User:Egm6322.s09.bit.sahin/r7

We know that $$\lambda^2-\lambda(a+c)+(ac-b^2)=0$$

and

$$\left (\lambda-\lambda_1  \right )\left (\lambda-\lambda_2   \right )=\lambda^{2}-\left ( \lambda_{1}+ \lambda_{2}\right )\lambda+\lambda_{1}\lambda_{2}=0$$

therefore

$$\lambda_{1}\lambda_{2}=ac-b^{2}$$

So,

Case#1: $$\lambda_1 \lambda_2 < 0$$ (hyperbola) or $$\lambda_1 \lambda_2 > 0$$ (ellipses)

Case#2: $$\lambda_1 \lambda_2 = 0$$(parabolas)

Section 2
$$\mathbf{J}=\begin{bmatrix} \phi _{x} &\phi _{y} \\ \psi _{x}&\psi_{y} \end{bmatrix}$$, $$\mathbf{J^{T}}=\begin{bmatrix} \phi _{x} &\psi _{x} \\ \phi _{y}&\psi_{y} \end{bmatrix}$$ and $$\mathbf{A}=\begin{bmatrix} a &b \\ b &c \end{bmatrix}$$

$$\mathbf{\bar{A}}=\mathbf{JAJ^T}=\begin{bmatrix} \bar{a} &\bar{b} \\ \bar{b} &\bar{c} \end{bmatrix} $$

Substituting these matrices into second equation

$$\mathbf{\bar{A}}=\begin{bmatrix} \phi _{x} &\phi _{y} \\ \psi _{x}&\psi_{y} \end{bmatrix} \begin{bmatrix} a &b \\ b &c \end{bmatrix} \begin{bmatrix} \phi _{x} &\psi _{x} \\ \phi _{y}&\psi_{y} \end{bmatrix}$$

$$\mathbf{\bar{A}}=\begin{bmatrix} a\left (\phi _{x} \right )^{2}+2b\phi _{x}\phi _{y}+c\left (\phi _{y}  \right )^{2} & a\phi _{x}\psi _{x}+b\phi _{x}\psi _{y}+b\psi _{x}\phi _{y}+c\phi _{y}\psi _{y} \\ a\phi _{x}\psi _{x}+b\phi _{x}\psi _{y}+b\psi _{x}\phi _{y}+c\phi _{y}\psi _{y} & a\left ( \psi _{x} \right )^{2}+2b\psi _{x}\psi _{y}+c\left ( \psi _{y} \right )^{2} \end{bmatrix}$$

Therefore,

$$\bar{a}=a\left (\phi _{x} \right )^{2}+2b\phi _{x}\phi _{y}+c\left (\phi _{y}  \right )^{2}$$

$$\bar{b}=a\phi _{x}\psi _{x}+b\phi _{x}\psi _{y}+b\psi _{x}\phi _{y}+c\phi _{y}\psi _{y}$$

$$\bar{c}=a\left ( \psi _{x} \right )^{2}+2b\psi _{x}\psi _{y}+c\left ( \psi _{y} \right )^{2}$$

Section 3
 Hyperbolic PDEs 

Method of characteristics applied in two different places

1) Finding canonical forms (Form 1 by hyperbolic PDEs) by setting

$$\bar{a}=\bar{c}=0$$

2) Solving Form 1 of hyperbolic PDEs

Setting $$\bar{a}=\bar{c}=0$$

Remembering that

$$\bar{a}=a\left (\phi _{x} \right )^{2}+b\phi _{x}\phi _{y}+c\left (\phi _{y}  \right )^{2}$$

and assuming that $$\phi _{x}$$ and $$\phi _{y}$$ are proportional

$$\phi _{x}=k\phi _{y}$$

substituting second equation to the first one, we have

$$\bar{a}=\left ( \phi _{y} \right )^{2}\left ( ak^{2}+bk+c \right )=0$$

the solution of $$\left ( ak^{2}+bk+c \right )=0$$ is

$$k=\frac{-b\pm \sqrt{\Delta }}{2a}$$, $$\Delta =b^{2}-4ac$$

so

$$k_{+}:=\frac{-b+ \sqrt{\Delta }}{2a}$$ and $$k_{-}:=\frac{-b- \sqrt{\Delta }}{2a}$$

it can be written that

$$d\phi =\phi _{x}dx+\phi _{y}dy$$ (2)

from the proportionality $$\phi _{x}=k\phi _{y}$$

$$0=\phi _{x}-k\phi _{y}$$ (3)

comparing Eq(2) and Eq(3) gives

$$\frac{d\phi }{0}=\frac{dx}{1}=\frac{dy}{-k}$$