Systems with Persistent Memory: the Observation Inequality Problems and Solutions


 Arnold Robertson
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1 Chapter 6 Systems with Persistent Memory: the Observation Inequality Problems and Solutions Facts that are recalled in the problems wt) = ut) + 1 c A 1 s ] R c t s)) hws) + Ks r)wr)dr ds. 6.1) w = w + hw + Kt r)wr)dr + Ft) = w + hw + K w + Ft) w) = u, w ) = v, w = f on Γ, w = on \Γ. 6.2) Theorem 6.1. Let u L 2 ), v H 1 ), F L 1,T ;L 2 )), f L 2,T ;L 2 Γ )). The function w C,T ;L 2 )) C 1,T ;H 1 )) solves Eq. 6.2) if and only if the following equality holds ϕ,w t) + + ϕt),v + ϕ 1 wt)dx = ϕ t)u dx + Γ f s)γ 1 ϕt s))dγ ds + ϕt τ)fτ)dxdτ 6.3) for every solution 1 ϕ of ϕ = ϕ + hϕ + Kt r)ϕr)dr, ϕ) = ϕ D), ϕ ) = ϕ 1 D) and ϕ = on. 6.4) ϕ x), χt) + ϕ 1 x)χx,t)dx 6.5) 1 the equation is the same as??). We rewrite and give a special number to stress that the initial conditions belong to D) and the affine term is zero. 61
2 62 Problems of Chapter 6 Theorem 6.2. Let. Let ϕ L 2 ), ϕ 1 H 1 ) have compact support in. Let ϕt) be the corresponding solution of??). Assume that the observation inequality holds at time T for the wave equation on and that ϕt) = on \, for t,t ]. Then the initial conditions ϕ and ϕ 1 which are zero in \ ) have the following additional regularity: ϕ H 1 ), ϕ 1 L 2 ) so that we have also ϕ C,T ];H 1)) C1,T ];L 2 )). Furthermore, there exists a constant K = K T such that ) ϕ 2 H 1) + ϕ 1 2 L 2 ) K ϕ 2 C,T ;L 2 )). 6.6) Theorem 6.3. Let Γ =. There exists a time T 1 such that system 6.2) is controllable. The time T 1 is the control time of the associated wave equation in any region 1 which contains. For every T T 1 there exist m > and M such that ) ) m ϕ 2 H 1) + ϕ 1 2 L 2 ) γ 1 ϕ 2 dγ dt M ϕ 2 H Γ,T ] 1) + ϕ 1 2 L 2 ). The Problems 6.1. Let w and ϕ solve u = u xx for x >, respectively with initial conditions wx,) =, w x,) =, ϕx,) = ϕ x) D,), ϕ x,) = ϕ 1 x) D,) and boundary conditions w,t) = f t) H 2,), ϕ,t) =. Justify the integration by parts of the left hand side of the equality and show that T ϕx,t s) w x,s) w xx x,s) ] dxds = ϕ x)w x,t ) + ϕ 1 x)wx,t ) ] dx = + ϕ x,t s) f s)ds. Compare with Problem 4.2 and with formula 6.3) and explain the sign of the right hand side Use 6.1) and Picard iteration to prove the direct inequality for system 6.1) In the sufficiency part of the proof of Theorem 6.1, the function t χt), χt)) has been defined. This function takes values in L 2 ) H 1 ) see formula 6.5)). Use the fact that the right hand side of 6.3) is a continuous function of time and prove that χt), χt)) C,T ];L 2 ) H 1 )) This and the next problem show that the inverse inequality is an observability property. Let us consider the problem
3 Problems of Chapter 6 63 ϕ = ϕ xx, x,1), { ϕx,) = ϕ x) C 1,1), ϕ x,) = ϕ,t) = ϕ1,t) =. 6.7) Prove the existence of initial conditions ϕ such that ϕ x 1,t) for t,t ], provided that T < 1 and prove that if ϕ x 1,t) for t,1] then ϕ = Do the same as in Problem 6.4 when the initial conditions are ϕx,) =, ϕ x,) = ϕ 1 x) C,1) Use the results in Problem 2.4. Show that the inverse inequality does not hold for the wave equation if =,π),π) and Γ =,1) {} Let ϕ and ϕ 1 and T be as in Theorem 6.2. Prove the existence of M such that the following holds for every t, T ]: ϕ t) L 2 ) M ϕ C,T ];L 2 )) Let ϕ H 1), ϕ 1 L 2 ). Use the Volterra integral equation 6.1) to represent the solutions of??). For simplicity let c = 1 and h =. Choose any T >. Consider the set K of the solutions such that ϕ H 1 ) 1, ϕ 1 L 2 ) 1. Prove that K is bounded in C,T ];L 2 )) and equicontinuous The set K in Problem 6.8 is both bounded and equicontinuous in C,T ];L 2 )). It is known that boundedness and equicontinuity of a subset C,T ];H) implies compactness if dimh <. Prove that this is not true if dimh = in spite of this negative result the special set K in Problem 6.8 is relatively compact, i.e. any sequence has convergent subsequences in C,T ];L 2 )), since its elements take value in H 1 ), which is compactly embedded in L 2 ), see 1, p. 266]) Both for the system with memory and for the associated wave equation, study the following problem. Let controllability holds at a certain time T with the control acting on Γ. Let H T be the subset of ϕ,ϕ 1 ) H 1 ) L2 ) such that the corresponding solution of 6.1) satisfies for a fixed number M, T γ 1 ϕ 2 dg T M. Γ Prove that H T is closed and decide whether it is also bounded. Discuss the role of the assumption of controllability in this problem The derivation of the inverse inequality for the wave equation uses conservation of energy. So, Theorem 6.3 might suggest conservation of energy for the system with memory. Examine that system ϕ = e αt s) ϕs)ds 6.8) with zero boundary conditions) and show that the quantity
4 64 Problems of Chapter 6 ϕ t) 2 + ϕ 2 dx is not constant along the solutions. The Solutions Solution of Problem 6.1 The equality can be justified as in Problem 4.2, where however we used f ) = and f ) =. It has an interest to rederive the equality without using these conditions. Let H + = { f C 2,)) such that t T = T f implies f t) = }. The space H + is dense in H 2,) and so the required formula can be proved assuming f H +. The formula is then extended by continuity in fact it can be extended also to square integrable controls f ). We integrate by parts and we get T = T T = ϕx,t s)w x,s)dsdx = ϕ x)w x,t ) + ϕ 1 x)wx,t ) ] dx + ϕx,t s)w xx x,s)dxds = T T ϕ x,t s) f s)ds + ϕ xx x,t s)wx,s)dxds in the second line we used ϕ,t) = ). We equate and we get T ϕ x,t s) f s)ds = ϕ x,t s)wx,s)dsdx, ϕ x)w x,t ) + ϕ 1 x)wx,t ) ] dx. We examine the first integral on the right hand side. We note that wx,t) = f t x)ht x), w x,t) = f t x)ht x) + f t x)δt x). Note that the numbers T here and T f in the definition of H + ) are not the same, so that the last term is not zero in general. So, the first integral takes the form
5 Problems of Chapter 6 65 = = T ϕ x)w x,t )dx = ϕ x) f T x)ht x)dx + ϕ x) f T x)δt x) = ϕ x) f T x)dx + ϕ T ) f ) = = ϕ T ) f ) + ϕ ) f T ) + T ] d = dx ϕ x) f T x)dx T ] d dx ϕ x) because ϕ D,) and so ϕ ) =. It is apparent from here that the transformation T ϕ ϕ x)w x,t )dx = f T x)dx + ϕ T ) f ) = d dx ϕ x) ] f T x)dx is a continuous functional on H 1,). Compare with Theorem 6.1. Solution of Problem 6.2 We prove the inequality with ϕ, ϕ 1 in D). The inequality can be extended to H 1) L2 ) by continuity. For simplicity, we confine ourselves to the case F =, c = 1. Let Lϕ)t) = hr t)ϕ + Kt s)r s)ϕds. Picard iteration asserts that s ϕt) = ut) + ha 1 Rt s)us)ds + A 1 R t s) Ks r)ur)dr ds A 1 L 2 A 1 L ) ) k u k= The function ut) solves 2 so that u = u, u) = ϕ, u ) = ϕ ), u = on. 6.9) ut) = R + t)ϕ + A 1 R t)ϕ 1. Hence, ut) C,T ],doma) C 1,T ],doma ). The function u solves a wave equation and so the direct inequality holds for u: ) γ 1 u 2 L 2 G T ) ϕ M 2 H 1) + ϕ 1 2 L 2 ). 2 for consistency with the notations in 6.1) here we use u even if the boundary conditions are zero.
6 66 Problems of Chapter 6 Equality??) shows that, when ϕ and ϕ 1 belong to D), γ 1 ϕ = D Aϕ. So, γ 1 A 1 L 2 A 1 L ) )) k u = D L 2 A 1 L ) ) k u k= k= and this depends continuously on ϕ H 1) and ϕ 1 L 2 ). Let us consider γ 1 A 1 Lu) ) = hγ 1 A 1 R t s)us)ds+γ 1 A 1 s note that Kt) is scalar valued). We consider the first integral which is Kt s r)r r)us)dsdr γ 1 A 1 R t s)r + s)ϕ ds + γ 1 A 1 R t s)a 1 R s)ϕ 1 ds. The second integral is a continuous function of ϕ 1. To study the first integral we note that D R t s)r s)ϕ 1 ds, R t s)r + s)ϕ ds = 1 4 tr t)ϕ. This is easily seen because, using the definitions of R t) and R + t), we have Hence the first integral is R τ)r + t) = 1 4 R t + τ) R + t τ)]. t γ 1 A 1 R t)ϕ ) 6.1) which depends continuously on ϕ H 1 ), using the direct inequality of the wave equation. The last term in 6.1) is treated analogously. Solution of Problem 6.3 Use the following properties of a Banach space: separation property, consequence of HahnBanach Theorem): if x 1 x 2 belong to a Banach space B then there exists χ B such that x 1, χ = x 2, χ. consequence of the Closed Graph Theorem) if x n then there exists χ B such that x n, χ.
7 Problems of Chapter 6 67 Identify the dual space of L 2 ) with itself. The dual space of B = L 2 ) H 1 ) is L2 ) H 1 ) and in turn L 2 ) H 1 ) ) = L 2 ) H 1 ) this is a consequence of the fact that H 1 ) and so also H 1 ) are Hilbert spaces). These properties have been used in Section By contradiction, if χt), χt)) is not continuous on,t ] then there exists a sequence t n t in,t ]) such that either χt n ), χt n )) or χt n ), χt n )) χ, χ ) χt ), χt ). The first case is impossible since the right hand side of 6.3) is bounded on,t ] for every ϕ and ϕ 1. Also the second case is impossible. In fact, the separation property would imply the existence of an element ϕ,ϕ 1 ) L 2 ) H 1 ) = L 2 ) H 1 ) ) such that the left hand side of 6.3) is not a continuous function of time while equality to the right hand side shows that it is continuous. Solution of Problem 6.4 As in Problem 4.1, we extend ϕ x) to a function ϕ defined on 1,2) which is odd respect to and respect to 1. I.e. for x,1) we require ϕ x) = ϕ x), ϕx) = ϕ2 x). Then, for t,1], the unique solution of 6.7) is ϕx,t) = 1 ϕx +t) + ϕx t) ]. 6.11) 2 Let { exp 1 ϕ x) = xx 1/n) if < x < 1/n if 1/n) < x < ) Let us fix any T < 1 and consider a time t < T and a number n > n where n satisfies T + 1/n < 1. Then we have ϕx,t) = if x < T + 1/n < 1. So we have also ϕ x 1,t) for t,t + 1/n] in spite of the fact that ϕ x). If ϕ x 1,t) for every t,1] then we have also D x ϕ1 + t) = D x ϕ1 t) and so D x ϕx) is odd respect to 1; but, it is also even, since it is the derivative of an odd function. Hence it is zero. Solution of Problem 6.5 We denote ψ the solution with initial conditions ψx,) = and ψ x,) = ψ 1 x) possibly not zero. An example is easily constructed using the solution ϕ found in Problem 6.4. Let ψx,t) = ϕ x,t). Then ψ1, t) = for t, 1 1/n) and ψ solves the string equation on, 1) with conditions ψx,) = ϕ x,) =, ψ x,) = ϕ xx x,). Let now ψ be a solution such that ψ x 1,t) = for t 1. We prove ψ 1 =. A solution of the problem for t,1] with initial conditions ψx,) =, ψ x,) = ψ 1 x)) is constructed as follows: we extend ψ 1 as an odd function ψx) respect to zero and respect to 1. The solution is
8 68 Problems of Chapter 6 ψx,t) = 1 2 x+t x t ψs)ds. If ψ x 1,t) = then we have ψ1 +t) ψ1 t) = and so ψ is both even and odd respect to 1. Hence it is zero. Solution of Problem 6.6 Let the initial conditions be Then, ϕx,y,t) = 2 π ϕx,y,) = ϕ x,y), ϕ x,y,) =. n,m=1 ) ϕ n,m cos n 2 + m 2 t sinnxsinmy. The coefficients ϕ n,m are the Fourier coefficients of ϕ x,y). We recall: and we note that: γ 1 ϕx,,t) = ϕ 2 H 1 Q) = 2 π n,m=1 So, the inverse inequality would be m n 2 + m 2) ϕn,m 2 n,m=1 T n,m=1 m 2 ϕ 2 n,m = π 4 m 2 ϕ 2 n,m { n,m=1 n 2 + m 2) ϕ 2 n,m 6.13) ) mϕ n,m )cos n 2 + m 2 t sinnx ) ) cos n m 2 t dt T + sin2t n 2 + m 2 2 n 2 + m 2 }. ] π sin 2 nx ) dx = It is clear that this inequality cannot hold for every sequence {ϕ nm } for which the series 6.13) converges. For example, for every fixed M let us consider the sequence {ϕ n,m } such that ϕ M 2,M = 1, ϕ n,m = otherwise. Then the inverse inequality would imply the existence of m > and T > such that the following inequality holds for every M: m M 4 + M 2) T + 1)π M 2, 4
9 Problems of Chapter 6 69 This is clearly impossible. A second example is as follows: let ) ϕ n,m x,y,t) = sin t n 2 + m 2 sinnxsinmy. This is a solution which corresponds to the initial conditions ϕ x,y) =, ϕ 1 x,y) = n 2 + m 2 sinnxsinmy. Substituting in the inverse inequality we find the following inequality, which should hold for every n and every m: m n 2 + m 2) π π T π ) sin 2 nxsin 2 mydxdy n 2 sin 2 t n 2 + m 2 sin 2 mydy. The integrals can be easily computed and it is to see that the inequality can hold only if m =. Hence, the system is not controllable whathever the time T, when the active part of the boundary is one side of the square. Solution of Problem 6.7 We use formula 6.1) that we rewrite as s ] ϕt) = R + t)ϕ +A 1 R t)ϕ 1 +A 1 R t s) hϕs) + Ks r)ϕr)dr ds. It follows from Theorem 6.2 that ϕ H 1) = doma, ϕ 1 L 2 ) so that ϕ C,T ];L 2 )) the proof is similar to the proof of Theorem??, where the condition ϕ doma and ϕ 1 doma are used to compute a second derivative, not needed here). So we have s ] ϕ t) = R t)a ϕ + R + t)ϕ 1 + R + t s) hϕs) + Ks r)ϕr)dr ds. Hence, ] ϕ t) M ϕ H 1 ) + ϕ 1 L 2 ) + ϕ C,T ];L 2 ). The required inequality follows from 6.6). Solution of Problem 6.8 The solution ϕ solves the Volterra integral equation ϕt) = R + t)ϕ + A 1 R t)ϕ 1 + A 1 s R t s) Ks r)ϕr)dr ds. Gronwall inequality shows boundedness of K. A sufficient condition for equicontinuity is boundedness of the set of the derivatives. Computing the derivative we see that s ϕ t) = R t)a ϕ + R + t)ϕ 1 + R + t s) Ks r)ϕr)dr ds.
10 7 Problems of Chapter 6 Hence the set of the derivatives of the elements of K is bounded in L 2 ). Solution of Problem 6.9 We must exibit a sequence of functions in C,T ];H), with dimh =, which is bounded and equicontinuous, but which does not have convergent subsequences. Let H be any Hilbert space and let {e n } be an orthonormal basis. We consider the constant functions ϕ n t) e n. The sequence {ϕ n } is bounded every element has norm 1) and it is equicontinuous since every element is constan). But, for every n and m we have So, there is no convergent subsequence. ϕ n ϕ m C,T ];H) = 2. Solution of Problem 6.1 Let {ϕ n,ϕ 1n )} be a sequence in H T which is convergent to ϕ,ϕ 1 ) H 1 ) L2 ). We must prove that it converges to an element of H T. Let ϕ n be the solution which correspond to the initial conditions ϕ n,ϕ 1n ) and ϕ be the one whose initial data are ϕ,ϕ 1 ). The direct inequality shows that ϕ,ϕ 1 ) γ 1 ϕ is continuous, and the norm is continuous too. Hence γ 1 ϕ 2 L 2 G T ) = lim γ 1ϕ n 2 L 2 G T ) M. This proves that H T is closed. The set H T is not bouded in general, without the assumption of controllability. For example, in the case of Problem 6.4 with T < 1 it contains every multiple of ϕ,). The set is bounded if the system is controllable at time T with controls acting on Γ, thanks to the inverse inequality. So, the set H T is bounded if the system is exactly controllable at time T. These facts hold both for the system with and without memory. Solution of Problem 6.11 Eq. 6.8) is a generalized) telegraph equation, ϕ = ϕ + αϕ Multiply both the sides with ϕ and integrate on. Using ϕ = on we get 1 d 2 dt Integrate both the sides on,t ] and get Hence ϕ t) 2 + ϕt) 2] dx = ϕ t) 2 + ϕt) 2] dx = α ϕ t) 2 dx. ϕ1 2 + ϕ 2] dx + α ϕ s) 2 dxds.
11 References 71 ϕ t) 2 + ϕt) 2] dx ϕ1 2 + ϕ 2] dx. We have ϕ t) 2 + ϕt) 2] dx ϕ1 2 + ϕ 2] dx. when α which is the case encountered in applications. So, in practice, the memory term introduces a dissipation. References 1. Dunford, N., Schwartz, J.T.: Linear operators part I: general theory. Wiley Classic Library Edition, John Wiley and Sons, New York 1988)
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