Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

4.4: A Method of Riemann

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    Riemann's method provides a formula for the solution of the following Cauchy initial value problem for a hyperbolic equation of second order in two variables. Let

    $${\mathcal S}:\ \ x=x(t), y=y(t),\ \ t_1\le t \le t_2,$$

    be a regular curve in \(\mathbb{R}^2\), that is, we assume \(x,\ y\in C^1[t_1,t_2]\) and \(x'^2+y'^2\not=0\). Set


    where \(a,\ b\in C^1\) and \(c,\ f\in C\) in a neighborhood of \({\mathcal S}\). Consider the initial value problem


    where \(f\in C\) in a neighbourhood of \({\mathcal S}\) and \(u_0,\ p_0,\ q_0\in C^1\) are given.

    We assume:

    1. \(u_0'(t)=p_0(t)x'(t)+q_0(t)y'(t)\) (strip condition),
    2. \({\mathcal S}\) is not a characteristic curve. Moreover assume that the characteristic curves, which are lines here and are defined by \(x=const.\) and \(y=const.\), have at most one point of intersection with \({\mathcal S}\), and such a point is not a touching point, i. e., tangents of the characteristic and \({\mathcal S}\) are different at this point.

    We recall that the characteristic equation to (\ref{riem1}) is \(\chi_x\chi_y=0\) which is satisfied if \(\chi_x(x,y)=0\) or \(\chi_y(x,y)=0\). One family of characteristics associated to these first partial differential of first order is defined by \(x'(t)=1,\ y'(t)=0\), see Chapter 2.

    Assume \(u,\ v\in C^1\) and that \(u_{xy},\ v_{xy}\) exist and are continuous. Define the adjoint differential expression by


    We have




    From (\ref{riem5}) it follows for a domain \(\Omega\in\mathbb{R}^2\)

    2\int_\Omega\ (vLu-uMv)\ dxdy&=&\int_\Omega\ (-P_y+Q_x)\ dxdy\nonumber\\
    &=&\oint\ Pdx+Qdy,

    where integration in the line integral is anticlockwise. The previous equation follows from Gauss theorem or after integration by parts:

    $$\int_\Omega\ (-P_y+Q_x)\ dxdy=\int_{\partial\Omega}\ (-Pn_2+Qn_1)\ ds,$$

    where \(n=(dy/ds,-dx/ds)\), \(s\) arc length, \((x(s),y(s))\) represents \(\partial\Omega\).

    Assume \(u\) is a solution of the initial value problem (\ref{riem1})-(\ref{riem4}) and suppose that \(v\) satisfies

    $$Mv=0\ \ \mbox{in}\ \Omega.$$

     Riemann's method, domain of integration

    Figure 4.4.1: Riemann's method, domain of integration

    Then, if we integrate over a domain \(\Omega\) as shown in Figure 4.4.1, it follows from (\ref{riem6}) that

    2\int_\Omega\ vf\ dxdy=\int_{BA}\ Pdx+Qdy+\int_{AP}\ Pdx+Qdy+\int_{PB}\ Pdx+Qdy.

    The line integral from \(B\) to \(A\) is known from initial data, see the definition of \(P\) and \(Q\).



    it follows

    \int_{AP} Pdx+Qdy&=&-\int_{AP} \left((uv)_x+2u(bv-v_x)\right)\ dx\\
    &=&-(uv)(P)+(uv)(A)-\int_{AP}\ 2u(bv-v_x)\ dx.

    By the same reasoning we obtain for the third line integral

    \int_{PB} Pdx+Qdy&=&\int_{PB} \left((uv)_y+2u(av-v_y)\right)\ dy\\
    &=&(uv)(B)-(uv)(P)+\int_{PB} 2u(av-v_y)\ dy.

    Combining these equations with (\ref{riem6}), we get

    2v(P)u(P)&=&\int_{BA} (u_xv-v_x+2buv)\ dx-(u_yv-v_yu+2auv)\ dy\nonumber\\
    &&+u(A)v(A)+u(B)v(B)+2\int_{AP}u(bv-v_x)\ dx\nonumber\\
    &&+2\int_{PB}u(av-v_y)\ dy-2\int_\Omega fv\ dxdy.

    Let \(v\) be a solution of the initial value problem, see Figure 4.2.2 for the definition of domain \(D(P)\),

    Definition of Riemann's function

    Figure 4.4.2: Definition of Riemann's function

    Mv&=&0\ \ \mbox{in}\ D(P)\\
    bv-v_x&=&0\ \ \mbox{on}\ C_1\\
    av-v_y&=&0\ \ \mbox{on}\ C_2\\

    Assume \(v\) satisfies (\ref{riem9})-(\ref{riem12}), then

    2u(P)&=&u(A)v(A)+u(B)v(B)-2\int_\Omega\ fv\ dxdy\\
    &&=\int_{BA}(u_xv-v_x+2buv)\ dx-(u_yv-v_yu+2auv)\ dy,

    where the right hand side is known from given data.

    A function \(v=v(x,y;x_0,y_0)\) satisfying (\ref{riem9})-(\ref{riem12}) is called Riemann's function.

    Remark. Set \(w(x,y)=v(x,y;x_0,y_0)\) for fixed \(x_0,\ y_0\). Then (\ref{riem9})-(\ref{riem12}) imply

    w(x,y_0)&=&\exp\left(\int_{x_0}^x\ b(\tau,y_0)\ d\tau\right)\ \ \mbox{on}\ C_1,\\
    w(x_0,y)&=&\exp\left(\int_{y_0}^y\ a(x_0,\tau)\ d\tau\right)\ \ \mbox{on}\ C_2.

    Example 4.4.1:

    \(u_{xy}=f(x,y)\), then a Riemann function is \(v(x,y)\equiv 1\).

    Example 4.4.2:

    Consider the telegraph equation of Chapter 3

    $$\varepsilon \mu u_{tt}=c^2\triangle_xu-\lambda\mu u_t,$$

    where \(u\) stands for one coordinate of electric or magnetic field.


    $$u=w(x,t)e^{\kappa t},$$

    where \(\kappa=-\lambda/(2\varepsilon)\), we arrive at


    Stretching the axis and transform the equation to the normal form we get finally the following equation, the new function is denoted by \(u\) and the new variables are denoted by \(x,y\) again,


    with a positive constant \(c\). We make the ansatz for a Riemann function

    $$v(x,y;x_0,y_0)=w(s),\ \ s=(x-x_0)(y-y_0)$$

    and obtain


    Substitution \(\sigma=\sqrt{4cs}\) leads to Bessel's differential equation

    $$\sigma^2 z''(\sigma)+\sigma z'(\sigma)+\sigma^2 z(\sigma)=0,$$

    where \(z(\sigma)=w(\sigma^2/(4c))\). A solution is


    which defines a Riemann function since \(J_0(0)=1\).

    Remark. Bessel's differential equation is


    where \(n\in\mathbb{R}^1\). If \(n\in{\mathbb N}\cup\{0\}\), then solutions are given by Bessel functions. One of the two linearly independent solutions is bounded at 0. This bounded solution is the Bessel function \(J_n(x)\) of first kind and of order \(n\), see [1], for example.