Be sure this exam has 8 pages including the cover
The University of British Columbia
MATH 257/316
Midterm Exam II – August 2015
Name
Signature
Student Number
Course Number
This exam consists of 3 questions with maximum score 50 points. No notes nor calculators.
Question
Points
1
30
2
20
3
3
Total:
50
Score
1. Each candidate should be prepared to produce his library/AMS card upon request.
2. Read and observe the following rules:
No candidate shall be permitted to enter the examination room after the expiration of one half hour, or to leave
during the first half hour of the examination.
Candidates are not permitted to ask questions of the invigilators, except in cases of supposed errors or ambiguities
in examination questions.
CAUTION - Candidates guilty of any of the following or similar practices shall be immediately dismissed from the
examination and shall be liable to disciplinary action.
(a) Making use of any books, papers or memoranda, other than those authorized by the examiners.
(b) Speaking or communicating with other candidates.
(c) Purposely exposing written papers to the view of other candidates. The plea of accident or forgetfulness
shall not be received.
3. Smoking is not permitted during examinations.
August 2015
(4 points)
Math 257/316 Midterm 2
Page 2 of 8
1. Consider the following initial boundary value problem for the wave equation:

utt = uxx + e−t cos(4x), 0 < x < π, t > 0,




BC : ux (0, t) = 2t, ux (π, t) = 3t, t > 0,



 IC : u(x, 0) = 1, ut (x, 0) = 1 x2 , 0 < x < π.
2π
(P-I)
(a) Determine a simple function w(x, t) that satisfies the inhomogeneous boundary
conditions in the problem (P-I).
Answer. Let w(x, t) = ax2 + bx be such that wx (0, t) = 2t and wx (π, t) = 3t, then
wx (x, t) = 2ax + b, which implies that
b = 2t,
So we get
a=
and
t
,
2π
Therefore, we have
w(x, t) =
(6 points)
2aπ + b = 3t.
and b = 2t.
t 2
x + 2tx.
2π
(b) Let u(x, t) = w(x, t) + v(x, t), determine the corresponding boundary value problem for
v(x, t) (Do NOT solve v(x, t)).
Answer.
Since u = w + v, then
0 = uxx + e−t cos(4x) − utt
= vxx + wxx + e−t cos(4x) − vtt − wtt
t
= vxx + + e−t cos(4x) − vtt
π
vx (0, t) = vx (π, t) = 0 By the construction of w(x, t) in part (a)
v(x, 0) = u(x, 0) − w(x, 0)
= 1−0
= 1
vt (x, 0) = ut (x, 0) − wt (x, 0)
1 2
1 2
x −
x − 2x
=
2π
2π
= −2x.
Then v(x, t) satisfies the following boundary value problem:

t


vtt = vxx + + e−t cos(4x), 0 < x < π, t > 0,


π
BC : vx (0, t) = vx (π, t) = 0, t > 0,




IC : v(x, 0) = 1, ut (x, 0) = −2x, 0 < x < π.
(1)
August 2015
(20 points)
Math 257/316 Midterm 2
Page 3 of 8
(c) Use the method of eigenfunction expansions to solve for v(x, t), and write down the
solution u(x, t) to the problem (P-I).
(
vtt = vxx , 0 < x < π, t > 0,
Answer. First, let’s look at the problem
, let
BC : vx (0, t) = vx (π, t) = 0, t > 0.
v(x, t) = X(x)T (t) be a non-zero separated solution, then X(x)T 00 (t) = X 00 (x)T (t), that
is,
X 00 (x)
T 00 (t)
−
=−
= λ.
X(x)
T (t)
So we get λt = λx = 0, that is, λ is a constant. Since vx (0, t) = vx (π, t) = 0, then
X 0 (0) = X 0 (π) = 0. Hence X(x) satisfies:
( 00
X + λX = 0,
X 0 (0) = X 0 (π) = 0.
(2)
For the eigenvalue problem (2), we know that the eigenvalues areSλ = n2 , and the
corresponding eigenfunctions are X(x) = C cos(nx), with n ∈ N {0}.
∞
a0 X
+
an cos(nx),
Now let’s compute the eigenfunction expansions of −2x, let −2x =
2
n=1
then
Z
2 π
(−2x) cos(nx) dx
an =
π 0

if n = 0,
 −2π,
n
=
 4 · 1 − (−1) , if n ≥ 1.
π
n2
Let v(x, t) =
∞
X
vn (t) cos(nx) be the solution to (1), then
n=0
∞
X
vn00 (t) cos(nx) = vtt = vxx +
n=0
= −
∞
X
t
+ e−t cos(4x)
π
n2 vn (t) cos(nx) +
n=0
t
+ e−t cos(4x)
π
vx (0, t) = vx (π, t) = 0
∞
X
n=0
∞
X
n=0
vn (0) cos(nx) = v(x, 0) = 1
vn0 (0) cos(nx) = vt (x, 0) = −2x
= −π +
∞
X
4 1 − (−1)n
·
cos(nx).
π
n
n=1
August 2015
Math 257/316 Midterm 2
Page 4 of 8
So we get
t
π
00
v4 (t) = −16v4 (t) + e−t
v000 (t) =
vn00 (t) = −n2 vn (t),
v0 (0) = 1
vn (0) = 0,
v00 (0)
for all n ≥ 1 but n 6= 4
for all n ≥ 1
= −π
4 1 − (−1)n
vn0 (0) =
·
, for all n ≥ 1.
π
n

 v 00 (t) = t ,
0
π
, then we get
For v0 (t), we need to solve

v0 (0) = 1, v00 (0) = −π.
v0 (t) =
For v4 (t), we need to solve
(
1 3
t − πt + 1.
6π
v400 (t) = −16v4 (t) + e−t ,
v4 (0) = 0, v40 (0) = 0.
, then we get
1
1
1
cos(4t) +
sin(4t) + e−t .
17
68
17
 00
2
 vn (t) = −n vn (t),
n , then we get
For n ≥ 1 but n 6= 4, we need to solve
 v (0) = 0, v 0 (0) = 4 · 1 − (−1) .
n
n
π
n
n
4 1 − (−1)
sin(nt).
vn (t) = ·
π
n2
v4 (t) = −
In summary, the solution to (1) is:
v(x, t) =
∞
X
vn (t) cos(nx)
n=0
=
1 3
1
1
1
t − πt + 1 + − cos(4t) +
sin(4t) + e−t cos(4x)
6π
17
68
17
∞
X 4 1 − (−1)n
+
·
sin(nt) cos(nx).
π
n2
n=1, n6=4
Hence the solution to (P-I) is:
u(x, t) = v(x, t) + w(x, t)
t 2
1 3
1
1
1 −t
x + 2tx +
t − πt + 1 + − cos(4t) +
sin(4t) + e
cos(4x)
=
2π
6π
17
68
17
∞
X
4 1 − (−1)n
+
·
cos(nt) cos(nx).
π
n2
n=1, n6=4
August 2015
Math 257/316 Midterm 2
Page 5 of 8
(20 points) 2. Find the solution u(r, θ) of following boundary condition problem for the Laplace’s equation
in an annulus region:

1
1


urr + ur + 2 uθθ = 0, 1 < r < 2, 0 < θ < 2π,


r
r
(P-II)
BC
:
u(1,
θ)
=
θ
+
1, u(2, θ) = 0, 0 ≤ θ ≤ 2π,




u(r, 0) = u(r, 2π), uθ (r, 0) = uθ (r, 2π), 1 ≤ r ≤ 2.
Answer.
First, let’s look at the problem:

1
1


urr + ur + 2 uθθ = 0,


r
r
1 < r < 2, 0 < θ < 2π,
BC : u(2, θ) = 0, 0 ≤ θ ≤ 2π,




u(r, 0) = u(r, 2π), uθ (r, 0) = uθ (r, 2π),
(3)
1 ≤ r ≤ 2.
Let u(r, θ) = R(r)Θ(θ) be a non-zero separated solution to (3), then
1
1
R00 (r)Θ(θ) + R0 (r)Θ(θ) + 2 R(r)Θ00 (θ) = 0.
r
r
Then we get
R00 (r) + 1r R0 (r)
Θ00 (θ)
= λ.
=
−
1
Θ(θ)
R(r)
r2
It’s easy to see that λr = λθ = 0, then λ is a constant. Since u(2, θ) = 0, u(r, 0) = u(r, 2π)
and uθ (r, 0) = uθ (r, 2π), then R(2) = 0, Θ(0) = Θ(2π) and Θ0 (0) = Θ0 (2π). So R(r) satisfies:

 R00 (r) + 1 R0 (r) − λR(r) = 0,
r
(4)

R(2) = 0.
And Θ(θ) satisfies:
(
Θ00 + λθ = 0,
Θ(0) = Θ(2π), Θ0 (0) = Θ0 (2π).
For the eigenvalue problem (5), we know that the eigenvalues are λ = n2 , and
S the
corresponding eigenfunctions are Θ(θ) = A cos(nθ) + B sin(nθ), with n ∈ N {0}.
S
For each n ∈ N {0}, when λ = n2 in (4), then
(
C[ln r − ln 2],
if n = 0,
R(r) =
n
n −n
C[r − 4 r ], if n ≥ 1.
So we can let the solution to (P-II) is:
u(r, θ) = a0 [ln r − ln 2] +
∞
X
[rn − 4n r−n ][an cos(nθ) + bn sin(nθ)].
n=1
(5)
August 2015
Math 257/316 Midterm 2
Page 6 of 8
Since u(1, θ) = θ + 1, then
θ + 1 = −a0 ln 2 +
∞
X
[1 − 4n ][an cos(nθ) + bn sin(nθ)].
n=1
∞
c0 X
+
[cn cos(nθ) + dn sin(nθ)] be the Fourier expansion of θ on [0, 2π], then
Let θ + 1 =
2
n=1
cn =
=
dn =
1
π
(
1
π
Z
2π
(θ + 1) cos(nθ) dθ
0
2π + 2, if n = 0,
0,
Z 2π
if n ≥ 1
(θ + 1) sin(nθ) dθ
0
2
= − ,
n
for all n ≥ 1.
So we know that
c0
=π+1
2
= 0, for all n ≥ 1
2
= − , for all n ≥ 1.
n
−a0 ln 2 =
[1 − 4n ]an
[1 − 4n ]bn
So we get
a0 = −
π+1
,
ln 2
an = 0,
and bn =
2
,
− 1)
n(4n
for all n ≥ 1.
In summary, the solution to (P-II) is:
∞
u(r, θ) = −
X
1
2
[ln r − ln 2] +
[rn − 4n r−n ] sin(nθ).
ln 2
n(4n − 1)
n=1
August 2015
Math 257/316 Midterm 2
Page 7 of 8
(3 points) 3. (Bonus Problem) Use the method of the characteristic coordinates to solve the following
initial value problem:
(
utt − utx = 2x, −∞ < x < ∞, t > 0,
(P-III)
IC : u(x, 0) = 0, ut (x, 0) = 0, −∞ < x < ∞.
Answer.
Notice that utt − utx
∂
∂x
∂
∂t
Then we have
∂ ∂
∂
=
u, let ξ = x and η = x + t, then
−
∂t ∂t ∂x
=
=
∂
∂ξ
∂
∂ξ
∂
∂
=
,
∂t
∂η
∂ξ
∂ ∂η
∂
∂
+
·
=
+
∂x ∂η ∂x
∂ξ ∂η
∂ξ
∂ ∂η
∂
·
+
·
=
.
∂t
∂η ∂t
∂η
·
and
So we get
utt − utx = −
∂
∂
∂
−
=− .
∂t ∂x
∂ξ
∂2u
= 2x = 2ξ.
∂η∂ξ
Let v = uη , then vξ = uηξ = −2ξ, which implies that uη = v = −ξ 2 + f (η). Hence
Z η
2
f (τ ) dτ + G(ξ) = −ξ 2 η + F (η) + G(ξ).
u = −ξ η +
0
So we get u(x, t) = −x2 (x + t) + F (x + t) + G(x). Since u(x, 0) = 0, then
−x3 + F (x) + G(x) = 0.
(6)
Since ut (x, t) = −x2 + F 0 (x + t) and ut (x, 0) = 0, then −x2 + F 0 (x) = 0, which implies that
x3
x3
F (x) =
+ C. By (6), then G(x) = x3 − F (x) = x3 −
− C. Therefore,
3
3
u(x, t) = −x2 (x + t) +
1
= xt2 + t3
3
(x + t)3
x3
+ C + x3 −
−C
3
3
Two useful integrals: If α 6= 0, then
Z
sin(αx) − αx cos(αx)
x sin(αx) dx =
+ C,
α2
and
Z
x cos(αx) dx =
αx sin(αx) + cos(αx)
+ C.
α2
Formula sheet - final exam
constant coefficients
ay 00 + by 0 + cy = 0
ar2 + br + c = 0
y = Aer1 x + Ber2 x
y = Aerx + Bxerx
eλx [A cos(µx) + B sin(µx)]
sin2 t + cos2 t = 1
sin t = 12 (1 − cos(2t))
cosh2 t − sinh2 t = 1
2
sinh t = 12 (cosh(2t) − 1)
Euler eq
ax2 y 00 + bxy 0 + cy = 0
ar(r − 1) + br + c = 0
y = Axr1 + Bxr2
y = Axr + Bxr ln |x|
xλ [A cos(µ ln |x|) + B sin(µ ln |x|)]
2
y1 (x) =
n=0
P∞
n+r1
an (x − x0 )
x→x0
where a0 = 1.
n=0
∞
X
y2 (x) = ay1 (x) ln(x − x0 ) +
n=0
∞
X
bn (x − x0 )n+r2 where b0 = 1.
bn (x − x0 )n+r2 for some b1 , b2...
Case 3: If r1 − r2 is a positive integer:
n=1
∞
X
bn (x − x0 )n+r2 where b0 = 1.
y2 (x) = y1 (x) ln(x − x0 ) +
Case 2: If r1 − r2 = 0:
y2 (x) =
The second linerly independent solution y2 is of the form:
Case 1: If r1 − r2 is neither 0 nor a positive integer:
x→x0
Regular singular point x0 : Rearrange (?) as:
(x − x0 )2 y 00 + [(x − x0 )p(x)](x − x0 )y 0 + [(x − x0 )2 q(x)]y = 0
If r1 > r2 are roots of the indicial equation:
r(r − 1) + br + c = 0 where
b = lim (x − x0 )p(x) and c = lim (x − x0 )2 q(x) then a solution of (?) is
Ordinary point x0 : Two linearly independent solutions of the form:
P∞
y(x) = n=0 an (x − x0 )n
Series solutions for y 00 + p(x)y 0 + q(x)y = 0 (?) around x = x0 .
ODE
indicial eq.
r1 6= r2 real
r1 = r 2 = r
r = λ ± iµ
Basic linear ODE’s with real coefficients
sin(α ± β) = sin α cos β ± sin β cos α
cos(α ± β) = cos α cos β ∓ sin β sin α.
sinh(α ± β) = sinh α cosh β ± sinh β cosh α
cosh(α ± β) = cosh α cosh β ± sinh β sinh α.
Trigonometric and Hyperbolic Function identities
Math 257-316 PDE
1
n=1
∞
X
bn sin(
nπx
),
L
bn =
2
L
0
Z
L
f (x) sin(
ODE: [p(x)y 0 ]0 − q(x)y + λr(x)y = 0, a < x < b.
BC:
α1 y(a) + α2 y 0 (a) = 0, β1 y(b) + β2 y 0 (b) = 0.
Hypothesis: p, p0 , q, r continuous on [a, b]. p(x) > 0 and r(x) > 0 for
x ∈ [a, b]. α12 + α22 > 0. β12 + β22 > 0.
Properties (1) The differential operator Ly = [p(x)y 0 ]0 − q(x)y is symmetric
in the sense that (f, Lg) = (Lf, g) for all f, g satisfying the BC, where (f, g) =
Rb
f (x)g(x) dx. (2) All eigenvalues are real and can be ordered as λ1 < λ2 <
a
· · · < λn < · · · with λn → ∞ as n → ∞, and each eigenvalue admits a unique
(up to a scalar factor) eigenfunction φn .
Rb
(3) Orthogonality: (φm , rφn ) = a φm (x)φn (x)r(x) dx = 0 if λm 6= λn .
(4) Expansion: If f (x) : [a, b] → R is square integrable, then
Rb
∞
X
f (x)φn (x)r(x) dx
f (x) =
cn φn (x), a < x < b , cn = a R b
, n = 1, 2, . . .
φ2 (x)r(x) dx
n=1
a n
Sturm-Liouville Eigenvalue Problems
nπx
) dx.
L
PDE: utt = c2 uxx , −∞ < x < ∞, t > 0 IC: u(x, 0) = f (x), ut (x, 0) = g(x).
R x+ct
1
SOLUTION: u(x, t) = 12 [f (x + ct) + f (x − ct)] + 2c
g(s)ds
x−ct
D’Alembert’s solution to the wave equation
Sf (x) =
For f (x) defined in [0, L], its cosine and sine series are
Z
∞
a0 X
nπx
2 L
nπx
Cf (x) =
+
an cos(
), an =
f (x) cos(
) dx,
2
L
L
L
0
n=1
Theorem (Pointwise convergence) If f (x) and f 0 (x) are piecewise continuous, then F f (x) converges for every x to 21 [f (x−) + f (x+)].
Parseval’s indentity
Z
∞
1 L
|a0 |2 X
|an |2 + |bn |2 .
+
|f (x)|2 dx =
L −L
2
n=1
Let f (x) be defined in [−L, L]then
Fourier series F f (x) is a 2L-periodic
P∞its function on R: F f (x) = a20 + n=1 an cos( nπx
) + bn sin( nπx
)
L
L
RL
R
1 L
nπx
where an = L1 −L f (x) cos( nπx
L ) dx and bn = L −L f (x) sin( L ) dx
Fourier, sine and cosine series