MATH323 2020 January Exam Solutions
February 3, 2020
1. (Lecture, homework based question)
(a) Inserting y = xn, y′ and y′′ into the equation we find n2 − n + 3n + 1 = 0. Solving for n
to obtain n1 = −1. The solution is
y(x) = C1x
−1 + C2x−1 lnx. [4 marks]
(b) Using the variation of parameters formula
yp(x) = L1(x)x
−1 + L2(x)x−1 lnx.
For finding L1(x) and L2(x) we have to solve
L′1
1
x2
+ L′2
lnx
x2
= 0,
−L′1
1
x2
+ L′2
(
1
x2
− lnx
x2
)
= 4
lnx
x2
.
[2 marks]
Adding the two equations
L′2
1
x2
=
4 lnx
x2
⇒ L′2 = 4 ln x.
L2 = 4
∫
lnxdx = 4(x lnx−
∫
dx) = 4(x lnx− x).
From the first equation
L′1 = −
L′2x lnx
x
= −L′2 lnx = −4 ln2 x.
L1 = −4
∫
ln2 xdx = −4
(
x ln2 x−
∫
x
2 lnx
x
dx
)
=
−4
(
x ln2 x− 2x lnx+ 2 ∫ x
x
dx
)
= −4 (x ln2 x− 2x lnx+ 2x) . [2 marks]
Thus the particular solution is
yp = −4
(
x ln2 x− 2x lnx+ 2x) 1
x
+ 4 (x lnx− x) lnx
x
yp = −4
(
ln2 x− 2 lnx+ 2)+ 4 (ln2 x− lnx) =
−4 ln2 x+ 8 lnx− 8 + 4 ln2 x− 4 lnx = 4 (ln x− 2) . [1 mark]
The general solution is y(x) = C1x
−1 + C2x−1 lnx+ 4 (lnx− 2) . [1 mark]
1
Boundary conditions y(1) = 0, y′(2) =
1
4
C1 − 8 = 0⇒ C1 = 8 .
y′ = −C1
x2
+ C2
(
1
x
x− lnx
)
x2
+ 4
1
x
⇒ 1
4
= −C1
4
+ C2
(1− ln 2)
4
+
4
2
,
1
4
= −2 + C21− ln 2
4
+ 2⇒ C2 = 1
1− ln 2
y(x) =
8
x
+
1
(1− ln 2)
lnx
x
+ 4 (lnx− 2) .
[2 marks]
(c) y1 = e
−x2 , y′1 = −2xe−x2 , y′′1 = −2e−x2 + 4x2e−x2 ,
y′′ − 1
x
y′ − 4x2y = e−x2
(
−2 + 4x2 − 1
x
(−2x)− 4x2
)
= 0 [1 mark]
Guess for the second solution: y2(x) = v(x)y1(x) = v(x)e
−x2 .
y′2(x) = v
′(x)e−x
2 − 2xv(x)e−x2
y′′2 = v
′′e−x
2 − 2xv′e−x2 + v
[
−2e−x2 + 4x2e−x2
]
− 2xv′e−x2 = e−x2 [v′′ − 4xv′ − 2v + 4x2v] .
[1 mark]
Substituting y2(x) and its derivatives into the differential equation:
y′′ − 1
x
y′ − 4x2y = e−x2
[
v′′ − 4xv′ − 2v + 4x2v − 1
x
(−2xv + v′)− 4x2v
]
=
= e−x
2
[
v′′ − 4xv′ − 2v + 4x2v + 2v − 1
x
v′ − 4x2v
]
= 0
v′′ − 4xv′ − 1
x
v′ = 0
[2 marks]
The equation is first order for w = v′(t).
w′ −
(
4x+
1
x
)
w = 0,
dw
dx
=
(
4x+
1
x
)
w ⇒ dw
w
=
(
4x+
1
x
)
dx
lnw =
4x2
2
+ lnx⇒ lnw = ln e2x2 + lnx⇒ w = xe2x2 . [2 marks]
v =
∫
xe2x
2
dx =
1
2
∫
e2x
2
dx2 =
1
4
∫
e2x
2
d(2x2) =
1
4
e2x
2
.
y2(t) = v(x)e
−x2 = 1
4
e2x
2
e−x
2
= 1
4
ex
2
. y(t) = C1e
−x2 + C2
1
4
ex
2
. [2 marks]
2
2. (Lecture, homework based question)
Solution. (a) The Euler-Lagrange equation is
∂F
∂y
− d
dx
(
∂F
∂y′
)
= 0.
The fact that a function satisfies the Euler–Lagrange equation and the boundary conditions
confirms that it is a critical function, and does not guarantee that it is the minimizer. The
nature of a critical function will be elucidated by the second derivative test (Legendre condition):
in order for y(x) to lead to a minimum of the functional J [y], the condition
∂ 2F
∂y′2
> 0
needs to be fulfilled for this function at all a ≤ x ≤ b. [3 marks]
(b) For the given functional
J [y(x)] =
∫ 2
1
(x2y′2 + 12y2 + 4yx4)dx
∂F
∂y
= 24y + 4x4,
∂F
∂y′
= 2x2y′
d
dx
(
∂F
∂y′
)
= 2x2y′′ + 4xy′.
The Euler-Lagrange equation is
2x2y′′ + 4xy′ − 24y − 4x4 = 0⇒ x2y′′ + 2xy′ − 12y = 2x4.
For the corresponding homogeneous equation substituting y = xn
x2n(n− 1)xn−2 + 2xnxn−1 − 12xn = 0
n2 − n+ 2n− 12 = 0 ⇒ n2 + n− 12 = 0
⇒ n1 = 3, n2 = −4, yCF = C1x3 + C2x−4
For a particular integral yp = αx
4, y′p = 4αx
3, y′′p = 12αx
2
Substituting into the equation x4(12α + 8α− 12α) = 2x4 ⇒ α = 1
4
, yp =
x4
4
⇒ y = C1x3 + C2x−4 + x
4
4
From BCs y(1) = −4, y(2) = 127 C1 + C2 + 14 = −4
8C1 +
C2
16
+ 4 = 127
⇒ 8C1 + 8C2 + 2 = −32
8C1 +
C2
16
+ 4 = 127
⇒ 8C2−C2
16
−2 = −32−127 = −159 ⇒ 8C2−C2
16
= −157 ⇒ 127C2 = −157×16
C2 = −2512
127
, C1 =
2512
127
− 1
4
− 4. y = C1x3 − 2512
127x4
+
x4
4
. [8 marks]
3
(c)
F = (x2y′2 + 12y2 + 4yx4),
∂F
∂y′
= 2x2y′,
∂2F
∂y′2
= 2x2 > 0, 1 ≤ x ≤ 2
The extremum is minimum. [3 marks]
(d) The solution of the Euler-Lagrange equation in this case is
y = C1x
3 + C2x
−4
From boundary conditions
C1 + C2 = 1
8C1 +
C1
16
= 8
⇒ C1 = 1, C2 = 0⇒ y(x) = x3.
J [x3] =
∫ 2
1
(x2y′2 + 12y2)dx =
∫ 2
1
(x2(3x2)2 + 12(x3)2)dx
=
∫ 2
1
(9x6 + 12x6)dx = 21
x7
7
|21 = 381
[3 marks]
For y(x) = 7x− 6 which satisfies the same boundary conditions y(1) = 1 and y(2) = 8
J [y] =
∫ 2
1
(x2y′2 + 12y2)dx
= 49
∫ 2
1
x2dx+ 12
∫ 2
1
(7x− 6)2dx = 49x
3
3
+
12
7
∫ 2
1
(7x− 6)2d(7x− 6)
=
49
3
(8− 1) + 12
7
(7x− 6)3
3
|21 =
49
3
7 +
12
7
(
(14− 6)3
3
− 1
3
)
= 406.333
This means that y = x3 is a minimizer for J[y]. [3 marks]
4
(Lecture, homework based question)
3. Solution. (a) If F solves the variational problem and G satisfies the constraint equation
then H = F +λG will also satisfy Euler-Lagrange equation for any λ. This leads to the solution
of the Euler-Lagrange equation for the functional H
∂H
∂y
− d
dx
(
∂H
∂y′
)
= 0 with H = F + λG. (1)
After finding a general solution of equation (1) the constants in the solution and the Langrange
multiplier can be determined from the boundary conditions y(a) = y1 and y(b) = y2 and the
constrained equation. [2 marks]
(b) We have F = y and G =
√
1 + y′2 and
H = F − λG = y − λ
√
1 + y′2.
Since H is independent of x the Euler-Lagrange equation is
H − y′∂H
∂y′
= c1 = const.
⇒ y − λ(1 + y′2) 12 + y
′λ2y′
2(1 + y′2)
1
2
= c1
⇒ y − λ
(1 + y′2)
1
2
= c1 ⇒ y − c1 = λ
(1 + y′2)
1
2
.
Introducing z = y − c1
⇒ z = λ√
1 + z′2
⇒
(
λ
z
)2
− 1 = z′2
⇒ dz
dx
=
√
λ2 − z2
z
⇒
∫
z√
λ2 − z2dz =
∫
dx = x− c2.
[2 marks]
Substituting z = λ cos θ, dz = −λ sin θdθ we have
x− c2 = −
∫
λ cos θ
λ sin θ
λ sin θ
dθ = −λ sin θ.
We have x− c2 = −λ sin θ, y − c1 = λ cos θ hence
(x− c2)2 + (y − c1)2 = λ2.
The solutions is a segment of a circle of radius |λ| and length 10. [4 marks]
From y(−4) = 0, y(4) = 0
(−4− c2)2 + c21 = λ2
(4− c2)2 + c21 = λ2 ⇒ c2 = 0, 16 + c21 = λ2
5
The solution becomes x = −λ sin θ, y − c1 = λ cos θ
⇒ x2 + (y − c1)2 = 16 + c21 . [2 marks]
Since dx = −λ cos θdθ, dy = −λ sin θdθ
From constraint
10 =
∫ 4
−4
√
1 + y′2dx =
∫ 4
−4
√
dx2 + dy2 =
∫ θ
−θ
λ
√
sin2 θ + cos2 θdθ =
∫ θ
−θ
λdθ.
⇒
∫ θ
−θ
λdθ = 2λθ = 2θ
√
16 + c21 = 10 . [2 marks]
In order to find c1 we need to express θ via c1.
Using one of the boundary conditions y(4) = 0
4 = −λ sin θ
−c1 = λ cos θ ⇒ tan θ = 4
c1
.
Thus
10 = 2
√
16 + c21 tan
−1 4
c1
⇒ c1 = 1.88163.
Thus if we want to maximise the enclosed area between a fixed length of a rope and a line
segment, then the rope is an arc of a circle. This is a circle centred at (0,−1.88163), with the
radius
√
16 + 1.881632 ≈ 4.420467. [3 marks]
(c) As c1 → 0, we will get a semi circle, with radious 4 and
l = 2
√
16 + c21 tan
−1
(
4
c1
)
→ 2× 4× pi
2
= 4pi ≈ 12.5664
and the area Amax =
pi42
2
= 8pi ≈ 25.132. [2 mark]
A rectangle rope of the same circumference, (−4 ≤ a ≤ 4), 2a+2h = 12.5664 has the height
h =
12.5664− 2× 4
2
= 2.2832 and hence the area A = 2ah = 18.2656 < 25.132.
Thus the area of a segment is maximum. [3 marks]
6
(Lecture, homework based question)
4. Solution. Writing the system in a matrix form
Aux +Buy = c ,
A=
(
y 0
0 y
)
, B=
(
0 1
1 0
)
, ux =
(
ux
vx
)
, uy =
(
uy
vy
)
.
det(λA−B) = det
(
λy −1
−1 λy
)
= 0⇒ λ2y2 − 1 = 0⇒ λ = ±1
y
⇒ hyperbolic.
[4 marks]
Characteristics :
dy
dx
=
1
y
⇒ ydy = dx,
⇒ y
2
2
= x+ c⇒ α = x− y
2
2
dy
dx
= −1
y
⇒ ydy = −dx
⇒ y
2
2
= −x+ c˜⇒ β = x+ y
2
2
Change of variables
ux = uααx + uββx = uα + uβ and the same for v
uy = uααy + uββy = y(uβ − uα) and the same for v.
Substituting into the original equations
(I) uα + uβ + vβ − vα = sinx
(II) − uα + uβ + vβ + vα = − cosx.
(I)+(II) 2(uβ + vβ) = sin x− cosx⇒ (uβ + vβ) = 1
2
(
sin
(
α + β
2
)
− cos
(
α + β
2
))
(I)-(II) 2(uα − vα) = sinx+ cosx (uα − vα) = 1
2
(
sin
(
α + β
2
)
+ cos
(
α + β
2
))
.
Integrating to find Riemann invariants
u+ v =
1
2
[
−2 cos
(
α + β
2
)
− 2 sin
(
α + β
2
)]
+ f(α)
u+ v + cos
(
α + β
2
)
+ sin
(
α + β
2
)
= f(α) (A)
u− v = 1
2
[
−2 cos
(
α + β
2
)
+ 2 sin
(
α + β
2
)]
+ g(β)
u− v + cos
(
α + β
2
)
− sin
(
α + β
2
)
= g(β) (B)
[6 marks]
7
Using boundary conditions u(x, 0) = sin x and v(x, 0) = cos x,
y = 0, α = x, β = x
u(x, 0) + v(x, 0) + cos x+ sinx = f(x)
u(x, 0)− v(x, 0) + cos x− sinx = g(x)
⇒ sinx+ cosx+ cosx+ sinx = f(x)⇒ f(α) = 2(sinα + cosα)
sinx− cosx+ cosx− sinx = g(x)⇒ g(β) = 0).
Finding u and v from (A) and (B)
2u = −2 cos
(
α + β
2
)
+ f(α) + g(β)⇒ u = − cos
(
α + β
2
)
+
f(α)
2
+
g(β)
2
2v = − sin
(
α + β
2
)
+ f(α)− g(β)⇒ v = − sin
(
α + β
2
)
+
f(α)
2
− g(β)
2
⇒ u = − cosx+ sin
(
x− y
2
2
)
+ cos
(
x− y
2
2
)
.
v = − sinx+ sin
(
x− y
2
2
)
+ cos
(
x− y
2
2
)
.
[6 marks]
(d) Checking equations
yux+vy = y sinx+y cos
(
x− y
2
2
)
−y sin
(
x− y
2
2
)
−y cos
(
x− y
2
2
)
+y sin
(
x− y
2
2
)
= y sinx
uy+yvx = −y cos
(
x− y
2
2
)
+y sin
(
x− y
2
2
)
−y cosx+y cos
(
x− y
2
2
)
−y sin
(
x− y
2
2
)
= −y cosx
Checking boundary conditions
u(x, 0) = − cosx+ sinx+ cosx = sinx,
v(x, 0) = − sinx+ sinx+ cosx = cosx.
[4 marks]
8
(Lecture, homework based question)
5. (a) A BVP or IBVP is said to be well posed if
i) a solution to the problem exists
ii) the solution is unique, and
iii) the solution depends continuously on the problem data. This means that a “small”
changes in initial or boundary functions (in appropriate norms) results in “small” changes in
the solution. If one of the above conditions is not satisfied, the PDE problem is said to be
ill-posed. [2 marks]
(b) 4uxx − utt + 3t = 0,⇒ A = 4, B = 0, C = −1, B2 − 4AC = 16 > 0(hyperbolic).
Hyperbolic PDE problems require boundary conditions on the part of the boundary and
Cauchy initial conditions. Here we have Neumann boundary conditions at x = 0 and x = 2
and Cauchy conditions for u and ut at t = 0. The problem is well posed in 0 < x < 2, t > 0.
[3 marks]
(c) The auxiliary equation y2µ2 − 2xyµ+ x2 = (yµ− x)2 = 0 has one real root yµ = x.
The PDE is therefore parabolic with just one characteristic equation
dy
dx
= −x
y
⇒ ξ = x2 + y2 = const. [3 marks]
(d) Setting ξ(x, y) = x2 + y2 the equations ξ = ξ(x, y) and η = η(x, y) define a non-singular
coordinate transformation provided
|J| =
∣∣∣∣∂(ξ, η)∂(x, y)
∣∣∣∣ = 2xηy − 2yηx 6= 0.
Choosing η = x, under this transformation
ux = uξξx + uηηx = 2xuξ + uη, uy = 2yuξ,
uxx = 2uξ + 2x(uξξξx + uξηηx) + uηξξx + uηηηx = 2uξ + 4x
2uξξ + 4xuξη + uηη,
uyy = 2uξ + 4y
2uξξ,
uxy = 4xyuξξ + 2yuξη.
whence
y2uxx − 2xyuxy + x2uyy =
2y2uξ + 4x
2y2uξξ + 4xy
2uξη + y
2uηη − 8x2y2uξξ − 4xy2uξη + 2x2uξ + 4x2y2uξξ
=
1
xy
[
x3(2yuξ) + y
3(2xuξ + uη)
]
.
⇒ y2uηη + 2x2uξ + 2y2uξ = 2x2uξ + y
2
x
uη + 2y
2uξ
It follows that the canonical form of the PDE is
uηη =
uη
η
or
uηη
uη
=
1
η
. [7 marks]
Denoting uη = v ⇒ vη
v
=
1
η
or
∫
dv
v
=
∫
dη
η
.
ln v = ln η + ln f(ξ)⇒ v = uη = ηf(ξ).
Integrating
u =
1
2
η2f(ξ) + g(ξ)⇒ u(x, y) = 1
2
x2f(x2 + y2) + g(x2 + y2).
9
(e) Applying the first boundary condition (u(1, y) = 0)
u(1, y) =
1
2
f(1 + y2) + g(1 + y2) = 0.
If t = 1 + y2 ⇒ g(t) = −1
2
f(t).
From the second boundary condition (u(0, y) = e−y
2
)
u(0, y) = g(y2) = e−y
2
,
and if t = y2 ⇒ g(t) = e−t.
⇒ f(t) = −2e−t.
⇒ u(x, y) = −x2e−(x2+y2) + e−(x2+y2) = (1− x2)e−(x2+y2). [5 marks]
10
(Lecture, homework based question)
6. Solution. (a)
w′(z) = 1− 1
z2
and w′′(z) =
2
z3
and w′(z) = 0 if and only if z = ±1 the transformation is not conformal at these two points.
Since the second derivative w′′(z) 6= 0 angles between lines intersecting at these point are
doubled in the w-plane.
[4 marks]
(b)
w =
(
z +
1
z
)
=
(
x+ iy +
1
x+ iy
)
=
(
x+ iy +
x− iy
x2 + y2
)
i.e., u =
(
x+
x
x2 + y2
)
= x
(
1 +
1
x2 + y2
)
,
v =
(
y − y
x2 + y2
)
= y
(
1− 1
x2 + y2
)
.
With polar co-ordinates, x = r cos θ, y = r sin θ :
u = cos θ
(
r +
1
r
)
, v = sin θ
(
r − 1
r
)
. (2)
For the circle |z| = 2 and z = 2eiθ
w = 2eiθ +
1
2
e−iθ =
(
2 +
1
2
)
cos θ + i
(
2− 1
2
)
sin θ = u+ iv (3)
u2(
2 +
1
2
)2 + v2(
2− 1
2
)2 = cos2 θ + sin2 θ = 1⇒ 4u225 + 4v29 = 1.
the origin-centred circle z = 2eiθ is mapped to an ellipse.
For the circle |z| = 1/2 and z = 1/2eiθ
w =
1
2
eiθ + 2e−iθ =
(
1
2
+ 2
)
cos θ + i
(
1
2
− 2
)
sin θ = u+ iv.
u2(
1
2
+ 2
)2 + v2(
1
2
− 2
)2 = cos2 θ + sin2 θ = 1⇒ 4u225 + 4v29 = 1
the origin-centred circle z = (1/2)eiθ is mapped to the same ellipse.
When r = 1 from (2) the ellipse degenerates to a line segment along the real u axes:
S = {w : Im w = 0,−2 ≤ Re w ≤ 2}.
As r varies between 1 and ∞, the ellipses sweep out the entire complex w-plane. Hence the
exterior of the circle z = 2eiθ in the z-plane is mapped to the exterior of the ellipse
4u2
25
+
4v2
9
= 1.
[9 marks]
11
(c)
z = reiθ, Φ = Re(A ln z) = ARe(ln r + iθ) = A ln r,
∂Φ
∂r
=
A
r
,
∂2Φ
∂r2
= −A
r2
,
∂Φ
∂θ
= 0⇒ ∂
2Φ
∂r2
+
1
r
∂Φ
∂r
+
1
r2
∂2Φ
∂θ2
= 0.
We can invert the Joukowski transformation to obtain z as a function of w
z± =
w ±√(w2 − 4)
2
. (4)
In (4) we obtain two possible inverses for the mapping both of which are conformal everywhere
except on the line segment S. The function z+(w) maps the exterior of S to the exterior of the
circle z = reiθ, while z−(w) maps the exterior of S to the interior of the circle.
Since the potential Φ is a harmonic function outside the circular region |z| = 2 in the z plane
and the mapping is conformal, the solution outside the ellipse E1 is given by
Φ = ARe ln
[
w +
√
(w2 − 4)
2
]
.
For w = 6i
Φ = ARe ln
[
6i +
√−36− 4
2
]
= ARe ln
(
(6 +
√
40)i
2
)
= ARe ln(3 +
√
10)i
= ARe ln(3 +
√
10)ei
pi
2 = ARe
(
ln(3 +
√
10) + ln ei
pi
2
)
= ARe
(
ln(3 +
√
10) + i
pi
2
)
= A ln(3 +
√
10).
[7 marks]
12
(Lecture, homework based question)
7. Solution.
(a) f(x) = e−a
2x2
f¯(ω) =
∫ ∞
−∞
e−a
2x2e−iωxdx
df¯(ω)
dω
=
∫ ∞
−∞
(−ix)e−a2x2e−iωxdx
=
i
2a2
∫ ∞
−∞
d
(
e−a
2x2
)
dx
e−iωx
=
i
2a2
[
e−a
2x2e−iωx
]∞
−∞
− ω
2a2
∫ ∞
−∞
e−a
2x2e−iωxdx
⇒ df¯
dω
= − ω
2a2
f¯ . (5)
[4 marks]
(b) Transformed to polar coordinates, x = r cos θ and y = r sin θ
(
f¯(0)
)2
=
∫ ∞
0
dr r
∫ 2pi
0
dθ e−a
2r2 = 2pi 12
∫ ∞
0
dr2 e−a
2r2 =
pi
a2
⇒ f¯(0) =
√
pi
a
[4 marks]
(c) Solving (5)
f¯ = Ae−ξ
2/4a2 ⇒ A = f¯(0) =
√
pi
a
.
f¯(ξ) =
√
pi
a
e−
ξ2
4a2 .
[3 marks]
(d) Fourier transform of the PDE
∂u¯
∂t
= −ω2c2u¯
u¯ = e−ω
2c2tu¯(ω, 0).
Inverse Fourier transform gives
u(x, t) =
1
2pi
∫ ∞
−∞
e−ω
2c2t+iωxu¯(ω, 0)dω ⇒ u(x, 0) = 1
2pi
∫ ∞
−∞
eiωxu¯(ω, 0)dω.
On the other hand from the initial condition
e−a
2x2 = g(x) = u(x, 0) =
1
2pi
∫ ∞
−∞
eiωxu¯(ω, 0)dω.
⇒ u¯(ω, 0) = g¯(ω)
Using
F{e−a2x2} =
√
pi
a
e−
ω2
4a2 , (6)
13
⇒ u(x, t) = 1
2a
√
pi
∫ ∞
−∞
e−
ω2
4a2 e−c
2ω2teiωxdω.
=
1
2pi
√
pi
a
∫ ∞
−∞
e−
ω2
4X2 eiωxdx,
where − 1
4X2
= − 1
4a2
− c2t ⇒ X = a√
1 + 4a2c2t
Thus for the last integral we can write
u(x, t) =
1
2pi
√
pi
a
∫ ∞
−∞
e−
ω2
4X2 eiωxdx =
X
a
1
2pi
√
pi
X
∫ ∞
−∞
e−
ω2
4X2 eiωxdx
Using (6) we can see that here we have the inverse transform of e−
ω2
4X2 . Thus
u(x, t) =
1√
1 + 4a2c2t
e−X
2x2 =
1√
1 + 4a2c2t
e
− a2x2
1+4a2c2t .
[9 marks]