14A (Methods)

x

. Since

θ_{xx} = - k^{2} θ

, the transformed heat equation is

\frac{\partial θ}{\partial t} = - D k^{2} θ, θ (k, 0) = Θ (k) .

Therefore

θ (k, t) = e^{- D k^{2} t} Θ (k) .

θ (x, t) = \frac{1}{4 π D t} \int_{- \infty}^{\infty} exp [- \frac{( x - ξ ) ^{2}}{4 D t}] Θ (ξ) d ξ .

b

The causal Green function for the heat operator is

G (x, t; ξ, τ) = H (t - τ) \frac{1}{4 π D ( t - τ )} exp [- \frac{( x - ξ ) ^{2}}{4 D ( t - τ )}],

where

H

is the Heaviside step function. It vanishes for

t < τ

, satisfies the homogeneous heat equation away from the source, and approaches

δ (x - ξ)

t ↓ τ

. Superposing the responses to all infinitesimal sources gives Duhamel principle:

θ_{f} (x, t) = \int_{0}^{t} \int_{- \infty}^{\infty} \frac{exp [ - \frac{( x - ξ ) ^{2}}{4 D ( t - τ )} ]}{4 π D ( t - τ )} f (ξ, τ) d ξ d τ .

The lower limit and causality give the required homogeneous initial data.

Let

g (x, t) = \frac{1}{4 π D t} exp (- \frac{x ^{2}}{4 D t}), t > 0.

Convolution with the initial Dirac delta function translates the heat kernel, while part (b) propagates the impulsive source from time one. Hence

θ_{c} (x, t) = g (x - 2 D, t) - A H (t - 1) g (x + 2 D, t - 1) .

(x, t) = (0, 2)

this becomes

θ_{c} (0, 2) = \frac{e ^{- 1/2}}{8 π D} - A \frac{e ^{- 1}}{4 π D} .

It vanishes exactly when

A = \frac{e ^{- 1/2}}{8 π D} \frac{4 π D}{e ^{- 1}} = \frac{e}{2} .

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