User:Egm6322.s09.bit.gk/r7

$$\frac{\mathrm{d} y}{\mathrm{d} x}=-k_{\pm }$$

Similarly,assume

$$\psi_x=k\psi_y$$

$$ak^2+bk+c=0$$

From $$\phi_x=k\phi_y$$  and  $$\psi_x=k\psi_y$$

We have

$$k=\frac{\phi_x}{\phi_y}$$

$$k=\frac{\psi_x}{\psi_y}$$

1-D Wave equation,we have ,

$$\hat{c}^2u_{xx}-u_{tt}=0$$ by looking at the units we have

$$\left [\hat{c}u_{xx} \right ]=\left [u_{tt}  \right ]=\left [\frac{L}{T^2}  \right ]$$

Therefore,

$$a=c^2,b=0,c=-1$$

$$\Rightarrow \xi =x-\hat{c}t$$ and $$\eta =\xi =x+\hat{c}t$$

By Scaling Method

$$\hat{c}^2u_{xx}-u_{tt}=0$$ $$\Rightarrow u_{xx}-u_{tt}=0$$

Coordinate Transformation

$$\begin{Bmatrix} \bar x\\ \bar y=t

\end{Bmatrix}\to \begin{Bmatrix} \xi \\ \eta

\end{Bmatrix}$$

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

The general form of a linear second order PDE in the matrix form is give as

$$\left[{\begin{matrix} \frac{\partial }{\partial x} &  \frac{\partial }{\partial x}  \\ \end{matrix}}\right]\left[{\begin{matrix} A& B \\ B & C \\ \end{matrix}}\right]\left[{\begin{matrix}

\frac{\partial u }{\partial x}   \\

\frac{\partial u}{\partial y}    \\ \end{matrix}}\right] +\left[{\begin{matrix} D &  E  \\ \end{matrix}}\right]\left[{\begin{matrix}

\frac{\partial u }{\partial x}   \\

\frac{\partial u}{\partial y}    \\ \end{matrix}}\right]+F(x,y)u+G(x,y)=0 $$

and for a second order non linear PDE the general equation in two independent variables is given as

$$F\left ( u,x,y,\frac{\partial u}{\partial x} ,\frac{\partial u}{\partial y},\frac{\partial^2 u}{\partial x^2},\frac{\partial^2 u}{\partial x\partial y},\frac{\partial^2 u}{\partial y^2}\right )=0$$

Let, $$p=\frac{\partial^2 u}{\partial x^2}$$; $$q=\frac{\partial^2 u}{\partial x\partial y}$$; $$r=\frac{\partial^2 u}{\partial y^2}$$

and denote

$$a=\frac{\partial F}{\partial p}$$

$$b=\frac{1}{2}\frac{\partial F}{\partial q}$$

$$c=\frac{\partial F}{\partial r}$$

Let the specific solution $$u=u(x,y)$$ be considered at any point (x,y)and the values of a,b,c be calculated and then let the determinant $$\delta $$ be calculated using $$ac-b^2$$

if

$$\delta < 0$$ then hyperbolic

$$\delta > 0$$ then elliptic

$$\delta = 0$$ then parabolic

In a nonlinear PDE the coefficients are not only dependent on the point$$(x,y)$$but also on the selection of the solution.Therefore it is not possible to know the nature of the eqution without knowing the specific solution.

Reference from Andreĭ Dmitrievich Poli͡a︡nin, Aleksandr Vladimirovich Manzhirov, 2006 p.653,

Given that

$$U_{\xi\eta}=0$$

and the solution is obtained by

$$U\left ( \xi,\eta \right )=F\left ( \xi \right )+G\left ( \eta \right )$$

To verify this ,we have to differentiate twice,Therefore

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