Mathematical Tools for Physics

10—Partial Differential Equations 286

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

1

f

df

dt

=D

1

g

d^2 g

dx^2

(7)

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 (8)

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 (9)

For negative realα, the hyperbolic functions become circular functions.

g(x) =Asin

√

−α/Dx+Bcos

√

−α/Dx and f(t) =eαt (10)

Ifα= 0then
g(x) =Ax+B, and f(t) = constant (11)

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 ( 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)