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Question

Solve the following heat equation on the semi-infinite rod by Fourier Transform $$u_t=ku_{xx}$$ where $x,t>0$ and $u_x(0,t) =0$ and $u(x,0)=\begin{cases} 1, & 0 < x <2 \\ 0, & 2\leq x \end{cases} $

My attempt

If we apply Fourier Trans. to both sides, we get

$ \frac{\partial \mathcal{F} \{U(w,t)\}}{\partial t} = -kw^{2} \mathcal{F} \{U(w,t)\}$

Solving the ODE, we have $U(w,t)=C(w)e^{-kw^2t}$

to find $C(w)$ we use the initial condt. $u(x,0)$

$C(w)=\mathcal{F} \{u(x,0)\}=\frac{1}{(2\pi)^{1/2}}\int_{-\infty}^{\infty} u(x,0) e^{-iwx} \ dx $

Could you help me to solve the rest? What is the solution $$u\left(x,t\right)=\mathrm{TF}^{-1}\{U(w,t)\}=\frac{1}{(2\pi)^{1/2}}\intop_{x\in\mathbb{R}}C(w)e^{-kw^2t}e^{iwx}dw$$?

(I can't calculate $C(w)$ since $u(x,0)$ is undefined on $-\infty<x<0$ )

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  • $\begingroup$ Possible duplicate of Solve Heat Equation using Fourier Transform (non homogeneous) $\endgroup$
    – Guy Fsone
    Oct 15, 2017 at 10:41
  • $\begingroup$ No, it is not fully duplicate. $u(x,0)$ is piecewise function in here. I can' t solve the question. $\endgroup$
    – HD239
    Oct 15, 2017 at 10:45
  • $\begingroup$ And what is the problem just plugging your piecewise function in the given solution. $\endgroup$
    – Guy Fsone
    Oct 15, 2017 at 10:48
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    $\begingroup$ There is no missing something. That is all. May be somebody will solve the question. So, we will learn how to solve. $\endgroup$
    – HD239
    Oct 15, 2017 at 11:00
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    $\begingroup$ @GuyFsone To be precise, OP is asking whether the problem can be solved with proper Fourier transform. :) One can argue that Fourier transform can't be used as a general name of complex Fourier transform, Fourier sine transfrom, Fourier cosine transform, etc. though. $\endgroup$
    – xzczd
    Oct 20, 2017 at 13:14

2 Answers 2

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Let $u_0(x) =u(x,0)$ then $$C(w)=\mathcal{F} \{u_0\}(w)=\frac{1}{2\pi}\int_{-\infty}^{\infty} u(x,0) e^{-iwx} \ dx$$ We also know the following $$ e^{-kw^2t} = \mathcal{F}\left(\frac{1}{\sqrt{2kt}} \displaystyle{e^{-\frac{x^2}{4kt}}}\right)(w)$$

Whence it follows that, $$U(w,t)= C(w)e^{-kw^2t}=\mathcal{F} \{u_0\}(w)\cdot\mathcal{F}\left(\frac{1}{\sqrt{2kt}} \displaystyle{e^{-\frac{x^2}{4kt}}}\right)(w)$$

Using the product rule for Fourier we have $$u\left(x,t\right)=\mathcal{F}^{-1}\{U(w,t)\} =\mathcal{F}^{-1}\left\{\mathcal{F} \{u_0\}\cdot\mathcal{F}\left(\frac{1}{\sqrt{2kt}} \displaystyle{e^{-\frac{x^2}{4kt}}}\right)\right\}\\= u_0\star\frac{1}{\sqrt{2kt}} \displaystyle{e^{-\frac{x^2}{4kt}}}~~~\text{convolution }\\= \frac{1}{\sqrt{2kt}}\int_\Bbb R u_0(y)e^{-\frac{(x-y)^2}{4kt}}dy \\=\frac{1}{\sqrt{2kt}}\int_{2}^{\infty} \underbrace{u_0(y)}_{0}e^{-\frac{(x-y)^2}{4kt}}dy +\frac{1}{\sqrt{2kt}}\int_{0}^{2} \underbrace{u_0(y)}_{1}e^{-\frac{(x-y)^2}{4kt}}dy \\+\frac{1}{\sqrt{2kt}}\int_{-\infty}^{2}\underbrace{u_0(y)}_{?}e^{-\frac{(x-y)^0}{4kt}}dy\\=\frac{1}{\sqrt{2kt}}\int_{0}^{2} e^{-\frac{(x-y)^2}{4kt}}dy +\frac{1}{\sqrt{2kt}}\int_{-\infty}^{2}\underbrace{u_0(y)}_{?-u_0~~is ~~not~~define}e^{-\frac{(x-y)^0}{4kt}}dy $$

YOU HAVE TO DEIFNE $u_0$ FOR $x<0$

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The separated solutions $X(x)T(t)$ must solve $$ \frac{T'(t)}{kT(t)} = -\lambda = \frac{X''(x)}{X(x)},\;\; X'(0)=0. $$ The solutions have the form $T(t)=Ce^{-\lambda t}$, $X(x)=D\cos(\sqrt{\lambda}x)$, and $\lambda > 0$ is natural in order to have boundedness in time and space of the separated solutions. Letting $\lambda=s^2$ gives a trial solution $$ u(x,t)=\int_{0}^{\infty}F(s)e^{-s^2t}\cos(sx)ds. $$ In order to satisfy the condition on $u(x,0)$, it is necessarily to determine a coefficient function $F$ such that $$ \int_{0}^{\infty}F(s)\cos(sx)ds=\chi_{[0,2]}(x), $$ which is solved by the inverse Fourier cosine transform: \begin{align} F(s) & = \frac{2}{\pi}\int_{0}^{\infty}\chi_{[0,2]}(x)\cos(sx)dx \\ & = \frac{2}{\pi}\int_{0}^{2}\sin(sx)dx= -\frac{2}{\pi}\frac{\cos(2s)-1}{s}. \end{align} Therefore, $$ u(x,t) = \frac{2}{\pi}\int_{0}^{\infty}e^{-s^2 t}\frac{1-\cos(2s)}{s}\cos(sx)ds. $$

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