Mathematical Tools for Physics - Department of Physics - University

10—Partial Differential Equations 244

If you want to find out if you have a solution, plug in:

∂T

∂t

=

κ

cρ

∂^2 T

∂x^2

is

df

dt

g=

κ

cρ

f

d^2 g

dx^2

Denote the constant byκ/cρ=Dand divide by the productfg.

1

f

df

dt

=D

1

g

d^2 g

dx^2

(10.8)

The left side of this equation is a function oftalone, nox. The right side is a function ofxalone with

not, hence the name separation of variables. Becausexandtcan vary quite independently of each

other, the only way that this can happen is if the two side are constant (the same constant).

1

f

df

dt

=α and D

1

g

d^2 g

dx^2

=α (10.9)

At this point, the constantαcan be anything, even complex. For a particular specified problem there

will be boundary conditions placed on the functions, and those will constrain theα’s. Ifαis real and

positive then

g(x) =Asinh

√

α/Dx+Bcosh

√

α/Dx and f(t) =eαt (10.10)

For negative realα, the hyperbolic functions become circular functions.

g(x) =Asin

√

−α/Dx+Bcos

√

−α/Dx and f(t) =eαt (10.11)

Ifα= 0then

g(x) =Ax+B, and f(t) = constant (10.12)

For imaginaryαthef(t)is oscillating and theg(x)has both exponential and oscillatory behavior in

space. This can really happen in very ordinary physical situations; see section10.3.

This analysis provides a solution to the original equation (10.3) valid for anyα. A sum of such

solutions for differentα’s is also a solution, for example

T(x,t) =A 1 eα^1 tsin

√

−α 1 /Dx+A 2 eα^2 tsin

√

−α 2 /Dx

or any other linear combination with variousα’s

T(x,t) =

∑

{α′s}

fα(t)gα(x)

It is the combined product that forms a solution to the original partial differential equation, not the
separate factors. Determining the details of the sum is a job for Fourier series.

Example

A specific problem: You have a slab of material of thicknessLand at a uniform temperatureT 0. Plunge

it into ice water at temperatureT= 0and find the temperature inside at later times. The boundary

condition here is that the surface temperature is zero,T(0,t) =T(L,t) = 0. This constrains the

separated solutions, requiring thatg(0) =g(L) = 0. For this to happen you can’t use the hyperbolic