# Canonical form of the hyperbolic equation

• MHB
• mathmari
In summary, the hyperbolic equation has the normal form of $A u_{\xi \xi}+ 2B u_{\xi \eta}+C u_{\eta \eta}=D$ and at the hyperbolic case, we can set $A=C=0$ by finding characteristics equations for which $\Phi (x,y)= \text{ constant }$. This ensures that the equation satisfies the condition for a hyperbolic equation and is a necessary step in solving the equation.
mathmari
Gold Member
MHB
Normal form of the hyperbolic equation

Hey!

I am looking at the following in my notes:

$$a(x,y) u_{xx}+2 b(x,y) u_{xy}+c(x,y) u_{yy}=d(x,y,u,u_x,u_y)$$

$$A u_{\xi \xi}+ 2B u_{\xi \eta}+C u_{\eta \eta}=D$$

$$A=a \xi_x^2+2b \xi_y \xi_x+c \xi_y^2 \ \ \ (*)$$
$$B=a \xi_x \eta_x +b \eta_x \xi_y+b\eta_y \xi_x+c \xi_y \eta_y$$
$$C=a \eta_x^2 +2b \eta_x \eta_y+c \eta_y^2 \ \ \ (**)$$

Hyperbolic case:
We will show that at the hyperbolic case we can set $A=C=0$.
The equations $(*)$ and $(**)$ have the general form:
$$a \Phi_x^2+2b \Phi_x \Phi_y+c \Phi_y^2=0$$
$$\Phi=\{ \xi, \eta \}$$

$$a (\frac{\Phi_x}{\Phi_y})^2+2b (\frac{\Phi_x}{\Phi_y})+c=0$$

$$\frac{\Phi_x}{\Phi_y}=\frac{1}{a} ( -b \pm \sqrt{b^2-ac})$$

We are looking for characteristics equations for which $\Phi (x,y)= \text{ constant }$.

$$\frac{\partial{\Phi}}{\partial{x}} dx+\frac{\partial{\Phi}}{\Phi{y}} dy=0$$

$$\frac{dy}{dx}=- \frac{\Phi_x}{\Phi_y}$$

So
$$\frac{dy}{dx}=\frac{1}{a}(b \pm \sqrt{b^2-ac})$$

The canonical form of the hyperbolic equation is:
$$u_{\xi \eta}=D'(x,y,u,u_x,u_y)$$Why are we looking for characteristics equations for which $\Phi (x,y)= \text{ constant }$?? (Wondering)

Last edited by a moderator:
We are looking for characteristics equations for which $\Phi (x,y)= \text{ constant }$ because this will ensure that the equation is hyperbolic. This is because the equation $u_{\xi \eta}=D'(x,y,u,u_x,u_y)$ must be satisfied for a hyperbolic equation to hold. If the equation is satisfied then we know that the equation is hyperbolic.

## 1. What is the canonical form of a hyperbolic equation?

The canonical form of a hyperbolic equation is a standard form that is used to represent all hyperbolic equations. It is written as uxx - c2utt = 0 where c is a constant.

## 2. How is the canonical form of a hyperbolic equation derived?

The canonical form of a hyperbolic equation is derived by separating the highest order derivatives of the dependent variable and setting them equal to zero. This results in a form that is independent of the specific equation and can be used to solve a variety of hyperbolic equations.

## 3. What are the characteristics of a hyperbolic equation in canonical form?

A hyperbolic equation in canonical form has two independent variables, x and t, and two constants, c and a. It also has two distinct families of characteristic curves, one in the x-t plane and one in the x-t plane, which are used to determine the solution.

## 4. How is the solution to a hyperbolic equation in canonical form determined?

The solution to a hyperbolic equation in canonical form can be determined by using the method of characteristics. This involves finding the characteristic curves and using them to determine the solution at any given point in the domain.

## 5. What are some real-life applications of hyperbolic equations in canonical form?

Hyperbolic equations in canonical form are commonly used in the fields of physics and engineering to model wave propagation, such as in acoustics, electromagnetics, and fluid dynamics. They are also used in economics to model supply and demand in market systems.

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