 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-4/14c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

θ (x, t) = t^{- 1/2} e^{- x^{2} / (4 D t)},

θ_{t} = (- \frac{1}{2 t} + \frac{x ^{2}}{4 D t ^{2}}) θ

and

D θ_{xx} = D (- \frac{1}{2 D t} + \frac{x ^{2}}{4 D ^{2} t ^{2}}) θ = (- \frac{1}{2 t} + \frac{x ^{2}}{4 D t ^{2}}) θ .

Thus

θ_{t} = D θ_{xx}

Its total mass is found from the Gaussian integral:

\int_{- \infty}^{\infty} θ (x, t) d x = 2 π D .

The unit-mass solution converging to the Dirac delta function as

t ↓ 0

is therefore the heat kernel

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

θ (x, t) = \int_{- 1}^{1} \frac{1}{4 π D t} exp (- \frac{( x - y ) ^{2}}{4 D t}) d y .

With

u = (x - y) / (2 D t)

and the definition of the error function, this becomes

θ (x, t) = \frac{1}{2} [Erf (\frac{x + 1}{2 D t}) - Erf (\frac{x - 1}{2 D t})] .

a = 1

t = 0

the graph is the rectangle of height one on

[- 1, 1]

. For every

t > 0

it is smooth, positive and even, with its maximum at

x = 0

. As time increases the graph broadens and its maximum falls, while its total area remains

2

; pointwise it tends to zero as

t \to \infty

 New to topics? Read the docs here!