Paper 2, Section II, 16D

Methods | Part IB, 2014

The Fourier transform f~\tilde{f} of a function ff is defined as

f~(k)=∫−∞∞f(x)e−ikxdx, so that f(x)=12π∫−∞∞f~(k)eikxdk\tilde{f}(k)=\int_{-\infty}^{\infty} f(x) e^{-i k x} d x, \quad \text { so that } f(x)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \tilde{f}(k) e^{i k x} d k

A Green's function G(t,t′,x,x′)G\left(t, t^{\prime}, x, x^{\prime}\right) for the diffusion equation in one spatial dimension satisfies

∂G∂t−D∂2G∂x2=δ(t−t′)δ(x−x′)\frac{\partial G}{\partial t}-D \frac{\partial^{2} G}{\partial x^{2}}=\delta\left(t-t^{\prime}\right) \delta\left(x-x^{\prime}\right)

(a) By applying a Fourier transform, show that the Fourier transform G~\tilde{G} of this Green's function and the Green's function GG are

G~(t,t′,k,x′)=H(t−t′)e−ikx′e−Dk2(t−t′)G(t,t′,x,x′)=H(t−t′)4πD(t−t′)e−(x−x′)24D(t−t′)\begin{aligned} \tilde{G}\left(t, t^{\prime}, k, x^{\prime}\right) &=H\left(t-t^{\prime}\right) e^{-i k x^{\prime}} e^{-D k^{2}\left(t-t^{\prime}\right)} \\ G\left(t, t^{\prime}, x, x^{\prime}\right) &=\frac{H\left(t-t^{\prime}\right)}{\sqrt{4 \pi D\left(t-t^{\prime}\right)}} e^{-\frac{\left(x-x^{\prime}\right)^{2}}{4 D\left(t-t^{\prime}\right)}} \end{aligned}

where HH is the Heaviside function. [Hint: The Fourier transform F~\tilde{F} of a Gaussian F(x)=14πae−x24a,a=constF(x)=\frac{1}{\sqrt{4 \pi a}} e^{-\frac{x^{2}}{4 a}}, a=\mathrm{const}, is given by F~(k)=e−ak2.]\left.\tilde{F}(k)=e^{-a k^{2}} .\right]

(b) The analogous result for the Green's function for the diffusion equation in two spatial dimensions is

G(t,t′,x,x′,y,y′)=H(t−t′)4πD(t−t′)e−14D(t−t′)[(x−x′)2+(y−y′)2]G\left(t, t^{\prime}, x, x^{\prime}, y, y^{\prime}\right)=\frac{H\left(t-t^{\prime}\right)}{4 \pi D\left(t-t^{\prime}\right)} e^{-\frac{1}{4 D\left(t-t^{\prime}\right)}\left[\left(x-x^{\prime}\right)^{2}+\left(y-y^{\prime}\right)^{2}\right]}

Use this Green's function to construct a solution for t⩾0t \geqslant 0 to the diffusion equation

∂Ψ∂t−D(∂2Ψ∂x2+∂2Ψ∂y2)=p(t)δ(x)δ(y)\frac{\partial \Psi}{\partial t}-D\left(\frac{\partial^{2} \Psi}{\partial x^{2}}+\frac{\partial^{2} \Psi}{\partial y^{2}}\right)=p(t) \delta(x) \delta(y)

with the initial condition Ψ(0,x,y)=0\Psi(0, x, y)=0.

Now set

p(t)={p0=const for 0⩽t⩽t00 for t>t0p(t)= \begin{cases}p_{0}=\mathrm{const} & \text { for } \quad 0 \leqslant t \leqslant t_{0} \\ 0 & \text { for } \quad t>t_{0}\end{cases}

Find the solution Ψ(t,x,y)\Psi(t, x, y) for t>t0t>t_{0} in terms of the exponential integral defined by

Ei(−η)=−∫η∞e−λλdλE i(-\eta)=-\int_{\eta}^{\infty} \frac{e^{-\lambda}}{\lambda} d \lambda

Use the approximation Ei(−η)≈ln⁡η+CE i(-\eta) \approx \ln \eta+C, valid for η≪1\eta \ll 1, to simplify this solution Ψ(t,x,y)\Psi(t, x, y). Here C≈0.577C \approx 0.577 is Euler's constant.

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