ABSTRACT
We begin with the task of finding a solution u = u(x, t), x ∈ Rn, t > 0, of the initial value problem{
∂u ∂t (x, t) = (∆u)(x, t), x ∈ Rn, t > 0, u(x, 0) = f(x), x ∈ Rn, (6.1)
where ∆ is the Laplacian on Rn defined by
∆ =
∂2
and f ∈ S. The partial differential equation in (6.1) is known as the heat equation. The trick is to take the partial Fourier transform of u with respect to x. If we do this, then we get{
∂uˆ ∂t (ξ, t) + |ξ|2uˆ(ξ, t) = 0, ξ ∈ Rn, t > 0, uˆ(ξ, 0) = fˆ(ξ), ξ ∈ Rn. (6.2)
Thus, from the first equation in (6.2), we get
uˆ(ξ, t) = Ce−|ξ| 2t, ξ ∈ Rn, t > 0,
where C is an arbitrary constant, which depends on ξ. Using the initial condition for uˆ(ξ, 0) in (6.2), we get C = fˆ(ξ). Thus,
uˆ(ξ, t) = e−t|ξ| 2
fˆ(ξ), ξ ∈ Rn, t > 0. If we take the inverse Fourier transform with respect to ξ, then, by the second formula in Proposition 4.5 and the adjoint formula in Proposition 4.7, we get
u(x, t) = (2π)−n/2 ∫ Rn
eix·ξe−t|ξ| 2
fˆ(ξ) dξ
= (2π)−n/2 ∫ Rn
(Mxe −t|·|2)(ξ)fˆ(ξ) dξ
= (2π)−n/2 ∫ Rn
(T−x(e−t|·| 2
)∧)(y) f(y) dy
= (2π)−n/2 ∫ Rn
)∧(y − x) f(y) dy
=
kt(x− y) f(y) dy, x ∈ Rn, t > 0,
Analysis
kt(x) = (2π) −n Rn
eix·ξe−t|ξ| 2
dξ, x ∈ Rn, t > 0.
