hmk#11_s.pdf - ENM510 - Advanced Engineering Mathematics I...

ENM510 - Advanced Engineering Mathematics I (Practice)
Fall Semester, 2018 M. Carchidi
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Problem #1
Determine the speci fi c solution to the heat equation
∂ 2 u ( x, t )
∂ x
2
− 6 x =
∂ u ( x, t )
∂ t
for a one-dimensional rod between x = 0 and x = 1 for 0 < t , assuming the
insulated boundary conditions:
∂ u ( x, t )
∂ x
¯
¯
¯
¯
¯
x =0
= 0 and
∂ u ( x, t )
∂ x
¯
¯
¯
¯
¯
x =1
= 0
for 0 ≤ t , and the initial condition u ( x, 0) = 0 for 0 ≤ x ≤ 1 . Is your
solution unique? Explain.
Problem #2
Solve completely for u ( x, t ) in the region
R = { ( x, t ) | 0 ≤ x ≤ 1 , 0 ≤ t }
given the heat equation,
∂
2
u ( x, t )
∂ x 2
=
∂ u ( x, t )
∂ t
along with the boundary conditions: u (0 , t ) = 0 and u (1 , t ) = t 2 , and the
initial condition, u ( x, 0) = x.
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Problem #3
Consider the following boundary-value, initial-value problem in the region
R = { ( x, t ) | 0 ≤ x ≤ 1 , 0 ≤ t } , having partial di ff erential equation
∂ 2 u ( x, t )
∂ x
2
=
∂ u ( x, t )
∂ t
,
boundary conditions:
u (0 , t ) = 0 and
∂ u (1 , t )
∂ x
+
∂ u (1 , t )
∂ t
= 0 ,
and initial condition: u ( x, 0) = f ( x ) .
a.) (10 points) Show that the separation of variables method applied to this
problem does not lead to a regular Sturm-Liouville problem.
b.) (10 points) Continue with solving this problem using the method of sepa-
ration of variables anyway, and explain why the hard part of this problem
is fi tting the initial condition.
Problem #4
Solve for u ( x, t ) in the region
R = { ( x, t ) | 0 ≤ x ≤ 1 , 0 ≤ t }
given the partial di ff erential equation,
x
∂ 2 u ( x, t )
∂ x
2
−
∂ u ( x, t )
∂ x
= x
3
∂ u ( x, t )
∂ t
,
the boundary conditions:
u (0 , t ) = 1 , u (1 , t ) = 0
and the initial condition,
u ( x, 0) = 0 .
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2

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Problem #5
Solve completely for u ( x, t ) in the region
R = { ( x, t ) | 0 ≤ x ≤ 1 , 0 ≤ t }
given the equation,
x
2
∂ 2 u ( x, t )
∂ x
2
− 2 x
∂ u ( x, t )
∂ x
+ 2 u ( x, t ) = x
2
∂ 2 u ( x, t )
∂ t
2
+ 2 x
3
along with the boundary conditions:
∂ u (0 , t )
∂ x
= 1 and u (1 , t ) = 3
and the initial conditions:
u ( x, 0) = x
3
and
∂ u ( x, 0)
∂ t
= x
2
.
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3

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Solution to Problem #1
Using the approach of fi nding a u
e
( x ) so that
u
00
e
( x ) − 6 x = 0 or u
00
e
( x ) = 6 x yielding u
0
e
( x ) = 3 x
2
+ A
and
u
e
( x ) = x
3
+ Ax + B.