I'm working on a system of the following form:

$$(1) \,\,\,\,\,\ \begin{cases} u_{tt} + L_1u + L_2v= f, \\ \nabla u - \nabla v - \nabla v_t=0 \end{cases} $$

where $u(x,t)$ and $v(x,t)$ belong to suitable sobolev spaces (I work with weak solutions), $L_1$ and $L_2$ are elliptic operators in divergence form. I am looking for methods to show the uniqueness of solutions for this kind of problem. I have already consulted classical books (e.g. Evans) with no luck. I understand that the above problem is very particular but any book or paper dealing with PDE systems could be useful.

I am also interested in other type of problems of the following form:

$$(2) \,\,\,\,\,\ \begin{cases} u_{tt} + L_1u + L_2v= f, \\ div( \nabla u - \nabla v - \nabla v_t)=0 .\end{cases} $$ Also in this case I am looking for methods to prove uniqueness of solutions.

  • $\begingroup$ When you say $L_1$ and $L_2$ are elliptic, which sign do you mean? (If you write $L_1 = \mathrm{div}(a \cdot \nabla u)$, is $a$ positive definite or negative definite?) Are you solving the boundary value problem , the initial value problem, or an initial-boundary value problem? $\endgroup$ – Willie Wong Nov 21 '19 at 17:14
  • $\begingroup$ In particular, without boundary conditions, supposing that $L_2 v = 0$ has a non-trivial solution which we call $v_0$, then if you add $v \mapsto v + e^{-t} v_0$ this will also solve the equation. $\endgroup$ – Willie Wong Nov 21 '19 at 17:19
  • $\begingroup$ Hi @WillieWong . I am the author of the post but I had to write it as a guest because today I was having problems with the log in. The operators are of the form $L_i=-div(A_i\nabla u)$ with positive definite matrix. I have both boundary and initial conditions. $\endgroup$ – Ef_Ci Nov 21 '19 at 19:45
  • $\begingroup$ The fact that the equation for $v$ cannot be solved respect to the $t$ derivatives makes me think this is a problem of Sobolev type, i.e. (1) can be modeled as a Sobolev type equation: there are several monographs published on this topic. $\endgroup$ – Daniele Tampieri Nov 21 '19 at 21:25

Let's try energy estimates (I'll assume constant coefficients, but variable coefficients shouldn't give too much trouble).

Since you are interested in uniqueness, we take two different solutions to your linear system with the same initial and boundary conditions and subtract them. Denote still by $u$ and $v$ the corresponding differences, now the inhomogeneity $f$ drops out.

Let us rewrite the transport equation for $v$ as $$ \partial_t (e^{t} \nabla v) = e^{t} \nabla u $$ which we can integrate (and using that the initial data for the difference is 0) as $$ \nabla v(t) = \nabla \int_0^t e^{s-t} u(s) ds $$ hence $$ L_2 v(t) = L_2 \int_0^t e^{s-t} u(s) ds $$ Plugging this into the first equation, integrating by parts and assuming sensible boundary conditions, you should have the differential form of the energy identity looking like

$$ \frac{d}{dt} \int \frac12 |\partial_t u|^2 + \frac12 A_1(\nabla u, \nabla u) + A_2( \nabla \int_0^t e^{s-t} u(s) ds, \nabla u) dx = \int A_2(\nabla u, \nabla u) - A_2(\nabla \int_0^t e^{s-t} u(s) ds, \nabla u) dx $$

Denote by $E(t) = \sup_{s \in [0,t]} \int |\partial_t u(s)|^2 + A_1 (\nabla u(s), \nabla u(s)) dx$

We can estimate, depending on how $A_1$ compares with $A_2$ (in terms of ellipticity) $$ |\int A_2( \nabla \int_0^t e^{s-t} u(s) ds, \nabla u) dx| \leq C t E(t) $$

So we have that the energy estimates imply

$$ E(t) - Ct E(t) \leq C \int_0^t E(s) + C s E(s) ~ds $$

For $t > 0$ sufficiently small we can absorb $Ct E(t)$ on the left, and get

$$ E(t) \leq C' \int_0^t E(s) + C s E(s) ~ ds $$

and by Gronwall's lemma, since $E(0) = 0$, we have $E(t) \equiv 0$ (for all $t$ sufficiently small, with smallness depending only on the ellipticity of $A_1$ and $A_2$, so we can iterate and get that this holds for all times).

This implies that $u\equiv 0$, which implies that $e^t \nabla v \equiv 0$ so that $v$ is constant on every given time. If you have sane boundary conditions this also implie that $v \equiv 0$.

For the case that the equation satisfied by $v$ is $\triangle ( v_t + v - u) = 0$, you will need to do additional elliptic estimates, but I think the basic idea should still work.

| cite | improve this answer | |
  • $\begingroup$ Thanks. I will try to use these ideas. $\endgroup$ – Ef_Ci Nov 21 '19 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.