Introduction - Finite Elements: Theory, Fast Solvers, and Applications in Elasticity Theory - page 7

For a one-dimensional rod and Q = 0, with u = T this simplifies to

u t = σu xx , ( 1 . 12 )

where σ = κ/a . As before, we may assume the normalization σ = 1byan

appropriate choice of units.

Parabolic problems typically lead to initial-boundary-value problems .

We first consider the heat distribution on a rod of finite length . Then, in
addition to the initial values, we also have to specify the temperature or the heat
fluxes on the boundaries. For simplicity, we restrict ourselves to the case where
the temperature is constant at both ends of the rod as a function of time. Then,
without loss of generality, we can assume that

σ = 1 , = π and u( 0 ,t) = u(π, t) = 0 ;

cf. Problem 1.10. Suppose the initial values are given by the Fourier series expan-
sion

∞

u(x, 0 ) = a k sin kx, 0 <x<π.

k = 1

Obviously, the functions e − k 2 t sin kx satisfy the heat equation u t = u xx , and thus

∞

a k e − k 2 t sin kx, u(x, t) = t ≥ 0 ( 1 . 13 )

k = 1

is a solution of the given initial-value problem.

For an infinitely long rod, the boundary conditions drop out. Now we need
to know something about the decay of the initial values at infinity, which we
ignore here. In this case we can write the solution using Fourier integrals instead
of Fourier series. This leads to the representation

+∞

1
2 √ πt

e − ξ 2 / 4 t f(x − ξ)dξ, u(x, t) = ( 1 . 14 )

−∞

where the initial value f(x) : = u(x, 0 ) appears explicitly. Note that the solution at

a point (x, t) depends on the initial values on the entire domain, and the propagation
of the data occurs with infinite speed.