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4.4: A Method of Riemann

( \newcommand{\kernel}{\mathrm{null}\,}\)

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 R2, that is, we assume x, yC1[t1,t2] and x2+y20. Set

$$Lu:=u_{xy}+a(x,y)u_x+b(x,y)u_y+c(x,y)u,\]

where a, bC1 and c, fC in a neighborhood of S. Consider the initial value problem

Lu=f(x,y)u0(t)=u(x(t),y(t))p0(t)=ux(x(t),y(t))q0(t)=uy(x(t),y(t)),

where fC in a neighbourhood of S and u0, p0, q0C1 are given.

We assume:

  1. u0(t)=p0(t)x(t)+q0(t)y(t) (strip condition),
  2. 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 S, and such a point is not a touching point, i. e., tangents of the characteristic and S are different at this point.

We recall that the characteristic equation to (4.4.1) is χxχy=0 which is satisfied if χx(x,y)=0 or χy(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, vC1 and that uxy, vxy exist and are continuous. Define the adjoint differential expression by

$$Mv=v_{xy}-(av)_x-(bv)_y+cv.\]

We have

2(vLuuMv)=(uxvvxu+2buv)y+(uyvvyu+2auv)x.

Set

P=(uxvxxu+2buv)Q=uyvvyu+2auv.

From (???) it follows for a domain ΩR2

2Ω (vLuuMv) dxdy=Ω (Py+Qx) dxdy= 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 Ω.

Assume u is a solution of the initial value problem (4.4.1)-(4.4.4) 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 Ω as shown in Figure 4.4.1, it follows from (4.4.8) that

2Ω vf dxdy=BA Pdx+Qdy+AP Pdx+Qdy+PB Pdx+Qdy.

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

Since

$$u_xv-v_xu+2buv=(uv)_x+2u(bv-v_x),\]

it follows

APPdx+Qdy=AP((uv)x+2u(bvvx)) dx=(uv)(P)+(uv)(A)AP 2u(bvvx) dx.

By the same reasoning we obtain for the third line integral

PBPdx+Qdy=PB((uv)y+2u(avvy)) dy=(uv)(B)(uv)(P)+PB2u(avvy) dy.

Combining these equations with (4.4.8), we get

2v(P)u(P)=BA(uxvvx+2buv) dx(uyvvyu+2auv) dy+u(A)v(A)+u(B)v(B)+2APu(bvvx) dx+2PBu(avvy) dy2Ωfv 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  in D(P)bvvx=0  on C1avvy=0  on C2v(P)=1.

Assume v satisfies (4.4.15)-(4.4.18), then

2u(P)=u(A)v(A)+u(B)v(B)2Ω fv dxdy=BA(uxvvx+2buv) dx(uyvvyu+2auv) dy,

where the right hand side is known from given data.

A function v=v(x,y;x0,y0) satisfying (4.4.15)-(4.4.18) is called Riemann's function.

Remark. Set w(x,y)=v(x,y;x0,y0) for fixed x0, y0. Then (4.4.15)-(4.4.18) imply

w(x,y0)=exp(xx0 b(τ,y0) dτ)  on C1,w(x0,y)=exp(yy0 a(x0,τ) dτ)  on C2.

Example 4.4.1:

uxy=f(x,y), then a Riemann function is v(x,y)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.

Introducing

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

where κ=λ/(2ε), we arrive at

$$w_{tt}=\frac{c^2}{\varepsilon\mu}\triangle_xw-\frac{\lambda^2}{4\epsilon^2}.\]

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,

$$u_{xy}+cu=0,\]

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

$$sw''+w'+cw=0.\]

Substitution σ=4cs leads to Bessel's differential equation

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

where z(σ)=w(σ2/(4c)). A solution is

$$J_0(\sigma)=J_0\left(\sqrt{4c(x-x_0)(y-y_0)}\right)\]

which defines a Riemann function since J0(0)=1.

Remark. Bessel's differential equation is

$$x^2y''(x)+xy'(x)+(x^2-n^2)y(x)=0,\]

where nR1. If nN{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 Jn(x) of first kind and of order n, see [1], for example.

Contributors and Attributions


This page titled 4.4: A Method of Riemann is shared under a not declared license and was authored, remixed, and/or curated by Erich Miersemann.

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