Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose we have two waves $u_1$ and $u_2$ satisfying

$$ \begin{cases} \partial_{tt}u_i = \Delta u_i\\ u_i(0,x) = f_i(x) \\ \partial_t u_i(0,x) = g_i(x) \end{cases} $$ in $\mathbb{R}^+ \times \mathbb{R}^n$. Then $u = u_1 + u_2$ will be a solution of $$ \begin{cases} \partial_{tt}u = \Delta u\\ u(0,x) = f_1(x) + f_2(x)\\ \partial_t u(0,x) = g_1(x) + g_2(x) \end{cases} $$ i.e. the solution depends linearly on the initial data.

However, if we consider the energy of solutions $$ e(u;t) = \int \left(u_t(t,x)\right)^2 + \left|\nabla u(t,x) \right|^2 \,dx $$ it is pretty clear that this quantity doesn't depend linearly on the initial data. For example, if $f_1 = -f_2 =: f$ and $g_i \equiv 0$ (for simplicity), then $u \equiv 0$, so obviously its energy is $0$ as well. On the other hand, the energy of $u_1$ and $u_2$ are both $$ e(u_i;t) = \int \left|\nabla{f(x)}\right|^2\,dx > 0 $$ (by conservation of energy).

So, my question is: what kind of (if any) relationships do hold, or what kind of conditions are necc./suff. for interesting/useful conclusions to be drawn? For example, (as the beginning of an idea), if the supports of $f_1$ and $f_2$ are disjoint, then it seems we can (at least) conclude $$ e(u_1 + u_2;0) = \int \left| \nabla f_1 + \nabla f_2\right|^2\,dx \\ = \int \left| \nabla f_1\right|^2 + \left|\nabla f_2\right|^2\,dx = e(u_1;0) + e(u_2;0) $$ (it would be nice if this were satisfied for $t>0$ as well, but I don't know if I can conclude that just yet.) Maybe the statement is something like "if, at time $t$, the supports of the $u_i(t,\cdot)$'s are disjoint, then the total energy is the sum of the individual energies".

Anyway, this is of course a broad question, but thanks in advance for any thoughts you may have.

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

it would be nice if this were satisfied for $t>0$ as well

It is satisfied for $t>0$ (provided the initial supports were disjoint), because the energy is conserved by the wave equation. $$ e(u_1+u_2;t)= e(u_1+u_2;0) = e(u_1;0)+e( u_2;0) = e(u_1;t)+e( u_2;t) \tag{1}$$

Thus, additivity of energy for all times is equivalent to $e(u_1+u_2;0) = e(u_1;0)+e( u_2;0) $. In terms of the initial data, the latter says precisely that $$\int g_1g_2 + \int \nabla f_1\cdot \nabla f_2=0 \tag{2}$$

Thus, (1) holds if and only if (2) holds.

One can imagine situations in which (2) naturally holds withough the supports being disjoint: for example, the support of one wave is contained in a "flat part" of the other; and initial velocities are zero.

share|improve this answer
    
Great, thanks for the answer; though I wonder if you have any sense whether (2) is a necessary condition as well. Or further, if (1) holds at some $t>0$ does it imply either or both of (2) (with $\nabla f_i$ replaced by $\nabla u_i(t,\cdot)$ etc.)? Or even further, whether it could imply $\nabla f_1 \cdot \nabla f_2 = 0$ (and not just their integral). Thanks! –  BaronVT Nov 19 '13 at 5:18
    
@BaronVT Edited with iff condition. No, (1) does not imply a pointwise statement $\nabla f_1\cdot \nabla f_2=0$, since (1) holds whenever the integral condition (2) holds. There is nothing deep going on here, in my opinion. –  user103402 Nov 19 '13 at 5:53
    
Alright, thanks again for the discussion. –  BaronVT Nov 19 '13 at 15:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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