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Let $\mathbf c(t)$ be a flow line of a gradient field $\mathbf F = - \nabla V$. Prove that $V(\mathbf c(t))$ is a decreasing function of t.

Not sure where to begin here, although it might have to do with the gradient chain rule?

My attempt:

$$\mathbf c'(t) = \mathbf F(\mathbf c(t)) = -\nabla V(\mathbf c(t))$$

So for $\mathbf c'(t) > 0, \nabla V(\mathbf c(t)) < 0$ indicating that $V$ is decreasing. Is that right?

Sean Haugh
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  • What does this "$\mathbf c'(t) > 0, \nabla V < 0$" mean? – Raskolnikov Mar 07 '13 at 22:09
  • I edited my post, is it clearer now? – Sean Haugh Mar 07 '13 at 22:13
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    No. Let me be more clear: an inequality for vectors does not make any sense. There's no "natural" order on a vector space like there is on the reals. – Raskolnikov Mar 08 '13 at 22:19
  • That equation you wrote isnt any chain rule, its just the definition of a flow line. You have to use the chain rule. So for a function $V$ in 3 variables, each depending on $t$, write down the chain rule. Then use your equation. Then you are done. – wilsonw Mar 29 '20 at 03:01

2 Answers2

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I would start by picking $t_1$ and $t_2$ which define a segment of curve $c_{12}$ which starts at $c(t_1)$ and ends at $c(t_2)$. The integral

$$ \int_{c_{12}} \textbf{F}\cdot dr$$

will tell you the change in the function $-V(c(t)$ from $t_1$ to $t_2$. Prove that this is always positive, and you will prove that $V(c(t))$ is decreasing in t.

Specifically, when you expand the integral, you will be able to find the directional derivative of $-V$ in the direction of $c'(t)$. Argue that this must always be positive, so your integral must be positive.

If this isn't enough help, let me know in comments and I can expand my answer. I don't want to give it away if you are close.

Paul Orland
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  • Why does the integral tell you the change in the function $-V(c(t))$ and not $V(c(t))$? – Joe Z. Feb 26 '14 at 06:27
  • (I'm doing this exact homework question for calculus and while your hint helped quite a lot, I'm stuck on just that last portion.) – Joe Z. Feb 26 '14 at 13:40
  • Assume the coordinates you are working in are Euclidean $(x, y, z) $, and try expanding the dot product of $\mathbf F\cdot dr$ in these coordinates and in terms of $V$ and the partial derivatives. You then get a differential. – wilsonw Mar 29 '20 at 02:55
  • I think this generalizes to higher dimensions easily. – wilsonw Mar 29 '20 at 02:56
  • Or you can go physicsy and think this $V$ as the potential, and that integral as the work done. – wilsonw Mar 29 '20 at 02:58
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I'd like to share a trick to solve this problem.

Let $f:M\subset\mathbf{R}^n\longrightarrow\mathbf{R}$ be a function defined on $M$, let $x\in M, X$ be an arbitrary vector on the tangent space $T_xM$, then the gradient field will induce a vector field, namely $$\bar{X}(x)=(df)_xX=\left<\mathrm{grad}_xf,X\right>$$

Then, let $\phi^s(x)$ be the flow line induced by $-\mathrm{grad}_xf$, i.e.

$$-\mathrm{grad}_{\phi}f=\phi'$$

Now if we take $X=-\mathrm{grad}_{\phi^s(x)}f$, by the definition of the tangent map:

$$\frac{d(f(\phi^s))}{ds}=-\Vert \mathrm{grad}_{\phi^s}f \Vert^2$$

Thus it is decreasing along the flow line.

Actually, the result holds for any Riemannian Manifold $M$.

Z. Liu
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