5.2 Three-Dimensional Heat Equation 301
Again using Fourier’s law, we obtain the boundary condition
κ∂∂un(P,t)+hu(P,t)=hT(P,t) forPonB′. (9)As an example, we set up the three-dimensional problem for a solid in the
form of a rectangular parallelepiped. In this case, Cartesian coordinates are
appropriate, and we may describe the regionRby the three inequalities 0<
x<a,0<y<b,0<z<c. Assuming no generation inside the object, we have
the partial differential equation
∂^2 u
∂x^2 +∂^2 u
∂y^2 +∂^2 u
∂z^2 =1
k∂u
∂t,^0 <x<a,^0 <y<b,^0 <z<c,^0 <t.
(10)
Suppose that on the faces atx=0anda, the temperature is controlled, so
the boundary condition there is
u( 0 ,y,z,t)=T 0 , u(a,y,z,t)=T 1 , 0 <y<b, 0 <z<c, 0 <t.
(11)
Furthermore, assume that the top and bottom surfaces are insulated. Then the
boundary conditions atz=0andcare
∂u
∂z(x,y,^0 ,t)=^0 ,∂u
∂z(x,y,c,t)=^0 ,^0 <x<a,^0 <y<b,^0 <t.
(12)
(The outward normal directions on the top and bottom are the positive and
negativez-directions, respectively.) Finally, assume that the faces aty=0and
aty=bare exposed to a fluid at temperatureT 2 , so they transfer heat by
convection there. The resulting boundary conditions are
−κ∂∂yu(x, 0 ,z,t)+hu(x, 0 ,z,t)=hT 2 ,κ∂u
∂y(x,b,z,t)+hu(x,b,z,t)=hT 2 ,0 <x<a, 0 <z<c, 0 <t. (13)Finally, we add an initial condition,
u(x,y,z, 0 )=f(x,y,z), 0 <x<a, 0 <y<b, 0 <z<c. (14)
A full, three-dimensional problem is complicated to solve, so we often look
for ways to reduce it to two or even one dimension. In the example problem
of Eqs. (10)–(14), we might eliminatezby finding the temperature averaged
over the interval 0<z<c,
v(x,y,t)=^1 c∫c0u(x,y,z,t)dz.