26 August 2005
Mathematical Physics I: What is a differential equation?

by George E. Hrabovsky, President MAST

A Note to the Reader

You will notice that I ask numerous questions and number them. I have no intention of answering these questions in sequence. The numbering is a bookkeeping device so that I can keep track of things. The point of this column is to investigate what heat is, and I will only go so far afield from that goal. This means that some questions will go unanswered. This is reasonable, and it allows for future projects based on those unanswered questions. Feel free to attempt to answer these questions for yourself.

Where We Have Been

Last time, we explored the nature of work and related kinetic energy to force. This raised the question of what a differential equation is.

Session 8: What is a differential equation?

Before delving into the meat, lets discuss the garnish. In order to do physics we need to make calculations. This involves the use of equations. In an equation two numerical quantities are said to be equal,

A = B .

We can resort to algebraic means to derive expressions for specific quantities in complicated equations.

Since most quantities in the real world change with respect to some other quantity, we need to account for those changes in our equations. When we study minute changes, we must use differentials and apply the rules of calculus. Such equations are differential equations, such as the one we had before

K v = 1/2F x .

Here we have an equation that claims that the quantity of the product of kinetic energy and a minute change in velocity is the same as the quantity of the product of one half the force producing the kinetic energy and the corresponding small quantity of distance traveled. So, then the question becomes, how do we solve a differential equation?

51. How do we solve a differential equation?

How do we solve a differential equation?

Unlike mathematicians we will proceed with the standard attitude of the physicist, that mathematics is there as a tool for us to solve physical problems. As such, we do not need to understand all of the intricate details of a method to use it. Of course, there is a great danger that we will use a technique in an inappropriate way. We will know quickly enough if we do this and get a wildly inaccurate answer (or at least one that falls outside of the error bounds).

Examining our differential equation, we note that we can manipulate it to get an actual derivative (see my earlier column, "What is a derivative?" for details).

K v/x = 1/2F .

We need to find a way to get the function that became the derivative. In other words we need to reverse the process of differentiation. If we have a derivative, we can change our notation,

x/t = x ' (t) .

Reversing the process of differentiation, we can write

/tX(t) = x ' (t)

where X(t) is called the antiderivative. The process of finding the antiderivative involves asking the question, "What function produced this derivative?" To be completely general, we add a constant (since the derivative of a constant is 0 this will never change the value of the derivative). This process is called indefinite integration and is written

X(t) = ∫x ' (t) t + c

where c is called the constant of integration.

So how does this work in practice? Let us use a simple example,

x ' (t) = t .

What derivative produces this as an answer? If we think about it, the only thing that comes to mind is the power rule,

x ' (t^n) = n t^(n - 1) .

In our case we are missing both the n and the n - 1 power. The power is easy enough to figure out, we only need to solve the equation

n - 1 = 1

FormBox[RowBox[{n, =, RowBox[{1 + 1, =, 2.}]}], TraditionalForm]

So, n = 2. That means that we need to divide our expression for t by 2,

x(t) = t^2/2 .

To complete the indefinite integration we add a constant,

x(t) = ∫x ' (t) t + c = t^2/2 + c .

We can use this technique to directly solve our differential equation by indefinitely integrating both sides.

∫K v = ∫ 1/2F x .

We will examine the left-hand side first,

∫K v .

What derivative produces this result? Since K does not appear to be dependent on v, we have a constant as a derivative; hence the only differentiation rule that leaves a constant is the constant multiple rule.

/v (K v) = K,

so,

∫K v = K∫ v = K v + c_1

where c_1 is the constant of integration. We can use the fundamental theorem of calculus, which can be used to define another kind of integral (the definite integral). This integral allows us to remove the constant of integration by considering the independent dummy variable v ' from some initial condition v_0 to some later condition v

∫_v_0^vKv ' = ∫K v - ∫K v_0 = K v + c_1 - K v_0 - c_1 = K v - K v_0 .

We now look at the right hand side,

∫F x .

This looks like the left-hand side with different symbols,

∫F x = F∫ x = F x + c_2 .

Applying the fundamental theorem of calculus again gives us,

∫_x_0^xFx ' = ∫F x - ∫F x_0 = F x + c_2 - F x_0 - c_2 = F x - F x_0 .

The solution to the differential equation is then

K v - K v_0 = F x - F x_0 .

This is a really simple differential equation. There are lots of other differential equations and methods to solve them.

52. What other kinds of differential equations are there?

53. What other methods are there to solve differential equations?

Created by Mathematica  (February 28, 2005)

   
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