Uniform Exponential Stability of Galerkin Approximations for a Damped Wave System

Abstract

We consider the numerical approximation of a linear damped wave system modeling the propagation of pressure waves in a pipeline by Galerkin approximations in space and appropriate time-stepping schemes. By careful energy estimates, we prove exponential decay of the physical energy on the continuous level and uniform exponential stability of semi-discrete and fully discrete approximations obtained by mixed finite element discretization in space and certain one-step methods in time. The validity and limitations of the theoretical results are demonstrated by numerical tests.

Appendix

6.1.1 Auxiliary Results

We start with proving a generalized Poincaré inequality.

Lemma 6.12

Let a ∈ L ²(0, 1) and $\bar a = \int a dx \ne 0$ . Then for any u ∈ H ¹(0, 1) we have

$$\displaystyle \begin{aligned} \|u\|{}_{L^2(0,1)} \le \tfrac{1}{\pi}\big( 1 + \tfrac{1}{\bar a} \|a-\bar a\|{}_{L^2(0,1)} \big) \ \|\partial_x u\|{}_{L^2(0,1)} + \tfrac{1}{\bar a} \big| \int_0^1 a u \; dx\big|, \end{aligned} $$

Proof

Let $\bar u = \int _0^1 u$ be the average of u and ∥⋅∥ the norm of L ²(0, 1). Then

$$\displaystyle \begin{aligned} \|u\| \le \|u-\bar u\| + \|\bar u\| \le \tfrac{1}{\pi} \|\partial_x u\| + \|\bar u\|, \end{aligned} $$

where we used the standard Poincaré inequality. To bound the last term, observe that

$$\displaystyle \begin{aligned} \bar u = \tfrac{1}{\bar a} \int_0^1 \bar a u \; dx = \tfrac{1}{\bar a} \int_0^1 (\bar a -a) \; u + a u \; dx = \tfrac{1}{\bar a} \int_0^1 (\bar a -a) \; (u-\bar u) + a u \; dx. \end{aligned} $$

Application of the triangle, Cauchy Schwarz, and Poincaré inequalities yields

$$\displaystyle \begin{aligned} \|\bar u\| \le \tfrac{1}{\bar a} \|\bar a - a\| \|u - \bar u\| + \tfrac{1}{\bar a} |\int_0^1 a u \; dx| \le \tfrac{\|\bar a - a\|}{\bar a \pi} \|\partial_x u\| + \tfrac{1}{\bar a} |\int_0^1 a u \; dx|. \end{aligned} $$

The assertion of the lemma now follows by combination of the two estimates. □

An application of this lemma to solutions of the damped wave system yields the following estimate which will be used several times below.

Lemma 6.13

Let (u, p) be smooth solution of (6.3)–(6.5) and 0 < a ₀ ≤ a(x) ≤ a ₁ . Then

$$\displaystyle \begin{aligned} \|u(t)\|{}_{L^2(0,1)} \le \tfrac{a_1}{a_0} \|\partial_t p(t)\|{}_{L^2(0,1)} + \tfrac{1}{a_0} \big| \|\partial_t u(t)\|. \end{aligned} $$

Proof

Using the bounds for the parameter, we obtain from the previous lemma that

$$\displaystyle \begin{aligned} \|u\|{}_{L^2(0,1)} \le \tfrac{a_1}{a_0} \|\partial_x u\|{}_{L^2(0,1)} + \tfrac{1}{a_0} \big| \int_0^1 a u \; dx\big|. \end{aligned} $$

Note that this estimate holds for any function u ∈ H ¹(0, 1). Using the mixed variational characterization of the solution, we further obtain

$$\displaystyle \begin{aligned} |\int_0^1 a u \; dx| = |(au, 1)| = |-(\partial_t u, 1) + (p, \partial_x 1)| = |(\partial_t u, 1)| \le \|\partial_t u\|. \end{aligned} $$

Note that the boundary condition on the pressure was used implicitly here. □

6.1.2 Proof of the Theorem 6.1 for k = 1

To establish the decay estimate for the energy $E^1(t) = \frac {1}{2} \big (\|\partial _t u(t)\|{ }^2 + \|\partial _t p(t)\|{ }^2 \big )$, let us define the modified energy

$$\displaystyle \begin{aligned}E_\epsilon^1(t) = E^1(t) + \epsilon (\partial_t u(t),u(t)). \end{aligned}$$

We assume that (u, p) is a classical solution of (6.3)–(6.5), such that the energies are finite. As a first step, we will show now that for appropriate choice of 𝜖, the two energies E ¹ and $E^1_\epsilon$ are equivalent.

Lemma 6.14

Let $|\epsilon | \le \frac {a_0}{ 4 +2 a_1}$ . Then

$$\displaystyle \begin{aligned} \tfrac{1}{2} E^1(t) \le E^1_\epsilon(t) \le \tfrac{3}{2} E^1(t). \end{aligned}$$

Proof

We only have to estimate the additional term in the modified energy. By the Cauchy-Schwarz inequality and the estimate of Lemma 6.13, we get

$$\displaystyle \begin{aligned} (\partial_t u(t),u) \le \|\partial_t u(t)\| \|u(t)\| \le \tfrac{1}{a_0} \|\partial_t u(t)\|{}^2 + \tfrac{a_1}{a_0} \|\partial_t u(t)\| \|\partial_t p(t)\|. \end{aligned}$$

Using Young’s inequality to bound the last term yields

$$\displaystyle \begin{aligned} |(\partial_t u(t),u(t))| \le \tfrac{2+a_1}{2a_0} \|\partial_t u(t)\|{}^2 + \tfrac{a_1}{2a_0} \|\partial_t p(t)\|{}^2 \le \tfrac{2+a_1}{2a_0} \big(\|\partial_t u(t)\|{}^2 + \|\partial_t p(t)\|{}^2 \big). \end{aligned}$$

The bound on 𝜖 and the definition of E ¹(t) further yields $|\epsilon (\partial _t u(t), u(t))| \le \frac {1}{2} E^1(t)$, from which the assertion of the lemma follows via the triangle inequality. □

We can now establish the exponential decay for the modified energy.

Lemma 6.15

Let $0 \le \epsilon \le \tfrac {2a_0^3}{8 a_0^2+ 4 a_0^2 a_1 + 2 a_0 a_1 +a_1^4}$ . Then

$$\displaystyle \begin{aligned} E^1_\epsilon(t) \le e^{-2 \epsilon (t-s)/3} E^1_\epsilon(s). \end{aligned}$$

Proof

To avoid technicalities, let us assume that the solution is sufficiently smooth first, such that all manipulations are well-defined. By the definition of the modified energy and the energy identity given in Lemma 6.3, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \tfrac{d}{dt} E^1_\epsilon(t) &\displaystyle &\displaystyle = \tfrac{d}{dt} E^1(t) + \epsilon \tfrac{d}{dt} (\partial_t u(t),u(t)) \\ &\displaystyle &\displaystyle \le - a_0 \|\partial_t u(t)\|{}^2 + \epsilon \tfrac{d}{dt} (\partial_t u(t),u(t)). \end{array} \end{aligned} $$

The last term can be expanded as

$$\displaystyle \begin{aligned} \epsilon \tfrac{d}{dt} (\partial_t u(t),u(t)) = \epsilon \|\partial_t u(t)\|{}^2 + \epsilon (\partial_{tt} u(t), u(t)). \end{aligned} $$

(6.19)

Using the fact that (u, p) as well as (∂ _t u, ∂ _t p) solve the variational principle, we can estimate the last term by

$$\displaystyle \begin{aligned} \begin{array}{rcl} (\partial_{tt} u(t), u(t)) &\displaystyle &\displaystyle = (\partial_t p(t), \partial_x u(t)) - (a \partial_t u(t), u(t)) \\ &\displaystyle &\displaystyle = -(\partial_t p(t), \partial_t p(t)) - (a \partial_t u(t), u(t)) \\ &\displaystyle &\displaystyle \le -\|\partial_t p(t)\|{}^2 + a_1 \|\partial_t u(t)\| \|u(t)\|. \end{array} \end{aligned} $$

Using Lemma 6.13 to bound ∥u(t)∥ and Young’s inequality, we further get

$$\displaystyle \begin{aligned} \begin{array}{rcl} (\partial_{tt} u(t), u(t)) &\displaystyle &\displaystyle \le -\|\partial_t p(t)\|{}^2 + \tfrac{a_1}{a_0} \|\partial_t u(t)\|{}^2 + \tfrac{a_1^2}{a_0} \|\partial_t u(t)\| \|\partial_t p(t)\| \\ &\displaystyle &\displaystyle \le -\tfrac{1}{2} \|\partial_t p(t)\|{}^2 + \big( \tfrac{a_1}{a_0} + \tfrac{a_1^4}{2a_0^2} \big) \|\partial_t u(t)\|{}^2. \end{array} \end{aligned} $$

Inserting this estimate in (6.19) then yields

$$\displaystyle \begin{aligned} \frac{d}{dt} E^1_\epsilon(t) \le -\big(a_0 - \epsilon (1+\tfrac{a_1}{a_0} + \tfrac{a_1^4}{2 a_0^2})\big) \|\partial_t u(t)\|{}^2 - \tfrac{\epsilon}{2} \|\partial_t p(t)\|{}^2. \end{aligned}$$

The two factors are balanced by the choice $\epsilon = \frac {2 a_0^3}{3 a_0^2 + 2 a_0 a_1 + a_1^4}$. In order to satisfy also the condition of the Lemma 6.14, we enlarge the denominator by $5a_0^2 + 4 a_0^2a_1$, which yields the expression for 𝜖 stated in the lemma. In summary, we thus obtain

$$\displaystyle \begin{aligned} \tfrac{d}{dt} E^1_\epsilon(t) \le -\epsilon E^1(t) \le -\tfrac{2\epsilon}{3} E^1_\epsilon(t). \end{aligned}$$

The result for smooth solutions now follows by integration. The general case is obtained by smooth approximation and continuity; cf. the proof of Lemma 6.3. □

Combination of the previous estimates yields the assertion of Theorem 6.1 for k = 1.

6.1.3 Proof of Theorem 6.1 for k = 0 and k ≥ 2

We will first show how the estimate for k = 0 can be deduced from that for k = 1. Let u ₀, p ₀ ∈ L ²(0, 1) be given and consider the following stationary problem

$$\displaystyle \begin{aligned} \begin{array}{rcl} \partial_x \bar p + a \bar u &\displaystyle = u_0, \qquad \mbox{in } (0,1), \\ \partial_x \bar u &\displaystyle = p_0, \qquad \mbox{in } (0,1), \end{array} \end{aligned} $$

with boundary condition $\bar p=0$ on {0, 1}. Using Lemma 6.2, we readily obtain

Lemma 6.16

Let 0 < a ₀ ≤ a(x) ≤ a ₁ . Then there exists a unique strong solution $(\bar u,\bar p) \in H^1(0,1) \times H_0^1(0,1)$ and $\|\bar u\|{ }_{H^1(0,1)} + \|\bar p\|{ }_{H^1(0,1)} \le C \big ( \|u_0\| + \|p_0\| \big )$.

Let us now define

$$\displaystyle \begin{aligned} U(t) = \int_0^t u(s) ds - \bar u \qquad \mbox{and} \qquad P(t) = \int_0^t p(s) ds - \bar p. \end{aligned}$$

Then (U, P) is classical solution of the damped wave system (6.3)–(6.5) with initial values $U(0)=-\bar u$ and $P(0)=-\bar p$. Applying Theorem 6.1 for k = 1 to (U, P) yields

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \|u(t)\|{}^2 + \|p(t)\|{}^2 = \|\partial_t U(t)\|{}^2 + \|\partial_t P(t)\|{}^2 \\ &\displaystyle &\displaystyle \le C e^{-\alpha (t-s) } \big(\|\partial_t U(s)\|{}^2 + \|\partial_t P(s)\|{}^2 \big) = C e^{-\alpha (t-s)} \big( \|u(s)\|{}^2 + \|p(s)\|{}^2 \big). \end{array} \end{aligned} $$

This yields the assertion of Theorem 6.1 for k = 0. The estimates for k ≥ 2 follow by simply applying the estimate for k = 0 to the derivatives $(\partial _t^k u, \partial _t^k p)$.

6.1.4 Proof of the Theorem 6.3

Let us start with considering the case k = 1. To establish the decay estimate for the energy $E_h^{1,n}=\frac {1}{2}(\|d_\tau u_h^n\|{ }^2+\|d_\tau p_h^n\|{ }^2)$, let us define the modified energy

$$\displaystyle \begin{aligned} E_{h,\epsilon}^{1,n}= E_h^{1,n}+\epsilon(d_\tau u_h^n,u_h^{n,\theta}). \end{aligned}$$

As before, the two energies $E_h^{1,n}$ and $E_{h,\epsilon }^{1,n}$ are equivalent for appropriate choice of 𝜖.

Lemma 6.17

Let $|\epsilon |<\frac {a_0}{4+2a_1}$ . Then

$$\displaystyle \begin{aligned} \frac{1}{2} E_h^{1,n}\le E_{h,\epsilon}^{1,n}\le \frac{3}{2} E_h^{1,n}. \end{aligned}$$

The proof of this assertion follows almost verbatim as that of Lemma 6.14. With similar arguments as on the continuous level, we can then also establish the exponential decay estimate for the modified energy $E_{h,\epsilon }^{1,n}$.

Lemma 6.18

Let $\frac {1}{2}<\theta \le 1$ , $0<\epsilon \le \epsilon _0=\frac {2a_0^3}{8a_0^2+4a_0^2a_1+3a_0a_1+4a_1^4}$ , and 0 < τ ≤ τ ₀ with

$$\displaystyle \begin{aligned} \tau_0=\tfrac{1}{\epsilon_0}\tfrac{\theta-\frac{1}{2}}{\tfrac{5}{4}\theta^2+\frac{a_1}{2a_0}\theta^2+\frac{(1-\theta)^2}{4}+\tfrac{\theta(1-\theta)}{2}}. \end{aligned}$$

Then there holds $E_{h,\epsilon }^{1,n} \le e^{-\epsilon (n-m)\tau /3} E_{h,\epsilon }^{1,m}$ for all m ≤ n.

Note that the maximal step size τ ₀ only depends on a ₀, a ₁, and the choice of θ. Moreover, observe that the condition θ > 1∕2 is required here to make τ ₀ positive.

Proof

Following the arguments of the proof of Lemma 6.15, we start with

$$\displaystyle \begin{aligned} \begin{array}{rcl} d_\tau E_{h,\epsilon}^{1,n} &\displaystyle &\displaystyle = d_\tau E_{h}^{1,n} + \epsilon d_\tau (d_\tau u_h^n, u_h^{n,\theta}) \\ &\displaystyle &\displaystyle \le - a_0 \|d_\tau u_h^{n,\theta}\|{}^2 - (\theta-\tfrac{1}{2}) \tau \big(\|d_{\tau\tau} u_h^{n}\|{}^2 + \|d_{\tau\tau} p_h^{n}\|{}^2 \big) + \epsilon d_\tau (d_\tau u_h^n, u_h^{n,\theta}). \end{array} \end{aligned} $$

The last term can be expanded as

$$\displaystyle \begin{aligned} d_\tau (d_\tau u_h^n, u_h^{n,\theta}) = (d_\tau u_h^n, d_\tau u_h^{n,\theta}) + (d_{\tau\tau} u_h^n, u_h^{n-1,\theta}). \end{aligned}$$

Using $d_\tau u_h^n=d_\tau u_h^{n,\theta }+(1-\theta )\tau d_{\tau \tau }u_h^n$, the first term of the above expression yields

$$\displaystyle \begin{aligned} (d_\tau u_h^n, d_\tau u_h^{n,\theta}) \le 2\|d_\tau u_h^{n,\theta}\|{}^2 + \tfrac{(1-\theta)^2\tau^2}{4}\|d_{\tau\tau}u_h^n\|{}^2. \end{aligned}$$

To estimate the second term, we use the fact that besides $(u_h^n,p_h^n)$ also $(d_\tau u_h^n,d_\tau p_h^n)$ satisfies Eqs. (6.14) and (6.15). This implies

$$\displaystyle \begin{aligned} \begin{array}{rcl} (d_{\tau\tau} u_h^n, u_h^{n-1,\theta}) &\displaystyle =&\displaystyle (d_\tau p_h^{n,\theta},\partial_x u_h^{n-1,\theta}) -(ad_\tau u_h^{n,\theta},u_h^{n-1,\theta}) \\ &\displaystyle =&\displaystyle -(d_\tau p_h^{n,\theta},d_\tau p_h^{n-1})-(ad_\tau u_h^{n,\theta},u_h^{n-1,\theta}). \end{array} \end{aligned} $$

Using that $d_\tau p_h^{n-1} = d_\tau p_h^{n,\theta } - \theta \tau d _{\tau \tau } p_h^n$, we see that

$$\displaystyle \begin{aligned} \begin{array}{rcl} -(d_\tau p_h^{n,\theta},d_\tau p_h^{n-1}) &\displaystyle &\displaystyle = -\|d_\tau p_h^{n,\theta}\|{}^2 + \theta\tau (d_\tau p_h^{n,\theta}, d_{\tau\tau} p_h^n) \\ &\displaystyle &\displaystyle \le -\frac{3}{4} \|d_\tau p_h^{n,\theta}\|{}^2 + \theta^2\tau^2 \|d_{\tau\tau} p_h^n\|{}^2. \end{array} \end{aligned} $$

A discrete version of Lemma 6.13 allows us to bound

$$\displaystyle \begin{aligned} \|u_h^{n-1,\theta}\| \le \tfrac{1}{a_0} \|d_\tau u_h^{n-1}\| + \tfrac{a_1}{a_0} \|d_\tau p_h^{n-1}\|. \end{aligned}$$

The remaining term in the above estimate can then be treated by

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle -(ad_\tau u_h^{n,\theta},u_h^{n-1,\theta}) \le \|d_\tau u_h^{n,\theta}\| \big( \tfrac{a_1}{a_0}\|d_\tau u_h^{n-1}\| +\tfrac{a_1^2}{a_0}\|d_\tau p_h^{n-1}\|\big) \\ &\displaystyle &\displaystyle \qquad \le \|d_\tau u_h^{n,\theta} \| \big( \tfrac{a_1}{a_0} \|d_\tau u_h^{n,\theta}\| + \theta \tau \tfrac{a_1}{a_0} \|d_{\tau\tau} u_h^n\| + \tfrac{a_1^2}{a_0} \|d_\tau p_h^{n,\theta}\| + \theta \tau \tfrac{a_1^2}{a_0}\|d_{\tau\tau} p_h^n\|\big), \end{array} \end{aligned} $$

where for the last step, we used the same expansion of $p_h^{n-1}$ as above and a similar formula for $u_h^{n-1}$. Via Youngs inequalities and basic manipulations, we then obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} -(ad_\tau u_h^{n,\theta},u_h^{n-1,\theta}) &\displaystyle &\displaystyle \le (\tfrac{3a_0a_1+4a_1^4}{2a_0^2})\|d_\tau u_h^{n,\theta}\|{}^2 +\tfrac{1}{4}\|d_\tau p_h^{n,\theta}\|{}^2 \\ &\displaystyle &\displaystyle \qquad \qquad +\tfrac{1}{4}\theta^2\tau^2\|d_{\tau\tau}p_h^n\|{}^2 +\tfrac{a_1}{2a_0}\theta^2\tau^2\|d_{\tau\tau}u_h^n\|{}^2. \end{array} \end{aligned} $$

In summary, we thus arrive at

$$\displaystyle \begin{aligned} \begin{array}{rcl} (d_{\tau\tau} u_h^n, u_h^{n-1,\theta}) &\displaystyle &\displaystyle \le \tfrac{3a_0a_1+4a_1^4}{2a_0^2}\|d_\tau u_h^{n,\theta}\|{}^2 -\tfrac{1}{2}\|d_\tau p_h^{n,\theta}\|{}^2 \\ &\displaystyle &\displaystyle \qquad \qquad +\tfrac{5}{4}\theta^2\tau^2\|d_{\tau\tau}p_h^n\|{}^2+\tfrac{a_1}{2a_0}\theta^2\tau^2\|d_{\tau\tau}u_h^n\|{}^2. \end{array} \end{aligned} $$

Putting all estimates together, we finally obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} d_\tau E_{h,\epsilon}^{1,n} &\displaystyle &\displaystyle \le -\big(a_0-\epsilon \tfrac{4a_0^2 + 3a_0a_1+4a_1^4}{2a_0^2}\big)\|d_\tau u_h^{n,\theta}\|{}^2 -\tfrac{\epsilon}{2}\|d_\tau p_h^{n,\theta}\|{}^2 \\ &\displaystyle &\displaystyle \,\quad -\big(\theta-\tfrac{1}{2}-\epsilon\tau\tfrac{2 a_1 \theta^2 + a_0 (1-\theta)^2}{4a_0}\big)\tau\|d_{\tau\tau}u_h^n\|{}^2 -\big(\theta-\tfrac{1}{2}-\epsilon\tau\tfrac{5\theta^2}{4}\big)\tau\|d_{\tau\tau}p_h^n\|{}^2. \end{array} \end{aligned} $$

By the particular choice of τ ₀, we may estimate the terms in the second line from above by $-\frac {\epsilon }{2} \theta (1-\theta ) \tau ^2 \big ( \|d_{\tau \tau }u_h^n\|{ }^2 + \|d_{\tau \tau }p_h^n\|{ }^2\big )$. The two factors in the first line are balanced by the choice $\epsilon =\frac {2a_0^3}{5a_0^2+3a_0a_1+4a_1^4}$. In order to satisfy also the condition of Lemma 6.17, we enlarge the denominator by $3a_0^2+4a_0^2a_1$, and obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} d_\tau E_{h,\epsilon}^{1,n} &\displaystyle \le -\epsilon_0 \big\{ \tfrac{1}{2} \big( \|d_\tau u_h^{n,\theta}\|{}^2 + \|d_\tau p_h^{n,\theta}\|{}^2\big) + \theta (1-\theta) \tfrac{\tau^2}{2} \big( \|d_{\tau\tau}u_h^n\|{}^2+\|d_{\tau\tau}p_h^n\|{}^2\big) \big\} \\ &\displaystyle = -\epsilon_0 \big( \theta E_h^{1,n} + (1-\theta) E_h^{1,n-1} \big) \le -\epsilon \big( \theta E_h^{1,n} + (1-\theta) E_h^{1,n-1} \big) \end{array} \end{aligned} $$

for 𝜖 ≤ 𝜖 ₀. By equivalence of the energies stated in Lemma 6.17, this leads to

$$\displaystyle \begin{aligned} E_{h,\epsilon}^{1,n} \le \tfrac{1-\frac{2}{3} \epsilon (1-\theta) \tau}{1+\frac{2}{3}\epsilon \theta \tau} E_{h,\epsilon}^{1,n-1} \le(1-\tfrac{\epsilon\tau}{3})E_{h,\epsilon}^{1,n-1} \le e^{-\epsilon \tau/3} E_{h,\epsilon}^{1,n-1}, \end{aligned}$$

where we used that $\frac {2}{3}\epsilon \theta \tau \le \frac {2}{3}\epsilon _0 \theta \tau _0 \le 1$ in the second step, which follows from the definition of τ ₀. The assertion of the Lemma now follows by induction. □

Remark 6.2

Let us emphasize that the assertion of Lemma 6.18 holds true also for the choice $\theta =\frac {1}{2} + \lambda \tau$ with λ sufficiently large, but independent of τ.

Using the equivalence of the discrete energies stated in Lemma 6.17, we readily obtain the proof of Theorem 6.3 for the case k = 1. The result for k = 0 and k ≥ 2 follows from the one for k = 1 with the same arguments as on the continuous level.