User:Egm6322.s09.bit.la.hw7

=Summary : Differences Between conics and PDE's=

Transformation of Coordinates

$$\begin{Bmatrix}\bar{x}\\ \bar{y}\end{Bmatrix} = \mathbf{J} \begin{Bmatrix}x\\ y\end{Bmatrix},\begin{Bmatrix}x\\ y\end{Bmatrix}=J^{-1} \begin{Bmatrix}\bar{x}\\ \bar{y}\end{Bmatrix}$$

Conics : $$\mathbf{\bar{A}}=\mathbf{J}^{-T}\mathbf{A}\mathbf{J}^{-1}$$

PDE's : $$\mathbf{\bar{A}}=\mathbf{J}\mathbf{A}\mathbf{J}^{T}$$

Invariance:

For Conics : $$det\mathbf{\bar{A}}=(det\mathbf{A})(\frac{1}{det\mathbf{J}})^2$$

For PDE's: $$det\mathbf{\bar{A}}=(det\mathbf{A})(det\mathbf{J})^2$$

Conics - $$\mathbf{A}=V\mathbf{\Lambda}V^T $$ Eigenvalue problem : Rotation

And the transformation of coordinates in this case is:

$$\begin{Bmatrix}\bar{x}\\ \bar{y}\end{Bmatrix} = \mathbf{V}^T$$

where:

$$=\mathbf{J_\alpha }\mathbf{V}^{-1}=\mathbf{V}^T$$

$$\mathbf{\bar{A}}=\mathbf{\Lambda }=\mathbf{V^T}A\mathbf{V}$$

where:

$$\mathbf{V}=\mathbf{J_\alpha }^T=\mathbf{J_\alpha }^{-1}$$

And Translating $$\mathbf{J_\alpha }^{-T}=\mathbf{J_\alpha }$$

$$\bar{\bar{x}}=\bar{x}-r$$

$$\bar{\bar{y}}=\bar{y}-s$$

After that Scaling:

$$\xi =\bar{\bar{x}}/dx$$

$$\eta=\bar{\bar{y}}/dy$$

where:

$$dx=\sqrt{\left |\lambda _1 \right |}$$

$$dy=\sqrt{\left |\lambda _2 \right |}$$

Applying the Form 1 - Assuming hyperbola

$$\xi^2-\eta ^2=\pm 1$$

First Step:

$$\begin{Bmatrix}\bar{\xi} \\ \bar{\eta}\end{Bmatrix}=\mathbf{J}_\gamma \begin{Bmatrix}\xi \\ \eta\end{Bmatrix}$$

$$\begin{Bmatrix}\xi \\ \eta\end{Bmatrix}=\mathbf{J}_\gamma^{-1} \begin{Bmatrix}\bar{\xi} \\ \bar{\eta}\end{Bmatrix}$$

Second Step:

$$\mathbf{\bar{B}}=\mathbf{J}_\gamma^{-T}\mathbf{B}\mathbf{J}_\gamma^{-1}$$

where: $$\mathbf{B}=\begin{bmatrix} 1 &0 \\ 0 &-1 \end{bmatrix}$$

This is a rotation in hyperbola cases

Going to Form 2 (Canonical Form):

$$\pm \bar{\xi} \bar{\eta}=1$$

Scaling again as the same processes described above:

$$\begin{Bmatrix}\xi \\ \eta\end{Bmatrix}=\begin{bmatrix}1/dx &0 \\ 0 &1/dy \end{bmatrix}\begin{Bmatrix}\bar{\bar{x}} \\ \bar{\bar{y}}\end{Bmatrix}$$

For the 2nd order linear PDE's:

General Transformation of Coordinates:

$$\mathbf{\bar{A}}=\mathbf{J}_\alpha\mathbf{A}\mathbf{J}_\alpha^{T}$$

$$\begin{Bmatrix}\partial x \\ \partial y\end{Bmatrix}=\mathbf{J}_\alpha^{T} \begin{Bmatrix}\bar{\partial x} \\ \bar{\partial y}\end{Bmatrix}$$

Particularize to Eigenvalue problem:

$$\mathbf{\bar{A}}=\mathbf{\Lambda }=\mathbf{V^T}A\mathbf{V}$$

Again where: $$\mathbf{V}=\mathbf{J_\alpha }^{T}$$

$$\lambda _1u_{\bar{x}\bar{x}}+\lambda _2u_{\bar{y}\bar{y}}=h(\bar{x},\bar{y},u,u_{\bar{x}},u_{\bar{y}})$$

where :

$$\lambda _1=\lambda _x;\lambda _2=\lambda _y$$

Assuming a hyperbolic PDE's:

$$\lambda _1>0;\lambda _2<0$$

$$\begin{Bmatrix}\xi \\ \eta\end{Bmatrix}=\begin{bmatrix}1/\sqrt{\left |\lambda _1 \right |} &0 \\ 0 &1/\sqrt{\left |\lambda _2  \right |} \end{bmatrix}\begin{Bmatrix}\bar{x} \\ \bar{y}\end{Bmatrix}$$

Where:

$$\mathbf{J}_\beta =\begin{bmatrix}1/\sqrt{\left |\lambda _1 \right |} &0 \\ 0 &1/\sqrt{\left |\lambda _2  \right |} \end{bmatrix}$$

And Scaling:

$$u_{\xi \xi }-u_{\eta \eta}=h$$ Form 1

Find: $$\begin{Bmatrix}\bar{\xi} \\ \bar{\eta}\end{Bmatrix}=\mathbf{J_\gamma }\begin{Bmatrix}\xi \\ \eta\end{Bmatrix}$$

This transformation was inspired from the conics form

Continuing:

$$\mathbf{\bar{B}}=\mathbf{J}_\gamma\mathbf{B}\mathbf{J}_\gamma^{T}$$

and

$$u_{\bar{\xi}\bar{\eta}}=h $$

new
Find

$$u_{\xi \xi }+u_{\eta \eta }=g(\xi ,\eta ,u,u_{\xi },u_{ \eta })$$

Using the PDE notation: $$\lambda_1u_{xx}+\lambda_2u_{yy}=g$$

Where the Jacobian = $$J_B=\begin{bmatrix} \frac {1}{\sqrt {\lambda_1}} & 0 \\ 0 & \frac {1}{\sqrt {\lambda_2}}\end{bmatrix}$$

The Jacobian is an orthogonal matrix and then:

$$\left \lfloor \partial_{\xi} \partial_{\eta} \right \rfloor J_B \begin{bmatrix} u \lambda_1 & 0 \\  0 & u \lambda_2\end{bmatrix} J_B\begin{bmatrix} \partial_{\xi} \\ \partial_{\eta}\end{bmatrix}=g $$

substituting in the equation above:

$$ \left \lfloor \partial_{\xi} \partial_{\eta} \right \rfloor \begin{bmatrix} \frac {1}{\sqrt {\lambda_1}} & 0 \\ 0 & \frac {1}{\sqrt {\lambda_2}}\end{bmatrix} \begin{bmatrix} u \lambda_1 & 0 \\  0 & u \lambda_2\end{bmatrix} \begin{bmatrix} \frac {1}{\sqrt {\lambda_1}} & 0 \\ 0 & \frac {1}{\sqrt {\lambda_2}}\end{bmatrix}\begin{bmatrix} \partial_{\xi} \\ \partial_{\eta}\end{bmatrix}=g$$

Solving it we find the expected result:

$$u_{\xi \xi }+u_{\eta \eta }=g$$