l2 stabilization of coupled viscous burgers' equations · 1 3 u3(1;t) saad qadeer &...

L2 Stabilization of Coupled Viscous Burgers’Equations

Saad Qadeer & Jean-Baptiste Sibille

UC Berkeley

May 8th, 2014

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Objective of the project

Goal: stabilize the system:

ut(x , t)− ε1uxx(x , t) + u(x , t)ux(x , t) = 0, x ∈ [0, 1]u(0, t) = 0

ux(1, t) = U(t)vt(x , t)− ε2vxx(x , t) + v(x , t)vx(x , t) = 0, x ∈ [1, 1+ D]

v(1, t) = qu(1, t)vx(1+ D, t) = W (t)

whereε1, ε2,D > 0, q ∈ R.U,W are the control inputs to the system.

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Motivation

Interconnected roads example

Figure: Real life example of the coupled Burgers’ equations

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First approach: with only one equationBack to the coupled problem

Current/Future Work: the inviscid case

1 First approach: with only one equationProblem FormulationLyapunov Stability & Controller DesignSimulation

2 Back to the coupled problemLyapunov Stability & Controller design: coupled caseSimulation

3 Current/Future Work: the inviscid case

Saad Qadeer & Jean-Baptiste Sibille L2 Stabilization of Coupled Viscous Burgers’ Equations

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Problem FormulationLyapunov Stability & Controller DesignSimulation

Table of Contents





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Problem Formulation

We first consider the system


ux(1, t) = U(t)


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Lyapunov Function

Let’s define the Lyapunov function (L2 norm of u):

V (t) =12

∫ 1

0u(x , t)2dx

And take its derivative:

V̇ (t) = ε1U(t)u(1, t)− ε1∫ 1

0u2x (x , t)dx − 1

3u3(1, t)


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Controller Design

We choose a controller:

U(t) = −c(u(1, t) + u(1, t)3), with c ≥ 16ε1

which leads to:

V̇ (t) = −ε1∫ 1

0u2xdx − ε1c(u(1, t) + u3(1, t))u(1, t)− 1

3u3(1, t)

= −ε1∫ 1

0u2xdx − ε1c

[1+

u(1, t)3ε1c

+ u2(1, t)]

︸︷︷︸≥0 with this choice of c

u2(1, t)

≤ −ε1∫ 1

0u2xdx


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V̇ (t) ≤ −ε1∫ 1

0u2xdx

≤ −ε1∫ 1

0u2(x , t)dx ∵ Poincaré’s inequality

which leads to

V̇ (t) ≤ −2ε1V (t)

Hence

Exponential Stability

V (t) ≤ V (0) exp(−2ε1t)


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Figure: Simulation using finite differences with ∆x = 0.1,∆t = 0.005


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Lyapunov Stability & Controller design: coupled caseSimulation

Table of Contents





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Recall the system

Goal: stabilize the system:


ux(1, t) = U(t)vt(x , t)− ε2vxx(x , t) + v(x , t)vx(x , t) = 0, x ∈ [1, 1+ D]

v(1, t) = qu(1, t)vx(1+ D, t) = W (t)

whereε1, ε2,D > 0, q ∈ R.U,W are the control inputs to the system.


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Lyapunov Function

Let’s define the Lyapunov function

V (t) =12

∫ 1

0u(x , t)2dx +

12

∫ 1+D

1v(x , t)2dx

We have:

V̇ (t) = −ε1∫ 1

0ux(x , t)2dx − ε2

∫ 1+D

1vx(x , t)2dx

+ u(1, t)(ε1U(t) +

13(q3 − 1)u(1, t)2 − ε2qvx(1, t)

)+ v(1+ D, t)

(ε2W (t)− 1

3v(1+ D, t)2

)


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Controller Design

We choose the control laws as

U(t) =ε2ε1

qvx(1, t)−13ε1

(q3 − 1)u(1, t)2

W (t) =13ε2

v(1+ D, t)2 − 2D

v(1+ D, t)

We get

V̇ (t) = −ε1∫ 1

0ux(x , t)2dx − ε2

∫ 1+D

1vx(x , t)2dx

+ u(1, t)(ε1U(t) +

13(q3 − 1)u(1, t)2 − ε2qvx(1, t)

)+ v(1+ D, t)

(ε2W (t)− 1

3v(1+ D, t)2

)


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Controller Design

We choose the control laws as

U(t) =ε2ε1

qvx(1, t)−13ε1

(q3 − 1)u(1, t)2

W (t) =13ε2

v(1+ D, t)2 − 2D

v(1+ D, t)

We get

V̇ (t) = −ε1∫ 1

0ux(x , t)2dx − ε2

∫ 1+D

1vx(x , t)2dx

+ u(1, t)(ε1U(t) +

13(q3 − 1)u(1, t)2 − ε2qvx(1, t)

)+ v(1+ D, t)

(ε2W (t)− 1

3v(1+ D, t)2

)Saad Qadeer & Jean-Baptiste Sibille L2 Stabilization of Coupled Viscous Burgers’ Equations

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V̇ (t) = −ε1∫ 1

0ux(x , t)2dx − ε2

∫ 1+D

1vx(x , t)2dx

−2ε2D

v(1+ D, t)2

≤ −ε1∫ 1

0u(x , t)2dx − ε2

D2

∫ 1+D

1v(x , t)2dx

In the case D ≤ 1, we use −1D2 ≤ −1 to get

V̇ (t) ≤ −ε1∫ 1

0u(x , t)2dx − ε2

∫ 1+D

1v(x , t)2dx

≤ −2min{ε1, ε2}(12

∫ 1

0u(x , t)2dx +

12

∫ 1+D

1v(x , t)2dx

)= −2min{ε1, ε2}V (t)


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V̇ (t) = −ε1∫ 1

0ux(x , t)2dx − ε2

∫ 1+D

1vx(x , t)2dx

−2ε2D

v(1+ D, t)2

≤ −ε1∫ 1

0u(x , t)2dx − ε2

D2

∫ 1+D

1v(x , t)2dx


V̇ (t) ≤ −ε1∫ 1

0u(x , t)2dx − ε2

∫ 1+D

1v(x , t)2dx

≤ −2min{ε1, ε2}(12

∫ 1

0u(x , t)2dx +

12

∫ 1+D

1v(x , t)2dx

)

= −2min{ε1, ε2}V (t)


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V̇ (t) = −ε1∫ 1

0ux(x , t)2dx − ε2

∫ 1+D

1vx(x , t)2dx

−2ε2D

v(1+ D, t)2

≤ −ε1∫ 1

0u(x , t)2dx − ε2

D2

∫ 1+D

1v(x , t)2dx


V̇ (t) ≤ −ε1∫ 1

0u(x , t)2dx − ε2

∫ 1+D

1v(x , t)2dx

≤ −2min{ε1, ε2}(12

∫ 1

0u(x , t)2dx +

12

∫ 1+D

1v(x , t)2dx

)= −2min{ε1, ε2}V (t)


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In the case D ≥ 1, we use −1D2 ≥ −1 to get

V̇ (t) ≤ − ε1D2

∫ 1

0u(x , t)2dx − ε2

D2

∫ 1+D

1v(x , t)2dx

≤ − 2D2 min{ε1, ε2}

(12

∫ 1

0u(x , t)2dx +

12

∫ 1+D

1v(x , t)2dx

)= − 2

D2 min{ε1, ε2}V (t)

In the end:

V̇ (t) ≤ −2 min{ε1, ε2}max{1,D2}

V (t)


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V̇ (t) ≤ − ε1D2

∫ 1

0u(x , t)2dx − ε2

D2

∫ 1+D

1v(x , t)2dx

≤ − 2D2 min{ε1, ε2}

(12

∫ 1

0u(x , t)2dx +

12

∫ 1+D

1v(x , t)2dx

)

= − 2D2 min{ε1, ε2}V (t)

In the end:

V̇ (t) ≤ −2 min{ε1, ε2}max{1,D2}

V (t)


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V̇ (t) ≤ − ε1D2

∫ 1

0u(x , t)2dx − ε2

D2

∫ 1+D

1v(x , t)2dx

≤ − 2D2 min{ε1, ε2}

(12

∫ 1

0u(x , t)2dx +

12

∫ 1+D

1v(x , t)2dx

)= − 2

D2 min{ε1, ε2}V (t)

In the end:

V̇ (t) ≤ −2 min{ε1, ε2}max{1,D2}

V (t)


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Hence

Exponential Stability

V (t) ≤ V (0) exp(−2 min{ε1, ε2}

max{1,D2}t)


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Figure: Simulation using finite differences with ∆x = 0.1,∆t = 0.005. Thered lines represents x = 1.


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Figure: Plot of log(V (t)) against t and the best fit line.Theoretical slope ≤ −0.25. Simulated = −0.78.


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Table of Contents





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Problem formulation

ut(x , t) + u(x , t)ux(x , t) = 0, x ∈ (0, 1)u(0, t) = U0(t)u(1, t) = U1(t)

Issue: Even for smooth boundary conditions, the solution doesn’talways exist in a classical sense, but rather in a weak sense, withweak boundary conditions. Hence the solution can exhibit shocks.


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Equilibrium

Most general form of equilibrium for this equation:

Figure: Inviscid case equilibrium.


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Conclusion

We achieved stabilization of the coupled Burgers’ equation.Not shown in this presentation:

Equilibria analysis for the viscous case.Constant control in the viscous case.Linearized form of the Burgers’ equation (heat equation).


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Questions?

Figure: McDonald’s approves Burgers’ equations


l2 stabilization of coupled viscous burgers' equations · 1 3 u3(1;t) saad qadeer &...

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