1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

198 Chapter 2 The Heat Equation

u( 0 ,t)= 0 , 0 <t,

u(x, 0 )=f(x), 0 <x,

can be expressed as

u(x,t)=√^1

4 πkt

∫∞

0

f(x′)

[

exp

(

−(x′−x)^2

4 kt

)

−exp

(

−(x′+x)^2

4 kt

)]

dx′.

Hint: Start from the problem of this section with initial condition

u(x, 0 )=fo(x), −∞<x<∞,

wherefois the odd extension off.ThenuseEq.(7),andsplittheinterval

of integration at 0.

5.Verify by differentiating that the function

u(x,t)=√^1

4 kπt

exp

[

−x

2

4 kt

]

is a solution of the heat equation

∂^2 u

∂x^2 =

1

k

∂u

∂t,^0 <t, −∞<x<∞.

What can be said aboutuatx=0? att= 0 +?Whatislimt→ 0 +u( 0 ,t)?

Sketchu(x,t)forvariousfixedvaluesoft.

6.Suppose thatf(x)is an odd periodic function with period 2a. Show that

u(x,t)defined by Eq. (7) also has these properties.

7.If f(x)=1 for allx, the solution of our heat conduction problem is

u(x,t)=1. Use this fact together with Eq. (7) to show that

1 =

1

√

4 πkt

∫∞

−∞

exp

[−(x′−x) 2

4 kt

]

dx′.

8.Solve the problem that follows using Eq. (7).

∂^2 u

∂x^2 =

1

k

∂u

∂t, −∞<x<∞,^0 <t,

u(x, 0 )=

{ 1 , x>0,

− 1 , x<0.

9.Can Exercise 8 be solved in the form of Eq. (4)? Note that

2

π

∫∞

0

sin(λx)

λ

dλ=

{ 1 , 0 <x,

− 1 , x<0.