Hyperbolic Equations

The quadratic form, Eq.(6.12), determined by the coefficients $A$ , $B$ , and $C$ of the given p.d.e. can be factored into two ordinary differential equation

$$A\,dy = (B+\sqrt{B^2-AC})\,dx~\quad~\textrm{and}~\quad~A\,dy = (B- \sqrt{B^2-AC})\,dx\,.$$

These are the equations for the two families of characteristic curves of the given p.d.e.

Their significance, we recall, is this: if the boundary line coincides with one of them, then specifying Cauchy data on it will not yield a unique solution. If, however, the boundary line intersects each family only once, then the Cauchy data will yields a unique solution.

This point becomes particularly transparent if one introduces the curvilinear coordinates $\lambda$ and $\mu$ relative to which the given p.d.e. assumes its standard form, Eq.(6.13). We shall consider the hyperbolic case by assuming that

$$B^2(x,y)-A(x,y)C(x,y)>0$$

throughout the $(x,y)$ domain.

We shall demand the new coordinates $\lambda$ and $\mu$ - the characteristic coordinates - have the property that their isograms (``loci of points of constant values'') contain the characteristic lines $(x(s),y(s))$ , i.e.,

$$\lambda (x(s),y(s)) =\textrm{const}~\quad~\textrm{and}~\quad~\mu (x(s),y(s)) =\textrm{const}$$

for all $s$ . This implies that

$$\lambda_x\frac{dx}{ds}+\lambda_y\frac{dy}{ds}=0~\quad~\textrm{and}~\quad~ \mu_x\frac{dx}{ds}+\mu_y\frac{dy}{ds}=0$$

where, as usual

$$\lambda_x = \frac{\partial\lambda}{\partial x}\,,\textrm{etc.}$$

Substituting these equations into Eq.(6.12), the equation for the characteristic directions, one obtains

$$A\left(\frac{\partial\lambda}{\partial x}\right)^2+2B\frac{\parti... ...lambda}{\partial y} +C\left(\frac{\partial\lambda} {\partial y}\right)^2 = 0\,.$$ (614)

An equation with the same coefficients is obtained for the other function $\mu (x,y)$ . The two solutions $\lambda (x,y)$ and $\mu (x,y)$ are real valued functions. Their isograms, the characteristics of the hyperbolic equation, give us the new curvilinear coordinate system

$$\lambda =\lambda (x,y)~~\qquad~~\mu = \mu (x,y)\,.$$

The partial derivatives of the given differential equation are now as follows

$$\frac{\partial^2\psi}{\partial x^2} = \frac{\partial^2\psi}{\partial \lambda^2}(\lambda_x)^2+2\frac{\pa... ...partial\mu} \lambda_x\mu_x+\frac{\partial^2\psi}{\partial\mu^2}(\mu_x)^2+\cdots$$

$$\frac{\partial^2\psi}{\partial x\partial y} = \frac{\partial^2\psi} {\partial\lambda^2}\lambda_x\lambda_y+\frac... ...a_x\mu_y+\mu_x\lambda_y)+\frac{\partial^2\psi}{\partial \mu^2}\mu_x\mu_y+\cdots$$

$$\frac{\partial^2\psi}{\partial y^2} = \frac{\partial^2\psi}{\partial \lambda^2}\lambda^2_y+2\frac{\part... ...tial\mu} \lambda_y\mu_y +\frac{\partial^2\psi}{\partial\mu^2}\mu^2_y +\cdots\,.$$

Here $+\cdots$ refers to additional terms involving only the first partial derivatives of $\psi$ . Inserting these expressions into the given p.d. equation, one obtains

$$[A\lambda^2_x+2B\lambda_x\lambda_y+C\lambda^2_y]\frac{\partial^2\psi} {\partial\lambda^2} + [2A\lambda_x\mu_x+B(\lambda_x\mu_y+\mu_x\lambda_y)+ 2C\lambda_y\mu_y]\frac{\partial^2\psi}{\partial\lambda\partial\mu}$$

$$+ [A\mu^2_x+2B\mu_x\mu_y+C\mu^2_y]\frac{\partial^2\psi}{\partial\mu^2}$$

$$= \Phi '\left(\lambda ,\mu ,\psi ,\frac{\partial\psi}{\partial\lambda}\,,~ \frac{\partial\psi}{\partial\mu}\right)\,.$$ (615)

It follows from Equation 6.14 that the coefficients of $\psi_{\lambda\lambda}$ and $\psi_{\mu\mu}$ vanish. Solving for $\frac{\partial^2\psi}{\partial\lambda\partial\mu}$ yields Equation 6.13, the hyperbolic equation in normal form.

The coordinates $\lambda$ and $\mu$ , whose surfaces contain the characteristic lines, are called the characteristic coordinates or null coordinates of the hyperbolic equation.

These coordinates are important for at least two reasons. First of all, they are boundaries across which a solution can be nonanalytic. If $\lambda(x,y)=\lambda_0$ is one of the isograms (``locus of points where $\lambda$ has constant value'') of the solution to Eq.(6.14), then the first term of the p.d. Eq.(6.15)

$$[A\lambda^2_x+2B\lambda_x\lambda_y+C\lambda^2_y]\frac{\partial^2\psi} {\partial\lambda^2}=finite$$

even if $\frac{\partial^2\psi}{\partial\lambda^2}\rightarrow\infty$ as $\lambda\to \lambda_0$ . In other words, there are solutions to Eq.(6.15) for which the first derivative $\frac{\partial\psi}{\partial\lambda}$ has a discontinuity across the characteristic $\lambda(x,y)=\lambda_0$ . Similarly, there exist solutions to Eq.(6.15) whose first derivative $\frac{\partial\psi}{\partial\mu}$ has a discontinuity across $\mu(x,y)=\mu_0$ whenever $\mu (x,y)$ satisfies Eq.(6.14) with $\lambda$ replaced by $\mu$ .

Secondly, these coordinates depict the history of a moving disturbance. The simple string illustrates the issue involved.

Example: The Simple string The governing equation is

$$\frac{\partial^2\psi}{\partial z^2} - \frac{1}{c^2}~\frac{\partial^2\psi} {\partial t^2} = 0\,.$$

Its characteristic coordinates are the ``retarded'' and the ``advanced'' times

$$\lambda =ct-z~\quad~\textrm{and}~\quad~\mu =z+ct$$

and its normal form is

$$\frac{\partial^2\psi}{\partial\lambda\partial\mu} = 0\,.$$

The solution is

$$\psi = f(\lambda )+g(\mu )$$

where $f$ and $g$ are any functions of $\lambda$ and $\mu$ .

Next consider the initial value data at $t=0$ :

$$\psi_0(z) \equiv \psi (t=0,z) = f(-z)+g(z)~~\qquad~~\qquad~~~~~~\textrm{\lq\lq initial~amplitude''}$$

$$V_0(z) \equiv \left.\frac{\partial\psi (t,z)}{\partial t}\right\vert _{t=0} = \left.\frac{\partial \lambda}{\partial t}~\frac{\partial\psi}{\pa... ...al\psi}{\partial\mu} \right\vert _{\mu =z}~~\qquad\textrm{\lq\lq initial~velocity''}$$

$$= c f'(-z) +c g'(z)\,.$$

These equations imply

$$f(\lambda ) = \frac{1}{2}\psi_0 (-\lambda )+\frac{1}{2c}\int^{-\lambda}_0 V_0(z')dz'$$

$$g(\mu ) = \frac{1}{2} \psi_0(\mu )+\frac{1}{2c}\int^\mu_0 V_0(z') dz'\,.$$

Consider the intersection of the two families of characteristics with the boundary line $t=0$ as in the figure below.

Figure 6.1: Characteristic coordinate lines $\mu$ and $\lambda$ as determined by the wave equation for a simple string.

Note that $f$ is constant along the $\lambda$ characteristics (i.e., where $\lambda =$ constant), while $g$ is constant along the $\mu$ characteristics. It follows that if $f$ is known on the boundary segment $RS$ , then $f$ is known along all the $\lambda$ -characteristics intersecting $RS$ . Similarly, if $g$ is known along $RS$ , then $g$ is known along all the $\mu$ -characteristics intersecting $RS$ . And this is precisely the case because the Cauchy data on $RS$ determine the values of both $f$ and $g$ on that segment.

Being the sum of the two functions, the solution to the wave equation is

$$\psi (z,t) = f(ct-z)+g(ct+z)$$

$$= \frac{1}{2}\psi_0(z-ct)+\frac{1}{2}\psi_0(z+ct)+ \frac{1}{2c}\int_{z-ct}^{z+ct}V_0(z')dz'$$ (616)

Thus one sees that any disturbance on a string consists of two parts: one propagating to the right the other to the left. The propagation speeds are $\pm c$ , the slopes of the characteristics relative to the $t$ -$z$ coordinate system. The idiosyncratic aspect of the simple string is that these two parts do not change their shape as they propagate along the string.

A general linear hyperbolic system does not share this feature. However, what it does share with a simple string is that its solution is uniquely determined in the common region traversed by the two sets of characteristics which intersect $RS$ . In fact, the Cauchy data on $RS$ determine a unique solution $\psi(z,t)$ at every point in the region $PRQS$ . This is why it is called the domain of dependence of $RS$ . To justify these claims it is neccessary to construct this unique solution to a general second order linear hyperbolic differential equation.