Math 401-501/502 Solutions — Assignment 9

  1. Find the Fourier transform of the funtion $f(x)$. You may use table A.2 (p. 314) in the text.
    1. $f(x) = \left\{\begin{array}{cc} e^{-x}& x \ge 0\\ 0 & x<0 \end{array} \right.$

      Solution. Do this diectly using the the definition of the FT. \[ \begin{align} \hat f(\omega) &=\mathcal F [f](\omega) \\ &= (2\pi)^{-1/2}\int_{-\infty}^\infty f(x)e^{i\omega x}dx = (2\pi)^{-1/2}\int_{-\infty}^0 0\cdot dx + (2\pi)^{-1/2}\int_0^\infty e^{-x+i\omega x}dx \\ &= (2\pi)^{-1/2}\int_0^\infty e^{-(1-i\omega) x}dx \\ &= -(2\pi)^{-1/2}\frac {e^{-(1-i\omega) x}}{1-i\omega} \bigg|_0^\infty = (2\pi)^{-1/2}(1-i\omega)^{-1} \end{align} \]

    2. $f(x) = \left\{\begin{array}{cc} xe^{-x}& x \ge 0\\ 0 & x<0 \end{array} \right.$

      Solution. Do this directly from the definition or use the identity derived in class, namely $\mathcal F[x f(x)]= -i\frac{d\hat f}{d\omega}(\omega)$. We can use the latter method because the $f$ here is $x\cdot f(x)$ from the previous problem. \[ \hat f(\omega) = \mathcal F[x\cdot (\text{previous}\ f)]= -i\frac{d }{d\omega}\bigg((2\pi)^{-1/2}(1-i\omega)^{-1}\bigg)= (2\pi)^{-1/2}(1-i\omega)^{-2} \]

  2. Show that if $F(\omega)$ is the Fourier transform of $f(x)$, then the Fourier transform of $F(x)$ is $f(-\omega)$. (Hint: this follows straight from the definition of the Fourier transform and inverse Fourier transform.) Use this and table A.2 to find the Fouier transform of $f(x) = \frac{\sin{2x}}{x}$.

    Solution. To do the first part, observe that we have \[ \mathcal F[f](\omega) = F(\omega) = \int_{-\infty}^\infty f(x)e^{i\omega x}dx \quad \text{and}\quad \mathcal F^{-1}[F](x) f(x)= \int_{-\infty}^\infty F(\omega)e^{-i\omega x}d\omega \] In the second integral, interchange the roles of $x$ and $\omega$. This gives us $f(\omega)=\int_{-\infty}^\infty F(x)e^{-i\omega x}dx$. In this formula, replace $\omega$ by $-\omega$ to get $f(-\omega)=\int_{-\infty}^\infty F(x)e^{i\omega x}dx=\mathcal F[F(x)]$. To find the Fourier transform of $f(x) = \frac{\sin{2x}}{x}$, look at Table A.2, #12, with $a=2$. After adjusting the constants, this is \[ \mathcal F\big[\underbrace{\sqrt{\frac{\pi}{2}}\,H(2-|x|)}_{f(x)}\big](\omega) = \underbrace{\frac{\sin{2\omega}}{\omega}}_{F(\omega)} \] By what was derived above, we have \[ \mathcal F\big[\underbrace{\frac{\sin{2x}}{x}}_{F(x)}\big] = \underbrace{\sqrt{\frac{\pi}{2}}H\big(2-\big|-\omega\big|\big)}_{f(-\omega)}= \sqrt{\frac{\pi}{2}}H(2-|\omega|). \]

  3. Use the convolution theorem and A.2, #7, to find $f*g$, if $f(x) = e^{-4x^2}$ and $g(x) =e^{-12x^2}$.

    Solution. By #7, $\hat f(\omega) = \frac{1}{2\sqrt{2}}e^{-\omega^2/16}$ and $\hat g(\omega) = \frac{1}{2\sqrt{6}}e^{-\omega^2/48}$. By the convolution theorem, \[ \mathcal F[f*g] = \hat f(\omega)\hat g(\omega)=\frac{1}{8\sqrt{3}} e^{-(1/48+1/16)\omega^2}=\frac{1}{8\sqrt{3}} e^{-\omega^2/12}, \] so $f*g(x) =\mathcal F^{-1}[ \frac{1}{8\sqrt{3}} e^{-\omega^2/12}]$. From #7, with $a^2=3$, we get $f*g(x) = \frac{1}{4\sqrt{2}} e^{-3x^2}$.

  4. Find the inverse Fourier transform of \[ \hat f(\omega) = \sin(\omega) e^{-2| \omega|} \]

    Solution. By A.2, #3, we have $\mathcal F[f(ax+b)]= \frac{1}{a}e^{-i(b/a)\omega}\hat f(\omega/a)$. If $a=1$, this says that $\mathcal F[f(x+b)]=e^{-ib\omega}\hat f(\omega)$, and so $f(x+b)=\mathcal F^{-1}[e^{-ib\omega}\hat f(\omega)]$. Apply this: \[ \begin{align} \mathcal F^{-1}[\sin(\omega) e^{-2| \omega|}] &= \frac{1}{2i}\mathcal F^{-1}[e^{i\omega}e^{-2| \omega|}] - \frac{1}{2i}\mathcal F^{-1}[e^{-i\omega}e^{-2| \omega|}\big)] \\ &=\frac{1}{2i} \mathcal F^{-1}[e^{-2| \omega|}](x-1) - \frac{1}{2i} \mathcal F^{-1}[e^{-2| \omega|}](x+1) \\ &= \frac{1}{2i} 2\sqrt{\frac{2}{\pi}}\frac{1}{(x-1)^2+2^2} - \frac{1}{2i} 2\sqrt{\frac{2}{\pi}}\frac{1}{(x+1)^2+2^2} \quad \text{-- by A.2, #10} \\ &= i\sqrt{\frac{2}{\pi}}\bigg(\frac{1}{(x+1)^2+4} - \frac{1}{(x-1)^2+4} \bigg). \end{align} \]

  5. Use the Fourier transform method for the infinite bar to solve this convective heat flow problem: \[ \frac{1}{k}\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}+ \alpha \frac{\partial u}{\partial x}, \ u(x,0)=f(x),\ -\infty < x < \infty, \ \alpha>0. \]

    Solution. Let $U(\omega,t)=\mathcal F_x[u(x,t)]$.

    1. Take the FT of the heat equation. \[ \mathcal F \big[\frac{1}{k}\frac{\partial u}{\partial t}\big] = \mathcal F\big[\frac{\partial^2 u}{\partial x^2}\big] + \mathcal F\big[\alpha\frac{\partial u}{\partial x}\big]. \]
    2. Use A.2, #1 and #2, along with the fact that "$t$" is is not involved in the FT in $x$, to get \[ \frac{1}{k}\frac{\partial U}{\partial t}(\omega,t) = -\omega^2 U(\omega,t) -i\alpha \omega U(\omega,t). \]
    3. Solve the previous equation. \[ \begin{align} U(\omega,t) &= e^{-kt\omega^2 - i\alpha k t \omega}U(\omega,0)\\ &= e^{-kt\omega^2 - i\alpha k t \omega}\hat f(\omega) \\ &= e^{-kt\omega^2} \big(e^{- i(\alpha k t) \omega}\hat f(\omega)\big)\\ &= \mathcal F\bigg[\frac{1}{2\sqrt{kt}} e^{-x^2/(4kt)}\bigg]\mathcal F[f(x+\alpha kt)] \end{align} \]
    4. Apply the Convolution Theorem. \[ u(x,t) = \frac{1}{\sqrt{4\pi kt}}\int_{-\infty}^\infty e^{-(x-z)^2/(4kt)} f(z+\alpha kt)dz. \]
  6. For $f(x) = \left\{\begin{array}{cc} 1 &0 \le x \le 1\\ 0 & x<0 \end{array}\right.$ and $\, k=\alpha =1$, find a solution to the previous problem in terms of the error function, erf(x).

    Solution. Put $\alpha=k=1$ in the solution: $u(x,t) = \frac{1}{\sqrt{4\pi t}}\int_{-\infty}^\infty e^{-(x-z)^2/(4t)} f(z+t)dz$

    1. Find limits on the integral. $f(z+t)=1$ if $0\le z+t\le 1$ or $-t\le z\le 1-t$. Outside of this, $f(z+t)=0$, so \[ u(x,t) = \frac{1}{\sqrt{4\pi t}}\int_{-t}^{1-t} e^{-(x-z)^2/(4t)}dz . \]
    2. Change the variable of integration to $s= (z-x)/\sqrt{4t}$. \[ u(x,t) = \frac{1}{\sqrt{4\pi t}}(\sqrt{4t})\int_{-(t+x)/\sqrt{4t}}^{(1-t-x)/\sqrt{4t}} e^{-s^2}ds = \frac{1}{\sqrt{\pi}}\int_{-(t+x)/\sqrt{4t}}^{(1-t-x)/\sqrt{4t}} e^{-s^2}ds. \]
    3. Use $\frac{d}{ds}\text{erf}(s) = \frac{2}{\sqrt{\pi}}e^{-s^2}$. \[ \begin{align} u(x,t) &= \frac12 \int_{-(t+x)/\sqrt{4t}}^{(1-t-x)/\sqrt{4t}} \frac{2}{\sqrt{\pi}}e^{-s^2}ds \\ &= \frac12 \int_{-(t+x)/\sqrt{4t}}^{(1-t-x)/\sqrt{4t}} \frac{d}{ds}\text{erf}(s)ds \\ &=\frac12 \text{erf}\big(\frac{1-t-x}{\sqrt{4t}}\big) - \frac12 \text{erf}\big(-\frac{t+x}{\sqrt{4t}}\big) \end{align} \]
  7. Solve this wave equation for an infinite string: \[ \frac{\partial^2 u}{\partial t^2} = 9\frac{\partial^2 u}{\partial x^2}, \ u(x,0) =0,\ \frac{\partial u}{\partial t}\!(x,0) = xe^{-x^2},\ - \infty < x < \infty, \ t\ge 0. \]

    Solution. Let $U(\omega,t)=\mathcal F_x[u(x,t)]$.

    1. Take the FT of the wave equation in $x$. Use the same properties that we did in problem 5. \[ \frac{\partial^2 U}{\partial t^2}(\omega,t) = -9\omega^2 U(\omega,t) \]
    2. Transform the initial conditions. Use A.2, #8 with $a=1$. \[ U(\omega,0) = \mathcal F[\underbrace{u(x,0)}_{0}]=0 \quad \text{and} \quad \frac{\partial U}{\partial t}(\omega,0)= \mathcal F[\frac{\partial u}{\partial t}(x,0)] = \mathcal F[xe^{-x^2}] = \frac{i }{2\sqrt{2}} \omega e^{-\omega^2/4} \]
    3. Solve $\frac{\partial^2 U}{\partial t^2}+9\omega^2 U = 0$ subject to the initial conditions $U(\omega,0)=0$, $\frac{\partial U}{\partial t}(\omega,0) = \frac{i }{2\sqrt{2}} \omega e^{-\omega^2/4}$. \[ U(\omega,t) = 0\cdot \cos(3\omega t)+ \frac{1}{3\omega}\frac{i }{2\sqrt{2}} \omega e^{-\omega^2/4}\sin(3\omega t) = \frac{i }{6\sqrt{2}} e^{-\omega^2/4}\sin(3\omega t) = \frac{1}{12\sqrt{2}}\big(e^{3i\omega t}e^{-\omega^2/4}-e^{-3i\omega t}e^{-\omega^2/4}\big) \]
    4. Use $f(x+b)=\mathcal F^{-1}[e^{-ib\omega} \hat f(\omega)]$ \[ \begin{align} u(x,t) &= \mathcal F^{-1}[U(\omega,t)] \\ &= \frac{1}{12\sqrt{2}}\big(\mathcal F^{-1}[e^{3i\omega t}e^{-\omega^2/4}]-\mathcal F^{-1}[e^{-3i\omega t}e^{-\omega^2/4}]\big) \\ &=\frac{1}{12\sqrt{2}}\big(\mathcal F^{-1}[e^{-\omega^2/4}](x-3t)-\mathcal F^{-1}[e^{-\omega^2/4}](x+3t) \end{align} \]
    5. Finish up by using A.2, #7, with $a=1$. \[ u(x,t) = \frac{1}{12}\big(e^{-(x-3t)^2} - e^{-(x+3t)^2}\big). \]

Updated 4/19/2013 (fjn)