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Name.
3150 Sample Exam2S2Ol3
Partial Differential Equations 3150
Sample Midterm Exam 2
Exam Date: Monday, 22 April 2013
Instructions: This exam is timed for 50 minutes. You will be given double time to complete the
exam. No calculators, notes, tables or books. Problems use only chapters 1 to 4 of the textbook. No
answer check is expected. Details count 3/4, answers count 1/4.
la. (CH3. Finite String: Fourier Series Solution)
(a) [75%] Display the series formula without derivation details for the finite string problem
I
j
u(x, t)
u(0,t)
u(L,t)
u(x,0)
ut(z,0)
c
u
2
(x, 1),
0,
0,
0,
g(x),
=
=
=
=
t > 0,
t>0,
t>0,
< L,
0 <
0<z<L.
(b) [25%] Display an explicit formula for the Fourier coefficients which contains the symbols L,
g(x).
7l
Lt
_c.•
j
‘141 )1
S4
( 0
(ic)
M’ pt
QI
j
( ii?TC)
Co.f’
‘)‘
I ‘7’-
‘Vhc,t)=
ç
(ii
1
fl fli
(
774..%
2er
(r7rct
)‘L
1I
-n
=
‘
1lfl2.c(o)
TC
..
c
t
y1rrC
—
—
I
4i-
c)
L)1
-
2-14 or#?3
(
2-L
h
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Name.
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3150 Sample Exam 2 S2013
lb. (CH3. d’Alembert’s Solution: Finite String)
Let
=
f
0.3z
0.3(1—x)
0<x0/5,
0.5<x< 1.
and define g(x)
=
0 on 0< x
1. Assumed is
dAlemberts solution to the vibrating string problem on 0 < x < 1, t> 0, which is the formula
u(x,t)
Assume L
=
1 and c
(a) [25%] Define
f,
=
(f(z
-
ct) + f(z + ct)) +
1
x+ct
L
gs)ds.
1/7r.
g as appropriate periodic extensions of
f
and g, respectively.
(b) [75%] Display a piecewise-defined formula for u(z, ir/3) on 0 <x < 1.
c
-
—peM
i,” os ‘
C
€.AL1c
rr P..
If
-,
(1÷2)
( (
,,
-
-=
j
z
-c—c
0
-
tc
‘
c
2
X
-
‘fl,
€.
-
f
C.
L (+C
c-&)
pr,
T’
±
± ((
;t::v.
2=
)
t
‘
P
-4’
-
-x
i
-
U..
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3150 Sample Exam 2 S2013
Name.
2. (CR3. Heat Conduction in a Bar)
Consider the heat conduction problem in a laterally insulated bar of length 1 with one end at
zero Celsius and the other end at 100 Celsius. The initial temperature along the bar is given by
function f(x).
u(0,t)
u(l,t)
()
o<<i, t>o,
=
Ut
U(X,0)
=
0,
100,
=
f(x),
=
t>0,
O<x<1.
(a) [25%] Find the steady-state temperature ui(z).
(b) [50%] Solve the bar problem with zero Celsius temperatures at both ends, but f(x) replacej
(x,t). The answer has Fourier coefficients in integral form,
2
Qr). Call the answer u
1
by f(x) —u
unevaluated, to save time.
(c) [25%] Explain why
(ta-)
u(x)
+ u2(x,t).
Cl’.
0
‘
A
U(x,t) =
QQ4t4.t
-it,
çjL-.?
7
1
C
(I)
(z.)
1
r’J U
n
2L*-=
-’L(,,c
2
c
C2
‘.it41D.
CAt”
+
CIO)C
(O),U)lO
L
(c
1A
qt)-
41
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