math 241 final exam fall 2013 · math 241 final exam fall 2013 name: instructor: wong pantev...
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Math 241 FINAL EXAMFall 2013
Name:
Instructor: Wong Pantev
Recitation Number and Day/Time:
Please turn off and put away all electronic devices. You may use both sides of a 8.5′′×11′′
sheet of paper for handwritten notes while you take this exam. No calculators, no coursenotes, no books, no help from your neighbors. Show all work. Please clearly markyour final answer. Remember to put your name at the top of this page. Good luck!
My signature below certifies that I have complied with the Universityof Pennsylvania’s Code of Academic Integrity in completing this exam-ination.
Your signature
Question Points Your
Number Possible Score
1 10
2 10
3 10
4 10
5 10
6 10
7 10
8 10
Total Score /80
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A Partial Table of Integrals∫ x
0
u cosnu du =cosnx+ nx sinnx− 1
n2for any real n 6= 0∫ x
0
u sinnu du =sinnx− nx cosnx
n2for any real n 6= 0∫ x
0
emu cosnu du =emx(m cosnx+ n sinnx)−m
m2 + n2for any real n,m∫ x
0
emu sinnu du =emx(−n cosnx+m sinnx) + n
m2 + n2for any real n,m∫ x
0
sinnu cosmu du =m sinnx sinmx+ n cosnx cosmx− n
m2 − n2for any real numbers m 6= n∫ x
0
cosnu cosmu du =m cosnx sinmx− n sinnx cosmx
m2 − n2for any real numbers m 6= n∫ x
0
sinnu sinmu du =n cosnx sinmx−m sinnx cosmx
m2 − n2for any real numbers m 6= n
Formulas Involving Bessel Functions
• Bessel’s equation: r2R′′ + rR′ + (α2r2 − n2)R = 0 – The only solutions of this which are bounded atr = 0 are R(r) = cJn(αr).
Jn(x) =∞∑k=0
(−1)k
k!(k + n)!
(x2
)n+2k
.
J0(0) = 1, Jn(0) = 0 if n > 0. znm is the mth positive zero of Jn(x).
• Orthogonality relations:
If m 6= k then
∫ 1
0
xJn(znmx)Jn(znkx) dx = 0 and
∫ 1
0
x(Jn(znmx))2 dx =1
2Jn+1(znm)2.
• Recursion and differentiation formulas:
d
dx(xnJn(x)) = xnJn−1(x) or
∫xnJn−1(x) dx = xnJn(x) + C for n ≥ 1 (1)
d
dx(x−nJn(x)) = −x−nJn+1(x) for n ≥ 0 (2)
J ′n(x) +n
xJn(x) = Jn−1(x) (3)
J ′n(x)− n
xJn(x) = −Jn+1(x) (4)
2J ′n(x) = Jn−1(x)− Jn+1(x) (5)
2n
xJn(x) = Jn−1(x) + Jn+1(x) (6)
• Modified Bessel’s equation: r2R′′ + rR′ − (α2r2 + n2)R = 0 – The only solutions of this which arebounded at r = 0 are R(r) = cIn(αr).
In(x) = i−nJn(ix) =∞∑k=0
1
k!(k + n)!
(x2
)n+2k
.
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Formulas Involving Associated Legendre and Spherical BesselFunctions
• Associated Legendre Functions: ddφ
(sinφ dg
dφ
)+(µ− m2
sinφ
)g = 0. Using the substitution x = cosφ,
this equation becomes ddx
((1− x2) dg
dx
)+(µ− m2
1−x2
)g = 0. This equation has bounded solutions only
when µ = n(n+ 1) and 0 ≤ m ≤ n. The solution Pmn (x) is called an associated Legendre function of
the first kind.
• Associated Legendre Function Identities:
P 0n(x) =
1
2nn!
dn
dxn(x2 − 1)n and Pm
n (x) = (−1)m(1− x2)m/2dm
dxmPn(x) when 1 ≤ m ≤ n
• Orthogonality of Associated Legendre Functions: If n and k are both greater than or equal to m,
If n 6= k then
∫ 1
−1
Pmn (x)Pm
k (x)dx = 0 and
∫ 1
−1
(Pmn (x))2 dx =
2(n+m)!
(2n+ 1)(n−m)!.
• Spherical Bessel Functions: (ρ2f ′)′+(α2ρ2−n(n+1))f = 0. If we define the spherical Bessel function
jn(ρ) = ρ−12Jn+ 1
2(ρ), then only solution of this ODE bounded at ρ = 0 is jn(αρ).
• Spherical Bessel Function Identity:
jn(x) = x2
(−1
x
d
dx
)n(sinx
x
).
• Spherical Bessel Function Orthogonality: Let znm be the m-th positive zero of jm.
If m 6= k then
∫ 1
0
x2jn(znmx)jn(znkx)dx = 0 and
∫ 1
0
x2(jn(znmx))2dx =1
2(jn+1(znm))2.
One-Dimensional Fourier Transform
F [u](ω) =1
2π
∫ ∞−∞
u(x)eiωxdx, F−1[U ](x) =
∫ ∞−∞
U(ω)e−iωxdω
Table of Fourier Transform Pairs
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Fourier Transform Pairs Fourier Transform Pairs(α > 0) (β > 0)
u(x) = F−1[U ] U(ω) = F [u] u(x) = F−1[U ] U(ω) = F [u]
e−αx2 1√
4παe−
ω2
4α
√π
βe−
x2
4β e−βω2
e−α|x|1
2π
2α
x2 + α2
2β
x2 + β2e−β|ω|
u(x) =
{0 |x| > α
1 |x| < α
1
π
sinαω
ω2
sin βx
xU(ω) =
{0 |ω| > β
1 |ω| < β
δ(x− x0)1
2πeiωx0 e−iω0x δ(ω − ω0)
∂u
∂t
∂U
∂t
∂2u
∂t2∂2U
∂t2
∂u
∂x−iωU ∂2u
∂x2(−iω)2U
xu −i∂U∂ω
x2u (−i)2∂2U
∂ω2
u(x− x0) eiωx0U1
2π
∫ ∞−∞
f(s)g(x− s)ds FG
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(1) 10 points Let u := u(x, t) be the temperature in a one-dimensional rod, and satisfy the followinginitial and boundary value problem:
∂u
∂t=∂2u
∂x2+ βx for 0 < x < 2, t > 0,
∂xu(0, t) = 0 and ∂xu(2, t) = −2,
u(x, 0) =8
πcos
π
4x,
where β is a constant. Denote the total thermal energy in the rod (0 < x < 2) by
E(t) :=
∫ 2
0
u(x, t) dx.
Answer the following questions.
(i) (2 Points) What is the physical meaning of the boundary condition ∂xu(0, t) = 0?
(ii) (3 Points) Verify thatdE
dt= −2 + 2β.
(iii) (3 Points) Using part (ii), compute E.
(iv) (2 Points) For which value of β does the limit limt→∞E(t) exist, and what is the limit?
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(2) 10 points Consider the following boundary value problem for a heat conduction in a rod:∣∣∣∣∣∣∣∣ut = 4uxx − 2u, on 0 < x < 1
u(0, t) = 0
u(1, t) = 0
(a) (5 points) Separate variables and determine all product solutions of this problem.
(b) (5 points) Find the solution for which the initial temperature is u(x, 0) = 3 sin(πx) + sin(3πx).
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(3) 10 points Let f(x) be a piecewise smooth function. Denote by g(x) the Fourier series of thefunction f(x) on the interval [−π, π], that is,
f(x) ∼∞∑n=0
an cosnx+∞∑n=1
bn sinnx =: g(x).
Decide whether the following statements are true or false. To obtain full credit, you must justifyyour answers.
(i) (2.5 points) If f is an odd function, then bn = 0 for all n = 1, 2, · · · . T / F
(ii) (2.5 points) If f is continuous, then g must be continuous. T / F
(iii) (2.5 points) If f is bounded, then g must be bounded. T / F
(iv) (2.5 points) For any given function f , the Fourier coefficient bn can be computed bybn = 2
π
∫ π0f(x) sinnx dx. T / F
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(4) 10 points Solve the damped wave equation
utt = uxx − ut, 0 < x < 1, t > 0,
with initial conditionsu(x, 0) = sin(2πx)
ut(x, 0) = 0
and boundary conditionsu(0, t) = 0
u(1, t) = 0.
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(5) 10 points Consider the eigenvalue problemd2φ
dx2+ (λ− x4)φ = 0
dφ
dx(0) =
dφ
dx(1) = 0.
Answer the following questions.
(i) (2 points) Is the above eigenvalue problem a regular Sturm-Liouville eigenvalue problem?
(ii) (5 points) Show that all eigenvalues λ are non-negative.
(iii) (3 points) Is λ = 0 an eigenvalue?
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(6) 10 points The displacement function u(r, θ, t) describing the vibration of a circular membraneD : 0 < r < 2, −π < θ < π satisfies the boundary value problem∣∣∣∣∣∣∣∣∣∣∣∣∣
utt = 9∇2u on D
u(2, θ, t) = 0
u(r, θ, 0) = 0
ut(r, θ, 0) = −J0
(z6
2r),
where J0(z) is the 0-th Bessel function of the first kind with roots z1, z2, . . . , zn, . . . .
(a) (7 points) Compute u(r, θ, t).
(b) (3 points) Find the value u(2, π
6, 1).
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(7) 10 points Let u := u(x, t) satisfy the following initial and boundary value problem
∂u
∂t=∂2u
∂x2+ x+ 2e−
π2
4t sin
πx
2for 0 < x < 1, t > 0
u(0, t) = 0
∂u
∂x(1, t) = t
u(x, 0) = 5 sin3πx
2.
Answer the following questions.
(ii) (8 Points) Find the solution u to the above initial and boundary value problem.
(iii) (2 Points) Prove or disprovelimt→∞
u(1, t) = 0.
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(8) 10 points Let u(x, t) be the solution of∣∣∣∣∣ ut = 2uxx, −∞ < x <∞, t > 0,
u(x, 0) = sin(3πx).
(a) (3 points) Use Euler’s formula to rewrite the initial data in terms of exponentials.
(b) (7 points) Use a Fourier transform in x to find u(x, t).
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