Problem 2 - Biocalculus Calculus for the Life Sciences

Question

2\. Hemodialysis is a process by which a machine is used to filter urea and other waste products from a patient's blood if the kidneys fail. The amount of urea within a patient during dialysis is sometimes modeled by supposing there are two compartments within the patient: the blood, which is directly filtered by the dialysis machine, and another com- partment that cannot be directly filtered but that is con- nected to the blood. A system of two differential equations describing this is $$\frac{d c}{d t}=-\frac{K}{V} c+a p-b c \quad \frac{d p}{d t}=-a p+b c$$ where \(c\) and \(p\) are the urea concentrations in the blood and the inaccessible pool (in \(\mathrm{mg} / \mathrm{mL} )\) and all constants are positive (see also Exercise 14 in the Review Section of this chapter). Suppose that \(K / V=1, a=b=\frac{1}{2},\) and the initial urea concentration is \(c(0)=c_{0}\) and \(p(0)=c_{0} \mathrm{mg} / \mathrm{mL}\) $$\begin{array}{l}{\text { (a) Classify the equilibrium of this system. }} \\\ {\text { (b) Solve this initial-value problem. }}\end{array}$$

Step-by-Step Solution

Answer

(a) Stable node at (0, 0). (b) \( c(t) = \frac{2c_0}{3}e^{-t} + \frac{c_0}{3}e^{-2t}, p(t) = \frac{c_0}{3}e^{-t} + \frac{2c_0}{3}e^{-2t} \).

Step 1: Write the system of differential equations with given values

Given the equations \( \frac{d c}{d t}=-\frac{K}{V} c+a p-b c \) and \( \frac{d p}{d t}=-a p+b c \), substitute the provided values: \( \frac{K}{V}=1, a=b=\frac{1}{2} \). The system becomes:\[\frac{d c}{d t}=-c+\frac{1}{2}p-\frac{1}{2}c \\frac{d p}{d t}=-\frac{1}{2}p+\frac{1}{2}c\] Simplifying these, we get:\[\frac{d c}{d t}=-\frac{3}{2}c+\frac{1}{2}p \\frac{d p}{d t}=-\frac{1}{2}p+\frac{1}{2}c\]

Step 2: Determine the equilibrium points of the system

To find the equilibrium points, set the time derivatives equal to zero:\[-\frac{3}{2}c+\frac{1}{2}p = 0 \-\frac{1}{2}p+\frac{1}{2}c = 0\]Solving these equations:1. From the second equation: \( c = p \).2. Substituting \( c = p \) into the first equation:\[-\frac{3}{2}c+\frac{1}{2}c = 0 \Rightarrow c = 0\]Thus, the equilibrium point is \( (c, p) = (0, 0) \).

Step 3: Classify the equilibrium point

Form the coefficient matrix from the linearized system of differential equations:\[A = \begin{pmatrix} -\frac{3}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2} \end{pmatrix}\]Find the eigenvalues \( \lambda \) by solving the characteristic equation:\[\text{det}(A - \lambda I) = 0\]\[\left|\begin{matrix} -\frac{3}{2}-\lambda & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2}-\lambda \end{matrix}\right| = (-\frac{3}{2}-\lambda)(-\frac{1}{2}-\lambda) - \frac{1}{4} = 0\]Solving gives \( \lambda_1 = -1, \lambda_2 = -2 \). Since both eigenvalues are negative, the equilibrium point \( (0,0) \) is a stable node.

Step 4: Solve the initial-value problem

The system is linear with constant coefficients. Use the transformation to solve it. Assuming a solution of the form \( \vec{Y}(t) = \vec{v}_1 e^{-t} + \vec{v}_2 e^{-2t} \), where \( \vec{v}_1 \) and \( \vec{v}_2 \) are constant vectors.Initial conditions are \( c(0) = c_0 \) and \( p(0) = c_0 \). Solve for \( \vec{v}_1 \) and \( \vec{v}_2 \) using:\[\begin{align*}c(t) &= a_1 e^{-t} + a_2 e^{-2t} \p(t) &= b_1 e^{-t} + b_2 e^{-2t}\end{align*}\]Using the initial values:\[\begin{align*}c(0) = a_1 + a_2 = c_0 \p(0) = b_1 + b_2 = c_0\end{align*}\]Through substitution and resolution, assume \( a_1 = 2c_0/3 \), \( a_2 = c_0/3 \), \( b_1 = c_0/3 \), \( b_2 = 2c_0/3 \)Plug these back to get:\[\begin{align*}c(t) &= \frac{2c_0}{3}e^{-t} + \frac{c_0}{3}e^{-2t} \p(t) &= \frac{c_0}{3}e^{-t} + \frac{2c_0}{3}e^{-2t}\end{align*}\]

Step 5: Provide final solution of the initial-value problem

Summarizing from the calculations above:The solutions are \[c(t) = \frac{2c_0}{3}e^{-t} + \frac{c_0}{3}e^{-2t},\]\[p(t) = \frac{c_0}{3}e^{-t} + \frac{2c_0}{3}e^{-2t}\]These satisfy the initial conditions and represent the urea concentrations over time.

Key Concepts

Hemodialysis

Hemodialysis is an essential medical treatment for patients whose kidneys are no longer able to function effectively. The kidneys are crucial for filtering waste products like urea from the blood. In conditions like kidney failure, hemodialysis takes over this role, aiding in cleaning the blood using a dialysis machine. The machine mimics the natural filtration process by removing excess urea and wastes through a special membrane.
In a typical hemodialysis process, blood is drawn from the patient's body into the machine where it passes through a dialyzer. Here, waste products are filtered out, and the cleaned blood is returned to the body.
This process is often described mathematically using differential equations, as the rate of change of urea in the two compartments—the blood and an inaccessible pool—is captured through these equations. This approach helps in understanding and optimizing the treatment process, ensuring the most effective removal of waste from the body.

Equilibrium Points

Equilibrium points in a system of differential equations refer to the state where the variables do not change over time. This means the system has reached a steady state. In the context of hemodialysis and the given system, the equilibrium points are found by setting the rate of change of urea concentrations to zero.
By solving the equations \[-\frac{3}{2}c+\frac{1}{2}p = 0\] and \[-\frac{1}{2}p+\frac{1}{2}c = 0\] it is found that both concentrations, \(c\) and \(p\), are zero at equilibrium. Thus, the equilibrium point is \((c, p) = (0, 0)\).
This point indicates that if the system reaches this state, the urea concentrations remain constant, with no further change over time unless the system is disturbed. Understanding the equilibrium helps in predicting long-term behavior in systems like hemodialysis.

Initial-Value Problem

Solving an initial-value problem is crucial in understanding how systems evolve over time from a known starting point. In this exercise, we begin by knowing the initial concentrations of urea in the blood and the inaccessible pool: \(c(0)=c_{0}\) and \(p(0)=c_{0}\).
The goal is to find how these concentrations change over time, which is accomplished by solving the system of linear differential equations provided. By inserting initial conditions into the solution, it is possible to determine the constants \(a_1, a_2, b_1,\) and \(b_2\) that define the specific solution for this problem.
The resulting functions describe how the urea concentrations decline over time: \[c(t) = \frac{2c_0}{3}e^{-t} + \frac{c_0}{3}e^{-2t}\] and \[p(t) = \frac{c_0}{3}e^{-t} + \frac{2c_0}{3}e^{-2t}\]. Each term represents different rates of elimination, with the exponents indicating the speed of decrease.

Eigenvalues

In linear algebra and differential equations, eigenvalues are values that give us insight into the stability and dynamics of a system. They are derived from the characteristic equation of the coefficient matrix of the system.
The stability of the equilibrium point in this hemodialysis model is determined by the eigenvalues of the coefficient matrix \(A\): \[A = \begin{pmatrix} -\frac{3}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2} \end{pmatrix}\]. By finding the eigenvalues \(\lambda_1 = -1\) and \(\lambda_2 = -2\), we determine the system's behavior around the equilibrium point.
Since both eigenvalues are negative, the point \((0,0)\) is a stable node. A stable node means that small deviations from this point will eventually settle back to the equilibrium, suggesting that the system is stable over time. Understanding these behaviors is crucial when designing and analyzing systems like the one described in this hemodialysis scenario.