Stabilization of Heat Equation
Mythily Ramaswamy
TIFR Centre for Applicable Mathematics, Bangalore, India
CIMPA Pre-School,
I.I.T Bombay
22 June - July 4, 2015
Mythily Ramaswamy
Stabilization of Heat Equation
4th July, 2015
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contents
1
2
3
4
5
6
Introduction
Stabilization- A model
Heat Equation - Interior Control
Weak Solutions
Projected Systems
Spectral Analysis
Equivalent Projected Systems
Stabilization of unstable projected system
finite dimensional system
Unique Continuation
Stabilization of full system
Heat equation - Boundary Control
Checking of Hautus condition
Control of minimal norm
Closed loop system
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Stabilization of Heat Equation
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Introduction
Stabilization- A model
One dimensional heat equation
The heat equation in (0, L) × (0, ∞) for L > 0
yt − yxx = uχO
in
y(0, t) = y(L, t) = 0
y(x, 0) = y0 (x)
in
(0, L) × (0, ∞),
in
(0, ∞),
(0, L),
Questions : For a given y0 ∈ L2 (0, L), is there a control u with support
in a subset O of (0, L) such that the solution
y ∈ L2 (0, ∞; H01 (0, L)) ∩ C([0, ∞); L2 (0, L)) decays exponentially in
time :
ky(t)k ≤ Ke−µt ky0 k, ∀ t ≥ 0?
At what rate µ ?
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Introduction
Stabilization- A model
Recall the eigenfunctions φk and eigenvalues λk of the homogeneous
problem
r
2
kπx
sin
, ∀ k ∈ N, ∀ x ∈ (0, L).
φk (x) =
L
L
k2 π2
.
L2
is a Hilbert basis of L2 (0, L).
λk = −
The family (φk )k∈N
If A =
d2
dx2
with domain D(A) = H 2 ∩ H01 , then
φk ∈ D(A),
Aφk = λk φk .
For the homogeneous problem, the solution is
y(x, t) =
∞
X
y0k e−
k2 π 2 t
L2
φk (x),
∀ x ∈ (0, L),
t ∈ (0, T ),
k=1
The function y ∈ L2 (0, T ; H01 (0, L)) ∩ C([0, T ]; L2 (0, L)).
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Stabilization of Heat Equation
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Heat Equation - Interior Control
Weak Solutions
Weak solution of heat equation
Definition (Weak solution of heat equation)
Let y0 ∈ L2 (0, L) and f ∈ L2 (0, T ; L2 (0, L)). Then y belonging in
W (0, T ; H01 (0, L), H −1 (0, L)) = L2 (0, T ; H01 (0, L) ∩ H 1 (0, T ; H −1 (0, L)),
is called a weak solution of the heat equation
yt − yxx = f
in
(0, L) × (0, T ),
y(0, t) = y(L, t) = 0
y(x, 0) = y0 (x)
in
in (0, T ),
(0, L),
if y(·, 0) = y0 (·) and for all φ ∈ H01 (0, L), y satisfies
d
dt
Mythily Ramaswamy
Z
L
Z
y(t)φ = −
0
L
Z
yx (t)φx +
0
Stabilization of Heat Equation
L
f (t)φ.
0
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Heat Equation - Interior Control
Weak Solutions
Theorem (Existence and uniqueness of the weak solution)
Let y0 ∈ L2 (0, L) and f ∈ L2 (0, T ; L2 (0, L)). The heat equation with
forcing term f and initial condition y0 , admits a unique weak solution y
and it satisfies
kykL∞ (0,T ;L2 (0,L)) +kykL2 (0,T ;H01 (0,L)) ≤ C(ky0 kL2 (0,L) +kf kL2 (0,T ;L2 (0,L)) ),
for some constant C > 0.
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Projected Systems
Spectral Analysis
We can always choose µ where µ is strictly between two eigenvalues.
Let us assume
0 ≤ −λ1 < · · · < −λN < µ < −λN +1 < · · · .
Define
E + = ⊕N
k=1 E(λk ),
E − = ⊕∞
k=N +1 E(λk ),
where E(λk ) = Ker(λk I + A), the eigenspace associated to λk , for
all k ∈ N.
We see that
L2 (0, L) = E + ⊕ E − .
Let PN be the orthogonal projection of L2 (0, L) onto E + , defined by
PN f =
N
X
(f, φk )L2 (0,L) φk ,
∀ f ∈ L2 (0, L).
k=1
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Projected Systems
Equivalent Projected Systems
Theorem
Let y be the solution of the heat equation with control u. Then PN y
satisfies
∂t PN y − ∂xx PN y = PN (uχO )
in
(0, L) × (0, ∞),
in (0, ∞),
PN y(0, t) = PN y(L, t) = 0
PN y(·, 0) = PN y0 (·)
in
(0, L),
and (I − PN )y satisfies
∂t (I − PN )y − ∂xx (I − PN )y = (I − PN )(uχO )
(I − PN )y(0, t) = (I − PN )y(L, t) = 0
(I − PN )y(·, 0) = (I − PN )y0 (·)
Mythily Ramaswamy
in
in
in
(0, L) × (0, ∞),
(0, ∞),
(0, L).
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Projected Systems
Equivalent Projected Systems
Proof
Let y be a solution of the heat equation.
We show that PN y satisfies the weak formulation
d
dt
Z
L
Z
PN y(t)φ = −
0
L
Z
PN yx (t)φx +
0
for all φ ∈ E + and so φ =
L
PN (uχO )φ,
0
PN
k=1 (φ, φk )L2 (0,L) φk .
L2 (0, L) is the orthogonal
It follows from the fact that
sum Of E +
and E − , and definition of PN .
Z
Z
d L
d L
y(t)φ =
PN y(t)φ,
In fact, we show
dt 0
dt 0
Z L
Z L
Z L
Z L
yx (t)φx =
PN yx (t)φx and
uχO φ =
PN (uχO )φ, for
0
0
0
0
P
+
all φ = N
k=1 (φ, φk )L2 (0,L) φk ∈ E .
Similarly, we can show for (I − PN )y.
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Stabilization of unstable projected system
finite dimensional system
The system for PN y is in E + for all t ≥ 0 and E + is the finite
dimensional space.
We study the controllability of the finite dimensional system for PN y.
Set finite dimensional control
u=
N
X
vk φk ∈ L2 (0, T ; E + ),
k=1
for some time T > 0.
Note that
P
PN (uχO ) = PN ( N
k=1 vk φk χO )
N X
N
Z
X
=
vj
φj φk φk .
j=1 k=1
Mythily Ramaswamy
O
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Stabilization of unstable projected system
finite dimensional system
We introduce B ∈ L(RN , E − ) such as
N X
N
Z
X
φj φk φk ,
Bv =
vj
O
j=1 k=1
for all v = (v1 , · · · , vN )T ∈ RN .
For this u, we write
∂t PN y − ∂xx PN y = Bv
PN y(0, t) = PN y(L, t) = 0
PN y(·, 0) = PN y0 (·)
(0, L) × (0, ∞),
in
in
in (0, ∞),
(0, L),
Setting PN y = yN , A+ = ∆PN ∈ L(E + ) and PN y0 = y0,N , the
above system can be written in E +
0
yN
= A+ yN + Bv,
yN (0) = y0,N ∈ E + .
Now we
want to write an equivalent system in RN . Let
PN
yN = k=1 yk φk ∈ E + and z = (y1 , · · · , yN )T ∈ RN .
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Stabilization of unstable projected system
finite dimensional system
We write
z 0 = Λz + Cv,
z(0) = z0 ,
where
Λ = diag(λ1 , · · · , λN ) ∈ RN × RN ,
Z
φj φk
C=
∈ RN × RN ,
O
1≤j≤N,1≤k≤N
z0 = (y0,N , φ1 )L2 (O) , · · · , (y0,N , φN )L2 (O)
Mythily Ramaswamy
Stabilization of Heat Equation
T
∈ RN .
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Stabilization of unstable projected system
Unique Continuation
Theorem
The matrix C is invertible if and only if the family {φk |O }1≤k≤N is linearly
independent in L2 (O).
Proof Since the matrix C is the Gram matrix in L2 (O) of the family
{φk |O }k∈N , the result follows.
Theorem
The family {φk |O }1≤k≤N is linearly independent in L2 (O).
Aim To show if
N
X
αk φk |O = 0 in L2 (O), then αk = 0,
∀1≤k≤N
k=1
and hence {φk |O }1≤k≤N is linearly independent in L2 (O).
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Stabilization of Heat Equation
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Stabilization of unstable projected system
Unique Continuation
Proof
Let
N
X
αk φk |O = 0 in L2 (O).
k=1
Denote f =
N
X
αk φk and we have f |O = 0.
k=1
Suppose f is in the form f = αk φk for some k ∈ {1, 2, · · · , N }.
q
Since φk (x) = L2 sin kπx
for all x ∈ (0, L), using this expression
L
we conclude that f = 0 on (0, L), if it is given that f |O = 0.
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Stabilization of unstable projected system
Now suppose, f =
N
X
Unique Continuation
αk φk .
k=1
Then we reduce it to the first case as follows. We have
[λ1 I + ∂xx ]f =
N
X
(λ1 − λk )αk φk ,
k=2
using[λ1 I + ∂xx ]φk = (λ1 − λk )φk . Thus we eliminate φ1 from the
right hand side.
Note that (λj − λk ) 6= 0, for j 6= k.
Similarly, to eliminate φ1 and φ2 ,
[λ2 I + ∂xx ][λ1 I + ∂xx ]f =
N
X
(λ2 − λk )(λ1 − λk )αk φk .
k=3
Repeating the above method finitely many times, we can reduce it in
the first case and derive the result.
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Stabilization of unstable projected system
Unique Continuation
Theorem
The system
z 0 = Λz + Cv,
z(0) = z0 ,
is exactly controllable at time T , for all T > 0.
Proof
By the above two theorems, we have C is invertible.
By checking Kalman rank condition, we show the system is
controllable.
From the above results, it follows
Theorem
For all y0,N ∈ E + , there exists a v ∈ L2 (0, T ; RN ) such that yN , the
solution of the finite dimensional system satisfies yN (T ) = 0 and control v
obeys
kv(t)kRN ≤ Cky0,N kL2 (0,L) ,
for some constant C > 0.
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Stabilization of full system
Theorem
For any µ > 0, there exists a constant K > 0 such that for all initial
condition y0 ∈ L2 (0, L), there exists a control u ∈ L2 (0, ∞; L2 (0, L)) such
that yy0 ,u , the solution of the heat equation satisfies
kyy0 ,u (t)kL2 (0,L) ≤ Ke−µt ky0 kL2 (0,L) , ∀ t ≥ 0.
Proof
Set control
( PN
k=1 vk (t)φk ,
u(t) =
0,
0 ≤ t ≤ T,
t > T,
where vk1≤k≤N is obtained from the previous theorem such that
PN y(T ) = yN (T ) = 0.
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Stabilization of full system
We have
kv(t)kRN ≤ Cky0,N kL2 (0,L) ,
where v = (v1 , · · · , vN )T .
We also have kPN y(T )kL2 (0,L) ≤ CkPN y0 kL2 (0,L) , for all 0 ≤ t ≤ T
and PN y(t) = 0 for all t ≥ T .
It yields
kPN y(t)kL2 (0,L) ≤ Ce−λN +1 t eλN +1 T kPN y0 kL2 (0,L) ,
Mythily Ramaswamy
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∀ t ≥ 0.
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Stabilization of full system
Now to estimate the stable part for (A− , D(A− )) where
D(A− ) = E − ∩ D(A) and A− y = (I − PN )yx x for all y ∈ D(A− ),
we write
−
(I − PN )y(t) = eA t (I − PN )y0 +
Z t
N
X
−
e(t−s)A (I − PN )
vk (s)φk |O ds.
0
k=1
−
Using keA t (I − PN )y0 kL2 (0,L) ≤ e−λN +1 t kPN y0 kL2 (0,L) and the
above estimates, we can show
k(I − PN )y(t)kL2 (0,L) ≤ Ke−µt ky0 kL2 (0,L) , ∀ t ≥ 0.
Combining the estimates for PN y and (I − PN )y, the theorem follows.
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Heat equation - Boundary Control
The heat equation in (0, 1) × (0, ∞) with a boundary control
yt − yxx = 0
y(0, t) = 0,
in
(0, 1) × (0, ∞),
y(1, t) = u(t)
y(x, 0) = y0 (x)
in
in
(0, ∞),
(0, 1),
where for all t ∈ (0, ∞), u(t) ∈ R, one dimensional control.
We want to write the above equation in a Hilbert space Y as
y 0 = Ay + Bu,
y(0) = y0 ∈ Y.
B is the control operator and U is a Hilbert space for controls.
Consider Y = L2 (0, 1) and U = R, one dimensional space.
d2
2
1
Recall A = dx
2 with domain D(A) = H (0, 1) ∩ H0 (0, 1) and
∗
∗
D(A) = D(A ) with A = A .
φk is an eigenfunction of A for eigenvalue λk , where
√
φk (x) = 2sin (kπx) , ∀ k ∈ N, ∀ x ∈ (0, 1).
λk = −k 2 π 2 .
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Heat equation - Boundary Control
Expression of B ∗
Aim To determine B and B ∗ by the weak formulation.
y is a weak solution of the heat equation with nonhomogeneous
boundary condition if and only if for all
ψ ∈ D(A∗ ) = H 2 (0, 1) ∩ H01 (0, 1),
d
dt
Z
1
Z
0
1
y(x, t)ψxx (x)dx − u(t)ψx (1).
y(x, t)ψ(x)dx =
0
y is a weak solution of the evolution equation if and only if for all
ψ ∈ D(A∗ ) = H 2 (0, 1) ∩ H01 (0, 1),
d
dt
Z
1
Z
y(x, t)ψ(x)dx =
0
Mythily Ramaswamy
1
y(x, t)Aψ(x)dx + (Bu(t), ψ)L2 (0,1) .
0
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Heat equation - Boundary Control
Comparing above two identities, we obtain
(Bu(t), ψ)L2 (0,1) = −u(t)ψx (1),
t ∈ (0, ∞)
Thus for all t ∈ (0, ∞), (u(t), B ∗ ψ)R = −u(t)ψx (1) and hence
B ∗ ψ = −ψx (1).
Hautus condition for stabilization:
With a prescribed decay ω > 0, (A + ωI, B) is stabilizable if and only
if for all unstable eigenvalues λk + ω > 0,
A∗ φ = λk φ,
B∗φ = 0
implies φ = 0.
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Stabilization of Heat Equation
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Heat equation - Boundary Control
Let choose ω = 10.
The only unstable eigenvalue of A + ωI is λ1 + ω = −π 2 + 10 > 0.
Unstable space spanned by φ1 , eigenfunction of (A + ωI) for
eigenvalue (λ1 + ω) and
E + = Rφ1 ,
E − = ⊕∞
k=2 Rφk ,
where E − is the stable eigenspace.
Let us define the projectors associated with this decomposition
P1 f = (f, φ1 )L2 (0,1) φ1 ,
(I − P1 )f =
∞
X
(f, φk )L2 (0,1) φk .
k=2
Thus
P1 Bu = (Bu, φ1 )L2 (0,1) φ1 = (u, B ∗ φ1 )R φ1 = −uφ01 (1)φ1 ,
∞
∞
X
X
(I − P1 )Bu =
(Bu, φk )L2 (0,1) φk = −u
φ0k (1)φk .
k=2
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Stabilization of Heat Equation
k=2
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Heat equation - Boundary Control
Checking of Hautus condition
Using the expression of φk , we have
√
φ0k (1) = 2πk(−1)k .
P
0
∗ 0
2
The series ∞
k=2 φk (1)φk converges in (D(A )) , but not in L (0, 1).
Checking of Hautus condition
Recall that only λ1 + ω > 0. Let A∗ φ = λ1 φ and B ∗ φ = 0.
Using A∗ = A, φ = cφ1 , for all c ∈ R.
√
Since B ∗ φ = 0, we get cφ0 (1) = − 2cπ = 0.
Hence c = 0 and so φ = 0.
Hautus condition is satisfied and (A + ωI, B) is stabilizable.
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Heat equation - Boundary Control
Control of minimal norm
Control of minimal norm
To determine control of minimal norm, we project the equation
y 0 = (A + ω)y + Bu onto E + .
Let P1 y = y1 φ1 . The equation for y1 is
√
y10 = (10 − π 2 )y1 − u(t)φ0 (1) = (10 − π 2 )y1 − u(t) 2π,
y1 (0) = (y0 , φ1 )L2 (0,1) .
The Bernoulli equation for the system is
p > 0,
√
2(10 − π 2 )p − (−π 2)2 p2 = 0.
Thus
p=
Mythily Ramaswamy
2(10 − π 2 )
.
2π 2
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Heat equation - Boundary Control
Control of minimal norm
The control of minimal norm obeys the feedback law
√ 2(10 − π 2 )
u(t) = −(−π 2)
y1 (t).
2π 2
y1 (·) satisfies the closed loop system
y10 = −(10 − π 2 )y1 ,
y1 (0) = (y0 , φ1 )L2 (0,1) .
Hence y1 (t) = (y0 , φ1 )L2 (0,1) e−(10−π
system is stable.
Mythily Ramaswamy
2 )t
, for all t ≥ 0 and so this
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Heat equation - Boundary Control
Closed loop system
The closed loop system for ŷ = e10t y is
ŷt − ŷxx = 0
ŷ(0, t) = 0,
(0, 1) × (0, ∞),
√
2)
(ŷ(t), φ1 )L2 (0,1)
ŷ(1, t) = (π 2) 2(10−π
2π 2
in
ŷ(x, 0) = y0 (x)
in
in
(0, ∞),
(0, 1).
Since the above system is stabilizable, the closed loop system for y is
yt − yxx = 0
y(0, t) = 0,
(0, 1) × (0, ∞),
√
2)
y(1, t) = (π 2) 2(10−π
(y(t), φ1 )L2 (0,1)
2π 2
in
y(x, 0) = y0 (x)
in
in
(0, ∞),
(0, 1),
and y satisfies
ky(t)kL2 (0,1) ≤ Ce−10t ky0 kL2 (0,1) ,
for some constant C > 0.
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Heat equation - Boundary Control
Closed loop system
A. Bensoussan, G. Da Prato, M. Delfour, S. K. Mitter, Representation
and control of infinite dimensional systems. Second edition. Systems
and Control: Foundations & Applications. Birkhäuser Boston, Inc.,
Boston, MA, 2007.
Lasiecka and Triggiani, Differential and Algebraic Riccati Equations
with Applications to Boundary/Point control Problems,
Springer-Verlag, 1991
J-L Lions, Optimal Control of systems governed by Partial Differential
Equations, Springer, 1971.
Jean-Pierre Raymond, Optimal Control of PDEs, FICUS Course
Notes, 2001.
Jean-Pierre Raymond, Optimal control and stabilisation of flow related
models, FICUS Course Notes, 2001.
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