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Consider the following wave-type equation, $$u_{tt}-\frac{2}{t}u_t-\Delta u=g(t,x)$$ where $(t,x)\in [\epsilon, 1]\times \mathbb{R}^3$ for some $\epsilon>0.$ Furthermore assume that $(u,u_{t})=(0,0)$ at $t=\epsilon.$ Then define the energy of the solution $u$ as follows, $$E(t) = E(u(t))=t^2 \int (u_t)^2+|\nabla u|^2 dx.$$ My goal is to derive a time decay estimates on the energy $E$.

As usual, we compute the time derivative of $E$. Thus, $$\frac{d}{dt}E(t) = \frac{2E}{t} + 2t^2 \int u_t u_{tt} + \nabla u\cdot \nabla u_t.$$ Using integration by parts we get, $$\frac{d}{dt}E(t) = \frac{2E}{t} + 2t^2 \int u_t \left(g+\Delta u + \frac{2}{t}u_t\right) - u_t\Delta u \\ =\frac{2E}{t} + 2t^2 \int u_t g + 4t \int (u_t)^2.$$ For the third term, we can simply control it as follows, $$4t \int (u_t)^2 \leq 4t\int (u_t)^2 + |\nabla u|^2\leq \frac{4E(t)}{t}.$$ For the second term, using the above estimate, we get $$ 2t^2 \int u_t g\leq 2t^2 \|u_t\|_{L^2}\|g\|_{L^2}\leq 2 t \sqrt{E(t)}\|g\|_{L^2}$$ which implies, $$\frac{dE}{dt}\leq \frac{6E}{t} + 2t \sqrt{E}\|g\|_{L^2}.$$

I am not sure how to proceed further. I would like to apply Gronwall's inequality but the right-hand side does not seem very nice. I started with $g\in L^2$, but perhaps I need more regularity for the function $g.$ Any ideas here will be much appreciated.

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  • $\begingroup$ Hang on, are you trying to solve backwards in time (with data at $t = 1$ and trying to show decay as $t\to 0$)? $\endgroup$ Oct 27, 2021 at 19:12
  • $\begingroup$ Yeah, that's right! $\endgroup$
    – Student
    Oct 27, 2021 at 20:24
  • $\begingroup$ Then your energy inequality is going the wrong way. If you integrate from $t = \epsilon$ to $t = 1$, your inequality gives $E(1) \leq E(\epsilon) + \frac{6E}{t} + \ldots$ whereas a useful equality will allow you to bound $E(\epsilon)$ in terms of $E(1)$. $\endgroup$ Oct 28, 2021 at 2:38
  • $\begingroup$ What do you exactly hope to accomplish? If you use $H(t) = \int |u_t|^2 + |\nabla u|^2$ the standard energy, you get $$ \frac{d}{dt} H(t) \geq - \frac{t}{4} \|g\|_{L^2}^2 $$ or that $$ H(1) + \int_\epsilon^1 \frac{t}{4} \|g\|_{L^2}^2 ~dt \geq H(\epsilon) $$ This gives boundedness of the standard energy and as $\epsilon \searrow 0$, and hence your weighted energy will decay like $\epsilon^2$. Are you looking somehow for more decay? $\endgroup$ Oct 28, 2021 at 3:10
  • $\begingroup$ @WillieWong not sure how you derived the lower bound. Could you please explain? $\endgroup$
    – Student
    Oct 28, 2021 at 11:42

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Define using $H(t) = \int (u_t)^2 + |\nabla u|^2 ~dx $ the standard energy. Taking the time derivative you find $$ \frac{d}{dt}H(t) = 2 \int u_t( g + \frac{2}{t} u_t)$$ Writing $\|\cdot \|$ for the $L^2$ integral, you have then $$ \frac{d}{dt} H(t) \geq - 2 \|u_t\|\cdot \|g\| + \frac{4}{t} \|u_t\|^2 \tag{A}$$ by Cauchy-Schwarz, and then by AM-GM you get $$ \frac{d}{dt} H(t) \geq - \frac{4}{t} \|u_t\|^2 - \frac{t}{4} \|g\|^2 + \frac{4}{t} \|u_t\|^2 = - \frac{t}{4} \|g\|^2 $$ So integrating you find $$ H(1) + \int_\epsilon^1 \frac{t}{4} \|g\|^2 ~dt \geq H(\epsilon) $$

So if you assume that $$\tag{*} \int_0^1 \int_{\mathbb{R}^3} |g|^2 t ~ dx~dt $$is finite, the above shows that $H(\epsilon)$ is uniformly bounded as $\epsilon \searrow 0$ and hence what you defined as the weighted energy $E(\epsilon)$ converges to zero at rate $\epsilon^2$.


If you only assume (*), then it is impossible to prove that the standard energy decays with any rate like $\epsilon^\beta$. This is because of the ODE examples

$$ y(t) = t^{1+\alpha}$$

with $\alpha\in (0,1)$. You find that

$$ \ddot{y} - \frac{2}{t} \dot{y} = (1+\alpha)(\alpha - 2) t^{\alpha - 1} =: g(t)$$

and this $g(t)$ satisfies $\int_0^1 t |g(t)|^2 ~dt < \infty$. So the energy ($(\dot{y})^2$) cannot be proven to go to zero as $t^\beta$ for any $\beta > 0$. By spatial truncation this ODE example can be upgraded to a bona fide solution of the wave equation you wrote down.


On the other hand, if instead of (*) you know something stronger about $g$, then some decay may be concluded.

For example: suppose you know that $\int_0^1 \|g\|^2 ~dt$ is bounded. Then starting from (A) you can instead find $$ \frac{d}{dt} H(t) \geq \left( \frac4t-1\right) \|u_t\|^2 - \|g\|^2 $$ This would give you $$ H(1) + \int_0^1 \|g\|^2 ~dt \geq H(\epsilon) + \int_{\epsilon}^1 \frac{3}t\|u_t\|^2 ~dt $$ This gives you an integrated decay of time component of the standard energy.

But this would depend on what you want to know and what you are willing to provide for (in terms of properties of $g$).

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  • $\begingroup$ Thank you very much for your detailed response, this is very helpful! $\endgroup$
    – Student
    Oct 28, 2021 at 16:58

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