Special inverse problems

Now we derive the operator for special inverse problems of partial differential equations. The derivation is in no way rigouros. We neglect subtleties such as the exact domains of definition. The essential tools are Gauss' integral theorem and Green's formula.

1. Impedance tomography.
Let , n = 2,3 be sufficiently regular, and let be a function in , the conductivity of the material in . We want to determine from measurements of currents and voltages , , on . The currents are usually applied at pairs of electrodes on .

The mathematical model is as follows. The potential satisfies

 

We normalize by requiring that the mean value on is zero. The measurements provide us with the values of

 

We have to determine from (3.1)-(3.2) for . We assume to be known on .

It is clear that (3.1) can be written in the form of (1.1) with , , , and playing the role of f. Also, (3.2) is a special case of (1.2) with which is simply with the (known) scalar product , and we have

where is the solution of (3.1). Hence, (2.2) reads

 

   

Proof: For any z, w we have

Choosing from (3.3) and z from (3.5) we obtain

Since on an integration by parts yields

or

This holds for each vanishing on , verifying (3.4).

In the light of the theorem, the iteration (1.4) proceeds as follows.

For

We see that each sweep of the iteration requires the solution of 2p boundary value problems of nearly identical shape. This is not only easy to program but also very cheap computationally, compared to other methods.

2. A parabolic inverse problem.
Consider the initial-boundary value problem

 

The source terms are assumed to be known, while the diffusion coefficient has to be determined from the knowledge of on :

 

Again we assume to be known on .

This problem is of the form (1.1)-(1.2). We have

The operator is given by

where is the solution of (3.6). According to (2.2), the Fréchet derivative is

 

  

Proof: For any z, w we have

For , z from (3.8)-(3.9) this reads

An integration by parts on the left hand side turns this into

hence the theorem.

The algorithm (1.4) is as follows: For

Solve

and

Put

Alternatively we may consider the boundary condition on instead of in (3.6), and on as data in (3.7). Then,

and

where now satisfies (3.8) with the boundary condition on . The adjoint is given by the same expression as above, but with the boundary condition z = g on .

3. Laser tomography: Stationary case

Here the governing equation is

 

where

with the exterior normal to at . The source term q is delta-like, and the sources s are on . is assumed to be known. We want to recover a, b from knowing u on for sources . Denoting by the solution of (3.10) for we have

 

The information on u on which we want to make use of is

 

Obviously, (3.11)-(3.12) is of the type of (1.1)-(1.2) with f = (a,b) and

The operator is given by

where is the solution to (3.11). The linearized operator is

where w solves

 

  

Proof: For any z, w we have

For z, w from (3.13)-(3.14) this implies

hence the theorem.

The algorithm is as follows:

For

Solve

and

Put

4. SPECT

In SPECT one has to compute the source distribution f from

 

 

where .

Here, is a bounded domain in , and are direction vectors . If a is known, then (3.15)-(3.16) reduces to a linear problem which can be solved by inverting the attenuated Radon transform. We consider the case in which a is unknown, too.

(3.15)-(3.16) is of the form (1.1)-(1.2) again, with (a,f) playing the role of f and

The operator is given by

where is the solution of (3.15). The linearized operator is

where solves

 

  

Proof: For any z, w we have

For z, w as in (3.13)-(3.17) this means

hence the theorem.

The algorithm is:
For

Solve

 

and

 

Put

The forward problem (3.19) can be solved by inverting the attenuated Radon transform, and (3.20) is the attenuated backprojection. Thus the algorithm is very similar to the ART algorithm in computerized tomography and in fact reduces to this algorithm if a is known, e.g. a = 0.

5. Ultrasound tomography: Time harmonic case
Here we have the Helmholtz equation

where with the scattered wave v satisfying the radiation condition

Here, f is a complex valued function which has to be recovered from knowing u for each direction outside the support of f. is a real parameter controlling the spatial resolution.

Let be a ball containing supp(f) in its interior. We rewrite the radiation condition as a boundary condition on , to wit

with a linear operator B on . Assume that the measurements are made for finitely many directions , and let be the solution for direction . Then,

 

and f has to be determined from

 

Obviously, (3.21)-(3.22) is of the form (1.1)-(1.2), with

The operator is

where solves (3.21). The linearized operator is given by , where is a solution of

 

  

Proof: For any z, w we have

For the functions z, from (3.24)-(3.23) this yields

or

6. Ultrasound tomography: Time resolved case
Here we have the wave equation

 

with initial and boundary conditions

 

with the exterior normal on . The source term q is like and the sources s sit on . The problem is to recover the sound speed c in from the measurement of u on , or a part thereof, for sources . We assume with f = 0 on . Again, this problem is obviously a special case of (1.1), (1.2) with

The linearized operator is given by

where is the solution of

 

with the solution to (3.25), (3.26) for and .

  

Proof: For arbitrary w, z we have

For from (3.27) and z as in (3.28) this reads

hence

If u is measured only on a part of , then Z has to be replaced by . With E the extension from to by zero, the corresponding adjoint is then given by .

7. Laser tomography in the diffusion approximation:

Time harmonic case.
Here the partial differential equation is

The modulating frequency is fixed, and is the measured data for the source .

Putting we obtain

 

The problem is to recover , a from the . It is clear that (3.29) is of the form (1.1-2). We have

being considered as an operator from into . We readily compute the operator , obtaining

where solves

 

Proof: The proof is done by verification. We compute

by way of Green's theorem. The integral over vanishes, and the integral over can be rewritten as

which, by the differential equations for z and , can be written as

Hence

8. Laser tomography in the diffusion approximation:

Time resolved case

In this case the governing equation reads

 

The source terms are assumed to be -functions at t=0 and . Now the invers problem is to determine both the diffusion coeffient and the absorption coefficient a(x) from the given data

As before this problem is of the form (1.1)-(1.2).

The operator is given by

where is the solution of (3.30). According to (2.2) we calculate the linearized operator to be

where solves the equation

 

 

Proof: For any z,w we have

For from (3.31)-(3.32) this reduces to

An integration by parts on the left hand side leads to

This proofs the theorem.