2.6 Example: Convection 171
w( 0 ,t)= 0 , hw(a,t)+κ∂w∂x(a,t)= 0 , 0 <t, (7)
w(x, 0 )=f(x)−v(x)≡g(x), 0 <x<a. (8)The solution forw(x,t)can now be found by the product method. On the
assumption thatwhas the form of a productφ(x)T(t), the variables can be
separated exactly as before, giving two ordinary differential equations linked
by a common parameterλ^2 :
φ′′+λ^2 φ= 0 , 0 <x<a,
T′+λ^2 kT= 0 , 0 <t.Also, since the boundary conditions are linear and homogeneous, they can be
translated directly into conditions onφ:
w( 0 ,t)=φ( 0 )T(t)= 0 ,κ∂w
∂x(a,t)+hw(a,t)=[
κφ′(a)+hφ(a)]
T(t)= 0.EitherT(t)is identically zero (which would makew(x,t)identically zero) or
φ( 0 )= 0 ,κφ′(a)+hφ(a)= 0.Combining the differential equation and boundary conditions onφ,weget
the eigenvalue problem
φ′′+λ^2 φ= 0 , 0 <x<a, (9)
φ( 0 )= 0 ,κφ′(a)+hφ(a)= 0. (10)The general solution of the differential equation is
φ(x)=c 1 cos(λx)+c 2 sin(λx).The boundary condition atx=0requiresthatφ( 0 )=c 1 =0, leavingφ(x)=
c 2 sin(λx). Now, at the other boundary,
κφ′(a)+hφ(a)=c 2(
κλcos(λa)+hsin(λa))
= 0.
Discarding the possibilitiesc 2 =0andλ=0, which both lead to the trivial
solution, we are left with the equation
κλcos(λa)+hsin(λa)= 0 , or tan(λa)=−κ
hλ. (11)