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mathmari

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MHB

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**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)

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