Dampings in PDE theory and stabilisation

Question

I always wondered about the subject of stabilisation of Pde by addding some dampings to the equation. Taking as example the wave equation with Dirichlet boundary condition $u_{tt}=u_{xx}$, the associated energy is conservative, $E(t)=E(0)$. So in order to make the energy decays to an equilibrium point we add some dampings( for example $u_t$ ) "frictional damping" to the equation, therefore the energy is exponentially stable. My quetion is: what is the influence of this new term on the physical model? is the derived model stays the same? thank you.

Answer 1

This answer still needs a little work, but since I received a message from our OP Gustav reminding me it has been a while since I promised him I'd respond, I'm posting what I've got so far. Sorry for the delay . . .

I'm not exactly sure what our OP Gustav has in mind when he refers to the "physical" and "derived" models. In what follows is a more or less standard mathematical "derivation" of energy conservation for the undamped wave equation (1), then a sort of "addendum" which shows how a damping term of the form $\rho u_t$ causes the energy to decrease over time.

We start with

$u_{tt} = u_{xx}, \tag 1$

and assume the domain of definition of $u(x, t)$ is $[-a, a] \times \Bbb R_{\ge 0}$ where $R_{\ge 0} = \{ r \in \Bbb R \mid r \ge 0 \}$; we also assume that $u(x, t)$ is fixed on the boundary of $[-a, a]$, so that $u(-a, t) = u(a, t)$ are both constant; the so-called Dirichlet boundary conditions.

We define the energy

$E = \displaystyle \int_{-a}^a (u_t^2 + u_x^2) \; dx; \tag 2$

then

$\dot E = \displaystyle \int_{-a}^a (\dot{(u_t)^2} + \dot{(u_x)^2}) \; dx, \tag 3$

whence

$\dot E = \displaystyle \int_{-a}^a (u_{tt}u_t + u_{xt}u_x) \; dx; \tag 4$

now,

$(u_t u_x)_x = u_{tx}u_x + u_t u_{xx}, \tag 5$

or

$u_{xt}u_x = (u_x u_t)_x - u_t u_{xx}; \tag 6$

then

$\displaystyle \int_{-a}^a u_{xt} u_t \; dx = \int_{-a}^a (u_x u_t)_x \; dx - \int_{-a}^a u_t u_{xx} \; dx; \tag 7$

now,

$\displaystyle \int_{-a}^a (u_x u_t)_x \; dx = (u_x(a, t) u_t(a, t) - u_x(-a, t) u_t(-a, t)) = 0, \tag 8$

by virtue of the assumption that $u(x, t)$ is fixed on the set $\{-a, a \} = \partial [-a, a]$, which implies $u_t(-a, t) = u_t(a, t) = 0$. (7) thus becomes

$\displaystyle \int_{-a}^a u_{xt} u_t \; dx = - \int_{-a}^a u_t u_{xx} \; dx, \tag 9$

and returning to (4) we find

$\dot E = \displaystyle \int_{-a}^a u_{tt}u_t \; dx + \int_{-a}^a u_{xt}u_x \; dx$ $= \displaystyle \int_{-a}^a u_{tt}u_t \; dx - \int_{-a}^a u_t u_{xx} \; dx = \int_{-a}^a u_t(u_{tt} - u_{xx}) \; dx = 0; \tag {10}$

this shows the derivation of the conservation of energy for (1); in the event that there is a $u_t$ term,

$u_{tt} + \rho u_t = u_{xx}, \tag{11}$

everything carries through exactly as in the above until equation (10) is reached; then we have, by (11)

$u_{tt} - u_{xx} = -\rho u_t; \tag{12}$

(10) is thus replaced by

$\dot E = \displaystyle \int_{-a}^a u_t (-\rho u_t) \; dx = -\rho \int_{-a}^a u_t^2 \; dx; \tag{13}$

with $\rho > 0$, (13) shows the energy is dissipated as long as $u_t \ne 0$.

Now, a few words on the physics associated with all this. The equation (1) is often taken to describe small vibrations of a string, in this case a string stretched 'twixt $-a$ and $a$ and fixed at those points; the string is taken to have unit linear mass density , and is also under unit horizontal tension. The typical physical interpretation of the terms in (1) is that $u_{tt}(x, t)$ represents the intertial or "$ma$" force per unit length in the vertical direction if we assume as we do the string is stretched out horizontally. As for $u_{xx}(x, t)$, it represents the vertical force on the string, per unit length, at the position $x$ and time $t$, induced by the horizontal tension through what may be interpreted as curvature effects--i.e., the variation in the slope $u_x(x, t)$ over the string. These considerations are part of the standard derivation of the wave equation (1) for vibratory string motion, and need not be discussed at great length here; the important fact for the present purposes is that both $u_{tt}(x, t)$ and $u_{xx}(x, t)$ represent forces per unit length and so the net force per unit length on and undamped or otherwise unforced string will be $u_{tt} - u_{xx} = 0$. A more detailed derivation of this model may be found in this wikipedia entry. Now in the light of this understanding we may consider the meaning and role of the term $\rho u_t$ which appears in (11). Recalling that both $u_{tt}$ and $u_{xx}$ are forces per unit length, we are for the sake of consistency required to interpret $\rho u_t(x, t)$ as a force per unit length acting at position $x$ and time $t$; the question then becomes, "what physical effect yields a force modelled by $\rho u_t$?" If we write (11) in the form

$u_{tt} = u_{xx} - \rho u_t, \tag{14}$

we see that for $\rho > 0$ the term $\rho u_t$ always acts in a direction opposite to $u_t$; that is, it will tend to decrease the velocity $u_t(x, t)$ of the string at position $x$; indeed, if the tension force $u_{xx} = 0$, (14) becomes

$u_{tt} = -\rho u_t \tag{15}$

which has the solution

$u_t(x, t) = u_t(x, t_0) e^{-\rho(t - t_0)}, \tag{16}$

showing that in the absence of the tension force $u_{xx}$, $u_t(x, t) \to 0$ as $t \to \infty$.