Mathematical Tools for Physics - Department of Physics - University

4—Differential Equations 77

Some of the numerical methods you will find in chapter 11 start from the ideas of these expansions,
but then develop them along different lines.
There is an iterative methods that of more theoretical than practical importance, but it’s easy
to understand. I’ll write it for a first order equation, but you can rewrite it for the second (or higher)
order case by doing the same thing as in Eq. (4.26).

x ̇=f(x,t) with x(t 0 ) =x 0 generates x 1 (t) =

∫t

t 0

dt′f(x 0 ,t′)

This is not a solution of the differential equation, but it forms the starting point to find one because

you can iterate this approximate solutionx 1 to form an improved approximation.

xk(t) =

∫t

t 0

dt′f

(

xk− 1 (t′),t′

)

, k= 2, 3 , ... (4.27)

This will form a sequence that is usually different from that of the power series approach, though the
end result better be the same. This iterative approach is used in one proof that shows under just what

circumstances this differential equationx ̇=fhas a unique solution.

4.5 Trigonometry via ODE’s

The differential equationu′′=−uhas two independent solutions. The point of this exercise is to derive

all (or at least some) of the standard relationships for sines and cosinesstrictly from the differential
equation.The reasons for spending some time on this are twofold. First, it’s neat. Second, you have
to get used to manipulating a differential equation in order to find properties of its solutions. This is
essential in the study of Fourier series as you will see in section5.3.

Two solutions can be defined when you specify boundary conditions. Call the functionsc(x)and

s(x), and specify their respective boundary conditions to be

c(0) = 1, c′(0) = 0, and s(0) = 0, s′(0) = 1 (4.28)

What iss′(x)? First observe thats′satisfies the same differential equation assandc:

u′′=−u =⇒ (u′)′′= (u′′)′=−u′, and that shows the desired result.

This in turn implies thats′is a linear combination ofsandc, as that is the most general solution to

the original differential equation.

s′(x) =Ac(x) +Bs(x)

Use the boundary conditions:

s′(0) = 1 =Ac(0) +Bs(0) =A

From the differential equation you also have

s′′(0) =−s(0) = 0 =Ac′(0) +Bs′(0) =B

Put these together and you have

s′(x) =c(x) And a similar calculation shows c′(x) =−s(x) (4.29)

What isc(x)^2 +s(x)^2? Differentiate this expression to get

d

dx

[c(x)^2 +s(x)^2 ] = 2c(x)c′(x) + 2s(x)s′(x) =− 2 c(x)s(x) + 2s(x)c(x) = 0

This combination is therefore a constant. What constant? Just evaluate it atx= 0and you see that

the constant is one. There are many more such results that you can derive, but that’s left for the
exercises, problem4.21et seq.