Mathematical Tools for Physics

10—Partial Differential Equations 289

The differential equation for the temperature is still Eq. ( 3 ), and I’ll assume that the temperature inside
the material approachesT= 0far away from the surface. Separation of variables is the same as before, Eq. ( 8 ),
but this time I know the time dependence at the surface. It’s typical in cases involving oscillations that it is easier
to work with complex exponentials than it is to work with sines and cosines. For that reason I specify that the
surface temperature isT 1 e−iωtinstead of a cosine. I understand that at the end of the problem I’ll take the real
part of the result and throw away the imaginary part. The imaginary part corresponds to solving the problem for
a surface temperature ofsinωtinstead of cosine. It’s easier to solve the two problems together then either one
separately. (The minus sign in the exponent ofe−iωtis arbitrary; you could use a plus instead.)
The equation ( 8 ) says that the time dependence that I expect is

1

f

df

dt

=α=

1

e−iωt

(

−iωe−iωt

)

=−iω

The equation for thex-dependence is then

D

d^2 g

dx^2

=αg=−iωg

This is again a simple exponential solution, sayeβx. Substitute and you have

Dβ^2 eβx=−iωeβx, implying β=±

√

−iω/D (15)

Evaluate this as
√
−i=

(

e−iπ/^2

) 1 / 2

=e−iπ/^4 =

1 −i

√

2

Letβ 0 =

√

ω/ 2 D, then the solution for thex-dependence is

g(x) =Ae(1−i)β^0 x+Be−(1−i)β^0 x (16)

Look at the behavior of these two terms. The first has a factor that goes ase+xand the second goes ase−x.
The temperature at large distances is supposed to approach zero, so that says thatA= 0. The solutions for the
temperature is now
Be−iωte−(1−i)β^0 x