The Model and Preliminaries Problems and Solutions

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1 Chapter The Model and Preliminaries Problems and Solutions Facts that are recalled in the problems The wave equation u = c u + G(x,t), { u(x,) = u (x), u (x,) = v (x), u(x,t) = f (x,t) x Γ, u(x,t) = x Ω \Γ (.) and its solutions u(t) = R + (c t)u + A R (c t)v + A R (c (t s))g(s)ds c c c A R (c (t s))d f (s)ds, (.) u (t) = c A R (c t)u + R + (c t)v + R + (c (t s))g(s)ds c A R + (c (t s))d f (s)ds. (.3) The equations with persistent memory: w (x,t) = cw (x,t) + c w(x,t) + M(t s) w(x,s)ds + F(x,t), (.4) w (x,t) = cw(x,t) + N(t s) w(x,s)ds + H(x,t). (.5) Eq. (.5) is the HEAT EQUATION WITH MEMORY. In particular, when Ω is an interval, we get: θ = N(t s)θ xx (s)ds + H(t), θ(,t) = f (t), θ(π,t) =. (.6) This equation is obtained when the Fourier law for the flux q(x,t) is replaced by

2 Problems of Chapter The cosine formula: q(x,t) = k N(t s)θ x (x,s)ds, (.7) R + (t)r + (τ) = (R +(t) + R + (τ)) t,τ IR. (.8) Fourier expansion for the cosine operator R + (t) ( n= We introduce also the operators R (t) = Integration by parts: α n ϕ n ) = n= (α n cosλ n t)ϕ n (x). (.9) [ e A t e A t], S(t) = A R (t), t IR. R + (t s)u (s)ds = = u(t) R + (t)u() + c A Finally, we recall: R (t s)u (s)ds = = R (t)u() + c A R (t s)u(s)ds, (.) R + (t s)u(s)ds. (.) Theorem.. Let u doma, v doma, G D(Ω (,T )) and f D(Γ (,T )). Then, the function u(t) in (.) is a regular solution of Eq. (4.) and the following equality holds for every t: u (t) = c A(u(t) D f (t)) + G(t). (.) The Problems.. Consider the following system of two ordinary differential equations: u = 3u + w, w = 3w + u. Let the initial conditions be u() = u, u () =, w() = w, w () =. Represent the solution as the operator S(t) called SINE OPERATOR.

3 Problems of Chapter 3 ( ) ( ) u(t) u = R(t). w(t) w Prove that R(t) is a cosine operator in the sense that it verifies equality (.8)... Show that the family of the operators R + (t) defined by (R + (t)ϕ)(x) = (ϕ(x +t) + ϕ(x t)) (.3) is a cosine operator in L (IR), in the sense that equality (.8) holds for this family of operators. Decide whether it is possible to represent this cosine operator with a Fourier type expansion, as in (.9)..3. Show that the series representation (.9) for the cosine operator can be obtained by separation of variables. Write it explicitly in the case Ω = (,π)..4. Write the series (.9) which represents the cosine operator of the problem u = u xx + u yy (x,y) Q = (,π) (,π), u = on Q ( Q is not of class C but the results we have seen extend to this case)..5. Prove the integration by parts formulas (.) and (.)..6. Prove Theorem. when f =, G = but the initial conditions are not zero; and do the same when control and initial conditions are zero, but G..7. Prove that under the condition of Theorem., and if furthermore u and v belong to D(Ω) then the function u(t) is of class C ([,T ];L (Ω))..8. Let x (,π) and u = u(x,t) while w = w(t) depends only on the time. Assuming zero initial conditions, discuss the time at which the effect of the input f affects the output y in the case of the systems w = w + f (t) u = u xx, u(,t) = f (t), u(π,t) = w = w + / /3 u(x,s)dx y(t) = w(t), u tt = u xx + (/3,/) (x)w(t) u(,t) = u(π,t) = y(t) = 3/4 u(x,t)dt The function (/3,/) (x) is the characteristic function of (/3,/)..9. The notations u and w are as in Problem.8 while x (,7). Let y(t) = w(t) IR be the output of the following system (α > is a real parameter): 6 u = u xx + (,α) (x) f (t), w = w + u(x, t)dx 3 (u = u(x, t) with x (, 7)) and conditions

4 4 Problems of Chapter { u(,t) =, u(7,t) = f (t), u(x,) =, u (x,) =, w() =. Study at what time the effect of the external input f (t) will influence the observation y(t) and specify whether the time depends on α... On a region Ω, consider the problem w = w + w(s)ds, w() = u L (Ω), w = on Ω. (.4) This equation is not of the same type as those studied in this paper (it is called a Jeffrey, or a Colemann-Gurtin model. Of course in general the memory kernel is not a constant). Its solution cannot be obtained using cosine operators. Prove that Eq. (.4) can be reduced to a Volterra integral equation using the semigroup e At. It is a fact that the controllability properties of Eq. (.4) are very different from those of Eq. (.5), see [,, 3, 4]... Let Ω = (,) and consider the heat equation with memory (.6) with H = and N(t) (hence, an integrated version of the string equation). Assume zero initial condition and θ(,t) = f (t), θ(,t) =. Use (.7) with initial time t = and k =. Compute the flux q(t) on t (,) and show that the function t q(t) belongs to C([,];L (,)). Let either T = or T =. Study whether the pair (θ(t ),q(t )) can be controlled to hit any target (ξ,η) L (,) L (,) using a control f L (,T ). The Solutions Solution of Problem. Let ξ = u + w and η = u w. Then ξ and η solve ξ = ξ, η = 4η with initial condition So we have ξ () = u + w, ξ () =, η() = u w, η () =. ξ (t) = (u + v )cos t, η(t) = (u w )cost. We compute w(t) = (/)(ξ η) and u(t) = (/)(ξ + η) and we find [ ] [ ] u(t) u = R(t) w(t) where w

5 Problems of Chapter 5 ( R(t) = cos ) ( t + cost cos ) t cost (cos ) ( t cost cos ). t + cost The cosine formula of trigonometry easily shows that R(t) satisfies the (operator) cosine formula (.9). Solution of Problem. We compute [ ] R + (t)(r + (τ)ϕ) = R + (t) (ϕ(x + τ) + ϕ(x τ)) = [{ϕ(x +t + τ) + ϕ(x t + τ)} + {ϕ(x τ +t) + ϕ(x τ t)}]. 4 It is easily seen that this is equal to R +(τ +t)ϕ + R (t τ)ϕ = [ϕ(x +t + τ) + ϕ(x t τ) + ϕ(x +t τ) + ϕ(x t + τ)]. 4 So, the cosine formula holds. Expression (.3) represents the solution of the problem u = u xx x IR with conditions u(x,) = ϕ(x) L (IR), u (x,) =. Hence, R(t) is the cosine operator generated by A in L (IR), where dom(a) = H (IR), Au = u xx. It is easy to see that the operator A does not have any eigenvalue so that R + (t) cannot be represented using a Fourier expansion. Solution of Problem.3 Inner product of both the sides of u = u with ϕ n (x) shows that u n (t) = u(x,t)ϕ n (x)dx Ω solves u n = λn u n, { un () = Ω u (x)ϕ n (x)dx = u n,, u n() = Ω v (x)ϕ n (x)dx = v n,. So we have u n (t) = u n, cosλ n t + v n, λ n sinλ n t, u(t) = R + (t)u + A R (t)v = = n= In the special case Ω = (,π) we have ϕ n (x)[u n, cosλ n t] + ϕ n (x) [v n, sinλ n t]. (.5) λ n ϕ n (x) = n= π sinnx

6 6 Problems of Chapter and we get the Fourier expansion of the solution u(x,t): [ u(x, t) = π u n, cosnt sinnx + ] n= n v n, sinnt sinnx [ = π u n, {sinn(x +t) + sinn(x t)} ] n v n, {cosn(x t) cosn(x +t)}. n= Solution of Problem.4 The formula is the same as (.5), as soon as noted that the eigenvalues are the numbers λ nm = ( n + m ) and the eigenfunctions are ϕ nm = π sinnxsinmy where n and m are natural (strictly positive). Note that different eigenfunctions may correspond to the same eigenvalue. For example, the eigenfunctions of 3 are π sinxsin3y, π sin3xsiny. There exist also simple eigenvalues. For example 8 has solely the (normalized) eigenfunction π sin3xsin3y. Solution of Problem.5 It is known from semigroup theory that the following formula holds when A generates any C -semigroup and ξ C : e A (t s) ξ (s)ds = ξ (t) e A t ξ () + A e A (t s) ξ (s)ds (.6) (see [5, p. 7]). Note that this formula makes sense because when ξ C the last integral takes values in doma. The integration by parts follows from this formula and the definitions of R + (t) and R (t) (note that R () = ). Solution of Problem.6 When f = and F = we must study the function u (t) = R + (c t)u + c A R (c t)v = R + (c t)u +R (c t)ξ, ξ = c A v hence with both u and ξ in doma = doma. This proves that both R + (c t)u and R (c t)ξ take values in doma, are twice differentiable and solve u = Au. When u =, v = and f = we have to study u (t). A new analysis is not needed since we can write

7 Problems of Chapter 7 the same form as u 3 (t). u(t) = c A R (c s) c A G(t s)ds, Solution of Problem.7 The function u 3 (t) is clearly of class C when f D( Ω (,T )) since the computations of the derivatives in the proof of Theorem. can be iterated as many times as we whish. The argument in Problem.6 proves that u C. The function u (t) is of class C because D(Ω) doma k for every k. Solution of Problem.8 One of the equations is an ordinary differential equation: it does not introduce any delay. Instead signals propagate with velocity in the space variable x. In the case of the system on the left, y(t) will be affected by the input y as soon as we have u(x,s) for some values of x (/3,/) and this is the case if s > /3. Let us consider the system on the right. The affine term (/3,/) (x)w(t) is affected by the input f at every time t > but it enteres u on the interval (/3,/). It needs a time t > /4 before u affects the integral in the definition of y(t): the hyperbolic part of the system introduces a time lag h = /4 from the input and the output. Solution of Problem.9 If α then the time lag from the input and the outpit is. If α (,3) then the time lag is 3 α. If α > 3 then there is no time lag. Solution of Problem. This equation can be considered as a perturbation of the heat equation w = w + F, F(t) = w(s)ds. The semigroup representation of the solution of the heat equation is w(t) = e At u + e A(t s) F(s)ds where A = with domain H (Ω) H (Ω). So we have, formally, w(t) = e At u + = e At u = e At u s e A(t s) A d ds ea(t s) w(r)dr + s w(r)dr dsds = w(r)dr dsds = e A(t s) w(s)ds. (.7) These formal computations lead to an integral equation of Volterra type for w(t) (in which no unbounded operator appears). The solution of Eq. (.4) is by definition the unique mild solution of this Volterra integral equation. When u D(Ω) we have u doma k for every k and so e At u is of class C ([,);L (Ω)). Hence also the solution w is of class C and the integration

8 8 Problems of Chapter by parts formula (.6) shows that the integration by parts is correct in this special case. For every T >, the solution w C([,T ];L (Ω)) depends continuously on u L (Ω) and this justifies the definition of the mild solution. Solution of Problem. We first study the system on the time interval [,]. In this case θ(x,t) = f (t x)h(t x) so that q(x,t) = if x > t. Otherwise we have q(x,t) = x = x f (s x)h(s x)ds = x x f (s)ds = f (t x) = θ(x,t). x f (s x)ds = So, the flux and the temperature coincide when < x < t and they cannot be controlled to independent targets: the flux q(x,) and the temperature θ(x,) are not independent. Consequently, also the flux, as the temperature, belongs to C([,T ];L (,)) for T. Now we consider the system on the interval of time (,) and we compute the solution. The function f (t x)h(t x) solves the wave equation and also the boundary condition at x = also when t [,] but it is not zero for x = when t >. We force this second boundary condition by choosing θ(x,t) = f (t x)h(t x) f (t + x)h(t + x), t. Note that the second addendum is equal zero when t [,] and that both the boundary conditions are satisfied for t [,]. Now we compute the flux. We compute first So, θ(x,s)ds = H(t x) = H(t x) x x f (s x)ds H(t + x) f (r)dr H(t + x) +x f (r)dr. x q(x,t) = H(t x) f (t x) + H(t + x) f (t + x) f (s + x)ds = because x +x δ(t x) f (r)dr + δ(t + x) f (r)dr = (compare the remarks in the solutions of Problem.4). Lebesgue theorem on the continuity of the shift in L p spaces shows that q C([,T ];L (,)) for every T.

9 References 9 Now we compare θ(x,) and q(x,). We have θ(x,) = f ( x) f (x), q(x,) = f ( x) + f (x). We can impose θ(x, ) = ξ (x) and q(x, ) = η(x) (for x (, )) by choosing f (t) = [η(t) ξ (t)] t (,), f (t) = [η( t) + ξ ( t)] t (,). References. Guerrero, S., Imanuvilov, O.Y.: Remarks on non controllability of the heat equation with memory. ESAIM Control Optim. Calc. Var. 9, 88-3 (3). Halanay, A., Pandolfi, L.: Lack of controllability of the heat equation with memory. Systems Control Lett. 6, 999- () 3. Halanay, A., Pandolfi, L.: Lack of controllability of thermal systems with memory. In print, Evol. Equ. Control Theory. 3 n. 3 4, ArXiv: 4. Halanay, A., Pandolfi, L.: Approximate controllability and lack of controllability to zero of the heat equation with memory, submitted, ArXiv: 5. Pazy, A.: Semigroups of linear operators and applications to partial differential equations. Applied Mathematical Sciences 44. Springer-Verlag, New York (983)

10 Problems of Chapter