Document 11271536

advertisement
3150 Sample Final Exam S2013
Name.
Partial Differential Equations 3150
Sample Final Exam
Final Exam Date: Thursday, 2 May 2013
Instructions: This exam is timed for 120 minutes. You will be given 30 extra minutes to complete
the exam. No calculators, notes, tables or books. Problems use only chapters 1, 2, 3, 4, 7 of the
textbook. No answer check is expected. Details count 3/4, answers count 1/4.
Problems 1,2,3: Selected problem types from the first two midterm exams. See the sample exams
and exam solutions for midterms 1 and 2, for specific examples.
4. (CH4. Poisson Problem)
Solve for i(x, y) in the Poisson problem
(
‘
+u
u(x,0)
u(x, y)
p_
Cfr- V
41
(c-k tL &
=
sin(rx) sin(iry), 0
=
4sin(7rr),
0
< x < 1, 0 <y < 1,
0 <x < 1,
on the other 3 boundary edges.
46
p1efe
‘r
f2
i
7L IQ I
Q__
3
-
, rj
Le1
S/
)
1
(
(Y( b
A
A.o
L
0
FlJ
7
=
c
ept-4D V)
0
-fLA
2 çi
U
bo
42.
4 r1i
l(
L(
txty
C
V44
fr)
6
-
-
I
A— c/n hrlr)
rL4
z
prour Q(tc
—
—
(
Yni
;)1-
(mit)
2
(ii
‘((=
.
then staple.
Use this
1
n
o-4
th’tc
=
A4
((i-))
.j aáA t
c<. ckci-4 &4-D
T
1
M
2
‘fl=,
Name.
KE
/
3150 Exam 3 F2009
5. (CH7. Fourier Transform: Infinite Rod)
Let f(x) = pulse(x, —2,2), that is, f(x) = 1 for x
S < cc.
2 and
<
f(x)
=
0 elsewhere on —oc
Aar
(a) [75%] Solve the insulated rod heat conduction problem
f
u(x,t)
=
u(x,0)
=
ft)
=
FT
L
3
5ac;k’c
1 —cc<x<oo, t>0,
(x,t),
1
u
—cc <x < cc.
f(s),
(b) [25%] Evaluate the limit of u(x,t) at x
cc and x
=
<
X
()
7.Lt,
—cc.
‘41
LflQr
(L P-Ilk]
q
de
(Lo Fi
ç4
r3 /j
.
r-t
=
—
C
F’
7i
(ti feft
ôcr /
(&)
(x)
()_
2.
-c) 5-$)c
I
\......L.
e
-cs/(2’
; =(/(J;-))
2h.
2
:2
(S
X2..
-
(ev
1-
Use this page to start your solution. Attach extra pages as needed, then staple.
rç(co)
0
j
Er((—X’)
“-I
Name.
3150 Exam 3 F2009
1’..
{
6. (CH4. Heat Equation and Gauss’ Heat Kernel)
Solve the insulated rod heat conduction problem
< x <
tLt(X,t)
=
x,t),
u
(
1
—00
v(x,0)
=
f(x),
—1<x<1,
,
t > 0,
2o Ix<1.
otherwise
0
2 dz, and Fourier
Hint: Use convolutions, the heat kernel, the error function erf(x) =
’ e_z
0
f
transform rules to solve the problem. The answer is expressed in terms of the error function.
—=
t
1
l
l
.( .
C
ç
00
=
(
(
c
ç
20
-
2o
x)
_?>j
e
dc
Lef
x—l
;_
Use this page to start your solution. Attach extra pages as needed, then staple.
Download