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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:
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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
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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)
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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)
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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.