Section 7.9
The Nonhomogeneous Heat Equation
133
Solutions to Exercises 7.9
1.
Proceed as in Example 1 with
c
=1
/
2. Equation (3) becomes in this case
u
(
x, t
)=
2
√
2
t
e

x
2
/t
*
δ
1
(
x
)
=
1
√
πt
e

(
x

1)
2
/t
,
since the eﬀect of convolution by
δ
1
is to shift the function by 1 unit to the right
and multiply by
1
√
2
π
.
5.
We use the superposition principle (see the discussion preceeding Example 4). If
φ
is the solution of
u
t
=
1
4
u
xx
+
δ
0
,
u
(
x,
0) = 0 and
ψ
is the solution of
u
t
=
1
4
u
xx
,
u
(
x,
0) =
U
0
(
x
), then you can check that
φ
+
ψ
is the solution of
u
t
=
1
4
u
xx
+
δ
0
,
u
(
x,
0) =
U
0
(
x
). By Examples 1,
φ
(
x, t
2
√
t
√
π
e

x
2
/t

2

x

√
π
Γ
±
1
2
,
x
2
t
²
and by Exercise 20, Section 7.4,
ψ
(
x, t
1
2
erf
±
x
√
t
²
.
9.
Apply Theorem 2 with
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk

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