 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/14/d/solution

Put

D = σ^{2} > 0

and

ξ = x - x_{0}

. Logarithmic differentiation of the positive Gaussian function gives

ϕ_{t} = (- \frac{1}{2 t} + \frac{ξ ^{2}}{4 D t ^{2}}) ϕ, ϕ_{x} = - \frac{ξ}{2 D t} ϕ .

Differentiating once more in space gives

ϕ_{xx} = (- \frac{1}{2 D t} + \frac{ξ ^{2}}{4 D ^{2} t ^{2}}) ϕ .

Multiplying by

D

proves

ϕ_{t} = D ϕ_{xx} = σ^{2} ϕ_{xx} (t > 0) .

The assumption

σ \neq = 0

ensures both the normalization and exponential denominator are defined. This is the heat kernel, with diffusion coefficient

D

and centre

x_{0}

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