Chapter 1
d2 y
− y = 2 + x2 + ex .
1.Find the general solution of
dx2
Solution:
m2 = 1 ⇒ m = ±1. Thus the complementary solution is
Y = Aex + Be−x .
We also have
1
1
2 = −2 ,
x2 = −(1 + D2 + · · ·)x2 = −x2 − 2.
2
−1
D −1
1
1
1
ex = ex
1 = ex 2
1
D2 − 1
(D + 1)2 − 1
D + 2D
1
1
xex
= ex
1 = ex
1=
.
D(D + 2)
D(2)
2
Therefore, the general solution of the differential equation is
xex
y = Aex + Be−x − 4 − x2 +
.
2
D2
2. Find the general solution of the ordinary differential equation
d4 y
+ 4y = sin 2x + 2x + ex .
dx4
Solution:
m4 = −4 =⇒ m = 1 + i, −1 + i, −1 − i, 1 − i.
Thus the complementary solution is
Y (x) = ex (a cos x + b sin x) + e−x (c cos x + d sin x).
A particular solution is
sin 2x x ex
1
yp (x) = 4
[sin 2x + 2x + ex ] =
+ + .
D +4
20
2
5
The general solution is
sin 2x x ex
y(x) = ex (a cos x + b sin x) + e−x (c cos x + d sin x) +
+ + .
20
2
5
3. Find the general solution of the ordinary differential equation
d4 y
− y = 3e2x + ex + 2x.
dx4
Solution:
m4 − 1 = 0 ⇒ m = ±1, ±i.
Thus
Y = aex + be−x + c cos x + d sin x.
We also have
1
yP = 4
(3e2x + ex + 2x)
D −1
e2x
1
=
+
ex − (1 + D4 + · · ·)2x
5
(D − 1)(D + 1)(D2 + 1)
1
e2x
1
1 x
+
e − 2x
5
(D − 1) 4
e2x
1 1
e2x
ex x
=
+ ex
− 2x =
+
− 2x.
5
D4
5
4
Therefore,
xex
e2x
+
− 2x.
y = aex + be−x + c cos x + d sin x +
5
4
=
4. Find the general solution for
d3 y
− y = 2 + cos x.
dx3
Solution:
d3 Y
− Y = 0. Thus
dx3
mx
3
2inπ/3
Y = e , m = 1 ⇒ mn = e
.
√
√
The three roots are: m0 = 1, m1 = (−1 + i 3)/2, m2 = (−1 + i 3)/2.
We get
√
√
Y = Aex + e−x/2 [B cos(x 3/2) + C sin(x 3/2)].
The particular solution is
1
1
(2 + cos x) = −2 −
yp = 3
cos x
D −1
D+1
D−1
− sin x − cos x
= −2 − 2
cos x = −2 −
.
D −1
−1 − 1
We have
√
√
sin x + cos x
y = Aex + e−x [B cos(x 3/2) + C sin(x 3/2)] − 2 −
.
2
The complementary solution satisfies
5. Consider the equation of motion
dx
··
·
·
x + ax + x = b cos t, x ≡
, a > 0.
dt
which describes a harmonic oscillator under a linear damping force as well
as an external harmonic force of the same frequency. Find the general solution
of this equation.
Solution:
p
m2 +am + 1 = 0 ⇒ m1 = −a/2 + a2 /4 − 1,
m2 = −a/2 −
p
a2 /4 − 1.
p
If a < 2, the two roots above are −a/2 ± iω, where ω = 1 − a2 /4, and the
complementary solution X is given by
X = e−at/2 (c cos ωt + d sin ωt), where c and d are constants.
We note the motion is a damped oscillation with the angular fequency ω,
which is smaller than unity–the angular frequency when a = 0.
p
If a > 2, the two roots are −a/2 ± a2 /4 − 1. The complementry solution
is
√ 2
√ 2
X = e−at/2 (ce a /4−1t + de− a /4−1t ).
2
Both solutions above describe motions vanishing as t → ∞ without oscillations.
If a = 2,
X = e−t (c + dt).
Both solutions vanish as t → ∞ without oscillations.
The particular solution is
1
1
b sin t
xp = 2
b cos t =
b cos t =
.
D + aD + 1
aD
a
The general solution is
b sin t
x=X+
.
a
6. Show that the particular solution of a linear differential equation with
constant coefficients can always be expressed by an integral.
Solution:
Let the equation be
dn−1 y
dy
dn y
+ an−1 n−1 + · · · + a1
+ a0 y = f (x),
n
dx
dx
dx
where am , m = 0, 1 · ·· n, are constants.
We have
Y = c1 em1 x + c2 em2 x + · · · + cn emn x ,
where mj , j = 1, 2 · · · n are the roots of the algebric equation
mn + an−1 mn−1 + · · · + a1 m + a0 = 0.
We assume here that they are distinct. If the algebraic equation has multiple
roots, only slight modifications are needed.
The particular solution is given by
1
yp = n
f (x).
n−1
D + an−1 D
+ · · · + a1 D + a0
We shall write
1
1
.
=
n
n−1
D + an−1 D
+ · · · + a1 D + a0
(D − m1 )(D − m2 ) · · · (D − mn )
By partial fraction, we get
1
n
n−1
D + an−1 D
+ · · · + a1 D + a0
1
1
1
1
=
+
(m1 − m2 ) · · · (m1 − mn ) D − m1
(m2 − m1 ) · · · (m2 − mn ) D − m2
+···+
1
1
.
(mn − m1 ) · · · (mn − mn−1 ) D − mn
Since
R x −mx0
1
1
f (x) = emx [e−mx f (x)] =emx
e
f (x0 )dx0 ,
D−m
D
we have
3
0
yp (x) =
Rx
0
em1 (x−x )
em2 (x−x )
[
+
+·
(m1 − m2 )(m1 − m3 ) · · · (m1 − mn ) (m2 − m1 )(m2 − m3 ) · · · (m2 − mn )
0
··+
emn (x−x )
] f (x0 )dx0 .
(mn − m1 )(mn − m2 ) · · · (mn − mn−1 )
Chapter 2
1.As we know, the sum of two complex numbers can be represented as that of
two vectors in a two dimensional space. How about the product of two complex
numbers?
Solution:
Let
α ≡ a1 + ia2 , β ≡ b1 + ib2 .
These two complex numbers can be reprsented by the vectors
→
→
→ →
→
→
v ≡ a1 e1 + a2 e2 , u ≡ b1 e1 + b2 e2 .
→
→
where e1 and e2 are unit vectors in the direction of the x-axis and the y-axis,
respectively.
We have
α∗ β = (a1 − ia2 )(b1 + ib2 ) = (a1 b1 + a2 b2 ) + i(a1 b2 − a2 b1 ).
→
→
The real part of α∗ β is the scalar product of v and u ; the imaginary part
→
→
∗
of α β is the cross product of v and u .
2. Find the real part and the imaginary part of eiz as well as those of cos z.
For what values of z is the real part of eiz equal to cos z?
Solution:
eiz = eix−y = e−y (cos x + i sin x) = u + iv,
e−iz = e−ix+y = ey (cos x − i sin x).
Thus
eiz + e−iz
cosz =
= cos x cosh y − i sin x sinh y.
2
If u is equal to cos z, then
e−y cos x = cos x cosh y − i sin x sinh y.
The real part and the imaginary part of the equation above are:
(e−y − cosh y) cos x = 0 and sin x sinh y = 0.
If y 6= 0, these two equalities give sin x = cos x = 0, which is not possible.
Both equalities are satisfied if
y = 0.
Thus the real part of eiz is equal to cos z only if z is real.
3. For what constant value (or values ) of a is the function u = x + ax3
the real part of an analytic function? Find the imaginary part of this analytic
function. Express this analytic function as a function of z.
4
Solution:
uxx = 6ax , uyy = 0 .
Thus ∇2 u = 0 only if a = 0, in which case
u = x.
The conjugate of x, denoted as v, satisfies
vy = ux = 1 and vx = −uy = 0.
We get
v = y + c,
and
u + iv = z + ic,
where c is a constant.
4. Find the roots of (z + 1)8 = (z 2 − 1)8 .
Solution:
The equation above gives
(z 2 − 1) = ei2πn/8 (z + 1), n = 0, 1, 2 · ··, 7 ,
or
(z + 1)(z − 1 − ei2πn/8 ) = 0, n = 0, 1, 2 · ··, 7.
Thus the roots are
z = −1
and
z = 1 + ei2πn/8 , n = 0, 1, 2 · ··, 7 .
5. Find the Laurent series expanded around z = 0 for
1
.
f (z) =
z(z − 2)
Specify the region in which each of these series is valid.
Solution:
In the region |z| < 2 , we have
1
1
z
zn
f (z) = −
= − (1 + + · · · + n + · · ·) .
2z(1 − z/2)
2z
2
2
And in the region |z| > 2 , we have
1
1
1
2
2n
f (z) = 2
= 2 (1 + + · · · + n + · · ·).
z 1 − 2/z
z
z
z
6.Evaluate the following integrals:
R ∞ (1 + cos πx)dx
a. −∞
,
(1 − x2 )
b.
R ∞ sin2 x
dx,
−∞ 1 + x2
5
c.
R∞
−∞
1 + 3x2
dx.
(1 + x2 )(4 + x2 )
Solutions:
a. We get
R ∞ (1 + eiπx )dx
J ≡ −∞
=0
(1 − x2 )
convergent integral.) Thus
I = Re J = 0.
b. I =
Let
by closing contour upstairs. ( Note that J is a
R ∞ sin2 x
1 R ∞ e2ix − 2 + e−2ix
dx
=
−
dx.
−∞ 1 + x2
4 −∞
1 + x2
R ∞ e2ix − 1
dx.
−∞ 1 + x2
By clsoing contour upstairs, we get
e−2 − 1
= −π(1 − e−2 ).
J = 2πi
2i
Thus
I = − Re J/2 = π(1 − e−2 )/2.
J≡
c. By closing contour upstairs, we get
R∞
−2
−11
1 + 3x2
dx = 2πi[
+
] = 7π/6.
−∞ (1 + x2 )(4 + x2 )
(2i)(3) (−3)(4i)
7. Evaluate the following integrals:
R 2π
1
a. 0
dθ.
3 − 2 sin θ
R∞
1
b. −∞ ( 2
)3 dx.
x +1
R ∞ cos(πx/2)
dx.
c. −∞
x2 − 1
a. LetIz = eiθ . We have
dz
I=−
,
2 − 3iz − 1
z
c
where c is a unit circle with its center at the origin. The integrand above
has two simple√poles:
√
z1 = i(3 + 5)/2, z2 = i(3 − 5)/2.
Since z2 is the pole inside c,
2π
2π
√
I = 2πi Residue(z2 ) = −
=√ .
(3 − 5) − 3
5
b. Closing the contour upstairs, we get
I = 2πiRes(i).
Since the integrand has a third-order pole at i, we get
6
1
6
3
1
d2
=
=
lim 2
.
2 z→i dz (z + i)3
(2i)5
16i
3π
We have I =
.
8
Res(i) =
c. Since both the numerator and the denominator vanish at x = ±1, the
integrand is entire. By deforming the contour, we get
1 R eizπ/2 + e−izπ/2
I=
dz,
2 c
z2 − 1
where c passes around the points z = ±1 from above. By closing the contour
upstairs, we find the first term in the integrand equal to zero. We get
1 R e−izπ/2
i
i
I=
dz = −πi(− +
) = −π.
c
2
2
z −1
2 −2
8. Evaluate the following integrals with the use of the Cauchy residue theorem.
R∞
dx
,
a. 0
(9 + x2 )(4 + x2 )
b.
R ∞ sin2 x
dx,
−∞ x2
Solution:
1 R∞
dx
.
2 −∞ (9 + x2 )(4 + x2 )
Closing the contour upstairs, we get
1
1
+
] = π/60.
I = πi[
(6i)(−5) (5)(4i)
a. I =
b. Since the integrand is an entire function, we deform the contour and get
1 R e2iz − 2 + e−2iz
I=− c
dz,
4
z2
where c is a contour in the upper half-plane joining −∞ and ∞. The first
two terms in the integrand give zero. We get
2πi
de−2iz
I=
lim
= π.
4 z→0 dz
9. Evaluate the following integrals:
R∞
sin cx
a. −∞
dx,
(x + a)2 + b2
where a is a real number while b and c are both real and positive.
R∞
x2
b. −∞ 4
dx.
x +1
1/z
R e
c. C
dz, where c is a circle around the origin in the counterclockwise
z
direction.
7
Solution:
eicx
dx = Im[2πiRes(−a + ib)]
−∞ (x + a)2 + b2
e−iac e−bc
πe−bc sin ac
= Im[2πi
]=−
.
2ib
b
b. Closing the contour upstairs, we get
I = (2πi)[Res(eiπ/4 )+Res(−e−iπ/4 )].
Since
z2
z2
Res(−e−iπ/4 )= 3 ]z=− exp(−iπ/4) ,
Res(eiπ/4 )= 3 ]z=exp(iπ/4) ,
4z
4z
√
1
1
−2i sin(π/4)
I = (2πi)[ iπ/4 +
] = (2πi)
= π 2/2.
−iπ/4
4
4(e
) 4(−e
)
a. I = Im
R∞
c. We have
R 1
R e1/z
dz = C (1 + z + z 2 /2 + · · ·)dz.
I= C
z
z
By the Cauchy residue theorem, we get
I = 2πi,
10. Evaluate the integral I ≡
R∞
0
x
dx.
1 + xn
Solution: R
Let J ≡ C
z
dz,
1 + zn
where C is the contour of a pie with two of its boundaries the infinite ray of
θ = 0 and that of θ = 2π/n. Then
J = (1 − e4iπ/n )I .
By the Cauchy residue theorem, we have
2πi
J = i(n−2)π/n .
ne
Thus
π
I=
.
n sin(2π/n)
11. Find the branch points of the function f (z) = (z 4 − 1)1/4 . Draw a set of
branch cuts for this function to make this function single-valued in the complex
plane.
Solution:
The branch points of this function are at
z = 1, i, −1, −i .
As z → ∞, f (z) → z(1 − z −4 /4 + · · ·).Thus z∞ is not a branch point of the
function. We draw straight lines joining 1 to i , i to −1 , and −1 to −i , and
the function is single-valued in the cut-plane.
1 + z 1/3 1
)
in the ex1−z
1 + z2
tended z-plane. Draw branch cut (or cuts) to make this function single-valued.
12. (a) Locate the singularities of the function (
8
(b) Evaluate the integral
R 1 1 + x 1/3 dx
(
.
)
−1 1 − x
1 + x2
Solution:
1 + z 1/3 1
(a) The function (
has simple poles at z = ±i and branch
)
1−z
1 + z2
points at ±1.
Let z = 1/t. Then the function is
t + 1 1/3 t2
(
)
.
t−1
t2 + 1
This shows that t = 0, or z = ∞, is not a branch point of the function.
Therefore, drawing a finite branch cut from −1 to 1 makes the function singlevalued.
(b) I
Let
1 + z 1/3 1
)
dz,
J= (
1 + z2
c 1−z
where c is the “toilet seat” contour wrapping around the branch cut.
1 + z 1/3
We choose (
)
to be real and positive when z is infinitesimally above
1−z
1 + z 1/3
)
at a point infinitesimally below the
the branch cut. The value of (
1−z
branch cut can be obtained as follows. We start at a point infinitesimally above
the branch cut and go around z = −1 to reach the point infinitesimally below the
starting point. Since the path does not enclose the point z = 1, the argument of
1 + z 1/3
(1−z) does not change. The argument of (1+z) changes by 2π. Thus (
)
1−z
2πi/3
changes by a phase factor e
.
From this consideration we get
J = (1 − e2πi/3 )I.
Note that the infinitesimal circles around z = ±1 contribute nothing to J,
as −1/3 → 0 when → 0.
Since the integrand vanishes like 1/z 2 at infinity, we may close the contour
by adding an infinite circle to the toilet seat contour. By the Cauchy residue
theorem, we get
1 + z 1/3
1
1 + z 1/3
1
J = 2πi{[(
) ]z=i + [(
) ]z=−i
}.
1−z
2i
1−z
−2i
1 + z 1/3
To evaluate (
) ]z=i , we note that this factor is real and positive at
1−z
the origin infinitesimally above the branch cut. As we move from z = 0 to z = i,
the argument of (z + 1) increases by π/4, while that of (z − 1) decreases by π/4.
Thus we have
1 + z 1/3
eiπ/4
(
) ]z=i = ( −iπ/4 ) = eiπ/6 .
1−z
e
1 + z 1/3
Similarly, since (
)
at the origin infinitesimally below the branch cut
1−z
2πi/3
is equal to e
, we have
1 + z 1/3
e−iπ/4
[(
) ]z=−i = e2πi/3 ( iπ/4 )1/3 = e2πi/3 e−iπ/6 .
1−z
e
9
Therefore,
J = π(eiπ/6 − e2πi/3 e−iπ/6 ).
We get
e−iπ/3 eiπ/6 − eiπ/3 e−iπ/6
eiπ/6 − e2πi/3 e−iπ/6
=
π
I=π
1 − e2πi/3
e−iπ/3 − eiπ/3
sin(π/6)
π
=π
=√ .
sin(π/3)
3
13. (a) Locate and classify the singularities of the function (z + 2)−1 (1 −
z )
in the extended z-plane. Draw branch cut (or cuts) to make the function
single-valued.
(b) Evalaute the integral
R1
dx
.
I = −1
(x + 2)(1 − x2 )1/2
2 −1/2
Solution:
(a) The function (z + 2)−1 (1 − z 2 )−1/2 has a simple pole at −2 and branch
points at ±1. This function is single-valued in the plane cut from −1 to 1.
(b) Let
I
dz
J=
,
2 1/2
c (z + 2)(1 − z )
where c is the closed contour wrapping around the branch cut in the clockwise
direction. We note that
J = 2I.
We may view this contour as the one enclosing the cut-plane in the counterclockwise direction. Since the integrand vanishes like z −2 as z → ∞, the only
singularity enclosed by the contour is at z = −2. Therefore, the Cauchy residue
theorem gives
2π
2πi
2πi
J = 2πi Res(−2) =
= √ =√ .
(1 − 4)1/2
i 3
3
Thus
π
I=√ .
3
14. Let f (z) be analytic at z0 and f (z0 ) 6= 0. Prove log f (z) does not have
a branch point at z0 .
Solution:
If f (z0 ) 6= 0, f (z) has no zero in a sufficiently small neighborhood of z0 .
Therefore, the argument of f (z) does not change as we go around a closed
contour in this neighborhood. As a result, the value of log f (z) does not change
as we go around this closed contour. Therefore, z0 is not a branch point of f (z).
15. Evaluate the integral
.
.
R∞
0
dx
, where 0 < Re a < 1.
xa (1 + x)
10
Solution:
The function f (z) =
1
has a branch point at z = 0. We choose the
z a (1 + z)
branch cut of this function to be the positive x-axis.
Let
R
dz
J= c a
,
z (1 + z)
where c wraps around the positive x-axis in the clockwise direction. The
contour goes around the singularity at the origin via a circle of radius .
The integral of f (z) over a circle of radius R is of the order of
R
,
(Ra )(R)
which vanishes as R → ∞. ( This is because a > 0.) Therefore, we may close
the contour by adding an infinite circle to c. By the Cauchy residue theorem,
1
J = 2πiRes(-1)= 2πi iπa .
e
Now the contour c had three sub-contours:
1. Sub-contour c1 , the positive x-axis. The integral over c1 is I.
2. Sub-contour c2 , the circle of radius around the origin. The integral over
c2 is proportional to
,
a
which vanishes as → 0. (This is because a < 1.)
3. Sub-contour c3 , which is the x-axis below the branch cut. The integrand
on c3 is equal to
1
1
= e−2πia a
.
(e2iπ x)a (1 + x)
x (1 + x)
Thus the integral over c3 is equal to
−e−2πia I.
The minus sign above us due to the direction of the contour eing clockwise.
Thus we have
J = (1 − e−2πia )I.
We get
π
I=
.
sin πa
16. Find
Z ∞ the value of
Z ∞
ln x
1
π
I=
dx. Reminder:
dx =
.
5
5
5 sin(π/5)
0 1+x
0 1+x
Solution:
Let
I
ln z
J=
dz,
5
c1 + z
where c is the closed pie-shape contour with angle 2π/5 which encloses the
simple pole at eiπ/5 . By the Cauchy residue theorem, we get
iπ/5
2π 2 eiπ/5
2π 2 e−4iπ/5
J = 2πi 4iπ/5 = −
=
.
25
25
5e
To extract I from J, we have
11
J =I −e
2πi/5
Thus we have
(1−e2πi/5 )I =
Z
∞
0
2iπ 2 e2πi/5
ln r + 2iπ/5
2πi/5
dr
=
(1
−
e
)I
−
.
1 + r5
25 sin(π/5)
2π 2 iπ/5
eiπ/5 i
2π 2 iπ/5
i cos(π/5) − sin π/5
e
(1+
)=
e
(1+
)=
25
sin(π/5)
25
sin(π/5)
2π 2 iπ/5 i cos π/5
e
,
25
sin(π/5)
or
π 2 cos(π/5)
.
I=−
25 sin2 (π/5)
17. Evaluate I =
R∞
−∞
x
dx.
sinh x
Solution:
We note sinh z = (ez − e−z )/2 is finite if the real part of z is finite, regardless
z
of the vaue of the imaginary part of z. As a result,
does not vanish at
sinh z
all points of the upper infinite half circle. Consequently, we cannot close the
contour upstairs. For the same reason, we cannot close the contour downstairs.
Let’sIclose the contour in a differently way. We define
z
dz,
J=
sinh
z
c
where c is the boundary of the ( infinite) rectangle the horizontal sides of
which are the x-axis and the line z = x + iπ.
On the line z = x + iπ, the integrand is equal to
x + iπ
.
sinh(x + iπ)
Since
sinh(x + iπ) = − sinh x,
the integrand on this line is equal to
x + iπ
−
,
sinh x
which blows up at x = 0.
To avoid the singularity at z = iπ, we make the indentation of an infinitesimal half circle below iπ.
We shall denote the x-axis as c1 , the line z = x + iπ with |x| > as c2 , and
the indented half circle as c3 .
Since c encloses no singularity of the integrand,
J = 0.
Now the integral over c1 is I.
The integral over c2 is
R ∞ x + iπ
R∞
x
P −∞
dx = P −∞
dx = I,
sinh x
sinh x
as
R ∞ iπdx
P −∞
= 0.
sinh x
z
The residue of
at iπ is
sinh z
12
iπ
.
−1
Thus the integral over c3 is equal to
−iπ(−iπ) = −π 2 .
Finally, the integral over the two vertical boundary lines vanishes.
Thus we have
J = 2I − π 2 .
We get
I = π 2 /2.
2
18. Find the Fourier transform of the function f (x) = e−x , −∞ < x < ∞.
Solution:
2
The
transform of eR−x is
R ∞ Fourier
2
2
2
∞
dx e−ikx−x = e−k /4 −∞ dxe−(x+ik/2) .
−∞
By deforming the contour of integration to the line
z = −ik/2 + t,
where t is real,
R ∞ we get2 √
2
2
I = e−k /4 −∞ dte−t = πe−k /4 .
19. Find the Fourier transform of the function
x < ∞.
f (x) = (x2 +1)−1 , −∞ <
Solution:
f (x) = (x2 + 1)−1 ,
∼
R∞
e−ikx
f (k) = −∞ dx
.
1 + x2
If k > 0, we close contour downstairs and get
∼
e−k
f (k) = −2πi
= πe−k , k > 0.
−2i
If k < 0, we close contour upstairs and get
ek
= πek , k < 0.
2i
∼
Thus f (k) = πe−|k| , all k.
∼
f (k) = 2πi
20. (a) Let
F (x) = 1,
= 0,
−a < x < a,
|x|> a,
˜
where a is a positive constant. Find F (k), the Fourier transform of F (x).
(b) Show that the inversion formula holds in this example.
Solution:
13
˜
(a) F (k) =
Ra
−a
e−ikx dx =
e−ika − eika
2 sin(ka)
=
.
−ik
k
(b) We need to prove that
R ∞ ikx e−ika − eika dk
e
= 0,
−∞
−ik
2π
|x| > a,
= 1,
|x| < a.
e−ika − eika
Since
is an even function of k, the integral above is an even
−ik
function of x. Thus we only need to show that the above is true when x is
positive, or
R ∞ ikx e−ika − eika dk
e
= 0,
x > a,
(I)
−∞
−ik
2π
= 1,
0 < x < a.
(II)
We deform the contour to the upper half plane, avoiding the origin k = 0.
If x > a, both (x + a) and (x − a) are positive. Thus we may close the
contour upstairs and get
R ikx e−ika − eika dk
e
= 0,
−ik
2π
and (I) is proven.
If 0 < x < a, (x + a) is positive while (x − a) is negative. Closing the contour
upstairs, we get
R ikx eika dk
= 0.
e
−ik 2π
We also get, by closing the contour downstairs,
R ikx e−ika dk
1
1
e
= (−2πi)( )( ) = 1.
−ik 2π
−i 2π
Thus (II) is proven.
21. Find the Fourier transform of
1
, −∞ < x < ∞ .
f (x) =
1 + x4
Solution:
The Fourier transform of
1
is
1 + x4
R ∞ e−ikx
dx
−∞ 1 + x4
The poles are located at
e±iπ/4 , e±3iπ/4 .
For k > 0 , we close the contour downstairs; and for k < 0 , we close the
contour upstairs.√ We then obtain the Fourier
transform of the function as
√
√
π −|k|(1+i)/ 2+iπ/4
e
+ c.c. = πe−|k|/ 2 cos(k/ 2 − π/4) .
2
22. Find the Fourier coefficient cn of the function f (θ) = eθ , −π << π. By
evauating the Fourier series for f (θ) at θ = π and at θ = −π, prove
14
π eπ + e−π
1
1
=
− .
2
π − e−π
1
+
n
2
e
2
n=1
∞
P
Solution:
Rπ
dθ
e−inθ eθ π
eπ − e−π
cn = −π e−inθ eθ
=
|−π = (−1)n
.
2π
2π(1 − in)
(1 − in)
Thus
∞
eπ − e−π X (−1)n einθ
.
eθ =
2π
1 − in
n=−∞
At θ = π and θ = −π, the Fourier seris has the same value
∞
eπ − e−π X
2
1
eπ − e−π
2
2
+···+
+ · · ·)
=
(1 +
+
2
2π
1
−
in
2π
1
+
1
1
+
2
1
+
n2
n=−∞
∞
X
1
eπ − e−π
=
(1 + 2
).
2π
1
+
n2
n=1
Since f (π) = eπ , f (−π) = e−π , the Fourier seris at θ = ±π is equal to the
average of f (π) and f (−π). Thus
∞
X
eπ + e−π
eπ − e−π
1
=
(1 + 2
),
2
2π
1
+
n2
n=1
or
∞
P
1
π eπ + e−π
1
=
− .
2
π
−π
2 e −e
2
n=1 1 + n
23. Find Fourier coefficient an for
1
, with a and b positive and b < a.
f (θ) =
a + b cos θ
Solution:
We have
1 Rπ
einθ dθ
.
a−n =
2π −π a + b cos θ
Note that
an = a∗−n .
Let z ≡ Ieiθ , then we have
I
1
z n−1 dz
1
z n dz
a−n =
=
.
−1
πi c [2a + b(z + z )]
πbi c [2az/b + (z 2 + 1)]
where c is the unit circle and n is non-negative. The integrand has simple
poles at
p
z = −a/b ± a2 /b2 − 1.
p
The pole inside the unit circle is located at −a/b+ a2 /b2 − 1.Using Cauchy’s
residue theorem, we get,
√ for n > 0,
(−1)n (a − a2 − b2 )n
√
a−n =
.
bn a2 − b2
Since a−n is real, we have
√
(−1)n (a − a2 − b2 )n
√
, n ≥ 0.
an = a−n =
bn a2 − b2
15
24. Use the Laplace transform to find the solution of
d2 y
+ y = δ(t − 1) + 2δ(t − 2)
dt2
dy
·
·
satisfying the initial conditions y(0) = y(0) = 0, where y ≡
.
dt
Solution:
Let L(s) be the Lapalce transform of y(s), then
(s2 + 1)L(s) = e−s + 2e−2s ,
or L(s) = (e−s + 2e−2s )/(s2 + 1).
Thus
R i∞ es(t−1) + 2es(t−2) ds
.
y(x) = −i∞
s2 + 1
2πi
By carrying out the integration, we get
y(x) = θ(t − 1) sin(t − 1) + 2θ(t − 2) sin(t − 2),
where
θ(τ ) = 1, τ > 0,
= 0, τ < 0.
25. Solve with the method of Laplace transform the initial-value problem
√
d2 y
+ y = x, x > 0,
2
dx
with the initial condition
y(0) = 1, y 0 (0) = 0.
Solution:
We have
−s + (s2 + 1)Y (s) = Γ(3/2)/s3/2 ,
where Y (s) is the Laplace transform of y(x). Thus
Γ(3/2)
s
+ 3/2 2
.
Y (s) = 2
s + 1 s (s + 1)
By performing Ran inverse Laplace transform, we get
x
y(x) = cos x + 0 x01/2 sin(x − x0 )dx0 .
The last integral is obtained by using the convolution theorem.
26. Let the Laplace equation ∇2 u(r, θ) = 0 be satisfied inside the disk r < 5,
where r and θ are the polar coordinates,and let the boundary condition be
u(5, θ) = sin2 θ.
Find u(r, θ).
Solution:
(eiθ − e−iθ )2
1 − cos 2θ
sin2 θ =
=
.
−4
2
Thus
1 − r2 cos 2θ/25
u(r, θ) =
.
2
16
Chapter 3
1. Find the solution u(x, y) of
ux + (x2 + y)uy = 0,
u(0, y) =
1
.
1 + y2
Solution:
The equation for the characteristic curves is given by
dx
dy
= 2
,
1
x +y
or
dy
= y + x2 .
dx
The complementary solution of the equation above is
Y = cex ,
and a particular solution is
1
yp = −
x2 = −(1 + D + D2 + · · ·)x2 = −(x2 + 2x + 2).
1−D
Thus the characteristic curves are given by
y = cex − (x2 + 2x + 2),
or
(y + x2 + 2x + 2)e−x = c.
Therefore, the general solution of the PDE is
u(x, y) = f ((y + x2 + 2x + 2)e−x ).
The initial condition gives
1
= f (y + 2),
1 + y2
or
1
.
f (y) =
1 + (y − 2)2
Therefore,
u(x, y) =
1 + [(y +
x2
1
.
+ 2x + 2)e−x − 2)]2
2. Find the solution u(x, y) of
ux + (e2x + y)uy = 0, u(0, y) = ey
Solution:
The equation for the characteristic curves is
dx
dy
= 2x
,
1
e +y
or y 0 = y + e2x ⇒ y = cex + e2x ⇒ (y − e2x )e−x = c.
Therefore, the general solution for thie PDE is
u(x, y) = f (ye−x − ex ).
From the initial condition, we get
17
ey = f (y − 1) ⇒ f (y) = ey+1 .
Therefore we have
−x
x
u(x, y) = eye −e +1 .
3. Solve
ut + xux = 0
2
with the initial condition u(x, 0) = e−x .
Solution:
The equation for the characteristic curves are
dx
dt
=
.
1
x
Thus the characteristic curves are given by
t = ln x + c0 ,
or
xe−t = c.
Therefore, the general solution of the equation is
u(x, t) = f (xe−t ).
To determine f , we set t = 0 and find that
2
f (x) = e−x .
Thus
2 −2t
u(x, t) = e−x e .
Chapters 4 and 5
∂2u ∂2u
+ 2 = u holds in the rectangle with vertices (0, 0),
∂x2
∂y
πy
(a, 0), (a, b), (0, b). At the right vertical side, u(a, y) = sin( ), 0 < y < b. The
b
function u vanishes at the other three sides of the rectangle. Find u.
1. The equation
Solution:
Since u vanishes at the two horizontal sides, we express u as a Fourier sine
series of y :
∞
P
nπy
u(x, y) =
An (x) sin(
).
b
n=1
Substituting u into the partial differential equation, we get
d2 An (x) n2 π 2 + b2
−
An (x) = 0.
dx2
b2 √
√
2
An (x) = cn cosh(x n π 2 + b2 /b) + dn sinh(x n2 π 2 + b2 /b).
Thus
18
√
√
nπy
[cn cosh(x n2 π 2 + b2 /b) + dn sinh(x n2 π 2 + b2 /b)] sin(
).
b
n=1
The boundary condition u(0, y) = 0 gives
∞
P
nπy
0=
cn sin(
),
b
n=1
or cn = 0.
πy
That u(a, y) = sin( ) gives
b
∞
√
P
nπy
πy
2
2
dn sinh(a n π + b2 /b)] sin(
) = sin( ),
b
b
n=1
or
dn = 0, n 6= 1, √
d1 = 1/(sinh(a π 2 + b2 /b).
We have
√
√
πy
u(x, y) = sinh(x π 2 + b2 /b)] sin( )/(sinh(a π 2 + b2 /b).
b
u(x, y) =
∞
P
∂2u
∂2u
= 2 , −∞ < x < ∞, t > 0, with the
2
∂x
∂t
initial conditions u(x, 0) = f (x), ut (x, 0) = 0.
a. Find the
R ∞equation satisfied by L(x, s), where
L(x, s) ≡ 0 dt e−st u(x, t).
b. Assuming that both f (x) and L(x, s) have Fourier transforms, find L(x, s)
in the form of a Fourier integral. (You are allowed to differentiate a Fourier
integral by differentiating its integrand.)
c. Find u(x, t).
Note: the Laplace transform of u00 (t) is equal to s2 L(s)−u0 (0)−su(0), where
L(s) is the Laplace trandform of u(t).
2. Consider the wave equation
Solution:
a. Let the Laplace transform of u(x, t) be L(s, x).We have, by multiplying
the equation above with e−st and integrate with respect to t from 0 to ∞,
d2 L(x, s)
− s2 L(x, s) = −sf (x).
dx2
b. Let
R ∞ dk ikx
e l(k, s)
L(x, s) = −∞
2π
and substitute it into the equation for L(x, s) above, we have
(−k 2 − s2 )l(k, s) = −sF (k),
where F (k) is the Fourier transform of f (x).Therefore,
R ∞ dk ikx sF (k)
sF (k)
l(k, s) = 2
⇒ L(x, s) = −∞
e
.
2
k +s
2π
k 2 + s2
c. We have
R ds st R ∞ dk ikx sF (k)
u(x, t) = c
e −∞
e
,
2πi
2π
k 2 + s2
where c is a vertical contour to the right of all singularities of L(x, s). By
closing contour in the s-plane to the left, we get
1 R ∞ dk ikx ikt
f (x + t) + f (x − t)
u(x, t) =
e (e + e−ikt )F (k) =
, t > 0.
−∞
2
2π
2
19
Note: It is easier to find the solution of this problem by making the Fourier
transform with respect to x first, which we have demonstrated in the textbook.
3. Consider the Laplace equation
∇2 u(r, θ) = 0
which holds inside a disk of radius 3 with the center at the origin. The
boundary condition is
∞ sin nθ
P
.
u(3, θ) =
n=1 n!
Find u(r, θ) in a closed form.
Hint: Make use of ex = 1 + x + x2 /2! + · · · + xn /n! + · · ·.
Solution:
∞ sin nθ r
∞ einθ − e−inθ r
P
1 P
u(r, θ) =
( )n =
( )n
3
2i n=0
n!
3
n=1 n!
−iθ r n
iθ r n
)
−
(e
)
(e
∞
1 P
3
3 = 1 [ereiθ /3 − ere−iθ /3 ]
=
2i n=0
n!
2i
1 r cos θ/3 ir sin θ/3
−ir sin θ/3
= e
[e
−e
] = er cos θ/3 sin(r sin θ/3).
2i
Note: u(x, y) is the imaginary part of the analytic function ez/3 .
··
4. Consider the Sch0dinger equation
∂2Ψ
∂Ψ
=− 2,
Ψ = Ψ(x, t),
i
∂t
∂x
which holds for all values of x and all positive values of t. Let the initial
value of Ψ(x, t) be
Ψ(x, 0) = f (x)
and let the boundary conditions for Ψ be
Ψ(±∞, t) = 0.
··
(a) Make a Fourier transform of the Sch0dinger equation with respect to x
and solve the resulting ordinary differential equation.
(b) Express the solution Ψ(x, t) satisfying the initial condition (B) in the
form of a Fourier integral.
Solution:
a. After making the Fourier transform, we have
∼
∼
∂ Ψ(k, t)
i
= k 2 Ψ(k, t).
∂t
The solution of the equation above is
∼
2
Ψ(k, t) = a(k)e−ik t .
Thus
R ∞ dk ikx−ik2 t
Ψ(x, t) = −∞
e
a(k).
2π
b. To satisfy the initial condition, we have
20
∼
a(k) = f (k).
Therefore
R ∞ dk ikx−ik2 t ∼
f (k).
Ψ(x, t) = −∞
e
2π
The answer above can be expressed as an integral over x0 :
R ∞ dk ikx−ik2 t R ∞
0
Ψ(x, t) = −∞
e
dx0 e−ikx f (x0 )
−∞
2π
r
R∞
1
(x − x0 )2 i
= −∞ dx0
exp[
] f (x0 ).
4iπt
4t
5. Find the eigenvalues and the eigenfunctions of the Sturm-LIouviile problem
d2 u
= −λu, 0 < x < L; u0 (0) = 0 and u(L) = 0.
dx2
Solution:
The general solution
of the√differential equation above is
√
u(x) = a cos( λx) +b sin( λx).
Taking the
above, we get
√ derivative
√ of the expression
√
u0 (x) = λ[−a sin( λx) +b cos( λx)].
0
The
√ boundary condition u (0) = 0 gives
b λ = 0 ⇒ either λ = 0 or b = 0.
If λ = 0, the solution of the differential equation is u = c1 +c2 x. This solution
vanishes identically as the boundary conditions u0 (0) = u(L) = 0 are imposed.
If λ 6= 0, we
√ have b = 0, or
u = a cos( λx).
The boundary
condition u(L) = 0 gives
√
a cos( λL) = 0.
Since a 6= 0 for a non-trivial solution, the eigenvalues of the non-trivial
solutions
are
√
λn L = (n + 1/2)π, n = 0, 1, 2, · · ·
or
(n + 1/2)2 π 2
λn =
, n = 0, 1, 2, · · ·.
L2
The eigenfunction
corresponding to the eigenvalue above is
r
(n + 1/2)πx
2
cos[
].
un (x) =
L
L
These eigenfunctions are orthonormal:
RL
un (x)um (x)dx = δnm .
0
6. Find the solution of uxx + uyy = 0, ∞ > y > 0 and 0 < x < 1,
with ux (0, y) = u(L, y) = 0, u(x, 0) = δ(x − 1/2) and u(x, ∞) = 0.
Solution:
We express u by the Fourier cosine series
∞
P
u(x, y) =
An (y) cos[(n + 1/2)πx],
n=0
which satisfies the boundary condition ux (0, y) = u(L, y) = 0.
21
Substituting this expression of u into the PDE, we get
d2 An (y)
= [(n + 1/2)π]2 An (y).
dy 2
The solution of the differential equation above is
An (y) = an e(n+1/2)πy + bn e−(n+1/2)πy .
The boundary condition at y = ∞ gives an = 0. Therefore
∞
P
u(x, y) =
bn e−(n+1/2)πy cos[(n + 1/2)πx].
n=0
The boundary condition at y = 0 requires
∞
P
δ(x − 1/2 ) =
bn cos[(n + 1/2]πx].
n=0
Thus we have
R1
1
bn = 0 δ(x − 1/2) cos[(n + 1/2)πx]dx = cos[(n + 1/2)π/2],
2
or
bn = 2 cos[(n + 1/2)π/2].
Therefore we have
∞
P
(n + 1/2)π
u(x, y) = 2
e−(n+1/2)πy cos[
] cos[(n + 1/2)πx].
2
n=0
7. Consider the vibration of the surface of a drum of unit radious. The
equation of motion is
∂2u
1 ∂ ∂u
1 ∂2u
=
(r
)
+
, r < 1, 0 ≤ θ ≤ 2π,
∂t2
r ∂r ∂r
r2 ∂θ2
where u(r, θ, t) is the displacement of the drum surface, and r and θ are the
polar coordinates. The boundary condition is
∂u
.
ur (1, θ, t) = 0, where ur ≡
∂r
The initial conditions are
u(r, θ, 0) = cos θ,
ut (r, θ, 0) = 0.
Find u(r, θ, t).
Solution:
Substituting u = T (t)R(r)Θ(θ) into the PDE, we get
T”
1 d
Θ”
=
(rR0 ) + 2 = −c.
T
rR dr
r Θ
d dR
n2
Thus T ” = −cT, Θ” = −n2 Θ,
(r
)−
R = −crR.
dr dr
r
Since Θ is required to satisfy Θ(θ + 2π) = Θ(θ), n is an integer.
The solution
√ of the equation for R is
R = Jn ( cr),
which satisfies the boundary condition of R(0) being finite, where Jn is the
Bessel function of order n. Let rm,n be the mth root of Jn0 , i.e., Jn0 (rm,n ) = 0,
m = 1, 2, · · ·
22
Then
P
cos(rm,n t)Jn (rm,n r)(Am,n cos nθ+Bm,n sin nθ) satifies the boudary
n,m
condition at r = 1 as well as the the initial condition ut (r, θ, 0) = 0.
To satisfy the initial condition u(r, θ, 0) = cos θ we require
P
Jn (rm,n r)(Am,n cos nθ + Bm,n sin nθ) = cos θ.
n,m
Thus all Fourier sine coefficients Bm.n = 0, while the only non-vanishing
Fourier
P cosine coefficients are Am,1 . These coefficients are determined from
J1 (rm,1 r)Am,1 = 1.
m
We get R
1
J1 (rm,1 r)rdr
Am,1 = R 10
[J1 (rm,1 r)]2 rdr
0
and
∞
P
u(r, θ, t) =
Am,1 cos(rm,1 t)Jn (rm,1 r) cos θ.
m=1
8. Consider the partial differential equation
1 ∂
∂
1
∂2
1
∂
∂
[ 2 r2
+ 2 2
+ 2
sin θ ]u = V (r, θ, φ)u,
r ∂r ∂r r sin θ ∂φ2
r sin θ ∂θ
∂θ
where
V3 (θ)
V2 (φ)
.
V (r, θ, φ) = V1 (r) + 2 2 +
r2
r sin θ
Perform separation of variables and reduce the equation into three ordinary
differential equations.
Solution:
Substituting
u = R(r)Θ(θ)Φ(φ)
into the PDE and multiplying the resulting equation by r2 , we get
1 d 2 dR
1 d2 Φ
1
1 1
d
dΘ
[
[
(r
) − r2 V1 (r)] +
− V2 (φ)] + [
( sin θ
)−
2
2
R dr
dr
Θ sin θ dθ
dθ
sin θ Φ dφ
V3 (θ)] = 0.
Note that the first bracket above is a function of r only, while the second
and the third brackets are functions of θ and φ only. Thus we have
1
1 d2 Φ
1 1 d
dΘ
1 d 2 dR
(r
) − r2 V1 (r) = − 2 [
− V2 (φ)] − [
(sin θ
)−
R dr
dr
Θ sin θ dθ
dθ
sin θ Φ dφ2
V3 (θ)] = a,
where a is a constant. The equation above contains two ordinary differential
equations:
d 2 dR
(r
) − [r2 V1 (r) + a]R = 0,
dr
dr
and
1 1 d
1
1 d2 Φ
dΘ
2 [ Φ dφ2 − V2 (φ)] + [ Θ sin θ dθ (sin θ dθ ) − V3 (θ)] + a = 0.
sin θ
23
Multiplying the equation above by sin2 θ, we get
1
1 d2 Φ
d
dΘ
− V2 (φ)] + [ sin θ (sin θ
) − sin2 θV3 (θ) + a sin2 θ] = 0.
[
Φ dφ2
Θ
dθ
dθ
We see that the first bracket above is a function of φ only, while the second
bracket above is a function of θ only. Thus we have
1
d
dΘ
1 d2 Φ
− V2 (φ) = − sin θ (sin θ
) + sin2 θV3 (θ) − a sin2 θ = b,
Φ dφ2
Θ
dθ
dθ
where b is a constant. Therefore we have the ordinary differential equations
d2 Φ
− [b + V2 (φ)]Φ = 0
dφ2
and
d
dΘ
sin θ [sin θ
] + [b + a sin2 θ − sin2 θV3 (θ)]Θ = 0.
dθ
dθ
V3 (θ)
V2 (φ)
(Indeed, it is not difficult to prove that V = V1 (r) + 2 2 + 2 is the
r
r sin θ
most general form of V for the equation in this problem to be separable.)
d2 u
6.
= −λu, 0 < x < L; u0 (0) = 0 and u(L) = 0.
dx2
The general solution
of the√differential equation is
√
u(x) = a cos( λx) +b sin( λx).
Taking the
above, we get
√ derivative
√ of the expression
√
u0 (x) = λ[−a sin( λx) +b cos( λx)].
0
The
√ boundary condition u (0) = 0 gives
b λ = 0 ⇒ either λ = 0 or b = 0.
If λ = 0, the solution of the differential equation is u = c1 +c2 x. This solution
vanishes identically as the boundary conditions are imposed.
If λ 6= 0, we
√ have b = 0, or
u = a cos( λx).
The boundary
condition u(L) = 0 gives
√
a cos( λL) = 0.
Since
a 6= 0, the eigenvalues of the non-trivial solutions are
√
λn L = (n + 1/2)π, n = 0, 1, 2, · · ·
or
(n + 1/2)2 π 2
λn =
, n = 0, 1, 2, · · ·.
L2
The eigenfunction
corresponding to the eigenvalue above is
r
2
(n + 1/2)πx
un (x) =
cos[
].
L
L
These eigenfunctions are orthonormal:
RL
un (x)um (x)dx = δnm .
0
3. uxx + uyy = 0, ∞ > y > 0 and 0 < x < 1;
ux (0, y) = 0, u(L, y) = 0
and
u(x, 0) = δ(x − 1/2) and u(x, ∞) = 0.
Let us express u by the Fourier cosine series
24
∞
P
u(x, y) =
An (y) cos[(n + 1/2)πx],
n=0
which satisfies the boundary condition at x = 0 and x = 1. The coefficient
An (y) satisfies the differential equation
d2 An (y)
= [(n + 1/2)π]2 An (y).
dy 2
The solution of the differential equation above is
An (y) = an e(n+1/2)πy + bn e−(n+1/2)πy .
The boundary condition at y = ∞ gives an = 0. Therefore
∞
P
u(x, y) =
bn e−(n+1/2)πy cos[(n + 1/2)πx].
n=0
The boundary condition at y = 0 requires
δ(x − 1/2 ) =
∞
P
bn cos[(n + 1/2]πx].
n=0
Thus we have
R1
1
bn = 0 δ(x − 1/2) cos[(n + 1/2)πx]dx = cos[(n + 1/2)π/2],
2
or
bn = 2 cos[(n + 1/2)π/2].
Therefore,
∞
P
(n + 1/2)π
] cos[(n + 1/2)πx].
u(x, y) = 2
e−(n+1/2)πy cos[
2
n=0
Chapter 6
1. Find the two independent solutions of
d2 y
dy
+x
+y =0
dx2
dx
in the form of Maclaurin series. One of the series can be summed into a
closed form. Find this closed form.
Solution:
Since x = 0 is an ordinary point of the equation, we put y =
P
d2 y
=
An n(n − 1)xn−2 ,
dx2
P
P
dy
+y =
An (n + 1)xn = An−2 (n − 1)xn−2 .
x
dx
We have
An−2
An = −
.
n
25
P
An xn . Thus
The recurrence formula above for n = 2m gives
A2(m−1)
A0
A2m = −
=⇒ A2m = (−1)m m .
2m
2 m!
Thus one of the solutions is
∞ (−x2 )m
P
2
y1 (x) =
= e−x /2 ,
m
m=0 2 m!
which is in a closed form. The recurrence formula above for n = 2m + 1
gives
A2(m−1)+1
A1 Γ(3/2)
= (−1)m m
.
2(m + 1/2)
2 Γ(m + 3/2)
Thus the second solution is
∞ (−x2 /2)m
P
y2 (x) = x
.
m=0 Γ(m + 3/2)
The general solution of the equation is
∞ (−x2 /2)m
P
2
.
y(x) = ae−x /2 + bx
m=0 Γ(m + 3/2)
A2m+1 = −
Note: We may take advantage of the fact that the Wronskian between y1
and y2 is a known function, and express y2 as an integral.
2. Find the Frobenius solutions of
3
dy
d2 y
+ y = 0,
x(1 − x) 2 +
dx
dx 4
the truncated series of which are good approximations of the solutions near
x = 0.
Solution:
point x = 0 is a regular singular point. Thus we try
y =
P a. The
An xn+s .
We have
P
P
d2 y dy
x 2+
=
An [(n+s)(n+s−1)+(n+s)]xn+s−1 =
An (n+s)2 xn+s−1 ,
dx
dx
P
d2 y
3
−x2 2 + y = − An [(n + s)(n + s − 1) − 3/4]xn+s
dx
4
P
P
= − An (n + s − 3/2)(n + s + 1/2)xn+s = − An−1 (n + s − 5/2)(n + s −
n+s−1
1/2)x
.
Thus we have
(n + s)2 An = (n + s − 5/2)(n + s − 1/2)An−1 .
The indicial equation is obtained by setting n in the equation above to zero.
We get
s2 = 0 =⇒ s1 = s2 = 0.
(n + s − 5/2)(n + s − 1/2)
An =
An−1
(n + s)2
[(n + s − 5/2) · · · (s − 3/2)][(n + s − 1/2) · · · (s + 1/2)
=⇒ An =
(n + s)2 · · · (1 + s)2
26
0
[Γ(s + 1)]2
Γ(n + s − 3/2)Γ(n + s + 1/2)
,
[Γ(n + s + 1)]2
Γ(s − 3/2)Γ(s + 1/2)
where we have set A0 to unity. Thus one of the solutions is obtained by
setting s = 0 :
P Γ(n − 3/2)Γ(n + 1/2) n
1
x .
y1 (x) =
Γ(−3/2)Γ(1/2)
[n!]2
The second solution is
∞
∂ X [(n + s − 5/2) · · · (s − 3/2)][(n + s − 1/2) · · · (s + 1/2) n+s
y2 (x) =
x
]s=0
∂s n=0
(n + s)2 · · · (1 + s)2
∞
X
[(n − 5/2) · · · (−3/2)][(n − 1/2) · · · (1/2)]
= (ln x)y1 (x) +
(n!)2
n=0
1
1
1
1
1
1
·[(
+···+
)+(
+···+
) − 2( + · · · + )]xn
n − 5/2
−3/2
n − 1/2
1/2
n
1
=
3.Consider the equation
1
dy
d2 y
− by = 0 ,
x 2 + ( − x)
dx
2
dx
where b is a constant.
a.Locate and classify the singular points of this equation in the extended
plane.
b.Find by the Frobenius method the two independent solutions of this equation, the point of expansion being the origin.
Solution:
1.
(a)The point x = 0 is a regular singular point.
Let x = 1/t ,
then we have
d dy
1
dy
t t2
− ( t2 − t)
− by = 0 ,
dt dt
2
dt
or
d2 y 3t/2 + 1 dy
b
+
− 3y = 0 .
2
2
dt
t
dt
t
Thus the point ∞ is an irregular singualr point of rank 1 .
(b) Let
P
y = an xn+s ,
where
a0 6= 0 , and a−1 = a−2 = · · · = 0 .
Then
P
P
−by = − ban xn+s = − ban−1 xn+s−1 ,
1 0 P1
y =
an (n + s)xn+s−1 ,
2
2
27
P
−xy 0 = − an−1 (n + s − 1)xn+s−1
and P
xy 00 =
an (n + s)(n + s − 1)xn+s−1
We then have
an (n + s)(n + s − 1/2) = an−1 (n + s − 1 + b)
Setting n = 0 in the equation above, we get
s = 0 or 1/2 .
For n 6= 0, we have
(n + s − 1 + b)
an−1
an =
(n + s)(n + s − 1/2)
Γ(n + s + b)Γ(s + 1)Γ(s + 1/2)
a0 .
=
Γ(s + b)Γ(n + s + 1)Γ(n + s + 1/2)
Setting s = 0 , we find that one of the solutions is
∞
P
Γ(n + b)
y1 =
xn .
n=0 Γ(n + 1/2)n!
Setting s = 1/2 , we find the other solution is
∞ Γ(n + 1/2 + b)
P
y2 =
xn+1/2 .
n=0 Γ(n + 3/2)n!
4. Consider the differential equation
y” + x2 y = 0 .
a.Find and classify the singular points of this equation in the extended complex plane.
b.Find one (any non-trivial one) of its solutions in the form of a series expanded around the origin.
Solution:
a.There are no singular points in the finite plane.
1
Let x = ,
t
then
d dy
d2 y
dy
y” = t2 t2
= t4 2 + 2t3
,
dt dt
dt
dt
and hence
d2 y 2 dy
1
+
+ 6y = 0 .
dt2
t dt
t
Thus the infinity is an irregular singular point of the differential equation.
b. Let
P
y = an xn
with the understanding that
a−1 = a−2 = · · · = 0 .
28
We have
P
y” = an n(n − 1)xn−2
and P
P
x2 y =
an xn+2 =
an−4 xn−2 .
Thus we have
an n(n − 1) = −an−4 ,
or
an−4
.
an = −
n(n − 1)
Thus
a4(m−1)
1
a0 Γ(3/4)
a4m = −
= (− )m
.
16m(m − 1/4)
16 m!Γ(m + 3/4)
Thus a solution is
∞
X
x4m
1
.
(− )m
16 m!Γ(m + 3/4)
m=0
5. Find the general solution of
xy” − y = 0
in two steps.
a. Find a solution of this ODE with the Frobenius method.
b. Find another independent solution of this ODE in the form of an infinite
series.
Solution:
6. Let
y” − (x4 − 3x−2 /16)y = 0.
a. Locate and classify all the singular points of this equation, finite or
infinite. Give the rank of each of the irregular singular points, if any.
b. Find the two independent solutions of this equation which are useful when
x is small.
7.a. Find with the Frobenius method one of the independent solutions in
the form of a series (expanded around the origin) for
d2 y
dy
3
x(1 − x) 2 +
+ y = 0.
dx
dx 4
b. Find the second independence solution of the equation.
29
30