Canonical form of hyperbolic equations

Theorem 4   Suppose that the equation

$$au_{xx} + b u_{xy} + c u_{yy} + d u_x + e u_y + fu = g.$$ (2.6)

is hyperbolic in a domain $D.$ Then there exists a coordinate system $(\xi, \eta)$ in which this equation has the canonical form

$$w_{\xi \eta} + l_{1}[w] = G(\xi, \eta),$$

where $w(\xi, \eta) = u(x(\xi, \eta), y(\xi, \eta)),$ $l_{1}$ is a first order linear differential operator, and $G$ is a function which depends on given equation.

Proof Under a generic change of coordinate system $(x,y) \rightarrow (\xi, \eta),$ the given PDE () changes to the PDE

$$Aw_{\xi \xi} + B w_{\xi \eta} + C w_{\eta \eta} + D w_{\xi} + E w_{\eta} + Fw = G.$$ (2.7)

where the new coefficients are given by ().

Since the discriminant $B^2 - 4AC > 0$ for hyperbolic equations, we set

$$A(\xi, \eta) = a \xi_{x}^{2} + b \xi_{x} \xi_{y} + c \xi_{y}^{2} =0$$ (2.8)

$$C(\xi, \eta) = a \eta_{x}^{2} + b \eta_{x} \eta_{y} + c \eta_{y}^{2} = 0$$ (2.9)

Note that the second equation for $\eta$ is actually same as the first equation for $\xi.$ Therefore we need to solve only one equation. Eq. () is in quadratic form and can be expressed as product of two linear terms as

$$\frac{1}{4a} \left [ 2a \xi_{x} - (-b + \sqrt{b^2 - 4ac}) \xi_{y} \right ] \left [ 2a \xi_{x} - (-b - \sqrt{b^2 - 4ac}) \xi_{y}\right ] = 0.$$

If $a \neq 0,$ we need to solve

$$2a \xi_{x} - (-b + \sqrt{b^2 - 4ac}) \xi_{y} =$$ 0 (2.10)

$$2a \xi_{x} - (-b - \sqrt{b^2 - 4ac}) \xi_{y} =$$ 0 (2.11)

To get a required non-singular transformation $(\xi(x,y), \eta(x, y)),$ we choose $\xi$ to be a solution of (), and $\eta$ to be a solution of ().

The auxiliary equation for () is :

$$\displaystyle \frac{dx}{2a} = \frac{dy}{ - (-b + \sqrt{b^2 - 4ac})} = \frac{d \xi}{0}$$

$\Rightarrow \xi =$   constant and

$$\frac{dy}{dx} = - \frac{ (-b + \sqrt{b^2 - 4ac})}{2a}.$$ (2.12)

Similarly the auxiliary equation for () is:

$$\displaystyle \frac{dx}{2a} = \frac{dy}{ - (-b - \sqrt{b^2 - 4ac})} = \frac{d \eta}{0}$$

$\Rightarrow \eta =$   constant and

$$\frac{dy}{dx} = - \frac{ (-b - \sqrt{b^2 - 4ac})}{2a}$$ (2.13)

Integrating () and (), we get the equations of characteristics $\xi(x,y)=c_1$ and $\eta(x,y) = c_2,$ which are called as the characteristics of the given hyperbolic PDE (). Now to obtain the canonical form of the given PDE (), carry out the change of coordinates with $\xi$ and $\eta$, which reduces () to

$$\hfill~w_{\xi \eta} = \phi(\xi, \eta, w, w_{\xi}, w_{\eta}).$$

Note For hyperbolic PDEs, the discriminant $b^2 -4ac > 0,$ and we have two independent family of real characteristics curves.

Example 2.4.1   Show that the equation

$$3 u_{xx} + 10 u_{xy} + 3 u_{yy} = 0,$$

is hyperbolic, and reduce it to its canonical form.

Solution Comparing the given PDE with the canonical form

$$au_{xx}+bu_{xy}+cu_{yy}++ d u_x + e u_y + fu -g=0.$$

One gets

$$a=3,~b=10~$$and$$~c=3.$$

The discriminant $\Delta=b^2-4ac>0.$ Therefore, the given equation is hyperbolic type. The corresponding characteristics are

$$\frac{dy}{dx}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{1}{3},$$

and

$$\frac{dy}{dx}=\frac{-b-\sqrt{b^2-4ac}}{2a}=3,~$$and

which implies that

$$y=\frac{x}{3},~$$and$$~y=3x+c.$$

Therefore,

$$\xi(x,y)=y-\frac{x}{3}~$$and$$~\eta(x,y)=y-3x.$$

Change of coordinates

$$u_x =-\frac{1}{3} u_{\xi}-3u_{\eta}~ ~u_y=u_{\xi}+u_{\eta},$$

$$u_{xx} =\frac{u_{\xi \xi}}{9}+9u_{\eta \eta}+u_{\xi \eta}+u_{\eta \xi},$$

$$u_{xy} =-\frac{u_{\xi \xi}}{3}-3u_{\eta \eta}-\frac{1}{3}u_{\xi \eta}-3u_{\eta \xi},$$

$$u_{yy} =u_{\xi \xi}+u_{\eta \eta}+u_{\xi \eta}+u_{\eta \xi},$$

$$3u_{xx}+10u_{xy}+3u_y=0u_{\xi \xi}+0u_{\eta \eta}+\frac{8}{3}u_{\xi \eta}-24u_{\eta \xi}.$$

Using $uu_{\eta \xi}=u_{\xi \eta}$

$$3u_{xx}+10u_{xy}+3u_y=-\frac{64}{3}u_{\xi \eta}.$$

Therefore, the canonical form is

$$u_{\xi \eta}=0.$$