The Ordinary Differential Equations Project

We will use linear systems of differential equations to illustrate how we can use systems of differential equations to model how subtances flow back and forth between two or more compartments. Suppose that we have two tanks ($A$ and $B$) between which a mixture of brine flows (Figure 3.2.1). Tank $A$ contains 300 liters of water in which 100 kilograms of salt has been dissolved and Tank $B$ contains 300 liters of pure water. Fresh water is pumped into Tank $A$ at the rate of 500 liters per hour, and brine is pumped into Tank $B$ from Tank $A$ at the rate of 900 liters per hour. Brine is also pumped back into Tank $A$ from Tank $B$ at the rate of 400 liters per hour, and an additional 500 liters of brine per hour is drained from Tank $B\text{.}$ All brine mixtures are well-stirred. If we let $x=x(t)$ be the amount of salt in Tank $A$ at time $t$ and $y=y(t)$ be the amount of salt in Tank $B$ at time $t\text{,}$ then we know that

We know that the salt concentrations in the two tanks are $x/300$ kilograms per liter and $y/300$ kilograms per liter. Thus, we can describe the rate of change in each tank with a differential equation,

Matrix notation gives us a convenient way of representing the $2\times 2$ system (3.2.1)–(3.2.1). If we let

then we can rewrite our system as

In other words, we can write our system as

where

A linear planar system

has an equilibrium solution at $(x_0,y_0)$ if

The following proposition tells us exactly where to find the equilibrium solutions of a linear system with constant coefficients.

Now let us attack the problem of finding all of the solutions of the system ${\mathbf{x}}^{\prime }=A\mathbf{x}\text{.}$ Suppose that we can find a nonzero vector ${\mathbf{v}}_0$ such that $A{\mathbf{v}}_0=\lambda {\mathbf{v}}_0$ for some real number $\lambda \text{.}$ In this case, the matrix $A$ just sends the vector ${\mathbf{v}}_0$ to a vector on the same line through the origin, $\lambda {\mathbf{v}}_0\text{.}$ This is a very special case of course; however, we claim that

is a solution for our linear system if we can find such a vector. To see that this is indeed the case, we compute

In other words, the key to solving a linear system ${\mathbf{x}}^{\prime }=A\mathbf{x}$ is to be able to find eigenvalues and eigenvectors for the matrix $A\text{.}$ We are now ready to state the results of our discussion in a theorem.

We say that the solution $\mathbf{x}(t)=e^{\lambda t}{\mathbf{v}}_0$ is a straight-line solution. The vector $e^{\lambda t}{\mathbf{v}}_0$ lies on the same line for each value of $t\text{.}$ Note that if ${\mathbf{v}}_0$ is an eigenvector for $A\text{,}$ then any nonzero multiple of ${\mathbf{v}}_0$ is also an eigenvector for $A\text{,}$

If ${\mathbf{x}}_1(t)$ and ${\mathbf{x}}_2(t)$ are solutions to the linear system ${\mathbf{x}}^{\prime }=A\mathbf{x}\text{,}$ then

Thus, for any two real numbers $c_1$ and $c_2\text{,}$

We state this result in the following theorem.

Revisiting the mixing problem that we posed at the beginning of this section, we have the following initial value problem,

If we write our system in matrix form, ${\mathbf{x}}^{\prime }=A\mathbf{x}\text{,}$ then

It is easy to check that we have eigenvalues $\lambda =-1$ and $\mu =-5$ with eigenvectors $\mathbf{u}=(2,3)$ and $\mathbf{v}=(-2,3)\text{,}$ respectively. Thus, we have two solutions to our system,

Since any linear combination of solutions is also a solution,

is a solution to our system. Using the initial values $x(0)=100$ and $y(0)=0\text{,}$ we can determine that $c_1=25$ and $c_2=-25\text{.}$ We now have the solution that we seek,

We can now proceed to the proof of the theorem. Suppose that we have a linear system ${\mathbf{x}}^{\prime }=A\mathbf{x}$ such that $A$ has a pair of distinct real eigenvalues, ${\lambda }_1$ and ${\lambda }_2\text{,}$ with associated eigenvectors ${\mathbf{v}}_1$ and ${\mathbf{v}}_2\text{.}$ By the Principle of Superposition, we know that

is a solution to the linear system ${\mathbf{x}}^{\prime }=A\mathbf{x}\text{.}$ To show that this is the general solution, we must show that we can choose $c_1$ and $c_2$ to satisfy a given initial condition ${\mathbf{x}}_0=\mathbf{x}(0)=(x_0,y_0)\text{.}$ By Lemma 3.2.7, we know that ${\mathbf{v}}_1$ and ${\mathbf{v}}_2$ form a basis for ${\mathbb{R}}^2\text{.}$ That is, we can write ${\mathbf{x}}_0$ as a linear combination of ${\mathbf{v}}_1$ and ${\mathbf{v}}_2\text{.}$ In other words, we can find $c_1$ and $c_2$ such that

It remains to show that $\mathbf{x}(t)=c_1{e}^{{\lambda }_{1}t}{\mathbf{v}}_1+c_2{e}^{{\lambda }_{2}t}{\mathbf{v}}_2$ is the unique solution to the system

Suppose that there is another solution $\mathbf{y}(t)$ such that $\mathbf{y}(0)={\mathbf{x}}_0\text{.}$ Then we can write

where

Since $\mathbf{y}(t)$ is a solution to our system of equations, we know that

On the other hand,

Consequently, we have two first-order initial value problems,

and

The solutions of these initial value problems are

respectively. Thus, $\mathbf{y}(t)=\mathbf{x}(t)\text{,}$ and proof our theorem is complete.