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Suppose

$$ \begin{cases}u_{tt} - \Delta u = 0 \quad \textrm{ in } \mathbb R \times \mathbb R^n \\ u(0,x) = f(x); \quad u_t(0,x)= g(x). \end{cases} $$

then, depending on the dimension $n$, we have a formula for $u$ in terms of $f$ and $g$; for $n = 3$, e.g.

$$ u(t,x) = C\int_{\partial B(x,t)} tg(y) + f(y) + Df(y)\cdot(y-x) \, dS(y) $$

and for $n = 2$ $$ u(t,x) = C\int_{B(x,t)} \frac{t^2g(y) + tf(y) + tDf(y)\cdot(y-x)}{\left(t^2 - |y-x|^2\right)^{1/2}} \, dy $$

(see, for instance, Evans PDE, p. 72).

In the case the wave equation is

$$ \begin{cases}u_{tt} - c^2(x) \Delta u = 0 \quad \textrm{ in } \mathbb R \times \mathbb R^n \\ u(0,x) = f(x); \quad u_t(0,x)= g(x). \end{cases} $$ or $$ \begin{cases}u_{tt} - \nabla\cdot(c^2(x)\nabla u) = 0 \quad \textrm{ in } \mathbb R \times \mathbb R^n \\ u(0,x) = f(x); \quad u_t(0,x)= g(x). \end{cases} $$

there are corresponding formulae (see this question). (EDIT: and to be concrete, let's just consider the divergence form equation (the second one))

As the answerer in that question notes, the complete form of the integral is hard to compute, but I'm not too worried about the full specifics, mostly just the sign of the coefficients in the formula in the case initial displacement is $0$.

Specifically, I am mainly interested in the case where $f \equiv 0$, and further, I just want to know how the non-negativity (or non-positivity) of $g$ influences the sign of $u$; e.g. in the example above, if $f \equiv 0$ and $g \geq 0$ for all $x$, then $u(t,x) \geq 0$ for all $t \geq 0$.

So then, my question is:

is there any reason to expect, for a general $c(x)$, that the coefficients of $f$ and $g$ (or just $g$) in the Kirchoff formula are either non-negative or non-positive?

Resources (beyond the one mentioned in the other question) would also be greatly appreciated. Thanks in advance.

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BaronVT
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  • Minor comment, but in those solutions for $u$ you want the balls you are integrating over to be $\partial B(x,|t|)$ instead of just $\partial B(x,t)$ since you are considering all time, right? – user308973 Oct 24 '16 at 18:41

1 Answers1

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Nonpositive coefficients of $f$ or $g$?

Clearly impossible. Given positive initial position or velocity, the solution will be positive somewhere, for some period of time.

Nonnegative coefficient of $f$?

Fails in all dimensions $n\ge 2$ for constant $c$. Place a positive bump function $f$ near $0$, and focus on the value $u(x,t)$ for a fixed $x$ away from the bump. The Kirchhoff formula for $u(x,t)$ involves some derivative of a certain integral average of $f$. This average is zero for small $t$, and tends to zero again as $t\to\infty$ (in odd dimensions, it becomes zero again). So the derivative of this average (of whatever order) will change its sign, and so will $u(x,t)$.

The case $n=1$ is different because no derivative of the average of $f$ is taken.

Nonnegative coefficient of $g$?

Fails in all dimensions $n\ge 4$ for constant $c$: same reasoning as in the previous paragraph. The cases $n\le 3$ are different because no derivative of the average of $g$ is taken.

variable $c$?

For variable $c$, nonnegativity of coefficients fails in all dimensions. The reason is that by sharply varying $c$, one can emulate boundary conditions, bouncing the wave back. You may want to look up "strong/sharp density gradient" in conjunction with "wave propagation/reflection". I offer the following illustration in $n=1$: an elastic string has density 1 g/cm on the interval $(-1,1)$ but something like $10^3$ kg/cm outside of it. You punch the string near $0$, putting it in motion (originally, in the positive direction). But the momentum you gave to the string is not enough to move its dense part by any appreciable amount. What happens: the solution on $(-1,1)$ is practically the same as for the string with fixed ends at $\pm 1$. Which means it will take on both positive and negative values, being a sum of harmonics.

Sharp gradient of density inhibits the transfer of energy. When two billiard balls of comparable density collide, much of the energy is exchanged. But hitting a stationary ball of mass 1000 kg with a ball of mass 10 g will not transfer much energy to the larger ball. In the wave example, the energy will mostly stay in the low-density region (whatever the number of dimensions): all it can do there is to convert from kinetic to potential, and then back -- hence its oscillatory behavior and changes of sign.

I would not want to calculate an explicit counterexample with variable coefficient; if you are not convinced, run a numerical experiment.

Post No Bulls
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  • This is a great answer; the string example is convincing. A clarification and a question though: first, for "nonnegative coeff. of $g$" your answer states this fails for constant $c$ in $n=2$. However, the Kirchoff formula I have in this case (I've edited it into the question) seems to indicate the opposite. (though I now fully understand the implications for $f$ and for $g$ with $n > 3$). – BaronVT Dec 23 '13 at 18:10
  • (To be concrete, from here, let $f \equiv 0$, $n= 2,3$, wave eqn. in diverg. form) Second, it is maybe natural to next ask whether small perturbations (in $C^\infty$ or some other norm) of constant speeds would still be non-negative, and more generally whether one can describe necessary or sufficient conditions to ensure this. (In particular, your counterexample has poor local energy decay, not sure if that is related).

    Of course, these are new questions, and I don't expect a full answer, but if you could point me at a reference, I'd be grateful. Thanks for all your help so far!

     – BaronVT Dec 23 '13 at 18:19
  • @BaronVT You are right about $n=2$; I edited the answer. As for small perturbations, I am skeptical. You see, in 3 dimensions the function $u(x,t)$ returns precisely to $0$ when the wave passes. I suspect that an arbitrary small perturbation may cause it to overshoot. Sorry, I don't know of any references. Try to work out the equation of spherical wave beginning with impulse $g=\delta$, with radially symmetric increasing density (hence, decreasing $c$). It's just an ODE after all. – Post No Bulls Dec 24 '13 at 03:18
  • Ok, thanks a lot, I'll accept the answer now. Though in $3$ dimensions, the wave only returns to $0$ for constant $c$; i.e. Huygen's principle doesn't apply generally. Though this is something I can try and work out on my own. – BaronVT Dec 24 '13 at 04:30
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    @BaronVT Yes, that's what I meant (without saying it): for constant $c$, the solution stays nonnegative just barely, which makes it likely that a perturbation of $c$ can push it over the edge. – Post No Bulls Dec 24 '13 at 04:32