1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

5.3 Two-Dimensional Heat Equation: Solution 303

EXERCISES

1. For the functionu(x,y,z,t)that satisfies Eqs. (10)–(14), show that
   ∫c

0

∂^2 u

∂z^2 dz=^0.

2. Find the initial and boundary conditions satisfied by the function

v(x,y,t)=^1 c

∫c

0

u(x,y,z,t)dz,

whereusatisfies Eqs. (10)–(14).

3. In Eqs. (15)–(18), suppose thatw(x,z,t)→W(x,z)ast→∞.Stateand
   solve the boundary value problem forW. (This problem is much easier
   than it appears, because there is no variation withz.)


4. Find the dimensions ofρ,c,κ,q,andg, and verify that the dimensions of
   the right and left members of the heat equation are the same.


5. Suppose the plate lies in the rectangle 0<x<a,0<y<b.Stateacomplete
   initial value–boundary value problem for temperature in the plate if: there
   is no heat generation; the temperature is held atT 0 alongx=aandy=0;
   the edges atx=0andy=bare insulated.

5.3 Two-Dimensional Heat Equation:

Double Series Solution

In order to see the technique of solution for a two-dimensional problem, we
shall consider the diffusion of heat in a rectangular plate of uniform, isotropic
material. The steady-state temperature distribution is a solution of the poten-
tial equation (see Exercise 6). Suppose that the initial value–boundary value
problem for the transient temperatureu(x,y,t)is

∂^2 u

∂x^2 +

∂^2 u

∂y^2 =

1

k

∂u

∂t,^0 <x<a,^0 <y<b,^0 <t,(^1 )

u(x, 0 ,t)= 0 , u(x,b,t)= 0 , 0 <x<a, 0 <t,( 2 )

u( 0 ,y,t)= 0 , u(a,y,t)= 0 , 0 <y<b, 0 <t,( 3 )

u(x,y, 0 )=f(x,y), 0 <x<a, 0 <y<b.( 4 )

This problem contains a homogeneous partial differential equation and ho-
mogeneous boundary conditions. We may thus proceed with separation of