Let $\Omega$ be an open subset of $\mathbb{R^n}$. Consider the linear wave equation $$\begin{cases} \dfrac{\partial^{2}}{\partial t^{2}}u\left(t,x\right)=\Delta u\left(t,x\right)-mu\left(t,x\right)+h\left(t,x\right), & t\geq0, \ x\in \Omega \\ u\left(0,x\right)=\phi\left(x\right)\mbox{ and }\dfrac{\partial}{\partial t}u\left(0,x\right)=\psi\left(x\right), & x\in\Omega. \end{cases}$$ where $m\in \mathbb{R}$ and $h:[0,+\infty)\times \Omega\to\mathbb{R}$.

Consider the Hilbert space $X=H_{0}^{1}\left(\Omega\right)\times L^{2}\left(\Omega\right)$ equipped with the scalar product $$\left\langle \left(\begin{array}{c} u\\ v \end{array}\right),\left(\begin{array}{c} w\\ z \end{array}\right)\right\rangle =\int_{\Omega}\left(\nabla u.\nabla w+muw+vz\right)dx.$$ If we assume that $h\in L^{\infty}([0,+\infty),L^2(\Omega))$, is there any sufficient condition on the parameters of the equation which assures the existence of a global bounded solution i.e. $$\sup_{t\geq 0}\left\|(u,u_t) \right\|_X<+\infty. $$

I am also interested if someone knows some reference which deals with bounded global solutions of linear non-autonomous equations where the differential operator generates an isometry group of operators, which is the case for the above wave equation or for example Schrödinger equations.

  • $\begingroup$ $m$ is a constant? $\endgroup$ – Jason Knapp Aug 5 '14 at 17:26
  • $\begingroup$ @JasonKnapp Yes, it is a constant. $\endgroup$ – user50618 Aug 5 '14 at 17:28
  • $\begingroup$ I'm sorry to do this to you, I have another clarification. No boundary conditions: deliberate? Also I assume the other gradient factor in the inner product is $w$, not $v$. $\endgroup$ – Jason Knapp Aug 5 '14 at 17:40
  • $\begingroup$ $H^1_0$ - homogeneous Dirichlet boundary conditions I suppose $\endgroup$ – daw Aug 5 '14 at 17:59
  • $\begingroup$ If $(\phi,\psi) \in X$ then the desired bound holds. If you know about existence of solutions, the bound can be obtained by testing the weak formulation with $u_t$ and $\Delta u$. Evans' book on pde is a good reference. $\endgroup$ – daw Aug 5 '14 at 18:01

Start with an example. So, look at $\Omega = [0,\pi]\subset \mathbb{R}^{1}$, where you have an orthonormal basis of eigenfunctions for $-\Delta=-\frac{d^{2}}{dx^{2}}$ given by $\{ e_{n}(x)=\sqrt{2/\pi}\sin(nx)\}_{n=1}^{\infty}$. To solve the equation in this case, write the solution as $$ u(x,t) = \sum_{n=1}^{\infty}a_{n}(t)e_{n}(x) $$ Then $$ a_{n}''(t)=-n^{2}a_{n}(t)-ma_{n}(t)+(h(t),e_{n}) $$ The solution of this equation, where $a_{n}(0)$, $a_{n}'(0)$ are known, involves convolution with functions $\exp(\pm i\sqrt{n^{2}+m}\,t)$, and is given by $$ a_{n}(t) = a_{n}(0)\cos(\sqrt{n^{2}+m}\,t) +a_{n}'(0)\frac{\sin(\sqrt{n^{2}+m}\,t)}{\sqrt{n^{2}+m}} \\ +\int_{0}^{t}(h(u),e_{n})\frac{\sin(\sqrt{n^{2}+m}\,(t-u))}{\sqrt{n^{2}+m}}\,du. $$ Regardless of the value of $m$, for sufficiently large $n$, this value has negligible effect on $a_{n}(t)$. So the issue is not the parameter $m$: it is the driving function $h$. And this is the simplest possible example. If any $a_{n}(t)$ grows without bound as $t\rightarrow\infty$, you're in trouble. And controlling the time behavior of $a_{n}$ for large enough (but fixed) $n$ basically has nothing to do with $m$--it has to do with the growth of $(h(u),e_{n})$ as a function of $u$. I think you're barking up the wrong tree, from both the point of view of Mathematics and of Physics. Why? Because you can't expect to push the system forever and not see consequences, unless that tapers off in a predictable manner for every mode of the system, or unless you have some damping (but you threw that term away.)

Note: I agree that you get boundedness for the conditions you stated on $h$. That appears to me to be correct, too. The above construction of solution can be carried out using the spectral resolution for $\Delta$ in the general case, too.

  • $\begingroup$ Yes, thank you! It is then the damping $-\gamma u_t$ which makes things bounded on the future. $\endgroup$ – user50618 Aug 6 '14 at 14:16
  • $\begingroup$ You're welcome. $\endgroup$ – DisintegratingByParts Aug 6 '14 at 14:24

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