121 Lecture 21: An application to ordinary

dierential equations

In this nal lecture we are going to use a contour integral to help solve an ODE.

This principle is, in fact, very general and can be applied to partial dierential

equations on Rm (recall that ODEs refer to dierential equations on R1). As awarm up to this you are encouraged to re-read Lecture 5: \ODEs and the FT".

21.1 The ODE

Here, we are going to see how to nd the general solution to the ODE

d4f

dx4+ f = g (21.1)

provided that

dkf

dxk! 0 and g ! 0 as jxj ! 1 for k = 0; 1; 2; 3: (21.2)

Thus the function g : R1 ! R1 is assumed to be given, and the objective is todetermine f : R1 ! R1.

21.2 Applying Fourier transform

The assumption (21.2) means that we can apply the Fourier transform to the ODE

(21.15). Recall that the Fourier transform bf() = F(f)() := R11 eitf(t) dt islinear F(f + g)() = F(f)() + F(g)() and

F(f 0)() = i F(f)() provided that f(x)! 0 as jxj ! 1; (21.3)and hence F(f (m))() = (i)m F(f)() provided f (k)(x) ! 0 as jxj ! 1 fork = 0; 1; : : : ;m 1: Applied to (21.15) linearity yields

F(f (4))() + F(f)() = F(g)():

and because of (21.2) and (21.3) this is equivalent to the algebraic equation

bf() = bg()p4()

;

where

p4() = 4 + 1:

2Since p4 has no real zeroes this is well dened and we can apply Fourier inversion

to obtain

f(x) =1

2

Z 11

eix bf() d = 12

Z 11

eixbg()p4()

d:

Substituting bg() = R11 eiyg(y) dy and rearranging yieldsf(x) =

Z 11

k(x; y) g(y) dy

where

k(x; y) =1

2

Z 11

ei(yx)

4 + 1d: (21.4)

It remains, then, to evaluate the integral on the right-side of (21.4).

21.3 Using the Cauchy residue theorem

To evaluate this real integral previous examples suggest that we may get somewhere

by evaluating the contour integral

IR() =

ZCR

eiz

z4 + 1dz

around the contour CR in the z-plane which consists of the portion of the real axis

from R to R, together with the semi-circular arc R parametrized by R() =Rei; 0 :We assume that

2 R1: (21.5)

For R > 1 f(z) = eiz

z4+1has two poles inside CR at z0 = e

i=4 and z1 = ei 3=4. The

Cauchy residue theorem says that

IR() = 2i(res(f; z0) + res(f; z1)); (21.6)

where the right-side refers to the residues. By a standard formula

res(f; zk) =eizk

4z3k:

So

res(f; z0) + res(f; z1) =ei p

2(1+i)

4z30+ei p

2(1+i)

4z31

= (ei34 e

i p2 + ei

4 e

i p2 ) e

p2

3= i2p2

sin

p2

+ cos

p2

e p

2

= i2sin

p2+

4

e p

2 :

Hence by (21.6)

IR() = e p

2 sin

p2+

4

: (21.7)

Notice that this holds for any R > 1.

On the other hand,

IR() =

Z RR

ei

4 + 1d +

ZAR

eiz

z4 + 1dz

with AR the arc component of the contour. Using the estimate (19.10) from online

Lecture notes 19, we haveZAR

eiz

z4 + 1dz

=Z 0

eiRei

R4ei4 + 1Rei d

R

Z 0

jeiRei jjR4ei4 + 1j d

= R

Z 0

eR sin()

jR4ei4 + 1j d:

By the triangle inequality jR4ei4 + 1j jjR4ei4j 1j = jR4 1j = R4 1 forR > 1. Since also 0 < eR sin() 1 for 0 we obtain thereforeZ

AR

eiz

z4 + 1dz

RR4 1 ! 0 as R!1 if 0: (21.8)We are therefore left with for 0

e p

2 sin

p2+

4

(21.7)= lim

R!1

ZCR

eiz

z4 + 1dz

= limR!1

Z RR

ei

4 + 1d + lim

R!1

ZAR

eiz

z4 + 1dz| {z }

=0 by (21.8)

=

Z 11

ei

4 + 1d:

That is,

provided 0Z 11

ei

4 + 1d = e

p2 sin

p2+

4

: (21.9)

4From (21.4) we want to apply this with = y x. But y x can take on any realvalue, so we need to extend (21.9) to the case 0. To do that we make use ofthe fact that we have only used two of the four poles of f . So we now evaluate

IDR() =

ZDR

eiz

z4 + 1dz

around the contour DR in the z-plane which consists of the portion of the real axis

from R to R, together with the semi-circular arc BR parametrized by R() =Rei; 2; (note this is positively oriented (anticlockwise) which meansthe direction along the [R;R]] is reversed!). We now assume that

0: (21.10)

We now repeat the above process with DR instead of CR. For R > 1 f(z) =eiz

z4+1

has two poles insideDR at z2 = ei5=4 and z3 = e

i 7=4. The Cauchy residue theorem

says that

IDR() = 2i(res(f; z2) + res(f; z3)): (21.11)

As before, res(f; zk) =eizk4z3k

so

res(f; z2) + res(f; z3) =ei p

2(1+i)

4(z2)3+ei p

2(1i)

4(z3)3

= ie

p2

2

cos

p2

1p2 sin

p2

1p2

= i

ep2

2sin

p

2+

4

:

Hence by (21.11)

IDR() = ep2 sin

p

2+

4

:

On the other hand,

IDR() = Z RR

ei

4 + 1d +

ZBR

eiz

z4 + 1dz

with BR the arc component of the contour in the lower-half plane. Note the minus

sign in front ofR RR is there to take account of the fact that we traverse the interval

[R;R] in reverse: by starting at R and ending at R. Then, similarly as before,we obtain

provided 0Z 11

ei

4 + 1d = e

p2 sin

p

2+

4

: (21.12)

5Exercise: Make sure you can derive (21.12)!

But since

jj = ; if 0;

; if 0;then we can write down (21.9) and (21.12) in one go as

For any 2 R1Z 11

ei

4 + 1d = e

jjp2 sin

jjp2+

4

: (21.13)

We now obtain from (21.4) and (21.13) that the general solution to (21.15) is given

by

f(x) =

Z 11

k(x; y) g(y) dy

with

k(x; y) =1

2e jxyjp

2 sin

jx yjp2

+

4

:

21.4 Exercises:

Using the same contours as above:

[1] Find the general solution to the ODE

d2f

dx2+ f = g (21.14)

provided that f; f 0 ! 0 and g ! 0 as jxj ! 1.

[1] Find the general solution to the ODE

d4f

dx4 2d

2f

dx2+ f = g (21.15)

provided that f (k) ! 0, k = 0; 1; 2; 3, and g ! 0 as jxj ! 1.

Solutions will be posted on the web page shortly.