3. Apply Green's Identity

One obtains the solution $\psi(u,v)$ to the given hyperbolic equation by expressing it in terms of its Cauchy data and in terms of a simple solution to the adjoint differential equation, namely

$$L^*\phi=0~,$$

subject to the boundary conditions

$$\frac{\partial \phi}{\partial u}-E\phi = 0~~~~~\textrm{along}~SQ$$ (624)

$$\frac{\partial \phi}{\partial v}-D\phi = 0~~~~~\textrm{along}~RQ~,$$ (625)

and

$$\phi(u,v)=1~~~~~\textrm{at}~(u,v)=(u_Q,v_Q)~.$$ (626)

Riemann noticed that it is often easier to solve this adjoint boundary value problem, and once one has solved it, the solution to the given problem is immediate.

Apply the fact that $\psi$ and $\phi$ satisfy their respective differential equations. Green's identity, Eq.(6.23) becomes

$$0=\int_{SQ} P_udv-P_vdu+\int_{QR} P_udv-P_vdu+\int_{RS} P_udv-P_vdu~.$$ (627)

The left hand side vanishes because $L^*\phi=0$ and because the given differential Eq.(6.17) has no source. Consider the first integral. The integration segment $SQ$ consists of

$$v =const.\equiv v_Q$$

$$u_S< u <u_Q$$

Consequently, one is left with

$$\int_{SQ} P_udv-P_vdu \stackrel{1}{=} -\int_{u_S}^{u_Q}P_v\vert _{v_Q}du$$

$$\stackrel{2}{=} -\frac{1}{2}\int\left[ \phi\frac{\partial\psi}{\partial u}- \psi\frac{\partial\phi}{\partial u}+ 2E\phi\psi \right]_{v_Q}du$$

$$\stackrel{3}{=} -\frac{1}{2}\int\left[ \frac{\partial(\phi\psi)}{\partial u}+ 2\psi\left(E\phi-\frac{\partial\phi}{\partial u}\right) \right]_{v_Q}du$$

$$\stackrel{4}{=} -\frac{1}{2} \left[\phi\psi\vert _{u_S}^{u_Q}\right]_{v_Q}$$

$$\stackrel{5}{=} -\frac{1}{2}\psi(u_Q,v_Q)+ \frac{1}{2}\phi(u_S,v_S)\psi(u_S,v_S)$$

$$\stackrel{6}{\equiv} -\frac{1}{2}\psi(Q)+ \frac{1}{2}\phi(S)\psi(S)~.$$

(Equality 1 uses the fact that $\int_{SQ}P_udv=0$ because $v$ is constant; 2 uses Eq.(6.22); 3 adds and subtracts $\partial \phi/\partial u$ ; 4 uses Eq.(6.24); 5 uses Eq.(6.26). ) Equality 6 introduces the short hand notation

$$\psi(Q)=\psi(u_Q,v_Q),~\phi(S)=\phi(u_S,v_S),~\textrm{etc.}$$

Similarly the second integral reduces to

$$\int_{QR} P_udv-P_vdu = \int_{v_Q}^{v_R}P_u\vert _{u_Q}dv$$

$$= -\frac{1}{2}\psi(Q)+ \frac{1}{2}\phi(R)\psi(R)~.$$

Consequently, the vanishing of the closed line integral, Eq.(6.27), implies

$$\psi(Q)=\frac{1}{2}\phi(S)\psi(S)+\frac{1}{2}\phi(R)\psi(R)+ \int_{RS} P_udv-P_vdu~.$$ (628)

This is Riemann's representation of the solution $\psi$ to the hyperbolic differential equation in terms of the given initial value data on the curve segment $RS$ . A function such as $\psi(u,v)$ establishes a quantitative relationship between two sets of measurements:

1. The quantity $\psi$ which typically expresses a measured amplitude or voltage and
2. the coordinates $(u,v)$ which, for a hyperbolic system, indirectly express measurements of time and place, namely
   

   $$u = t-z$$

   $$v = t+z$$

   
   in terms of the familiar laboratory space and time coordinates.

Thus it is necessary to express the solution, Eq.(6.28), in terms of these coordinates. Using

$$du = dt-dz$$

$$dv = dt+dz$$

and

$$\frac{\partial \psi}{\partial v}+\frac{\partial \psi}{\partial u} = \frac{\partial \psi}{\partial t}$$

$$\frac{\partial \psi}{\partial v}-\frac{\partial \psi}{\partial u} = \frac{\partial \psi}{\partial z}$$

one finds that with the help of Eqs.(6.21)-(6.22) that the solution is

$$\psi(t_Q,z_Q)=\frac{1}{2}\phi(R)\psi(R)+\frac{1}{2}\phi(S)\psi(S)+$$ (629)

$$\int_R^S\left\{ \left[ \frac{1}{2}\left( \phi\frac{\partial\psi}{... ...}-\psi\frac{\partial\phi}{\partial t} \right) +(D+E)\phi\psi \right]dz \right\}$$

Example: String Imbedded in an Elastic Medium

Let us illustrate the integration method with a simple string imbedded in an elastic medium. The governing equation is the Klein-Gordon wave equation in 1+1 dimensions,

$$0=\frac{\partial^2\psi}{\partial t^2} - \frac{\partial^2\psi} {\partial z^2}+k^2\psi \,.$$

Its solution is to satisfy at $t=0$ the initial value conditions

$$\psi(t=0,z) = \psi_0(z)$$ (630)

$$\frac{\partial \psi}{\partial t}(t=0,z) = V_0(z)~.$$ (631)

Here $\psi_0(z)$ and $V_0(z)$ are the given initial value data (``Cauchy data'') associated with this initial value problem.

The equation for the characteristic coordinate functions is

$$\left(\frac{\partial S}{\partial t}\right)^2 -\left(\frac{\partial S}{\partial z}\right)^2=0$$

Being a quadratic, this equation has two distinct real solutions

$$S(x,y)=\left\{ \begin{array}{c} u(t,z)=t-z\\ v(t,z)=t+z \end{array}\right.$$

Its characteristic coordinate functions are

$$u=t-z~\quad~\textrm{and}~\quad~ v=z+t$$ (632)

and its normal form is

$$L(\psi)=\frac{\partial^2\psi}{\partial\lambda\partial\mu}+\frac{k^2}{4} \psi = 0\,.$$

Thus one has $D=E=0$ , which means that the hyperbolic operator is formally self-adjoint. Consequently, the adjoint differential equation is

$$L^*(\phi)=\frac{\partial^2\phi}{\partial v\partial u}+\frac{k^2}{4} \phi = 0\, .$$

The adjoint boundary conditions are

$$\frac{\partial \phi}{\partial u} = 0~~~~~\textrm{along}~SQ:~v=v_Q$$ (633)

$$\frac{\partial \phi}{\partial v} = 0~~~~~\textrm{along}~RQ:~u=u_Q\,,$$ (634)

and

$$\phi(u,v)=1~~~~~\textrm{at observation point Q:}~(u,v)=(u_Q,v_Q)~.$$ (635)

Remark. One can draw a very useful conclusion from Eqs.(6.33)-(6.35). The solution $\phi$ to the hyperbolic problem adjoint to the given one under consideration is constant along the two characteristics through point $Q$ :

$$\phi(u,v_Q) = 1$$

$$\phi(u_Q,v) = 1~.$$

Note that the two points $R$ and $S$ in Figure 6.2 lie on these charateristics. Consequently,

$$\phi(R)=\phi(S)=1~.$$

This simplifies the solution to the solution, Eq.(6.29), to the given problem considerably.

The solution to the adjoint boundary boundary value problem is achieved by recalling that the Bessel function of order zero, $J_0(x)$ , satisfies

$$\frac{d^2J_0}{dx^2}+\frac{1}{x}\frac{dJ_0}{dx}+J_0=0~.$$

Letting

$$x=k\sqrt{uv}~,$$

one finds

$$\frac{\partial J_0(k\sqrt{uv})}{\partial v} = \frac{1}{2}k\sqrt{\frac{u}{v}} J_0'(k\sqrt{uv})$$

$$\frac{\partial^2 J_0(k\sqrt{uv})}{\partial u\,\partial v} = \frac{1}{4}k\frac{1}{\sqrt{vu}}J_0'(k\sqrt{uv})+ \frac{k^2}{4}J_0{''}(k\sqrt{uv})$$

$$= -\frac{k^2}{4}J_0(k\sqrt{uv})~.$$

Consequently,

$$\left[ \frac{\partial^2}{\partial v\,\partial u}+\frac{k^2}{4} \right]J_0\left( k\sqrt{(u_Q-u)(v_Q-v)}\right) = 0$$

Furthermore, note that

$$\phi(u,v)=J_0\left( k\sqrt{(u_Q-u)(v_Q-v)}\right)$$

satisfies the three required boundary conditions

$$\left. \frac{\partial \phi(u,v)}{\partial u}\right\vert _{v=v_Q} =$$ 0

$$\left. \frac{\partial \phi(u,v)}{\partial v}\right\vert _{u=u_Q} =$$ 0

$$\phi(u_Q,v_Q) = 1~,$$

and also

$$\phi(u,v_Q)=\phi(u_Q,v)=1~,$$

as required.

The solution, Eq.(6.29), to the given problem is determined by the initial value data, Eqs.(6.30)-(6.31) at $t=0$ . Substituting this data into the expression for this solution, taking note of the fact

$$R:~(t,z) = (0,z_Q-t_Q)$$

$$S:~(t,z) = (0,z_Q+t_Q)~,$$

introducing (with the help of Eq.(6.32))

$$\phi=J_0\left( k\sqrt{(u_Q-u)(v_Q-v)}\right)= J_0\left( k\sqrt{(t-t_Q)^2-(z-z_Q)^2}\right)$$

into the integrand, and setting $t=0$ , one finds that the solution is

$$\psi(t_Q,z_Q) = \frac{1}{2}\psi_0(z_Q-t_Q)+\frac{1}{2}\psi_0(z_Q+t_Q)+$$

$$\frac{1}{2}\int_{z_Q-t_Q}^{z_Q+t_Q} \left[ J_0\left( k\sqrt{(t-t_Q)^2-(z-z_Q)^2}\right)V_0(z)-\right.$$

$$~~~~~~~~~~~~~~~~\left.\psi_0(z)\frac{\partial }{\partial t} J_0\left( k\sqrt{(t-t_Q)^2-(z-z_Q)^2}\right) \right]_{t=0}dz$$

or in terms of standard variables,

$$\psi(t,z) = \frac{1}{2}\psi_0(z-t)+\frac{1}{2}\psi_0(z+t)+$$ (636)

$$\frac{1}{2}\int_{z-t}^{z+t} \left[ J_0\left( k\sqrt{(t'-t)^2-(z'-z)^2}\right)V_0(z')-\right.$$

$$~~~~~~~~~~~~~~~~\left.\psi_0(z')\frac{\partial }{\partial t'} J_0\left( k\sqrt{(t'-t)^2-(z'-z)^2}\right) \right]_{t'=0}dz'~.$$

Compare this result with Eq.(6.16) and observe the influence of the elastic medium on the propagation of a disturbance along the string:

In the absence of that medium an initial pulse separates into two pulses also highly localized in the same way. They move into opposite directions, but they don't change their shapes and amplitudes. The region between these pulses is a widening gap having zero amplitude.

However, the presence of an elastic medium ($k^2\ne 0$ ) changes all this. An initial pulse also separates into two pulses, but each one leaves a nonzero trailing wake which fills the widening gap between them with a space-time dependent amplitude. It decreases with time in a manner dictated by the behaviour of the Bessel function in the integrand of Eq.(6.36).

Equations (6.16) and (6.36) are in perfect harmony. Indeed, the first is the $k^2=0$ limit of the second. This is as it must be. It is a mathematical consequence of the fact that $J_0(0)=1$ and that $J_0'(0)=0$ in the integrand of Eq.(6.36).

Lecture 51⁶³

Footnotes

... 51⁶³

Presentation given 10/3/2006 at the OSU Electro Science Lab.