Notes on Compartment Models and Pharmacokinetics
October 1998

Compartment models describe the flow of a substance (chemicals, drugs, information) between the components of a larger system. In general, there are N components (the compartments) and they may be linked to each other in any way . The flow rates or exchange constants between linked compartments are generally specified, and each compartment may also have output and input to and from the outside world.

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As we saw in studying stirred tank reaction, the governing equation for x(t) is

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subject to the initial condition x(0)=a. The rate constant k>0 specifies how quickly the substance is removed from the compartment.

Laplace transforms can be used to solve this initial value problem. To anticipate what happens with more than one compartment, let's review the process. We will let tex2html_wrap_inline249 denote the Laplace transform of x(t). Taking tranforms of both sides of the ODE, we have

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Using the derivative property, we have

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Substituting x(0)=a and solving for X(s) gives us the transform of the solution:

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where a and F(s) are known.

We can now invert the transform to recover the solution x(t). We see that

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We use the inverse transform of the exponential function for the first term and the convolution theorem on the second term (some review may be needed!):

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The first term of the solution reflects the effect of the initial condition (which is transient and eventually vanishes); the second term reflects the effect of the input. This solution could also have been obtained using an integrating factor. But the Laplace transform methods generalizes to systems of first order linear ODEs.

Notice that if we take a=0, then the transform of the solution can be written

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The function G is called the transfer function because it relates the transform of the input, F(s) to the transform of the output, X(s).

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We can now derive the governing equations for the N-compartment system. Let tex2html_wrap_inline313 denote the mass of substance in compartment i at time t. Over a small increment of time tex2html_wrap_inline319 , the change in mass in compartment i is given approximately by

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for tex2html_wrap_inline327 . In a familiar maneuver, we move tex2html_wrap_inline313 to the left side of the equation, divide through by tex2html_wrap_inline319 , and take the limit as tex2html_wrap_inline333 . This leads to the ODEs

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for tex2html_wrap_inline327 .

We will let tex2html_wrap_inline337 and note that tex2html_wrap_inline339 . Let tex2html_wrap_inline341 be the N-vector tex2html_wrap_inline345 , and let tex2html_wrap_inline347 be the N-vector tex2html_wrap_inline351 . Then the system of N ODEs can be written in matrix-vector forms as

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where the tex2html_wrap_inline355 matrix A has entires tex2html_wrap_inline357 . We will assume the initial conditions have the form tex2html_wrap_inline359 .

How do we solve a system of linear, constant coefficient, nonhomogeneous ODEs? No method is particularly easy to carry out. But conceptually, Laplace transforms work well. We can take the Laplace transform of a vector of functions one component at a time. We will let tex2html_wrap_inline361 . Taking transforms across the system tex2html_wrap_inline363 , we get

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We now have to solve for tex2html_wrap_inline365 using matrix-vector operations. Let I be the tex2html_wrap_inline355 identity matrix and recall that tex2html_wrap_inline371 . Then

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Multiplying both sides by tex2html_wrap_inline373 gives u

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Now what about the inversion? There are several ways to proceed. One is to mimic the one-compartment case and use a matrix version of the convolution theorem. This requires making sense of the matrix exponential tex2html_wrap_inline375 , which can be done. However, we will take another path that may require more work, but leads to some results. For the moment let's take tex2html_wrap_inline377 so we can focus on the effect of the input. Then we have

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In analogy with the one-compartment case, the tex2html_wrap_inline355 matrix tex2html_wrap_inline381 is called the transfer matrix because it relates the input to the output of the system. The key is to invert this matrix. Recall a result from linear algebra which gives the inverse of a matric B as

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where adj(B) is the transpose of the matrix of cofactors of B and det(B) is the determinant of B. Therefore, we have

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Now recall det(sI-A) is the Nth degree characteristic polynomial of A. The roots (which we will assume to be distinct) are the eigenvalues, tex2html_wrap_inline399 , of A.

At this point it probably doesn't pay to proceed in generality. But if the matrix of cofactors, adj(sI-A), can be computed without too much effort, then a solution in closed for can be found. A few examples will illustrate the method.

Mon Nov 2 16:55:55 MST 1998