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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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2  
Have we just been assigned a task? –  JohnD Jan 1 '13 at 15:19
    
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  user53153 Jan 1 '13 at 22:24
    
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

1 Answer 1

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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