206 Chapter 2 The Heat Equation
the ordinary heat equation has this form if we takeA=1,E=− 1 /kand all
other coefficients equal to zero.
Some astute students will have wondered why we should seek solutions in
product form. The simplest answer is that in many cases it works. A more
subtle rationale is that of seeking solutions that are geometrically similar func-
tions ofxat different times. The idea of similarity — related to dimensional
analysis — has been most fruitful in the mechanics of fluids.
Chapter Review
See the CD for Review Questions.
Miscellaneous Exercises
Also see Review Questions on the CD.
In Exercises 1–16, find the steady-state solution, the associated eigenvalue
problem, and the complete solution for each problem.
- ∂
(^2) u
∂x^2 =
1
k∂u
∂t,0<x<a,0<t,
u( 0 ,t)=T 0 , u(a,t)=T 0 ,0<t,
u(x, 0 )=T 1 ,0<x<a.2.∂^2 u
∂x^2 −γ(^2) u=^1
k
∂u
∂t,0<x<a,0<t,
u( 0 ,t)=T 0 , u(a,t)=T 0 ,0<t,
u(x, 0 )=T 1 ,0<x<a.
- ∂
(^2) u
∂x^2
+r=^1
k
∂u
∂t
,0<x<a,0<t,
u( 0 ,t)=T 0 , u(a,t)=T 0 ,0<t,
u(x, 0 )=T 1 ,0<x<a (ris constant).
- ∂
(^2) u
∂x^2 =
1
k∂u
∂t,0<x<a,0<t,
u( 0 ,t)=T 0 ,∂u
∂x(a,t)=0,^0 <t,
u(x, 0 )=T 1 x
a ,0<x<a.- ∂
(^2) u
∂x^2 −γ
(^2) u=^1
k
∂u
∂t,0<x<a,0<t,