Renjith Thazhathethil

Partial Differential Equations

Assignment 5

  1. [5 points] Use Fourier transform to derive the Poisson formula for the upper half plane in 2-dimension.

  2. Fundamental Solution

    1. [5 points] Define the fundamental solution:

      ϕ(x,t)={1(4πt)n/2ex24t,xRn,t>0,0,xRn,t=0.\begin{equation*} \phi(x,t) = \begin{cases} \frac{1}{(4\pi t)^{n/2}} e^{-\frac{|x|^2}{4t}}, & x \in \mathbb{R}^n, t > 0, \\ 0, & x \in \mathbb{R}^n, t = 0. \end{cases} \end{equation*}

      Show thatϕ\phisatisfiesϕtΔϕ=0\phi_t - \Delta \phi = 0forxRn,t>0x \in \mathbb{R}^n, t > 0and that

      lim(x,t)(x0,0)ϕ(x,t)=0,for x00.\begin{equation*} \lim\limits_{(x,t)\to(x_0,0)} \phi(x,t) = 0, \quad \text{for } x_0 \neq 0. \end{equation*}
    2. [5 points] Show that

      Rnϕ(x,t)dx=1t>0.\begin{equation*} \int_{\mathbb{R}^n} \phi(x,t)dx = 1 \quad \forall t > 0. \end{equation*}
    3. [5 points] Forδ>0\delta > 0, show that

      limt0xy>δϕ(xy,t)dy=0.\begin{equation*} \lim\limits_{t\to0} \int_{|x - y| > \delta} \phi(x - y, t)dy = 0. \end{equation*}

  3. [5 points] Solve the heat equation:

    ut2ux2=0,xR,t>0,\begin{equation*} \frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} = 0, \quad x \in \mathbb{R}, t > 0, \end{equation*}

    with initial conditionu(x,0)=f(x)u(x,0) = f(x)using Fourier transform (assume appropriate conditions onff).

  4. [5 points] Consider the equation:

    ut2ux2=0,xR,t>0.\begin{equation*} \frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} = 0, \quad x \in \mathbb{R}, t > 0. \end{equation*}

    Find all solutions of the formu(x,t)=1tv(x2t)u(x,t) = \frac{1}{\sqrt{t}} v\left(\frac{x}{2\sqrt{t}}\right).

  5. [10 points] LetE(x,t,r)E(x,t,r)be the heat ball andE(1)=E(0,0,1)E(1) = E(0,0,1). Show that:

    E(1)y2s2dyds=4.\begin{equation*} \int\int_{E(1)} \frac{|y|^2}{s^2} dyds = 4. \end{equation*}

    Use an appropriate transformation to evaluate:

    E(r)y2s2dyds,\begin{equation*} \int\int_{E(r)} \frac{|y|^2}{|s|^2} dyds, \end{equation*}

    whereE(r)=E(0,0,r)E(r) = E(0,0,r).

  6. [5 points] Define:

    g(t)={e1tα,t>0,0,t0,\begin{equation*} g(t) = \begin{cases} e^{-\frac{1}{t^\alpha}}, & t > 0, \\ 0, & t \leq 0, \end{cases} \end{equation*}

    withα>1\alpha > 1, and consider:

    u(x,t)=k=0g(k)(t)(2k)!x2k.\begin{equation*} u(x,t) = \sum\limits_{k=0}^{\infty} \frac{g^{(k)}(t)}{(2k)!} x^{2k}. \end{equation*}

    Show that this provides infinitely many solutions to the heat equation with zero boundary conditions.

  7. [5 points] Find a sequence of solutionsunu_nof the one-dimensional heat equation:

    ut2ux2=0,x(0,2π),t>0,\begin{equation*} \frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} = 0, \quad x \in (0, 2\pi), t > 0, \end{equation*}

    with boundary conditionsu(0,t)=u(2π,t)=0u(0,t) = u(2\pi,t) = 0. Using separation of variables, construct a series solution and derive the condition ensuring thatu(x,0)=f(x)u(x,0) = f(x).

  8. Show that ifuusatisfies the heat equationutΔu=0u_t - \Delta u = 0inΩ×(0,T)\Omega \times (0,T), then the following maximum principle holds:

    supΩ×(0,T)u(x,t)=supΓTu(x,t).\begin{equation*} \sup\limits_{\Omega \times (0,T)} u(x,t) = \sup\limits_{\Gamma_T} u(x,t). \end{equation*}
  9. [5 points] Show that ifuusatisfies:

    utΔu=u,in Ω×(0,T),\begin{equation*} u_t - \Delta u = u, \quad \text{in } \Omega \times (0,T), \end{equation*}

    withu(x,0)=0u(x,0) = 0forxΩx \in \Omegaandu(x,t)=0u(x,t) = 0onΩ×[0,T]\partial\Omega \times [0,T], thenu(x,t)=0u(x,t) = 0inΩ×(0,T)\Omega \times (0,T).

  10. [5 points] Solve the heat equation:

    ut=uxx,x>0,t>0,\begin{equation*} u_t = u_{xx}, \quad x > 0, t > 0, \end{equation*}
    with initial and boundary conditions:
    u(x,0)=g(x),x>0,u(0,t)=0,t>0.\begin{equation*} u(x,0) = g(x), \quad x > 0, \quad u(0,t) = 0, \quad t > 0. \end{equation*}
    (Hint: Use an odd extension to rewrite the equation inR×(0,)\mathbb{R} \times (0,\infty).)