# Diffusion equation in an infinite medium

Consider the one dimensional diffusion equation in an infinite medium given by;

$$\alpha^2U_{xx}(x,t)-U_t(x,t)=-F(x,t), \quad -\infty\lt x \lt\infty \quad \mathrm{and} \quad 0\lt t\lt \infty,\\ U(x,0)=f(x), \\ \lim_{x\to \infty}U(x,t)=\mathrm{finite,} \quad\lim_{x\to -\infty}U(x,t)=\mathrm{finite,}$$

where $F(x,t)$ is a source term and $f(x)$ is the initial temperature distribution.

A) Determine the response of the system, using Laplace transform for

$$F(x,t)=\sin(2\pi t) \ \mathrm{and}\ f(x)=0 .$$

B) Determine the response of the system, using Fourier transform for:

$$F(x,t)=0 \ \mathrm{and}\ f(x)=\delta(x), \quad \mathrm{where}\ \delta(x) \ \mathrm{is\ the Dirac\ impulse\ function}.$$

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Have we just been assigned a task? – JohnD Jan 1 '13 at 15:19
I have solve the problem a bit. I have found a non-homogeneous ODE after taking the Laplace. But, putting the boundry conditions do not give reasonable results. – mud08 Jan 2 '13 at 7:43
math.utsa.edu/~gokhman/ftp//courses/notes/heat.pdf might be useful. – doraemonpaul Jan 2 '13 at 8:43

A)

Let $U(x,t)=V(x,t)-\dfrac{\cos2\pi t}{2\pi}$ ,

Then $U_x(x,t)=V_x(x,t)$

$U_{xx}(x,t)=V_{xx}(x,t)$

$U_t(x,t)=V_t(x,t)+\sin2\pi t$

$\therefore\alpha^2V_{xx}(x,t)-V_t(x,t)-\sin2\pi t=-\sin2\pi t$

$\alpha^2V_{xx}(x,t)-V_t(x,t)=0$

Let $V(x,t)=X(x)T(t)$ ,

Then $\alpha^2X''(x)T(t)-X(x)T'(t)=0$

$X(x)T'(t)=\alpha^2X''(x)T(t)$

$\dfrac{T'(t)}{\alpha^2T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-\alpha^2s^2\\X''(x)+s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-\alpha^2ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\dfrac{\pi xs}{L}&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore U(x,t)=\int_0^\infty C_1(s)e^{-\alpha^2ts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-\alpha^2ts^2}\cos xs~ds-\dfrac{\cos2\pi t}{2\pi}$

$U(x,0)=0$ :

$\int_0^\infty C_1(s)\sin xs~ds+\int_0^\infty C_2(s)\cos xs~ds-1=0$

$\int_0^\infty C_2(s)\cos xs~ds=1-\int_0^\infty C_1(s)\sin xs~ds$

$\mathcal{F}_{c,s\to x}\{C_2(s)\}=1-\mathcal{F}_{s,s\to x}\{C_1(s)\}$

$C_2(s)=\delta(s)-\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_1(s)\}\}$

$\therefore U(x,t)=\int_0^\infty C_1(s)e^{-\alpha^2ts^2}\sin xs~ds+\int_0^\infty\delta(s)e^{-\alpha^2ts^2}\cos xs~ds-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_1(s)\}\}e^{-\alpha^2ts^2}\cos xs~ds-\dfrac{\cos2\pi t}{2\pi}=\int_0^\infty C_1(s)e^{-\alpha^2ts^2}\sin xs~ds-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_1(s)\}\}e^{-\alpha^2ts^2}\cos xs~ds-\dfrac{\cos2\pi t}{2\pi}+1$

B)

Let $U(x,t)=X(x)T(t)$ ,

Then $\alpha^2X''(x)T(t)-X(x)T'(t)=0$

$X(x)T'(t)=\alpha^2X''(x)T(t)$

$\dfrac{T'(t)}{\alpha^2T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-\alpha^2s^2\\X''(x)+s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-\alpha^2ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\dfrac{\pi xs}{L}&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore U(x,t)=\int_0^\infty C_1(s)e^{-\alpha^2ts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-\alpha^2ts^2}\cos xs~ds$

$U(x,0)=\delta(x)$ :

$\int_0^\infty C_1(s)\sin xs~ds+\int_0^\infty C_2(s)\cos xs~ds=\delta(x)$

$\int_0^\infty C_2(s)\cos xs~ds=\delta(x)-\int_0^\infty C_1(s)\sin xs~ds$

$\mathcal{F}_{c,s\to x}\{C_2(s)\}=\delta(x)-\mathcal{F}_{s,s\to x}\{C_1(s)\}$

$C_2(s)=\dfrac{1}{\pi}-\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_1(s)\}\}$

$\therefore U(x,t)=\int_0^\infty C_1(s)e^{-\alpha^2ts^2}\sin xs~ds+\dfrac{1}{\pi}\int_0^\infty e^{-\alpha^2ts^2}\cos xs~ds-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_1(s)\}\}e^{-\alpha^2ts^2}\cos xs~ds=\int_0^\infty C_1(s)e^{-\alpha^2ts^2}\sin xs~ds-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_1(s)\}\}e^{-\alpha^2ts^2}\cos xs~ds+\dfrac{e^{-\frac{x^2}{4\alpha^2t}}}{2\alpha\sqrt{\pi t}}$

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