Problems of Chapter 2
129
Solution of Problem 2.1 Let ξ = u + w and η = u − w. Then ξ and η solve
ξ ′′ = −2ξ ,
η ′′ = −4η
ξ ′ (0) = 0 ,
η (0) = u0 − w0 ,
with initial condition
ξ (0) = u0 + w0 ,
η ′ (0) = 0 .
So we have
√
ξ (t) = (u0 + v0 ) cos 2t ,
η (t) = (u0 − w0 ) cos 2t .
We compute w(t) = (1/2)(ξ − η ) and u(t) = (1/2)(ξ + η ) and we find
[
]
[ ]
u(t)
u
= R(t) 0
w(t)
w0
where
)
 ( √
1
cos
2t
+
cos
2t
2
)
R(t) =  ( √
1
cos
2t
−
cos
2t
2
1
2
1
2
(
)
√
cos 2t − cos 2t
( √
) .
cos 2t + cos 2t
The cosine formula of trigonometry easily shows that R(t) satisfies the (operator)
cosine formula (2.9).
Solution of Problem 2.2 We compute
[
]
1
R+ (t) (R+ (τ )ϕ ) = R+ (t) (ϕ (x + τ ) + ϕ (x − τ ))
2
1
= [{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
1
1
1
R+ (τ +t)ϕ + R− (t − τ )ϕ = [ϕ (x + t + τ ) + ϕ (x − t − τ ) + ϕ (x + t − τ ) + ϕ (x − t + τ )] .
2
2
4
So, the cosine formula holds.
Expression (2.53) represents the solution of the problem
u′′ = uxx
x ∈ IR
with conditions u(x, 0) = ϕ (x) ∈ L2 (IR), u′ (x, 0) = 0. Hence, R(t) is the cosine operator generated by A in L2 (IR), where dom(A) = H 2 (IR), Au = uxx .
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 2.3 Inner product of both the sides of u′′ = ∆u with ϕn (x)
shows that
130
Problems of Chapter 2
∫
un (t) =
{
solves
u′′n
= −λn2 un ,
Ω
u(x,t)ϕn (x)dx
∫
un (0) = ∫Ω u0 (x)ϕn (x)dx = un,0 ,
u′n (0) = Ω v0 (x)ϕn (x)dx = vn,0 .
So we have
1
sin λnt ,
λn
u(t) = R+ (t)u0 + A −1 R− (t)v0 =
un (t) = un,0 cos λnt + vn,0
=
+∞
+∞
n=1
n=1
1
∑ ϕn (x) [un,0 cos λnt] + ∑ ϕn (x) λn [vn,0 sin λnt] .
(8.1)
In the special case Ω = (0, π ) we have
√
ϕn (x) =
2
sin nx
π
and we get the Fourier expansion of the solution u(x,t):
√
[
]
2 +∞
1
u(x,t) =
∑ un,0 cos nt sin nx + n vn,0 sin nt sin nx
π n=1
√
[
]
1 +∞
1
=
∑ un,0 {sin n(x + t) + sin n(x − t)} − n vn,0 {cos n(x − t) − cos n(x + t)} .
2π n=1
Solution of Problem 2.4 The formula is(the same) as (8.1), as soon as noted that the
eigenvalues are the numbers −λnm = − n2 + m2 and the eigenfunctions are
ϕnm =
2
sin nx sin my
π
where n and m are natural (strictly positive). Note that different eigenfunctions may
correspond to the same eigenvalue. For example, the eigenfunctions of −13 are
2
sin 2x sin 3y ,
π
2
sin 3x sin 2y .
π
There exist also simple eigenvalues. For example −18 has solely the (normalized)
eigenfunction
2
sin 3x sin 3y .
π
Solution of Problem 2.5 It is known from semigroup theory that the following
formula holds when A generates any C0 -semigroup and ξ ∈ C1 :
Problems of Chapter 2
∫ t
0
131
eA (t−s) ξ ′ (s)ds = ξ (t) − eA t ξ (0) + A
∫ t
0
eA (t−s) ξ (s)ds
(8.2)
(see [5, p. 107]). Note that this formula makes sense because when ξ ∈ C1 the last
integral takes values in dom A . The integration by parts follows from this formula
and the definitions of R+ (t) and R− (t) (note that R− (0) = 0).
Solution of Problem 2.6 When f = 0 and F = 0 we must study the function
u1 (t) = R+ (c0t)u0 +
1 −1
A R− (c0t)v0 = R+ (c0t)u0 +R− (c0t)ξ0 ,
c0
ξ0 =
1 −1
A v0
c0
hence with both u0 and ξ0 in dom A = dom A 2 . This proves that both R+ (c0t)u0 and
R− (c0t)ξ0 take values in dom A, are twice differentiable and solve
u′′ = Au .
When u0 = 0, v0 = 0 and f = 0 we have to study u2 (t). A new analysis is not needed
since we can write
u(t) = c0 A
∫ t
0
R− (c0 s)
1 −1
A G(t − s)ds ,
c20
the same form as u3 (t).
Solution of Problem 2.7 The function u3 (t) is clearly of class C∞ when f ∈
D(∂ Ω × (0, T )) since the computations of the derivatives in the proof of Theorem 2.2 can be iterated as many times as we whish. The argument in Problem 2.6
proves that u2 ∈ C∞ .
The function u1 (t) is of class C∞ because D(Ω ) ⊆ dom Ak for every k.
Solution of Problem 2.8
One of the equations is an ordinary differential equation: it does not introduce
any delay. Instead signals propagate with velocity 1 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) ̸= 0 for some values of x ∈ (1/3, 1/2) and this is the case if s > 1/3.
Let us consider the system on the right. The affine term 1(1/3,1/2) (x)w(t) is affected by the input f at every time t > 0 but it enteres u on the interval (1/3, 1/2).
It needs a time t > 1/4 before u affects the integral in the definition of y(t): the
hyperbolic part of the system introduces a time lag h = 1/4 from the input and the
output.
Solution of Problem 2.9 If α ≤ 2 then the time lag from the input and the outpit is
1. If α ∈ (2, 3) then the time lag is 3 − α . If α > 3 then there is no time lag.
Solution of Problem 2.10 This equation can be considered as a perturbation of the
heat equation
∫
w′ = ∆w + F ,
t
F(t) =
0
∆w(s)ds .
The semigroup representation of the solution of the heat equation is
132
Problems of Chapter 2
w(t) = eAt u0 +
∫ t
eA(t−s) F(s)ds
0
where A = ∆ with domain H 2 (Ω ) ∩ H01 (Ω ). So we have, formally,
w(t) = eAt u0 +
∫ t
eA(t−s) A
∫ s
0
w(r)dr dsds =
0
∫ t
∫
d A(t−s) s
= eAt u0 −
w(r)dr dsds =
e
0 ds
0
∫ t
∫ t
= eAt u0 −
eA(t−s) w(s)ds .
w(r)dr +
0
(8.3)
0
These formal computations lead to an integral equation of Volterra type for w(t) (in
which no unbounded operator appears). The solution of Eq. (2.54) is by definition
the unique mild solution of this Volterra integral equation.
When u0 ∈ D(Ω ) we have u0 ∈ dom Ak for every k and so eAt u0 is of class
∞
C ([0, +∞); L2 (Ω )). Hence also the solution w is of class C∞ and the integration by
parts formula (8.2) shows that the integration by parts is correct in this special case.
For every T > 0, the solution w ∈ C([0, T ]; L2 (Ω )) depends continuously on u0 ∈
2
L (Ω ) and this justifies the definition of the mild solution.
Solution of Problem 2.11
We first study the system on the time interval [0, 1]. In this case
θ (x,t) = f (t − x)H(t − x)
so that q(x,t) = 0 if x > t. Otherwise we have
q(x,t) = −
∂
∂x
=−
∫ t
0
∂
∂x
f (s − x)H(s − x)ds = −
∫ t−x
0
∂
∂x
∫ t
x
f (s − x)ds =
f (s)ds = f (t − x) = θ (x,t) .
So, the flux and the temperature coincide when 0 < x < t ≤ 1 and they cannot be
controlled to independent targets: the flux q(x, 1) and the temperature θ (x, 1) are
not independent.
Consequently, also the flux, as the temperature, belongs to C([0, T ]; L2 (0, 1)) for
T ≤ 1.
Now we consider the system on the interval of time (0, 2) and we compute the
solution. The function f (t −x)H(t −x) solves the wave equation and also the boundary condition at x = 0 also when t ∈ [0, 2] but it is not zero for x = 1 when t > 1. We
force this second boundary condition by choosing
θ (x,t) = f (t − x)H(t − x) − f (t − 2 + x)H(t − 2 + x) ,
0≤t ≤2.
Note that the second addendum is equal zero when t ∈ [0, 1] and that both the boundary conditions are satisfied for t ∈ [0, 2].
References
133
Now we compute the flux. We compute first
∫ t
0
θ (x, s)ds = H(t − x)
= H(t − x)
∫ t−x
0
∫ t
x
f (s − x)ds − H(t − 2 + x)
f (r)dr − H(t − 2 + x)
∫ t
∫ t−2+x
2−x
f (s − 2 + x)ds =
f (r)dr .
0
So,
q(x,t) = H(t − x) f (t − x) + H(t − 2 + x) f (t − 2 + x)
because
−δ (t − x)
∫ t−x
0
f (r)dr + δ (t − 2 + x)
∫ t−2+x
f (r)dr = 0
0
(compare the remarks in the solutions of Problem 1.4).
Lebesgue theorem on the continuity of the shift in L p spaces shows that q ∈
C([0, T ]; L2 (0, 1)) for every T ≤ 2.
Now we compare θ (x, 2) and q(x, 2). We have
θ (x, 2) = f (2 − x) − f (x) ,
q(x, 2) = f (2 − x) + f (x) .
We can impose θ (x, 2) = ξ (x) and q(x, 2) = η (x) (for x ∈ (0, 1)) by choosing
1
[η (t) − ξ (t)] t ∈ (0, 1) ,
2
1
f (t) = [η (2 − t) + ξ (2 − t)] t ∈ (1, 2) .
2
f (t) =
References
1. Guerrero, S., Imanuvilov, O.Y.: Remarks on non controllability of the heat equation with memory. ESAIM Control Optim. Calc. Var. 19, 288-300 (2013)
2. Halanay, A., Pandolfi, L.: Lack of controllability of the heat equation with memory. Systems
Control Lett. 61, 999-1002 (2012)
3. Halanay, A., Pandolfi, L.: Lack of controllability of thermal systems with memory. In print,
Evol. Equ. Control Theory. 3 n. 3 2014, ArXiv: http://arxiv.org/abs/1304.1386
4. Halanay, A., Pandolfi, L.: Approximate controllability and lack of controllability to zero of
the heat equation with memory, submitted, ArXiv: http://arxiv.org/abs/1404.2745
5. Pazy, A.: Semigroups of linear operators and applications to partial differential equations.
Applied Mathematical Sciences 44. Springer-Verlag, New York (1983)
134
Problems of Chapter 2