The following is a summary of some of the work I did with my advisor Graeme Milton; the results have been submitted to Communications in the Mathematical Sciences, and a preprint is posted here.
In organ transplant surgery one measure of the quality of the organ to be transplanted is the percentage of damaged/dead cells in the organ. It is likely not feasible for a surgeon to cut the organ open and examine it before giving it to the patient, so electrical impedance tomography (EIT) offers a cheap and non-invasive alternative for looking inside the organ to measure its quality. The idea behind EIT can be described in the following way.
The conductivity of a material essentially describes how well that material conducts electric current, according to Ohm's Law. Good conductors (such as metals) have high conductivities while poor conductors (such as plastics, rubbers, and wood) have low conductivities. Different biological tissues (such as dead and living organ cells) typically have different conductivities--EIT exploits this difference in conductivity to create an image of the interior of an organ. We begin by attaching several electrodes to the boundary of the organ in question; we then apply a small current through these electrodes. The current will flow through the organ (without damaging it, since the applied current is small--this is in contrast to a CAT scan or an X-ray, where the radiation used for imaging destroys some of the organ in the process); in particular the current will be more concentrated in the regions of the organ with highest conductivity. We then measure the corresponding voltage on the boundary of the organ, so in the end we know both the current and voltage on the boundary of the organ--using this measurement, we can estimate the percentage (by volume) of damaged or dead cells in the liver and thus determine whether or not it is healthy enough to be transplanted into the patient. Many people also use EIT to construct images of the interior of the organ, such as in real-time monitoring of lung function. Electrical impedance tomography can also be used in the non-destructive testing of materials to determine, for example, the quality of a beam or other support structure.
In our work, rather than finding an image of the conductivity distribution, we derive bounds on the volume fraction occupied by an inclusion in a body. In the context of the above example, the bounds we derive are estimates on the percentage of damaged/dead cells in an organ. The above figure is a generic illustration of our setting. The inclusion (e.g., dead/damaged cells) is colored blue, the body (healthy organ cells) in question is colored red, and the boundary of the body (boundary of the organ) is the black curve. We simplify the problem considerably by considering the organ to be composed of two tissues--dead/damaged cells and healthy cells. We also assume that the conductivity in each tissue is a known complex constant, possibly obtained from a separate experiment. (As described in our paper, much work has been done in the case when the conductivities are assumed to be real, but not as much work has been done when the conductivities are assumed to be complex. The complex conductivity problem is more realistic in the sense that it takes into account the fact that tissues dissipate energy.) In the figure the inclusion is represented by only one rather large region--however, it could be divided into several smaller parts scattered throughout the organ and our method will still work. We apply a single current to the boundary of the body and measure the corresponding voltage around the boundary. Using this measurement, we estimate the quantity
which is called the volume fraction of the inclusion (the percentage by volume of the body that is occupied by the inclusion).
It is probably easiest to understand our method as applied to a specific example, illustrated below. In this case the body under consideration is circular and colored red while the inclusion is the blue annulus (ring). In this case we apply a voltage to the boundary (the black circle) and measure the resulting current (above we considered applying a current on the boundary and measuring the corresponding voltage--either procedure will give similar results for our problem). Using our method, we are able to derive some elementary bounds, namely:
organ(I put organ in quotes because no organ is perfectly circular like in the picture to the right--the circular geometry is used to simplify some computations. Our work holds for essentially any other shape, not just circles). This means somewhere between 79.4% and 80.8% of the volume of the organ is made up of healthy cells.
Note that the only pieces of information we assume we have are our measurement of the voltage and current on the boundary and the values of the conductivities of the healthy and damaged/dead cells (likely obtained from a different experiment). Also, for the particular example under consideration, the true volume fraction of the inclusion is
i.e., the annulus (of damaged/dead cells) actually occupies 20% of the volume of the body (organ) in question (so our bounds of 19.2% and 20.6% are quite tight).
Although the elementary bounds are pretty good, we would like to know if we can do even better. It turns out that we can, at least sometimes. We begin by defining a test value f. We can show that for each test value there is an ellipse associated with the inclusion and another ellipse associated with the body. As f varies, the ellipses move and change size. We can show mathematically that a given value of f could be the true volume fraction f1 if the ellipses intersect or if one is inside the other. If the ellipses do not intersect and neither is inside the other, then that particular value of f cannot be the true volume fraction. This gives us the following procedure for deriving tighter bounds on the volume fraction.
- Choose a test value f somewhere between the elementary bounds 0.192 and 0.206 (we only need to consider test values in this interval instead of all possible values between 0 and 1 because we have already proven that the true volume fraction must be between 0.192 and 0.206);
- draw the 2 ellipses (one associated with the inclusion and the other associated with the body);
- if the 2 ellipses do not intersect and neither is inside the other, we can eliminate the current value of f--it cannot be the true volume fraction f1;
- otherwise f could be the true volume fraction.
In the figure below we plot the ellipses for values of f between 0.192 and 0.206. In particular, the ellipses in (a) correspond to f = 0.206 (where the red ellipse is just a single point, represented by the red dot); the value of f is decreased as we move from (a)-(h) until we reach f = 0.192 in (h) (where the blue ellipse is just a single point, represented by a blue dot). The red ellipse is associated with the body while the blue ellipse is associated with the annular inclusion. Notice that either one of the ellipses is inside the other or the ellipses intersect for all the values of f shown in the figure; this means that any of those values could be the true volume fraction, so we cannot eliminate any of them from consideration. Ultimately, we find that the bounds from the ellipses are the same as the elementary bounds: 0.192 and 0.206. Note: the black rectangle in the figure is related to the elementary bounds and is described in more detail in our paper.
So far, everything we have discussed holds in both 2 and 3 dimensions. In 2 dimensions, however, we can use additional computations to derive a second set of improved elementary bounds. For the annular case considered above, these improved elementary bounds are
In 2 dimensions, we can also construct 2 new ellipses that lead to even better bounds (at least in the annular case). In the figure below, we show a plot of the ellipses at various test values f: in (a) f = 0.202 and we decrease f to 0.198 as we move from (a)-(h) in the figure. The previous ellipses from the above figure are shown with dashed curves in this case. These new ellipses do not intersect and neither is in the other in (a) (where the red ellipse is a point), (b), (g), and (h) (where the blue ellipse is a point), so we can eliminate those test values from consideration. Any of the values in (c)-(f) could potentially be the true volume fraction. Therefore, the best bounds we obtain in this case are
