2. Hemodialysis is a process by which a machine is used to...

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 $p(0)=c_{0} \mathrm{mg} / \mathrm{mL}$ $$\begin{array}{l}{\text { (a) Classify the equilibrium of this system. }} \\ {\text { (b) Solve this initial-value problem. }}\end{array}$$

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 $p(0)=c_{0} \mathrm{mg} / \mathrm{mL}$
$$\begin{array}{l}{\text { (a) Classify the equilibrium of this system. }} \\ {\text { (b) Solve this initial-value problem. }}\end{array}$$

Key Concepts

Initial Value Problems

An initial value problem is a differential equation or a system of differential equations that is supplemented with specified values at a starting point, known as initial conditions. These problems are essential in modeling real-world scenarios, as the initial state of the system uniquely determines its subsequent behavior. Solving an initial value problem involves finding a function or set of functions that satisfy both the differential equations and the initial conditions.

Eigenvalue and Eigenvector Analysis

Eigenvalue and eigenvector analysis is a powerful method used to simplify and solve linear systems of differential equations. By finding the eigenvalues and corresponding eigenvectors of the coefficient matrix, one can decouple the system into independent modes that evolve exponentially. This approach not only facilitates the solution of the system but also provides insight into the qualitative behavior of its solutions.

Equilibrium Points

Equilibrium points refer to the states of a system where all derivatives vanish, meaning there is no change over time. Identifying these points is crucial in understanding the long?term behavior of a system. By setting the right?hand sides of the differential equations to zero and solving the resulting algebraic equations, one can determine the conditions for equilibrium.

Systems of Linear Differential Equations

This concept involves multiple interrelated differential equations that describe the evolution of several dependent variables with respect to an independent variable. Such systems are often represented in matrix form, which allows the use of linear algebra techniques for analysis and solution. They are widely used to model phenomena in various fields, including physics, biology, and economics, where interactions between components occur.

Stability Analysis

Stability analysis involves determining whether small perturbations or deviations from an equilibrium state will diminish or amplify over time. In the context of linear systems, this is often accomplished by examining the eigenvalues of the system's coefficient matrix. If all eigenvalues have negative real parts, the equilibrium is stable; if any have positive real parts, the equilibrium is unstable.

Transcript

00:01 So we have the following system and first we want to classify the equilibrium.

00:06 So in order to do so, we need to rewrite this system in matrix form.

00:12 So we have dcdt and dpdt, which we can just call dxdt as a vector.

00:30 And this is equal to some matrix times cp.

00:39 Because those are our variables.

00:42 And so let's fill in this matrix.

00:46 So let's rewrite d -c -d -t.

00:50 Since we're given that a and b are one -half and k over v is one, this is just equal to negative c plus one -half -p minus one -half -c, or one -half -p minus three -half -c.

01:12 So we can fill that in in our matrix here.

01:15 So p needs to be multiplied by one -half, and c needs to be multiplied by negative three -half.

01:23 We can rewrite our second equation also as negative one -half p plus one -half c.

01:31 So p needs to be multiplied by negative one -half, and c needs to be multiplied by positive one -half.

01:39 So this is our matrix, so we need to find eigenvalues and eigenvectors.

01:45 So we've done this a lot by this point.

01:49 So we just find the determinant of this matrix a minus lambda i, and we get the lambda 1 and 2 are equal to negative 1 plus or minus square to 2 over 2.

02:09 And then we solve the system, a minus lambda i times v equals 0 to get our eigenvectors, which come out to be 1 and 1 plus or minus square to 2.

02:33 So to classify the equilibrium, which is what this question is all about, our eigenvalues, let's just rewrite those as a reminder, we have that they are the same sign, which tells us that we have a node.

02:57 And what is their sign? their sign is negative.

03:02 So this tells us that we have a stable.

03:05 Equilibrium.

03:08 So we have a stable node.

03:15 Now for part b we need to solve the initial value problem.

03:19 So we know our general solution is equal to some constant times e to the lambda 1 t.

03:30 So we have negative 1 plus square root of 2 over 2 t times corresponding item vector 1 and 1 plus square root of 2 plus plus another constant, e to the lambda 2t, so we have minus, and corresponding eigen vector with a minus sign.

03:56 So in order to solve for the c1 and c2, we need to use the initial condition.

04:02 So using the initial condition, x of 0 is equal to c of 0 times p of 0, which are both given to be these value, c not.

04:27 So let's actually plug zero into our general solution.

04:30 We see that the exponentials go away.

04:33 So we get c1 times the first eigenvector plus c2 times the second eigenvector.

04:51 So this tells me that c1 plus c2 equals the c not value.

05:00 So this is just the c not is just the number...