Cauchy Problem and Characteristics

In order to compute the function $\psi (x,y)$ at points off the boundary curve, we resort to the Taylor series on two dimensions;

$$\psi (x,y) = \psi (\xi ,\eta )+(x-\xi )\frac{\partial\psi}{\partial x} +(y-\eta )\frac{\partial\psi}{\partial y}$$

$$+\frac{1}{2!}\left[ (x-\xi )^2\frac{\partial^2\psi}{\partial x^2}... ...l x\partial y} +(y-\xi )^2 \frac{\partial^2\psi}{\partial y^2}\right]+\cdots\,.$$

Here the derivatives are to be evaluated on the boundary.

The problem we are confronted with is this:

Determine all partial derivatives, starting with the first partials on up from the given Cauchy boundary conditions, the given boundary, and the given partial differential equation!

We shall do this first for the first derivatives.

From the Cauchy data we obtain two equations

$$\left. \begin{array}{ccl} \displaystyle \frac{d\psi (s)}{dn} &= &... ...\frac{d\psi}{dy} \end{array} \right\}~~\textrm{at}~~(x,y)=(\xi(s) ,\eta(s) )\,.$$ (610)

From these we obtain the first partial derivatives of $\psi$ evaluates on the boundary

$$\begin{array}{rcr} \displaystyle \left(\frac{\partial\psi}{\p... ...rac{d\eta}{ds} +\frac{d\psi}{ds}~\frac{d\xi}{ds}\,. \end{array}$$ (611)

The procurement of the second derivatives is more interesting. We differentiate the (known) first derivatives along the boundary. Together with the given p.d.e. we have

$$\frac{d}{ds}\left(\frac{\partial\psi}{\partial x}\right) = \frac{d\xi}{ds} ~\frac{\partial^2\psi}{\partial x^2} + \frac{d\eta}{ds}~\frac{\partial^2\psi} {\partial y\partial x}$$

$$\frac{d}{ds}\left(\frac{\partial\psi}{\partial y}\right) = ~~~~~~~~~~~~~\frac{d\xi} {ds}~\frac{\partial^2\psi}{\partial x\partial y} +\frac{d\eta}{ds}~\frac {\partial^2\psi}{\partial y^2}$$

$$\Phi = A\frac{\partial^2\psi}{\partial x^2}~~ + 2B\frac{\partial^2\psi} {\partial x\partial y} +C\frac{\partial^2\psi}{\partial y^2}\,.$$

The left hand sides of these three equations are known along the whole boundary. So are the coefficients of the three unknown partial derivatives on the right hand side. One can solve for these partial derivatives unless

$$\left\vert\begin{array}{ccc} \displaystyle \frac{d\xi}{ds} &\disp... ...d\xi}{ds} &\displaystyle\frac{d\eta}{ds}\\ A &2B &C\end{array}\right\vert = 0$$

or

$$A\left(\frac{d\eta}{ds}\right)^2 - 2B\frac{d\eta}{ds}~\frac{d\xi}{ds} + C \left(\frac{d\xi}{ds}\right)^2 =0\,.$$

If this determinant does not vanish, one can solve for the second derivatives evaluated on the boundary. Differentiating along the boundary yields

$$\frac{d}{ds}\psi_{xx} = \frac{d\xi}{ds}\psi_{xxx} + \frac{d\eta}{ds} \psi_{yxx}$$

$$\frac{d}{ds}\psi_{xy} = ~~~~~~~~~~~~~\frac{d\xi}{ds}\psi_{xxy} + \frac{d\eta}{ds} \psi_{yxy}$$

$$\Phi_x +\cdots = A\psi_{xxx}~~ +2B\psi_{xyx} +C\psi_{xyy}\,.$$

Subscripts refer to partial derivatives. The last equation was obtained differentiating the given p.d.e. with respect to $x$ . The left hand side contains only lower order derivatives, which are known on the boundary.

We see that one can solve for

$$\psi_{xxx}\,,~\psi_{yxx}\,,~\psi_{xyy}$$

on the boundary unless the determinant, the same one as before, vanishes. It is evident that one can continue the process of solving for the other higher order derivatives, provided the determinant of the system does not vanish. We are led to the conclusion that one can expand $\psi (x,y)$ in a Taylor series at every point of the boundary and that the coefficients of the series are uniquely determined by the Cauchy boundary conditions on the given boundary.

We must now examine the vanishing of the system determinant

$$A(x,y)\left(\frac{dy}{ds}\right)^2-2B(x,y)\frac{dy}{ds}~\frac{dx}{ds} +C(x,y) \left(\frac{dx}{ds}\right)^2=0$$ (612)

at every point of the domain of the partial differential equation.

Depending on the coefficients $A$ , $B$ , and $C$ , this quadratic form determines two characteristic curves, $\lambda(x,y)=const.$ and $\mu(x,y)=const.$ , through each point $(x,y)$ . We distinguish between three cases:

1. $AC-B^2>0$ : elliptic type in which the two characteristics $\lambda$ and $\mu$ are complex conjugates of each other.
2. $AC-B^2<0$ : hyperbolic type in which case for each $(x,y)$ the characteristics $\lambda$ and $\mu$ are real. They sre two curves intersecting at $(x,y)$ . As one varies $(x,y)$ one obtains two distinct families.
3. $AC-B^2=0$ : parabolic type in which there is only one family of characteristics.

These three cases imply three different types of differential equations. By utilizing the characteristic, one can introduce new coordinates relative to which a differential equation of each type assumes a standard normal form. Let the new coordinate surfaces be

$$\lambda (x,y)=\textrm{const}~~\qquad~~\mu (x,y)=\textrm{const}\,.$$

Then the coordinate transformation

$$u+iv=\lambda~\quad~\textrm{and}~\quad~u-iv=\mu$$

yields a normal form of the elliptic type,

$$\frac{\partial^2\psi}{\partial u^2} +\frac{\partial^2\psi}{\parti... ...,\frac{\partial\psi}{\partial u}\,,~\frac{\partial\psi} {\partial v}\right)\,.$$

By contrast the coordinate transformation

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

yields a normal form of the hyperbolic type,

$$\frac{\partial^2\psi}{\partial\lambda\partial\mu} = \Phi\left(\lambda ,\mu ,\psi ,\frac{\partial\psi}{\partial\lambda}\,,~\frac{\partial\psi}{\partial \mu}\right)\,.$$ (613)

Finally, the coordinate transformation

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

yields a normal form of the parabolic type,

$$\frac{\partial^2\psi}{\partial\lambda^2} = \Phi\left( x,\lambda ,... ...c{\partial\psi}{\partial x}\,,~\frac{\partial\psi}{\partial\lambda} \right)\,.$$

We recognize that elliptic partial differential equations express an equilibrium or a static potential phenomenon.

By introducing the standard coordinates

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

in terms of which

$$\lambda = \frac{1}{2}(t+z)~\quad~\textrm{and}~\quad~\mu =\frac{1}{2}(t-z)\,,$$

one finds that

$$\frac{\partial^2\psi}{\partial t^2} -\frac{\partial^2\psi}{\parti... ...,\frac{\partial\psi}{\partial t}\,,~\frac{\partial\psi} {\partial z}\right)\,,$$

the wave equation of a general vibrating string. We, therefore, recognize that a hyperbolic p.d. equation expresses the phenomenon of a propagating wave or disturbance.

Finally, a parabolic p.d. equation expresses a diffusion process. In fact, the two dimensional Laplace equation, the equation for a vibrating sting, and the heat conduction equation are the simplest possible examples of elliptic, hyperoblic, and parabolic equations.