# Example: Differential equations

## Understanding the situation

Sometimes, understanding how the rate of change of a quantity is determined can let us set up differential equations. Solving those equations can then provide us with a lot of critical information about a physical or biological situation.

Let's consider a biological example.

Suppose we have a small spherical tumor of radius $r$. We know the tumor produces a chemical that can be detected in the blood stream if there is enough of it. We also know that tumors are made up of cells that are each about the same size and that the tumor grows by the cells dividing. As a result, we expect the rate at which the number of cells and, therefore, the volume of the tumor increases should be proportional to the number of cells that we currently have. Let's see what this implies.

## Presenting a sample problem

Suppose at a time $t$, a tumor consists of a number of cells, $n$. Since $n$ changes with time, we will write the number of cells as $n(t)$. If in a small time $\Delta t$ the new number of cells produces is $\Delta n = \lambda \Delta t$, write a differential equation satisfied by the function $n(t)$ and solve it. Interpret the constant $\lambda$ physically.

## Solving this problem

The equation that is given is

$$\Delta n = \lambda \Delta t$$

If we write this as a ratio and let the time interval become small, we can change our ratio of $\Delta$'s to a derivative:

$$\frac{\Delta n}{\Delta t} = \lambda n \quad\quad \Rightarrow \quad \quad \frac{dn}{dt} = \lambda n$$

Whenever the growth rate of a quantity is proportional to the quantity, as here dn/dt is proportional to n, we get exponential growth — faster than any power. We can see how to solve this by introducing a variable to cancel the $\lambda$

$$z = \lambda t$$

Since we then have

$$t = z/\lambda \quad \Rightarrow \quad \frac{d}{dt} = \frac{d}{d(z/\lambda)}\quad \Rightarrow \quad \frac{d}{dt} = \lambda\frac{d}{dz}$$

With this substitution, our equation becomes

$$\lambda\frac{dn}{dz} = \lambda n$$

The $\lambda$ cancels and we get the equation

$$\frac{dn}{dz} = n$$

From calculus, we know that the function that is its own derivative is the exponential, $e^z$ so our solution is

$$n = Ae^{\lambda t}$$

Since when $t$ is equal to our starting time of 0, $n$ becomes $A$, we can see this is our starting value, which we might conveniently call $n_0$, so

$$n = Ae^{\lambda t}$$

where $A$ is any constant.

We can now put back the time using $z = \lambda t$ to get

$$n = n_0e^{\lambda t}$$

Now we are asked to interpret $\lambda$.  By doing a dimensional analysis of our differential equation, we can see that since

$$\bigg[\frac{dn}{dt}\bigg] = \frac{1}{T}$$

that $\lambda$ must have dimensions of

$$[\lambda] = \frac{1}{T}$$

So lets define the reciprocal of $\lambda$ as a time constant:

$$\lambda = 1/\tau$$

With this our result becomes:

$$n = n_0e^{ t/\tau}$$

We can see that when our time increases by an amount $\tau = 1/\lambda$ that our number grows by a factor of $e$ or 2.718...

Notice that in this analysis we have used a number of the tools in our mathematical tool belt including building equations from the physics,  repackaging, and dimensional analysis.

How this analysis of tumor growth can give more insight into the biology of tumors is explored in the problem: Tumor Growth in the link below.

Joe Redish 3/11/19

Article 282