mohsen tadi: active research areasmohsen tadi: active research areas • systems & control...

UC-Denver Mech. Engng.

Mohsen Tadi: Active Research Areas

• Systems & Control

Practical stabilizability of dyadic homogeneous

bilinear systems, Asian Journal of Control

Vol. 14(3) pp. 1-7 (2012)

• Stefan Problems (Melting, Solidification)

A four-step fixed-grid method for 1-D Stefan

problems, ASME Journal of Heat Transfer

Vol. 132, pp. 114502(1-4) (2010)

• Inverse Problems: (Topic of this talk)


Parameter ID. & Inverse Problems

Rate Constants in Reactions: k

OH +H2CO −→ H2O+HCO

x1 = [OH], x2 = [H2CO],

x3 = [H2O], x4 = [HCO]

x1 = −kx1x2

x2 = −kx1x2

x3 = kx1x2

x4 = kx1x2, , y(t) = x1

Problem: Recover k based on y(t).


Inverse Problems (some applications:)

• Various problems in chemical kinetics

• Nondestructive evaluation of materials

• Identification of land mines

• Identification of unhealthy (cancerous) tissues

• Identification of incoming directed energy


Continuous Systems

• Hyperbolic Systems (Wave propagation)

• Parabolic Systems (Diffusion)

• Elliptic Systems (Equilibrium)

Existing methods:

• Optimization methods (adjoint, gradient)

Work best for hyperbolic systems

• Newton method, Born approx., Kalman Filtering....

Recent methods:

• Quzi-Reversibility (Tadi-Klibanov)

• Convexification method (Klibanov)

• No-name yet (present talk)


Items discussed in the talk

• A Hyperbolic problem (gradient method)

• Present method (applicable to any dimensions)

◦ Present a number of parabolic problems

◦ Consider one parabolic problem

◦ Consider two elliptic problems

• Discuss open problems

• Conclusions & questions & remarks


Atmospheric Beam Propagation: Maxwell’s Eqns.

ǫ∂E

∂t= ∇× B,

∂B

∂t= −∇× E,

∇.(ǫE) = 0, D = ǫE, B = µH

E: Electric field B: Magnetic field

µ: permeability ǫ: Index of refraction

For atmospheric application: µ = 1, ǫ = ǫ(x)

ǫ(x)∂2E

∂t2= ∇2E + ∇.[E.∇(ln(ǫ(x)))]

ǫ(x) : domain property (Index of Refraction)

Unknown


Atmospheric Beam Propagation:

ǫ(x)∂2E

∂t2= ∇2E + ∇.[E.∇(ln(ǫ(x)))]

Problem 1: Compute ǫ(x) for the domain of

interest

Problem 2: Compute E(x), i.e. identify the

existance of a directed energy


Consider a 2-D domain

ǫ(x)∂2E∂t2

= ∇2E + ∇.[E.∇(ln(ǫ(x)))]

Absorbing B.C.

Absorbing B.C.

⇒Ab.B.C.

••Sensors

6

ℓ

Assume a function for ǫ:

ǫ(x) = 1 + 0.05sin(3πx)cos(4πy)

Recover ℓ based on sensor data u(t), v(t)

E = ui+ vj


0 0.2

0.4 0.6

0.8 1 0

0.2

0.4

0.6

0.8

1

-1.5

-1

-0.5

0

0.5

1

1.5u(x,y)

A Simple Laser Beam, u(x,y), w:1000

x

y

u(x,y)


0 0.2

0.4 0.6

0.8 1 0

0.2

0.4

0.6

0.8

1

-0.0006

-0.0004

-0.0002

0

0.0002

0.0004

0.0006

v(x,y)

A Simple Laser Beam, v(x,y)

x

y

v(x,y)


Gradient type algorithm: Define Cost

J =

∫ τ

0[u(t) − u]2 + [v(t) − v]2dt

u(t), v(t): Components of E from the

simulation

I. Assume a guess ℓ

II. Obtain forward responses u(t), v(t)

III. Obtain the gradient (Sensitivity) ∂J∂ℓ

IV. update according to ℓ = ℓ− ∂J∂ℓ

Go to II.


Robust evaluation of ℓ

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

0 1 2 3 4 5 6 7

\ell

Iterations

Convergence


Parabolic Inverse Problems

Diffusion coefficient

ut = (d(x)ux)x, u(0, x) = g0(x)

d(x)u(t,0) = f0(t)

ux(t,0) = q0(t)

u(t,1) = f1(t)

ux(t,1) = q1(t)

Absorption coefficient

ut = duxx − a(x)u, u(0, x) = g0(x)

a(x)u(t,0) = f0(t)

ux(t,0) = q0(t)

u(t,1) = f1(t)

ux(t,1) = q1(t)


Ill-Posed Parabolic problem

ut = uxx, x ∈ [0 : 1], u(0, x) =??

u(t,0) = f0(t)

ux(t,0) = q0(t)

u(t,1) = f1(t)

ux(t,1) = q1(t)


Absorption coefficient

ut = uxx − a(x)u, x ∈ [0 : 1], u(0, x) = 1

u(t, x) light intensity > 0, let u = ev, then

vt = vxx + v2x − a(x)

Let P = vt and obtain ∂P∂t

Pt = Pxx + 2Px

∫ t

0Pxdτ P(0, x) = a(x) =??

P(t,0) = f0(t)

Px(t,0) = q0(t)

P(t,1) = f1(t)

Px(t,1) = q1(t)

An Ill-posed parabolic problem


Ill-Posed Parabolic System

ut = uxx, u(0, x) =??

u(t,0) = f0(t)

ux(t,0) = q0(t)

u(t,1) = f1(t)

ux(t,1) = q1(t)

I : Assume u(0, x) = g0(x),obtain background

ut = uxx, u(0, x) = g0(x)

u(t,0) = f0(t) u(t,1) = f1(t)

II : Let e(t, x) = u− u, error field given by

et = exx, e(0, x) = u(0, x) − g0(x) =??

e(t,0) = 0

ex(t,0) = h0(t)

h0 = q0 − ux

e(t,1) = 0

ex(t,1) = h1(t)

h1 = q1 − ux


III: Obtain I.C. for the error

et = exx, e(0, x) = u(0, x) − g0(x) =??

e(t,0) = 0

ex(t,0) = h0(t)

e(t,1) = 0

ex(t,1) = h1(t)

IV: Consider two well-posed problems

i:)

et = exx, e(0, x) =??e(t,0) = 0 ex(t,1) = h1(t)

ii:)

et = exx, e(0, x) =??ex(t,0) = h0(t) e(t,1) = 0


Obtain finite-element formulations

i :e =∑

j φj(x)ηj(t), ii :e =∑

j ψj(x)δj(t)

let s⊤ = [η1, ..., ηn]⊤, r⊤ = [δ1, ..., δn]

⊤,

s = A1s + b1(t), r = A2r + b2(t)

i : s(t) = eA1ts(0) +

∫ t

0eA1(t−τ)b1(τ)dτ

ii : r(t) = eA2tr(0) +

∫ t


e(t, x) =∑nj=1 φjηj(t) =

∑nj=1ψjδj(t)

< φi, φj > ηj(t) =< φi, ψj > δj ⇒ s(t) = Br(t)

B = [< φi, φj >]−1[< φi, ψj >]


i : s(t) = eA1tBr(0) +

∫ t


ii : Br(t) = BeA2tr(0) + B

∫ t


Solve for r(0), G(t)r(0) = g(t)

G(t) =[

eA1tB − BeA2t]

,

g(t) = B

∫ t

0eA2(t−τ)b2(τ)dτ −

∫ t


G(t) is singular, but G(t)r(0) = g(t) true ∀ t

G(t1)G(t2)...βΓ

r(0) =

g(t1)g(t2)...

0

V: Obtain r(0), go to I and repeat.

t = T8t = T7t = T6t = T5t = T4t = T3t = T2t = T1

Number of Eigenvalues

Norm

alized

Eig

envalu

es

14121086420

1

0.01

0.0001

1e-06

1e-08

1e-10

1e-12

1e-14

1e-16

ActualComputed

X

u(0,x)

10.80.60.40.20

1

0.8

0.6

0.4

0.2

0

-0.2

Final Itr.Computed

Actual

X

u(0,x)

10.80.60.40.20

1

0.8

0.6

0.4

0.2

0

-0.2


Summary

• Applicable to multidimensional problems

• Signal-to-noise ration %8

• More accurate results than previous methods

• Fewer iterations

• Proof of convergence not-yet-done


Evaluation of ǫ(x)

start with ǫ(x)∂2E

∂t2= ∇2E + ∇.[E.∇(ln(ǫ(x)))]

For a periodic electric field E(t,x) = u(x)cos(ωt)

∇2u + ω2ǫ(x)u = −∇[u.∇(log(ǫ))]︸︷︷︸

α

(α ≈ 0) for atmospheric beam propagation

∇2u + ω2ǫ(x)u = 0, Helmholtz Eqn.

UC-Denver Mech. Engng.Consider a 2-D domain

∇2u+ ω2ǫ(x)u = 0

=======

collect un

Incident field u = cos(ω(xcos(θ) + ysin(θ)))

•Free to choose ω and θ

•Recover ǫ(x) based on the incident and the collected data

Other applications

Med. Imaging.

Land mine detec.


An Iterative Algorithm

I: Assume a guess ǫ, obtain background field

∇2u+ ω2ǫ(x)u = 0

II: The error field e = u− u, also, ǫ = ǫ+ h

∇2e+ ω2e+ ω2hu = 0

III: Linearize around the background field

∇2e+ ω2e+ ω2uh(x) = 0

IV: Obtain 2 Eqns. for 2 unkns. e(x) and h(x)

V: Solve for h(x), update ǫ(x) = ǫ(x) + h(x)

go to I.


III: Linearized Error Field

∇2e+ ω2ǫe+ ω2uh = 0

=======

en = f(y)

e = 0e = 0

e = 0

e = 0

IV: Obtain 2 Equations for e(x) & h(x)

∆e+ ω2ge+ ω2hu = 0 ∆e+ ω2ge+ ω2hu = 0e = 0 en = f(y)

problem I problem II

V: Eliminate e(x), Solve for h(x), update, go to I.



∇2u+ ω2ǫ(x)u = 0

=======

collect un

Incident field u = cos(ω(xcos(θ) + ysin(θ)))

• 9 incident angles: π18,

π16,

π15,

π13,

π11,

π8,π7,

π4,π3

• 4 incident frequences: 10π,15π,17π,19π

• Treat data for each frequency separately.

• Add noise, noise-to-signal ration 8%.

• Run 35 iterations for every incident frequency.


0 0.2

0.4 0.6

0.8 1 0

0.2

0.4

0.6

0.8

1

1

1.2

1.4

1.6

1.8

2

Exact function

x

y


0 0.2

0.4 0.6

0.8 1 0

0.2

0.4

0.6

0.8

1

0.8

1

1.2

1.4

1.6

1.8

2

Recovered Function

x

y


0

0.2

0.4

0.6

0.8

1 0 0.2

0.4 0.6

0.8 1

1

1.2

1.4

1.6

1.8

2

Exact function

x

y


0

0.2

0.4

0.6

0.8

1 0 0.2

0.4 0.6

0.8 1

0.8

0.9

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

Recovered Function

x

y


k = 19k = 17k = 15k = 10Exact

cross section:

x

g(x,0

.3)

10.80.60.40.20

2

1.8

1.6

1.4

1.2

1


k = 19k = 17k = 15k = 10Exact

cross section:

y

g(0.3

,y)

10.80.60.40.20

2

1.8

1.6

1.4

1.2

1


Cauchy Problem

Periodic field E(t,x) = u(x)cos(kt), ǫ(x) ≈ 1

∇2u + k2u = 0, Helmholtz Eqn.


∇2u+ k2u = 0

‖ ‖ ‖ ‖ ‖ ‖ ‖ ‖ ‖ ‖ ‖collect un

g(y) f(y)

Unknown condition


An iterative algorithm

1: Assume a guess for the unknow condition h(x)

2: Obtain the background field u(x)

3: Obtain the error field e(x), including the B.C.

4: Solve for the error field, the correction to h(x)

5: Update the assumed guess, repeat 1-5,

3: Combine two methods to solve for e(x)


3: Obtain the error field e(x)

1: assume a guess,⇒ 2: Obtain u(x) ⇒

3: Obtain B.C. for e(x) = u− u

∇2e+ k2e = 0

e = 0,ey = q(x)

e = 0 e = 0

e = h(x) =?


Method 1: Consider two problems for e(x):

Let L = ∆ + k2

Le = 0 Le = 0e = 0 e = 0

ey = q(x)

e = h(x)

e = 0

e = 0

e = h(x)

problem I problem II

◦ Obtain two equations

◦ Eliminate e, obtain an equation for h(x)

Method 2:A different method to obtain e(x, y)


3: Obtain the error field e(x)

1: assume a guess,⇒ 2: Obtain u(x) ⇒

3: Obtain B.C. for e(x) = u− u

∇2e+ k2e = 0

e = 0,ey = q(x)

e = 0 e = 0

e = h(x) =?

Assume: e(x, y) =∞∑

j=1

Aj(y)sin(jπx)


Orthogonality, apply field Eqn:

Aj(y) = 2

∫ 1

0e(x, y)sin(jπx)dx

d2

dy2Aj(y)+k2Aj(y) = 2(jπ)2

∫ 1

0e(x, y)sin(jπx)dx

d2Ajdy2

+ (k2 − (jπ)2)Aj = 0 if k > jπ

d2Ajdy2

+ k2Aj = 2(jπ)2∫ 10 esin(jπx)dx otherwise

For k > jπ ⇒ Exact solution (Sine, Cosine)

For k < jπ ⇒ Iterate for Aj(y)


Aj(0) = 0,dAj

dy|0 = 2

∫ 1

0ey(x,0)sin(jπx)dx

for k > jπ ⇒ exact solution

for k < jπ ⇒ set up an iteration

1 : Use Differential form to obtain e(x, y)

2 : Obtain the function on the right-hand-side

d2Ajdy2

+ k2Aj = 2(jπ)2∫ 10 esin(jπx)dx = gj(y)

3 : Use (Variation of parameter, numerical)

h(x) = e(x,1) =∑∞j=1Aj(1)sin(jπx)


3: error field e(x): Method 1

1) Use differential method to get h ⇒ e(x, y)

[P][h] =

q

00

2) With h, (e(x,1) = h), Obtain e(x, y), ⇒ gj(y)

3) Obtain a better estimate of e(x,1)

h = e(x,1) =∑∞j=1Aj(1)sin(jπx)

4) Use the new result as regularization

[

P

β2I

]

[h] =

q

00β2h


2-itractual

x

u(x,1

)

10.80.60.40.20

2

1.8

1.6

1.4

1.2

1

0.8

SNR=0.04, β = 45, β2 = 0.1, k = 26, Error:

104 → 1


2-itractual

x

u(x,1

)

10.80.60.40.20

2.2

2

1.8

1.6

1.4

1.2

1

0.8

SNR=0.04, β = 45, β2 = 0.1, k = 26, Error:

104 → 1


Inverse Problems:Open questions and future research work

• Proof of convergence

• Apply to actual data (joint work with bioengineering)

• Numerical methods for very high frequency

electromagnetic equations

◦ Helmholtz Eqn.

◦ Time-dependent hyperbolic system

• Singular perturbation approach for very high

frequency field Eqns.

• Exact solutions for EM equations (Green’s functions)


Recent relevant publications:•’An inverse problem for Helmholtz equation’,

Inverse Problems in Science & Engineering,

Vol. 19(6), pp. 839-854 (2011).•’Evaluation of Diffusion Coefficient Based on Multiple

Forward Problems’, Applied Mathematics and Computation,Vol. 216(12), pp. 3707-3717 (2010).

•‘A Computational Method for an Inverse Problem in

Optical Tomography’Discrete and Continuous Dynamical Systems-B,

Vol. 12(1), pp. 205-214 (2009).•‘Solution of an Ill-Posed Parabolic Equation Based on

Multiple Background Fields’,Advn. in Math. Sci. Journal,

Vol. 1, pp. 1-20 (2009).

Questions, Comments, Suggestions

mohsen tadi: active research areasmohsen tadi: active research areas • systems & control...

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