Renjith Thazhathethil

Partial Differential Equations

Assignment 6

  1. [5 points] Prove the characteristic parallelogram property for wave equations.

  2. [5 points] Solve the problem with two different characteristic speedsc1c_1andc2c_2:

    (tc1x)(tc2x)u=0,in R×(0,).\begin{equation*} \left(\frac{\partial}{\partial t} - c_1 \frac{\partial}{\partial x}\right)\left(\frac{\partial}{\partial t} - c_2 \frac{\partial}{\partial x}\right) u = 0, \quad \text{in } \mathbb{R} \times (0,\infty). \end{equation*}

    Analyze the casesc1c2c_1 \neq c_2andc1=c2c_1 = c_2, and derive D’Alembert’s formula as a special case. Discuss any potential loss of regularity in the one-dimensional case.

  3. [5 points] Integrate the wave equationuttuxx=f(x,t)u_{tt} - u_{xx} = f(x,t)in the characteristic triangleP(x,t)P(x,t),Q(xct,0)Q(x-ct,0),R(x+ct,0)R(x+ct,0)to derive a formula for the solution. (Hint: Use the identityuttuxx=(ut)t(ux)xu_{tt} - u_{xx} = (u_t)_t - (u_x)_x.)

  4. [5 points] Solve the wave equation in the first quadrant with non-homogeneous Dirichlet boundary condition:

    uttuxx=0,in (0,)×(0,)\begin{equation*} u_{tt} - u_{xx} = 0, \quad \text{in } (0,\infty) \times (0,\infty) \end{equation*}

    with initial and boundary conditions:

    u(x,0)=f(x),ut(x,0)=g(x),u(0,t)=h(t),t>0.\begin{equation*} u(x,0) = f(x), \quad u_t(x,0) = g(x), \quad u(0,t) = h(t), \quad t > 0. \end{equation*}

    Derive the formula forx<ctx < ctusing the parallelogram identity.

  5. [5 points] Solve the above equation with the Neumann non-homogeneous boundary condition whereu(0,t)=h(t)u(0,t) = h(t)is replaced byux(0,t)=h(t)u_x(0,t) = h(t).

  6. [5 points] Letc1,,ckc_1, \dots, c_kbe distinct positive real numbers. Show that the solution of the equation:

    (t2c12x2)(t2ck2x2)u=0,\begin{equation*} (\partial_t^2 - c_1^2 \partial_x^2) \dots (\partial_t^2 - c_k^2 \partial_x^2) u = 0, \end{equation*}

    can be written as:

    u(x,t)=j=0kuj(x,t),\begin{equation*} u(x,t) = \sum_{j=0}^{k} u_j(x,t), \end{equation*}

    where eachuju_jsatisfiest2ujcj2x2uj=0\partial_t^2 u_j - c_j^2 \partial_x^2 u_j = 0. This result also holds in higher dimensions.

  7. [5 points] Consider the casen=3n=3for:

    (t2c2x2)(t2c2x2)u=0,c>0.\begin{equation*} (\partial_t^2 - c^2 \partial_x^2)(\partial_t^2 - c^2 \partial_x^2) u = 0, \quad c > 0. \end{equation*}

    Given smooth initial datatju(x,0)=fj(x)\partial_t^j u(x,0) = f_j(x)forj=0,1,2,3j = 0,1,2,3, explicitly determine the solution.

  8. [5 points] Consider the wave equation:

    uttΔu=0,in Rn×(0,),\begin{equation*} u_{tt} - \Delta u = 0, \quad \text{in } \mathbb{R}^n \times (0,\infty), \end{equation*}
    with initial data:
    u(x,0)=ϕ(x),ut(x,0)=ψ(x).\begin{equation*} u(x,0) = \phi(x), \quad u_t(x,0) = \psi(x). \end{equation*}
    Verify that the solution is given by:
    u(x,t)=uϕ(x,t)+0tuψ(x,s)ds,\begin{equation*} u(x,t) = u_{\phi}(x,t) + \int_0^t u_{\psi}(x,s)ds, \end{equation*}
    and also by:
    u(x,t)=vψ(x,t)+tvϕ(x,t).\begin{equation*} u(x,t) = v_{\psi}(x,t) + \frac{\partial}{\partial t} v_{\phi}(x,t). \end{equation*}