\phi_t:M \to M$ of diffeomorphisms of $M$ to itself, it makes sense talk about the evolution of a hypersurface $S \subset M$; we simply define $S_t = \phi_t(S_0)$; since $\phi_t$ is a diffeomorphism for each $t$, it preserves certain differentiable relationships such as tangency and transverality, so $S_t = \phi_t(S_0)$ will be a hyper-surface in $M$ for each $t \in R$.

As to the velocity of the evolving hyper-surface $S_t = \phi_t(S_0)$, this concept is given rigorous mathematical meaning through the notion of vector field, which is of course deeply tied in with one-parameter families of diffeomorphisms such as $\phi_t$. Indeed, for any point $x \in M$ there is a curve of the form $\gamma(t) = \phi_t(x)$ passing through $x$, since $\phi_0(x) = x$. If we take $\dot{\gamma}(0) = \frac{d\phi_t(x)}{dt}|_{t = 0}$ we obtain the vector $\dot{\gamma}(0)$ at the point $x$; that is, $\dot{\gamma}(0) \in T_xM$, where $T_xM$ is the tangent space to $M$ at $x$. Since this can be done at any point of $M$, we see that each one-parameter family of diffeomorphisms $\phi_t$ of the manifold $M$ gives rise to a unique vector field $X_\phi$ on $M$. And of course it is well known this works the other way around: given a sufficiently smooth vector field $X$ on $M$, we obtain, by integrating the differential equation $\frac{d \gamma(t)}{dt} = X(\gamma(t))$ with initial condition $\gamma(0) = x$ for each $x \in M$, a one-parameter family of maps $\phi_t:M \to M$ given by the solution curves to $\frac{d \gamma(t)}{dt} = X(\gamma(t))$: $\phi_t(x) = \gamma(t)$ where $\gamma(0) = x$. We can thus interpret the vector field $X_\phi$ as the velocity field associated with $\phi_t$; that is, for each $x \in M$, $X_\phi(x)$, and we can think of the action of $\phi_t(x)$ as moving the point $x$ along the curve $\gamma(t) = \phi_t(x)$ in such a manner that the rate of change of the position of the point $x$ is given by $X_\phi(x)$. If we imagine the points as particles of dust and $X_\phi(x)$ as the wind velocity at the point $x$, then a dust particle initially situated at $x$ will move take the path $\gamma(t) = \phi_t(x)$ under the influence of this breeze, carried along by a wind $X_\phi(\gamma(t))$ at each point $\gamma(t)$ on its path. Of course, one must be careful not to push these physical analogies too far or hard, but in the present case I think it is a useful aid to the intuition.

I have belabored these two notions, the evolution of a hyper-surface and the velocity of a point under such evolution--because these are the key concepts I feel need to be given mathematical teeth, as it were, to carry this discussion forward, in a mathematical sense. Perhaps I am overdoing it, preaching to the choir, so to speak, but when I read a phrase such as "a hyper-surface $S_t$ evolves" I feel a need for more precision, since, as I mentioned above, there are many ways for a hyper-surface to evolve. In any event, I settled upon what I believe to be the most widely-used interpretation of this phrase, and that is that $S_t = \phi_t(S_0)$ for some one-parameter family of diffeomorphisms $\phi_t$. And under the action of $\phi_t$, each point $x \in S_t$ has velocity $X_\phi(x)$. So, within this view of things, I take the meaning of, "$S_t$ evolves with velocity only in the tangential direction" to be that $X_\phi(x)$ is in fact tangent to $S_t$ for all $x \in S_t$.

Well, I've probably written enough, if not too much, about what these terms mean; now let me try to put forth some answers:

If, as the OP says, "$S_t$ evolves with velocity only in the tangential direction", then we must have $X_\phi$ tangent to $S_t$ for all $t$; in particular, $X_\phi(x)$ must be tangent to $S_0$ for any $x \in S_0$. But this means the entire integral curve $\phi_t(x)$ lies in $S_0$ for any $x \in S_0$. There are a few ways to see this, I think, one of which is: use the fact that, since $X_\phi$ is tangent to $S_0$, the entire solution process, including the machinery of existence and uniqueness etc., can be deployed in terms of $S_0$ itself to obtain solutions curves $\gamma(t)$ satisfying $\dot{\gamma}(t) = X_\phi(\gamma(t))$ which lie entirely within $S_0$; but these $\gamma(t) \subset S_0$ are in fact the same curves obtained (for initial $\gamma(0) \in S_0$) when integrating the entire system $\dot{\gamma}(t) = X_\phi(\gamma(t))$ in the ambient space $M$; this follows from uniqueness of solutions. Since the curves $\gamma(t)$ initialized in $S_0$ stay in $S_0$, we have $S_t = \phi_t(S_0) \subset S_0$, so in fact $S_t = S_0$ since for $y \in S_0$, $\phi_{-t}(y) \in S_0$ (because the integral curves may be extended for all time in either direction), hence $y = \phi_t(\phi_{-t}(y)) \in S_t$. Here I have used the group property of the $\phi_t$, viz., $\phi_{s + t}(x) = \phi_s(\phi_t(x))$ which implies that $x = \phi_0(x) = \phi_t(\phi_{-t}(x))$ so that $\phi_t^{-1} = \phi_{-t}$ (This and other properties of $\phi_t$ are well-known and may be found in any reasonably complete discussion of vector fields and the associated diffeomorphisms.) Another way to see that $\phi_t(x)$ lies in $S_0$ for any $x \in S_0$ is to employ Will Jagy's wonderfully succinct technique, given in his answer to this question. It should be noted though, that if we are dealing with a submanifold codimension greater than one, more than one function such as Will's $f(x, y, z)$ will be required to specify $S_0$. But I will leave to the reader to further investigate this possibility on his/her own. Finally, as noted in Will's answer, completeness is required for any of these arguments to fly. Under such circumstances, we can answer the OP's lead question in the affirmative: $X_\phi$ tangential to $S_0$ implies $S_t = S_0$ for all $t$.

What does tangential velocity mean in one dimension? I take it here the OP michael_faber means $\dim M = 1$, since the case $\dim M \ge 2$, $\dim S = 1$ is basically covered by what has been written above. If $\dim M = 1$, a "hypersurface", having codimension one, if connected must be a single point $p$; all tangent vectors to such a point must vanish, since any curve $\gamma: I \to {p}$, $0 \in I \subset R$ an open interval,$\gamma(0) = p$, satisfies $\dot{\gamma}(t) = 0$ for $t \in I$; it doesn't go anywhere under the action of a "tangential" vector field, so it appears that $p_t = p_0$ for all $t \in I$, and since for such a "constant curve" we might as well take $I = R$, we have $p_t = p_0$ for all $t \in R$ as well.

SOME EXAMPLES:

1.) Consider, on the punctured plane $R^2 - \{0\}$, the vector field $X(r, \theta)$ the components of which, in polar coordinates, are given by

$X_r = r(1 - r)$,

$X_\theta = \omega_0$,

where $\omega_0$ is a constant. Then the flow $\phi_X$ of $X$ is given by the system of ordinary differential equations

$\dot{r} = r(1 - r)$,

$\dot{\theta} = \omega_0$.

In Cartesian coordinates, with $x = r\cos \theta$ and $y = r\sin \theta$, these become

$\dot{x} = \dot{r}\cos \theta - r\sin \theta \dot{\theta} = (1 - r)x - \omega_0y$,

$\dot{y} = \dot{r}\sin \theta + r\sin \theta \dot{\theta} = (1 - r)y + \omega_0x$,

where of course $r = \sqrt{x^2 + y^2}$. I have restricted attention to $R^2 - \{0\}$ in order to avoid altogether the question of whether or not $X$ is singular (i.e., nondifferentiable) at the origin. In any event it is clear, from the polar form of these equations, that for $0 < r < 1$ we have $\dot{r} > 0$, for $r > 1$ we have $\dot{r} < 0$, and for $r = 1$ we have $\dot{r} = 0$. Meanwhile, the variable $\theta$ changes at the constant angular rate $\omega_0$, completing on full revolution in time $\tau = \frac{2\pi}{\vert \omega_0 \vert}$. Thus the flow $\phi_X$ of $X$ spirals outward or inward toward the circle $r = 1$, depending on whether the initial value $r_0$ of $r$ satisfies $r_0 > 1$ or $r_0 < 1$. The circle $r = 1$ is carried into itself by $\phi_X$, so if we take $S_0 = \{(r, \theta) \in R^2 - \{0\} \vert r = 1\}$, then $S_t = S_0$ for all $t$. Similar remarks apply to any periodic orbit of any vector field; indeed, for any initial $x_0$ the curve $\phi_{Xt}(x_0)$ is carried into itself by the action of $\phi_{Xt}$, since $X_\phi$, satisfying as it does $X_\phi(\phi_{Xt}(x_0)) = \dot{\phi}_{Xt}(x_0)$, is everywhere tangent $\phi_{Xt}(x_0)$.

hypersurface evolving with tangential velocity

Answer 2

Completeness of the surface is required for this. And it is just the implicit function theorem. In a small neigborhood, a smooth surface in $\mathbb R^3$ is given by $f(x,y,z) = 0$ for some smooth function $f.$ Tangential means that the direction of travel of any specific point is perpendicular to the gradient of $f.$ So, a point travels along $\gamma(t)$ with $\gamma'(t) \perp \nabla f (\gamma(t)).$ However, the chain rule says that $$ \frac{d}{dt} f(\gamma(t)) = \nabla f (\gamma(t)) \cdot \gamma'(t),$$ and the tangential condition says that this dot product remains zero. That is, along the path of travel, $f$ remains constant, and the point remains on the original level set of $f.$

It's a Good Thing.