Math 316/202: Solutions to Assignment 7
10.8.6(a)
Using separation of variables, we write u(r, θ) = R(r)Θ(θ), where Θ(0) =
Θ(π) = 0. The Laplace equation in polar coordinates (equation 19) becomes R00 Θ +
1
r2 R00 + rR0 Θ00
1 0
R Θ + 2 RΘ00 = 0, so
+
= 0. The second term must be constant. The
r
r
R
Θ
boundary value problem Θ00 + λΘ = 0, Θ(0) = Θ(π) = 0 has solutions Θ(θ) = sin(nθ)
for positive integers n, with λ = n2 . Then the Euler equation r2 R00 + rR0 − n2 R = 0
has solutions R = c1 rn + c2 r−n . We must reject r−n because u(r, θ) must be bounded as
r → 0. So our basic solutions are un (r, θ) = rn sin(nθ), and we look for a series solution
of the form u(r, θ) =
is satisfied if
∞
X
∞
X
an rn sin(nθ). The remaining boundary condition u(a, θ) = f (θ)
n=1
an an sin(nθ) is the Fourier sine series of f (θ) on the interval [0, π]. Thus
n=1
2 Zπ
an = n
f (θ) sin(nθ) dθ.
πa 0
(b)
With f (θ) = θ(π − θ) we have
2
an = n
πa
θ cos (nθ) (−π + θ) sin (nθ) (−π + 2 θ)
cos (nθ)
−
−2
2
n
n
n3
i.e. 8/(πan n3 ) if n is odd and 0 if n is even. So u(r, θ) =
!π
4(1 − (−1)n )
=
πan n3
0
8r
8
sin θ + 3
πa
3π
3
r
a
sin 3θ +
8 r 5
sin 5θ + . . ..
53 π a
(c)
Here is u versus r for θ = π/8, π/4, 3π/8 and π/2 in red, green, blue and magenta
respectively. Note that u(π −θ) = u(θ). This was produced by the following Maple code:
u:= add(8/Pi/(2*n-1)^3*(r/2)^(2*n-1)*sin((2*n-1)*theta),n=1..40):
plot([seq(u,theta=[Pi/8,Pi/4,3*Pi/8,Pi/2])], r=0..2,
colour=[red,green,blue,magenta],labels=[r,’u’]);
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Here is u versus θ for r = 1/2, 1, 3/2 and 2 in red, green, blue and magenta respectively. This was produced by the following Maple code:
plot([seq(u,r=[1/2,1,3/2,2])], theta=0..Pi,
colour=[red,green,blue,magenta],labels=[theta,’u’]);
2
Here is a plot of the three-dimensional surface z = u(r, θ). It was produced by the
following Maple code:
plot3d([r*cos(theta),r*sin(theta),u], r=0..2, theta=0..Pi, axes=box,
scaling=constrained, orientation=[-45,50], shading=zhue);
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Here is a contour plot. It was produced by the following Maple code:
plots[contourplot]([r*cos(theta),r*sin(theta),u], r=0..2, theta=0..Pi,
axes=frame, scaling=constrained, labels=[‘‘,‘‘], view=[-2..2,0..2]);
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We use separation of variables with u(x, y) = X(x)Y (y), X 0 (0) = Y 0 (0) =
X 00
Y 00
0
00
00
Y (b) = 0. We must have X Y + XY = 0 so
=−
is constant. As we’ve seen
X
Y
00
0
in the heat equation, the equation Y + λY = 0 with Y (0) = Y 0 (b) = 0 has solutions
Yn (y) = cos(nπy/b) for nonnegative integers n, with λn = (nπ/b)2 . Then the equation
X 00 − (nπ/b)2 X = 0 has solutions X(x) = c1 exp(nπx/b) + c2 exp(−nπx/b). From the
boundary condition X 0 (0) = (nπ/b)(c1 − c2 ) = 0, so c1 = c2 . We can take c1 = c2 = 1/2,
so that X(x) = cosh(nπx/b). Thus we have the fundamental set of solutions un (x, y) =
cosh(nπx/b) cos(nπy/b) for nonnegative integers n, where u0 (x, y) = 1.
10.8.10(a)
(b)
Writing the formal series solution
∞
X
nπx
nπy
u(x, y) =
cn cosh
cos
b
b
n=0
we have
∞
X
nπ
nπa
nπy
ux (a, y) =
cn
sinh
cos
b
b
b
n=1
= f (y)
which must be the Fourier cosine series of f on the interval [0, b], except that there is
no constant term. Thus we must have
Z
b
0
nπa
2
sinh
cn =
nπ
b
f (y) dy = 0, and for n ≥ 1
−1 Z
0
5
b
nπy
f (y) cos
b
dy
while c0 is arbitrary (you can always add a constant to a solution and it remains a
solution).
10.8.11
The separation of variables is similar to 10.8.6, except that the boundary
conditions on Θ are Θ(2π) = Θ(0) and Θ0 (2π) = Θ0 (0) (periodic boundary conditions).
Solutions of Θ00 + λΘ = 0 with these boundary conditions are Θ = 1 for λ = 0 and
Θ = cos(nθ) and Θ = sin(nθ) for integers n ≥ 1. Again from the R equation we take
R(r) = rn . Thus we have the formal series solution
u(r, θ) = a0 +
∞
X
rn (an cos(nθ) + bn sin(nθ))
n=1
We then need
ur (a, θ) =
∞
X
nan−1 (an cos(nθ) + bn sin(nθ)) = g(θ)
n=1
This must be the
full Fourier series of g(θ) on the interval [0, 2π]. The constant term
Z 2π
must be 0, i.e.
g(θ) dθ = 0, and for n ≥ 1
0
Z 2π
Z 2π
1
1
g(θ) cos(nθ) dθ and bn =
g(θ) sin(nθ) dθ
an =
nπan−1 0
nπan−1 0
while a0 is arbitrary.
11.1.9
For λ = 0, y 00 = 0 has solutions y = c1 + c2 x; y(0) − y 0 (0) = c1 − c2 = 0 and
0
y(1) + y (1) = c1 + 2c2 = 0 imply c1 = c2 = 0, so 0 is not an eigenvalue.
If λ = −σ 2 < 0 with σ > 0, the solutions of the differential equation y 00 − σ 2 y = 0
are y = c1 eσx + c2 e−σx ;
y(0) − y 0 (0) = c1 + c2 − σ(c1 − c2 ) = (1 − σ)c1 + (1 + σ)c2 = 0
y(1) + y 0 (1) = c1 eσ + c2 e−σ + σ(c1 eσ − c2 e−σ ) = (1 + σ)eσ c1 + (1 − σ)e−σ c2 = 0
The determinant of the coefficient matrix of this system of two equations in two unknowns is (1 − σ)2 e−σ − (1 + σ)2 eσ . To have an eigenvalue, this must be 0. But
(1 + σ)2 > (1 − σ)2 ≥ 0 and eσ > e−σ , so the determinant < 0, and we have no
eigenvalue in this case.
If λ = σ 2 > 0 with σ > 0, the solutions of the differential equation are c1 cos(σx) +
c2 sin(σx);
y(0) − y 0 (0) = c1 − σc2
y(1) + y 0 (1) = c1 cos σ + c2 sin σ − c1 σ sin σ + c2 σ cos σ
= c1 (cos σ − σ sin σ) + c2 (sin σ + σ cos σ)
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The determinant of the coefficient matrix is
(sin σ + σ cos σ) + σ(cos σ − σ sin σ)
= 2σ cos σ + (1 − σ 2 ) sin σ
This is 0 if tan σ = −2σ/(1 − σ 2 ). Graphically, we see that there is one solution in every
interval nπ < σ < (n + 1/2)π for nonnegative integers n. The two smallest solutions are
approximately σ1 = 1.30654 and σ2 = 3.67319, corresponding to λ1 = σ12 = 1.70705 and
λ2 = σ22 = 13.49236 respectively. Since −2σ/(1 − σ 2 ) → 0 as σ → ∞, as n → ∞ we
have σn ≈ nπ and λn ≈ (nπ)2 .
11.1.16
XT 00 + cXT 0 + kXT = a2 X 00 T , so
T 00 + cT 0 + kT = λT and a2 X 00 = λX.
X 00
T 00 + cT 0 + kT
= a2
. The ODE’s are
T
X
E.7
The temperature u in the meatball should depend only on ρ (the distance from
!
2
2
the centre) and t (the time). As discussed in class, the PDE is ut = α uρρ + uρ .
ρ
The boundary conditions are u(ρ, 0) = 0, u(1, t) = 100, and u should be continuous at
ρ = 0. By taking v(ρ, t) = 100 − u(ρ, t), we have v(ρ, 0) = 100 and v(1, t) = 0, and v
satisfies the same PDE as u. We get the series solution
v(ρ, t) =
∞
X
an
n=1
ρ
sin(nπρ)e−(nπα)
where
ρv(ρ, 0) = 100ρ =
∞
X
2t
an sin(nπρ)
n=1
The coefficients in this Fourier sine series are an = 200(−1)n+1 /(nπ). Note that lim sin(nπρ)/ρ =
ρ→0
nπ, so
v(0, t) =
∞
X
200(−1)n+1 e−(nπα)
2t
n=1
and we want to find t when this is 100 − 71 = 29, where α2 = .0038. Using 20 terms
in the sum, and Maple’s fsolve function, we find v(0, t) = 29 when t ≈ 51.4056 seconds.
Actually, by this value of t if you take just one term of the series the temperature would
be accurate to within less than 0.1 degree. If you solve 200 exp(−(πα)2 t) = 29 you get
t = (πα)−2 ln(200/29) ≈ 51.4877 seconds, which is within about .08 seconds of the exact
answer.
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