The finite time turnpike phenomenon for optimal control problems

The ﬁnite-time turnpike phenomenon for optimal
control problems
Martin Gugat martin.gugat@fau.de
Chair of Applied Analysis (Alexander von Humboldt-Professur)
Friedrich-Alexander-Universität Erlangen-Nürnberg (FAU)

Outline
The Turnpike Phenomenon: What is it?
An example with L1
-tracking term
The finite-time turnpike phenomenon
An example with squared L2
-tracking term
The classical turnpike phenomenon
Literature about the turnpike phenomenon
Optimal L1
-Dirichlet control of a linear hyperbolic 2x2 system
Application: Gas transport through pipelines
Optimal Dirichlet L1
-control of the wave equation:
A finite horizon example
Optimal Neumann control of the wave equation:
Conclusion
Martin Gugat · FAU · The finite-time turnpike phenomenon for optimal control problems 17.10.2019 2

Inhalt
An example with L1
-tracking term
-tracking term
Optimal L1
Conclusion

The Turnpike Phenomenon
Wikipedia: The
New Jersey Turnpike
Creative-Commons-Lizenz

Inhalt
An example with L1
-tracking term
-tracking term
Optimal L1
Conclusion

An example with L1-tracking term
For T ≥ 1 and γ > 0 we consider the problem
(OC)T



min
u∈L2(0,T)
T
0
1
2|u(t)|2
+ γ |y(t)| dt subject to
y(0) = −1,
y (t) = y(t) + u(t).

(OC)T



min
u∈L2(0,T)
T
0
1
2|u(t)|2
y(0) = −1,
y (t) = y(t) + u(t).
The problem without initial conditions is
(S)



min
u∈L2(0,T)
T
0
1
2|u(t)|2
y (t) = y(t) + u(t).

(OC)T



min
u∈L2(0,T)
T
0
1
2|u(t)|2
y(0) = −1,
y (t) = y(t) + u(t).
The problem without initial conditions is
(S)



min
u∈L2(0,T)
T
0
1
2|u(t)|2
y (t) = y(t) + u(t).
The solution of (S) is the turnpike! Here the turnpike is zero, y(σ)
= 0 and u(σ)
= 0.

• Variation of constants yields y(t) = et
−1 +
t
0
e−τ
u(τ) dτ .

−1 +
t
0
e−τ
u(τ) dτ .
•
(OC)T min
u∈L2(0,T)
T
0
1
2|u(t)|2
+ γ et
1 −
t
0
e−τ
u(τ) dτ dt

−1 +
t
0
e−τ
u(τ) dτ .
•
(OC)T min
u∈L2(0,T)
T
0
1
2|u(t)|2
+ γ et
1 −
t
0
e−τ
u(τ) dτ dt
In the problem we have no terminal conditions!

−1 +
t
0
e−τ
u(τ) dτ .
•
(OC)T min
u∈L2(0,T)
T
0
1
2|u(t)|2
+ γ et
1 −
t
0
e−τ
u(τ) dτ dt
• Consider the time-horizon T as a parameter.
For τ ∈ (0, T], consider the parametric optimal control problem (OC)τ .
As long as y(τ) ≤ 0, we can get rid of the absolute value and thus of the
non-smoothness!

−1 +
t
0
e−τ
u(τ) dτ .
•
(OC)T min
u∈L2(0,T)
T
0
1
2|u(t)|2
+ γ et
1 −
t
0
e−τ
u(τ) dτ dt
non-smoothness!
• So for sufﬁciently small τ > 0, we consider
(OC)τ
min
u∈L2(0,τ)
τ
0
1
2u(t)2
+ γ et
1 −
t
0
e−τ
u(τ) dτ dt.

−1 +
t
0
e−τ
u(τ) dτ .
•
(OC)T min
u∈L2(0,T)
T
0
1
2|u(t)|2
+ γ et
1 −
t
0
e−τ
u(τ) dτ dt
non-smoothness!
• So for sufﬁciently small τ > 0, we consider
(OC)τ
min
u∈L2(0,τ)
τ
0
1
2u(t)2
+ γ et
1 −
t
0
e−τ
u(τ) dτ dt.
• Integration by parts allows to transform the objective function:
(OC)τ min
u∈L2(0,τ)
τ
0
1
2
u(t)2
+ γ 1 − eτ−t
u(t) dt + γ (eτ
− 1)

(OC)τ min
u∈L2(0,τ)
τ
0
1
2
u(t)2
+ γ 1 − eτ−t
u(t) dt + γ (eτ
− 1)

(OC)τ min
u∈L2(0,τ)
τ
0
1
2
u(t)2
+ γ 1 − eτ−t
u(t) dt + γ (eτ
− 1)
• Then the necessary optimality conditions imply
u(t) = −γ + γ eτ−t
.

(OC)τ min
u∈L2(0,τ)
τ
0
1
2
u(t)2
+ γ 1 − eτ−t
u(t) dt + γ (eτ
− 1)
• Then the necessary optimality conditions imply
u(t) = −γ + γ eτ−t
.
• For such a control
s
0
e−t
u(t) dt = 1 holds if γ = 1
cosh(s)−1.
Hence for τ = s we have y(s) = 0 and u(s) = 0 and for t > s we can continue
with u(t) = 0 and obtain the optimal control for (OC)T for all T ≥ s!

Lemma
For γ > 0, define s > 0 as the value where cosh(s) = 1
γ + 1.
Assume that T ≥ s. Define
û(t) = γ(es−t
− 1)+.
Then for the state ˆy generated by û for t ≥ s we have ˆy(t) = 0.
Moreover, for all t ∈ (0, T) we have ˆy(t) ≤ 0.
The control û is the unique solution of (OC)T .

Lemma
For γ > 0, define s > 0 as the value where cosh(s) = 1
γ + 1.
Assume that T ≥ s. Define
û(t) = γ(es−t
− 1)+.
Then for the state ˆy generated by û for t ≥ s we have ˆy(t) = 0.
Moreover, for all t ∈ (0, T) we have ˆy(t) ≤ 0.
The control û is the unique solution of (OC)T .
The optimal state is y(t) = et
−1 + γ es
−es−2t
2 + e−t
− 1
−
.
The optimal state for s = 1 The optimal control for s = 1

The optimal state and control for s = 2 and γ = 1
cosh(s)−1. :

For sufficiently large T, due to the L1
-norm of y that appears in the objective
function, the solution has a finite–time turnpike structure where the system is
steered to zero in the finite stopping time s.

For sufficiently large T, due to the L1
-norm of y that appears in the objective
function, the solution has a finite–time turnpike structure where the system is
steered to zero in the finite stopping time s.
This time s is independent of T and only depends on γ.
Both state and control remain at zero for all t ∈ (s, T).
~

You can think of the turnpike as the point of the highest bliss level.

Inhalt
An example with L1
-tracking term
-tracking term
Optimal L1
Conclusion

For T sufﬁciently large, we can add a terminal condition:
For T ≥ 1 and γ 0 we consider the problem
(FROMATOB)T



min
u∈L2(0,T)
T
0
1
2|u(t)|2
y(0) = −1, y(T) = 1, y (t) = y(t) + u(t).

(FROMATOB)T



min
u∈L2(0,T)
T
0
1
2|u(t)|2
y(0) = −1, y(T) = 1, y (t) = y(t) + u(t).
The problem decouples into (OC)s on [0, s] and
the problem without initial condition and starting time s
(END)s



min
u∈L2(s, T)
T
s
1
2|u(t)|2
y(T) = 1, y (t) = y(t) + u(t), t ∈ [s, T].

(FROMATOB)T



min
u∈L2(0,T)
T
0
1
2|u(t)|2
y(0) = −1, y(T) = 1, y (t) = y(t) + u(t).
(END)s



min
u∈L2(s, T)
T
s
1
2|u(t)|2
y(T) = 1, y (t) = y(t) + u(t), t ∈ [s, T].
For the starting time s if cosh(T − s) − 1 = 1/γ, the optimal control of (END)s is
u2(t) = γ (1 − es−t
)+.

(FROMATOB)T



min
u∈L2(0,T)
T
0
1
2|u(t)|2
y(0) = −1, y(T) = 1, y (t) = y(t) + u(t).
(END)s



min
u∈L2(s, T)
T
s
1
2|u(t)|2
y(T) = 1, y (t) = y(t) + u(t), t ∈ [s, T].
For the starting time s if cosh(T − s) − 1 = 1/γ, the optimal control of (END)s is
u2(t) = γ (1 − es−t
)+.
Then we have y(s) = 0 and y(t) = et−T
[1 − γ(es−T
−eT+s−2t
2 + eT−t
− 1)].

The solution of (OC)s reaches the turnpike after finite stopping time s = sstop
The solution of (END)s leaves the turnpike after the finite starting time s = sstart.
Hence if sstop ≤ sstart, the solutions can be glued together to solve (FROMATOB)T
if T is sufficiently large, e.g. 1
γ ∈ (0, cosh(T/2) − 1)
(then sstop ≤ T/2 ≤ sstart).

γ ∈ (0, cosh(T/2) − 1)
This is the finite-time turnpike situation:
From a given initial state, the optimal state is driven to the turnpike in finite time.
Then it stays on the turnpike for a finite time interval.
Wir fahr’n fahr’n fahr’n auf der Autobahn
Text of the Song by Kraftwerk

γ ∈ (0, cosh(T/2) − 1)
This is the finite-time turnpike situation:
From a given initial state, the optimal state is driven to the turnpike in finite time.
Then it stays on the turnpike for a finite time interval.
Text of the Song by Kraftwerk
If the prescribed terminal state is different from the turnpike,
the state finally leaves the turnpike to reach the target state.

An example
Let γ = 1
cosh(1)−1. Then sstop = 1 and sstart = T − 1.

An example
Let γ = 1
Assume that T is sufﬁciently large that cosh(T/2) − 1 1/γ, that is T 2.

An example
Let γ = 1
Assume that T is sufﬁciently large that cosh(T/2) − 1 1/γ, that is T 2.
Then for the optimal control and the optimal state we have
u(t) = 0, y(t) = 0 for all t ∈ [1, T − 1].
This is the ﬁnite-time turnpike situation.
On [0, T] the control u is continuous and the state is continuously differentiable.

Numerical examples for a nonautomous system
by Michael Schuster: y (t) = y(t) + etu(t)

Preprint
Available in the Arxiv preprint
M. GUGAT, M. SCHUSTER, E. ZUAZUA, The Finite-Time Turnpike Phenomenon
for Optimal Control Problems: Stabilization by Non-Smooth Tracking Terms,
arXiv:2006.07051
To appear in ”Stabilization of Distributed Parameter Systems: Design Methods and
Applications” edited by ALEXANDER ZUYEV and G. SKLYAR within the SEMASIMAI
Springer Series

Preprint
arXiv:2006.07051
Springer Series
• Contains also results for inﬁnite-dimensional linear systems with L∞
and L2
-norm
tracking terms.
• Obviously, the ﬁnite-time turnpike phenomenon can only occur for systems that
are exactly controllable in the sense that the turnpike is reachable.

Preprint
arXiv:2006.07051
Springer Series
• Contains also results for infinite-dimensional linear systems with L∞
and L2
-norm
tracking terms.
• Obviously, the finite-time turnpike phenomenon can only occur for systems that
are exactly controllable in the sense that the turnpike is reachable.
• However, this also happens for systems that are nodal profile exactly controllable
if the turnpike is described through the nodal profiles.

Inhalt
An example with L1
-tracking term
-tracking term
Optimal L1
Conclusion

An example with squared L2-tracking term
(OC)T



min
u∈L2(0,T)
T
0
1
2|u(t)|2
+ γ |y(t)|2
dt subject to
y(0) = −1, y (t) = y(t) + u(t), y(T) = 0.
Here we have the end condition y(T) = 0.

(OC)T



min
u∈L2(0,T)
T
0
1
2|u(t)|2
+ γ |y(t)|2
dt subject to
y(0) = −1, y (t) = y(t) + u(t), y(T) = 0.
The problem without initial and terminal conditions:
(S)



min
u∈L2(0,T)
T
0
1
2|u(t)|2
+ γ |y(t)|2
dt subject to
y (t) = y(t) + u(t).

(OC)T



min
u∈L2(0,T)
T
0
1
2|u(t)|2
+ γ |y(t)|2
dt subject to
y(0) = −1, y (t) = y(t) + u(t), y(T) = 0.
The problem without initial and terminal conditions:
(S)



min
u∈L2(0,T)
T
0
1
2|u(t)|2
+ γ |y(t)|2
dt subject to
y (t) = y(t) + u(t).
Here the turnpike is zero, that is y(σ)
= 0 and u(σ)
= 0.

(OC)T min
u∈L2(0,T)+:
T
0
e−τ u(τ) dτ=1
T
0
1
2
u(t)2
− γ u(t) y(t) dt +
1
2
γ.

(OC)T min
u∈L2(0,T)+:
T
0
e−τ u(τ) dτ=1
T
0
1
2
u(t)2
− γ u(t) y(t) dt +
1
2
γ.
• The necessary optimality conditions yield
u(t) = γ y(t) − γ e−t
t
0
ez
u(z) dz dt + λ e−t
where λ is the multiplier corresponding to the moment equation.

(OC)T min
u∈L2(0,T)+:
T
0
e−τ u(τ) dτ=1
T
0
1
2
u(t)2
− γ u(t) y(t) dt +
1
2
γ.
u(t) = γ y(t) − γ e−t
t
0
ez
• Using u = y − y, this yields y = (1 + 2γ) y.

(OC)T min
u∈L2(0,T)+:
T
0
e−τ u(τ) dτ=1
T
0
1
2
u(t)2
− γ u(t) y(t) dt +
1
2
γ.
u(t) = γ y(t) − γ e−t
t
0
ez
• Using u = y − y, this yields y = (1 + 2γ) y.
• With the end condition y(T) = 0 we obtain for ω =
√
1 + 2γ
y(t) =
sinh(ω(t − T))
sinh(ω T)
, u(t) =
ω cosh(ω(t − T)) − sinh(ω(t − T))
sinh(ωT)
.

The optimal state and control for γ = 1, ω =
√
3:

The optimal state and control for γ = 1, ω =
√
3:
Note that for the problem with time interval [0, 2T] and the end condition y(2T) = 1,
we can extend the control u(t) that generates the state
y(t) =
sinh(ω(t − T))
sinh(ω T)
.
Thus we have
y(2 T) =
sinh(ω T)
sinh(ω T)
= 1.
Since u satisﬁes the necessary optimality conditions (with a different multiplier λ)
this yields the solution of

Inhalt
An example with L1
-tracking term
-tracking term
Optimal L1
Conclusion

(OC)2T



min
u∈L2(0,2T)
2T
0
1
2|u(t)|2
+ γ |y(t)|2
dt subject to
y(0) = −1, y (t) = y(t) + u(t), y(2T) = 1.
This is the classical turnpike phenomenon:
Note that u(T) = ω
sinh(ω T) = 0.
So the control-state pair (u, y) never reaches the turnpike (u(σ)
, y(σ)
).
But we have u(t) − u(σ) 2
≤ 2 ω2
+1
sinh2
(ω T)
cosh2
(ω (T − t).
So locally around the middle of the time interval [0, 2T] at T, the distance to the
turnpike decreases exponentially fast with T!

Inhalt
An example with L1
-tracking term
-tracking term
Optimal L1
Conclusion

Remarks about the turnpike phenomenon
• Non-smooth tracking terms: The ﬁnite-time turnpike phenomenon is possible.

• Smooth tracking terms: The classical turnpike phenomenon.
1. Exponential turnpike result:
u(t) − uσ 2
+ y(t) − uσ 2
≤ C0 [exp(−µτ) + exp(−µ(T − τ))]
where C0 and µ 0 do not depend on T.
For linear pdes by PORRETTA and ZUAZUA (SICON 2013), and TRELAT, ZHANG,
ZUAZUA (SICON 2018 in Hilbert space)
For nonlinear ODEs TRELAT, ZUAZUA in Journal of Differential Equations, 2015

u(t) − uσ 2
+ y(t) − uσ 2
≤ C0 [exp(−µτ) + exp(−µ(T − τ))]
2. Another team: LARS GRüNE from Bayreuth,
KARL WORTHMANN and MANUEL SCHALLER from Ilmenau and
TIMM FAULWASSER from Dortmund.
3. ALEXANDER ZASLAVSKI (Technion). Many books!

u(t) − uσ 2
+ y(t) − uσ 2
≤ C0 [exp(−µτ) + exp(−µ(T − τ))]
• Weakest property: Measure turnpike property.

u(t) − uσ 2
+ y(t) − uσ 2
≤ C0 [exp(−µτ) + exp(−µ(T − τ))]
• Intermediate: Integral turnpike, see for example GUGAT, HANTE, (SICON 2019).
Boundary control for linear 2 × 2 hyperbolic systems.

u(t) − uσ 2
+ y(t) − uσ 2
≤ C0 [exp(−µτ) + exp(−µ(T − τ))]
• Intermediate: Integral turnpike, see for example GUGAT, HANTE, (SICON 2019).
Boundary control for linear 2 × 2 hyperbolic systems.
• Strongest: Exponential turnpike as above and for example
GUGAT, TRELAT, ZUAZUA, Optimal boundary control for the wave equation,
(Syst. Control Letters 2016).

Inhalt
An example with L1
-tracking term
-tracking term
Optimal L1
Conclusion

Hyperbolic 2 × 2 boundary control systems:
Gas transportation through pipelines
The gas pressure is increased at compressor stations.

Application: Gas transportation through pipelines
Here the system dynamics is described by
the isothermal Euler equations



ρt + qx = 0
qt + p + q2
ρ
x
= −
fg
2δ
q|q|
ρ − ρ g sin(α)
or a similar (linearized) model, see A. Osiadacz, M. Chazykowski, Comparison of
isothermal and non-isothermal pipeline gas ﬂow models, 2001.
See the results of DFG CRC 154:

Model for the ﬂow in a single pipe
For ideal gas in a horizontal pipe
we have
ρt + qx = 0
qt + a2
ρ + q2
ρ
x
= −1
2 θ q |q|
ρ .
• ρ: density
• q: ﬂow rate
• a: sound speed
• θ =
fg
δ
• fg: friction, δ: diameter

we have
ρt + qx = 0
qt + a2
ρ + q2
ρ
x
= −1
2 θ q |q|
ρ .
• ρ: density
• q: flow rate
• a: sound speed
• θ =
fg
δ
Slow flow
For slow flow, this is close to the linear
model
ρt + qx = 0
qt + a2
ρx = 0

we have
ρt + qx = 0
qt + a2
ρ + q2
ρ
x
= −1
2 θ q |q|
ρ .
• ρ: density
• q: flow rate
• a: sound speed
• θ =
fg
δ
Slow flow
model
ρt + qx = 0
qt + a2
ρx = 0
Slow flow: Wave equation
This in turn implies the wave equations
ρtt = a2
ρxx
qtt = a2
qxx

we have
ρt + qx = 0
qt + a2
ρ + q2
ρ
x
= −1
2 θ q |q|
ρ .
• ρ: density
• q: flow rate
• a: sound speed
• θ =
fg
δ
Slow flow
model
ρt + qx = 0
qt + a2
ρx = 0
Slow flow: Wave equation
This in turn implies the wave equations
ρtt = a2
ρxx
qtt = a2
qxx
The problem to control the flow
leads to problems of optimal boundary
control.

Inhalt
An example with L1
-tracking term
-tracking term
Optimal L1
Conclusion

L1-Optimal Dirichlet control of the wave equation
Exact norm minimal control, ﬁnite horizon
T ≥ 1
Let y0 ∈ L2
(0, 1), y1 ∈ H−1
(0, 1) and T ≥ 1
be given. Deﬁne (EC) :



min
T
0
|u1(t)| + |u2(t)| dt subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = u1(t), y(t, 1) = u2(t), t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1)
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).

T ≥ 1
Let y0 ∈ L2
(0, 1), y1 ∈ H−1
(0, 1) and T ≥ 1



min
T
0
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = u1(t), y(t, 1) = u2(t), t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1)
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).
In general the optimal controls are
not unique!

T ≥ 1
Let y0 ∈ L2
(0, 1), y1 ∈ H−1
(0, 1) and T ≥ 1



min
T
0
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = u1(t), y(t, 1) = u2(t), t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1)
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).
In general the optimal controls are
not unique!
Gugat, French-German-Spanish
Conf. on Opt., Avignon 2004
Let T ≥ 1. There exist solutions of
(EC) that are 2–periodic and



u1(t + j) = −u2(t + (j + 1)),
u1(t + (j + 1)) = −u2(t + j).
For k ∈ {1, 2, ..., }, t ∈ (0, 2),
t + 2k ≤ T, l ∈ {1, 2} we have
ul(t + 2k) = ul(t).
The set of all solutions can be
parametrized by the measurable
convex combinations (t ∈ (0, 1))
λ
(l)
j (t) ≥ 0,
j:t+2 j≤T
λ
(l)
j (t) = 1.
there exists

Example: Let y1 = 0.
Let
k = max{j ∈ {1, 2, 3, ...} : j ≤ T}
and
∆ = T − k ≥ 0.
(0, T) = (0, ∆)∪(∆, 1)∪(1, 1 + ∆)∪(2+∆, 2)∪(2, 2 + ∆)...∪((k−1)+∆, k)∪(k, k + ∆)
There are k + 1 red intervals and k black intervals!

Let
k = max{j ∈ {1, 2, 3, ...} : j ≤ T}
and
∆ = T − k ≥ 0.
(0, T) = (0, ∆)∪(∆, 1)∪(1, 1 + ∆)∪(2+∆, 2)∪(2, 2 + ∆)...∪((k−1)+∆, k)∪(k, k + ∆)
Then for t ∈ (0, ∆), an optimal control is given by
u1(t + j) = u2(t + j) = (−1)j y0(t)
2 (k + 1)
.

Let
k = max{j ∈ {1, 2, 3, ...} : j ≤ T}
and
∆ = T − k ≥ 0.
(0, T) = (0, ∆)∪(∆, 1)∪(1, 1 + ∆)∪(2+∆, 2)∪(2, 2 + ∆)...∪((k−1)+∆, k)∪(k, k + ∆)
Then for t ∈ (0, ∆), an optimal control is given by
u1(t + j) = u2(t + j) = (−1)j y0(t)
2 (k + 1)
.
For t ∈ (∆, 1), an optimal control is given by
u1(t + j) = u2(t + j) = (−1)j y0(t)
2 k
.

Example: y1 = 0.
The control action can be shifted between the different time periods!
Let again
k = max{j ∈ {1, 2, 3, ... : j ≤ T}, ∆ = T − k ≥ 0.
Deﬁne
d(t) =
(k + 1) if t ∈ (0, ∆),
k if t ∈ (∆, 1).

Example: y1 = 0.
Let again
k = max{j ∈ {1, 2, 3, ... : j ≤ T}, ∆ = T − k ≥ 0.
Deﬁne
d(t) =
(k + 1) if t ∈ (0, ∆),
k if t ∈ (∆, 1).
The set of all optimal controls has the following structure:
For pairs of measurable convex combinations l ∈ {1, 2}, t ∈ (0, 1)
λ
(l)
j (t) ≥ 0,
j:t+2 j≤T
λ
(l)
j (t) = 1
we obtain the optimal controls

Example: y1 = 0.
Let again
k = max{j ∈ {1, 2, 3, ... : j ≤ T}, ∆ = T − k ≥ 0.
Deﬁne
d(t) =
(k + 1) if t ∈ (0, ∆),
k if t ∈ (∆, 1).
The set of all optimal controls has the following structure:
For pairs of measurable convex combinations l ∈ {1, 2}, t ∈ (0, 1)
λ
(l)
j (t) ≥ 0,
j:t+2 j≤T
λ
(l)
j (t) = 1
we obtain the optimal controls
u1(t + 2j) = λ
(1)
2j (t)
y0(t)
2 d(t)
, u1(t + 2j + 1) = −λ
(2)
2j+1(t)
y0(t)
2 d(t)
,
u2(t + 2j) = λ
(2)
2j (t)
y0(t)
2 d(t)
, u2(t + 2j + 1) = −λ
(1)
2j+1(t)
y0(t)
2 d(t)
.

• The convex combinations (t ∈ (0, 1), l ∈ {1, 2})
λ
(l)
j (t) ≥ 0,
j:t+2 j≤T
λ
(l)
j (t) = 1.
determine the support of the corresponding optimal control.
It can be the whole interval [0, T].

λ
(l)
j (t) ≥ 0,
j:t+2 j≤T
λ
(l)
j (t) = 1.
• If for all t ∈ (0, 1), the λ
(l)
j (t) are equal for all j, we obtain periodic controls.
In fact, this yields λ
(l)
j (t) = 1
d(t).

λ
(l)
j (t) ≥ 0,
j:t+2 j≤T
λ
(l)
j (t) = 1.
(l)
(l)
j (t) = 1
d(t).
• If one of the λ
(l)
j (t) is equal to 1, the others must be equal to 0.
In this case, the support of the corresponding optimal control can be constrained
to a subinterval of [0, T] of length 1.

λ
(l)
j (t) ≥ 0,
j:t+2 j≤T
λ
(l)
j (t) = 1.
(l)
(l)
j (t) = 1
d(t).
• If one of the λ
(l)
j (t) is equal to 1, the others must be equal to 0.
In this case, the support of the corresponding optimal control can be constrained
to a subinterval of [0, T] of length 1.
• Thus we have optimal controls with support (0, 1).
These controls steer the system to rest at the time t = 1.

Adding a tracking-term in the goal function
Finite horizon T ≥ 1, γ 0. Deﬁne (P):



min
u1,u2∈L1(0,T)
T
0
|u1(t)| + |u2(t)| dt
+ γ
T
1
1
0
|y(t, x)| dx dt
subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = u1(t), y(t, 1) = u2(t), t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1),
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).




min
u1,u2∈L1(0,T)
T
0
|u1(t)| + |u2(t)| dt
+ γ
T
1
1
0
|y(t, x)| dx dt
subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = u1(t), y(t, 1) = u2(t), t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1),
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).
Solution of (P)
The nonsmooth problem (P) has a
unique solution.
The unique solution of (P) steers the
state to rest at time t = 1.
Then the control is switched off.




min
u1,u2∈L1(0,T)
T
0
|u1(t)| + |u2(t)| dt
+ γ
T
1
1
0
|y(t, x)| dx dt
subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = u1(t), y(t, 1) = u2(t), t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1),
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).
Solution of (P)
unique solution.
Here we have an extreme
(ﬁnite time) turnpike structure!




min
u1,u2∈L1(0,T)
T
0
|u1(t)| + |u2(t)| dt
+ γ
T
1
1
0
|y(t, x)| dx dt
subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = u1(t), y(t, 1) = u2(t), t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1),
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).
Solution of (P)
unique solution.
For t 1 we have
u1(t) = u2(t) = 0,
y(t, x) = 0, x ∈ (0, 1)




min
u1,u2∈L1(0,T)
T
0
|u1(t)| + |u2(t)| dt
+ γ
T
1
1
0
|y(t, x)| dx dt
subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = u1(t), y(t, 1) = u2(t), t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1),
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).
Solution of (P)
unique solution.
For t 1 we have
u1(t) = u2(t) = 0,
y(t, x) = 0, x ∈ (0, 1)
and yt(t, x) = 0.




min
u1,u2∈L1(0,T)
T
0
|u1(t)| + |u2(t)| dt
+ γ
T
1
1
0
|y(t, x)| dx dt
subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = u1(t), y(t, 1) = u2(t), t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1),
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).
Solution of (P)
unique solution.
For t 1 we have
u1(t) = u2(t) = 0,
y(t, x) = 0, x ∈ (0, 1)
and yt(t, x) = 0.
This is possible due to
exact controllability!

Optimal Dirichlet L1-control of the wave equation
Proof.
Problem (EC) has a unique solution where
the support of the controls is in (0, 1).

Proof.
This optimal control (u∗
1, u∗
2) steers the state
to rest at time t = 1.

Proof.
1, u∗
2) steers the state
Let ν(EC) denote the optimal value of (EC).

Proof.
1, u∗
2) steers the state
Let ν(P) denote the optimal value of (P).

Proof.
1, u∗
2) steers the state
Since the objective function of (EC) is ≤ the
objective function of (P), we have
ν(EC) ≤ ν(P).

Proof.
1, u∗
2) steers the state
ν(EC) ≤ ν(P).
The objective value of (P) for (u∗
1, u∗
2) is
ν(EC)+γ
T
1
1
0
|y(t, x)| dx dt = ν(EC) ≤ ν(P).
Hence ν(EC) = ν(P).

Proof.
1, u∗
2) steers the state
ν(EC) ≤ ν(P).
1, u∗
2) is
ν(EC)+γ
T
1
1
0
|y(t, x)| dx dt = ν(EC) ≤ ν(P).
Hence ν(EC) = ν(P). (u∗
1, u∗
2) solves (P).

Proof.
1, u∗
2) steers the state
ν(EC) ≤ ν(P).
1, u∗
2) is
ν(EC)+γ
T
1
1
0
|y(t, x)| dx dt = ν(EC) ≤ ν(P).
1, u∗
2) solves (P).
In fact, we can replace (P) with
the problem without terminal con-
straints.

Proof.
1, u∗
2) steers the state
ν(EC) ≤ ν(P).
1, u∗
2) is
ν(EC)+γ
T
1
1
0
|y(t, x)| dx dt = ν(EC) ≤ ν(P).
1, u∗
2) solves (P).
straints.
Deﬁne (P ):



min
u1,u2∈L1(0,T)
1
2
T
0
|u1(t)|2
+ |u2(t)|2
dt
+ γ
T
1
|y(t, 1)|2 + |yx(t, 1)|2 dt
subject to
y(0, x) = y0(x), yt(0, x) = y1(x),
x ∈ (0, 1)
y(t, 0) = u1(t), y(t, 1) = u2(t),
t ∈ (0, T)
ytt(t, x) = yxx(t, x),
(t, x) ∈ (0, T) × (0, 1).

Proof.
1, u∗
2) steers the state
ν(EC) ≤ ν(P).
1, u∗
2) is
ν(EC)+γ
T
1
1
0
|y(t, x)| dx dt = ν(EC) ≤ ν(P).
1, u∗
2) solves (P).
straints.
Define (P ):



min
u1,u2∈L1(0,T)
1
2
T
0
|u1(t)|2
+ |u2(t)|2
dt
+ γ
T
1
|y(t, 1)|2 + |yx(t, 1)|2 dt
subject to
y(0, x) = y0(x), yt(0, x) = y1(x),
x ∈ (0, 1)
y(t, 0) = u1(t), y(t, 1) = u2(t),
t ∈ (0, T)
ytt(t, x) = yxx(t, x),
(t, x) ∈ (0, T) × (0, 1).
Then if γ is sufficiently large, the
solution has the same structure as
for (P).
The proof can be based uponMartin Gugat · FAU · The finite-time turnpike phenomenon for optimal control problems 17.10.2019 40

Inhalt
An example with L1
-tracking term
-tracking term
Optimal L1
Conclusion

Optimal Neumann control of the wave equation
T ≥ 2
Let y0 ∈ H1
(0, 1) with y0(0) = 0, y1 ∈ L2
(0, 1)
and T ≥ 2 be given. Deﬁne (EC) :



min u 2
L2(0,T)
subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = 0, yx(t, 1) = u(t) , t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1)
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).

T ≥ 2
Let y0 ∈ H1
(0, 1) with y0(0) = 0, y1 ∈ L2
(0, 1)



min u 2
L2(0,T)
subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = 0, yx(t, 1) = u(t) , t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1)
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).
Gugat, Arab. J. Math. 2015 Open Access
Let T ∈ N be even. The unique
solution of (EC) is 4–periodic and
u(t) =



y0(1−t)−y1(1−t)
T , t ∈ (0, 1),
y0(t−1)+y1(t−1)
T , t ∈ (1, 2).
For k ∈ {1, 2, ..., (T − 2)/2},
t ∈ (0, 2) we have
u(t + 2k) = (−1)k
u(t).

T ≥ 2
Let y0 ∈ H1
(0, 1) with y0(0) = 0, y1 ∈ L2
(0, 1)



min u 2
L2(0,T)
subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = 0, yx(t, 1) = u(t) , t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1)
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).
Gugat, Arab. J. Math. 2015 Open Access
Let T ∈ N be even. The unique
solution of (EC) is 4–periodic and
u(t) =



y0(1−t)−y1(1−t)
T , t ∈ (0, 1),
y0(t−1)+y1(t−1)
T , t ∈ (1, 2).
For k ∈ {1, 2, ..., (T − 2)/2},
t ∈ (0, 2) we have
u(t + 2k) = (−1)k
u(t).
For t0 = 0 (Moving horizon) this
yields the well-known feedback
law
yx(t0, 1) = − 1
T−1 yt(t0, 1).




min
u∈L2(0,T)
T
0
(yx(t, 0))2
+ γ u2
(t) dt subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = 0, yx(t, 1) = u(t) , t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1),
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).




min
u∈L2(0,T)
T
0
(yx(t, 0))2
+ γ u2
(t) dt subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = 0, yx(t, 1) = u(t) , t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1),
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).
Solution of (P), Syst. Control Lett.
2016 (with E. TRÉLAT, E. ZUAZUA)
The unique solution of (P) is the
sum of 2 parts that grow/decay
exponentially.




min
u∈L2(0,T)
T
0
(yx(t, 0))2
+ γ u2
(t) dt subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = 0, yx(t, 1) = u(t) , t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1),
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).
exponentially. Choose zγ with
z2
γ + (2 + 4
γ ) zγ + 1 = 0.




min
u∈L2(0,T)
T
0
(yx(t, 0))2
+ γ u2
(t) dt subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = 0, yx(t, 1) = u(t) , t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1),
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).
z2
γ + (2 + 4
γ ) zγ + 1 = 0.
For t ∈ (0, 2) let
H(t) =



y0(1−t)−y1(1−t)
2 , t ∈ (0, 1),
y0(t−1)+y1(t−1)
2 , t ∈ [1, 2).




min
u∈L2(0,T)
T
0
(yx(t, 0))2
+ γ u2
(t) dt subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = 0, yx(t, 1) = u(t) , t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1),
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).
z2
γ + (2 + 4
γ ) zγ + 1 = 0.
For t ∈ (0, 2) let
H(t) =



y0(1−t)−y1(1−t)
2 , t ∈ (0, 1),
y0(t−1)+y1(t−1)
2 , t ∈ [1, 2).
For t ∈ (0, 2), k ∈ N0 and
t + 2k ≤ T:
u(t + 2k) = zk
γ − 1
zk
γ
1+zγ
1−zγ
H(t).




min
u∈L2(0,T)
T
0
(yx(t, 0))2
+ γ u2
(t) dt subject to
y(0, x) = y0(x), yt(0, x) = y1(x), x ∈ (0, 1)
y(t, 0) = 0, yx(t, 1) = u(t) , t ∈ (0, T)
ytt(t, x) = yxx(t, x), (t, x) ∈ (0, T) × (0, 1),
y(T, x) = 0, yt(T, x) = 0, x ∈ (0, 1).
z2
γ + (2 + 4
γ ) zγ + 1 = 0.
For t ∈ (0, 2) let
H(t) =



y0(1−t)−y1(1−t)
2 , t ∈ (0, 1),
y0(t−1)+y1(t−1)
2 , t ∈ [1, 2).
For t ∈ (0, 2), k ∈ N0 and
t + 2k ≤ T:
u(t + 2k) = zk
γ − 1
zk
γ
1+zγ
1−zγ
H(t).This is an exponential turnpike structure!

Inhalt
An example with L1
-tracking term
-tracking term
Optimal L1
Conclusion

Conclusion
• The turnpike result for L1
–control cost shows that for suff. large T 1 the optimal
control steers the system to rest in the minimal time t = 1.

Conclusion
• Aim: Extension to semilinear and quasilinear models

Conclusion
• Exploit for numerical computations.
Link with mixed integer programming.
For nonlinear pdes, using piecewise linear approximations of the nonlinearities
leads to mixed integer programs.

Conclusion
• Exploit for numerical computations.
Link with mixed integer programming.
For nonlinear pdes, using piecewise linear approximations of the nonlinearities
leads to mixed integer programs.
• Thank you for your attention!

Thank you for your attention!

The finite time turnpike phenomenon for optimal control problems

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