User:Egm6322.s09.mafia/HW7

 See my comments below. Egm6322.s09 09:37, 3 May 2009 (UTC)

=The Eigenvalue Problem, Cot'd=

--Egm6322.s09.xyz 19:30, 23 April 2009 (UTC)

Let $$ \mathbf{A} =

\begin{pmatrix} 2 & 5 \\ 5 & 7 \end{pmatrix}

$$

HW: Matlab to find Eigenvalues/Eigenvectors

--Egm6322.s09.xyz 19:52, 23 April 2009 (UTC)

Using Matlab, the eigenvalues are given as:

$$\lambda_1 =-1.0902$$ and $$\lambda_2 = 10.0902$$

OR $$ \mathbf{\Lambda} =

\begin{pmatrix} -1.0902 & 0 \\ 0 & 10.0902 \end{pmatrix}

$$

The corresponding eigenvectors are the columns of the following matrix, V:

$$ \mathbf{V}=

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

$$

MATLAB CODE: A = [2 5 ; 5 7]; [v,d] = eig(A); %where eigenvalues are given in matrix d, and, %the eigenvectors are the columns of matrix v %such that A*v = b*v*d

HW: Use Matlab to prove that $$\mathbf{V}^{-1} = \mathbf{V}^T$$

--Egm6322.s09.xyz 02:17, 24 April 2009 (UTC)

Recall that for the matrix given above, the eigenvector matrix is given as:

$$

\mathbf{V} =

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

$$

Using Matlab, the inverse of the eigenvector matrix is:

$$

\mathbf{V}^{-1} =

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

$$

MATLAB CODE for the inverse of a matrix, V inv(v);

The respective transpose matrix is given as:

$$

\mathbf{V}^T =

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

$$

MATLAB CODE for the transpose of a matrix, V v';

$$\therefore \mathbf{V}^{-1} = \mathbf{V}^T \surd$$ (verified)

HW: Use Matlab to show that $$\mathbf{V} \mathbf{\Lambda}\mathbf{V}^T = \mathbf{A}$$

--Egm6322.s09.xyz 02:19, 24 April 2009 (UTC)

Again, the matrix A is given as:

$$ \mathbf{A} =

\begin{pmatrix} 2 & 5 \\ 5 & 7 \end{pmatrix}

$$

From the information given above, it has already been shown that the eigenvalue (diagonal) matrix, given as $$\mathbf{\Lambda}$$ and the transpose of the eigenvector matrix, $$\mathbf{V}^T$$ are given as:

$$ \mathbf{\Lambda} =

\begin{pmatrix} -1.0902 & 0 \\ 0 & 10.0902 \end{pmatrix}

$$

and

$$ \mathbf{V}^T =

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

$$

where

$$

\mathbf{V} =

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

$$

Breaking the stated multiplication of the aformentioned matrices into two steps, one can see that:

$$

\mathbf{V} \mathbf{\Lambda} =

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

\begin{pmatrix} -1.0902 & 0 \\ 0 & 10.0902 \end{pmatrix}

$$

$$ \therefore

\mathbf{V} \mathbf{\Lambda} =

\begin{pmatrix} 0.9274 & 5.3047 \\ -0.5731 & 8.5832 \end{pmatrix}

$$

and

$$

\left( \mathbf{V} \mathbf{\Lambda} \right) \mathbf{V}^T =

\begin{pmatrix} 0.9274 & 5.3047 \\ -0.5731 & 8.5832 \end{pmatrix}

\begin{pmatrix} -0.8507 & 0.5257 \\ 0.5257 & 0.8507 \end{pmatrix}

$$

$$ \therefore

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

\begin{pmatrix} 2 & 5 \\ 5 & 7 \end{pmatrix}

\surd $$ verified

MATLAB Code for the combined matrix multiplication given above: A = [2 5; 5 7]; [v,d]=eig(A); %note: v = eigenvector matrix and d = diagonal eigenvalue matrix A_mult = v*d*v'; %Comparison of the matrix A to matrix A_mult %will yield A = A_mult

note: Given any matrix, attempt to change coordinates so to express that matrix as simple as possible. Best = diagonal form. Not all matrices are diagonalizable; best = Jordan canonical form. An important case is the real symmetric matrix

$$

\begin{bmatrix} \lambda_1     & \cdots & 0      \\ \vdots & \ddots & \vdots \\ 0     & \cdots & \lambda_n \end{bmatrix}

$$

Ref: pg. 34-1, eqn 2

$$ \left \lfloor x \; y \right \rfloor

\mathbf{V} \mathbf{\Lambda}\mathbf{V}^T

\begin{Bmatrix} x \\ y \end{Bmatrix}

+

\left \lfloor d \; e \right \rfloor

\begin{Bmatrix} x \\ y \end{Bmatrix}

+ f

= 0

$$

Definition

$$

\begin{Bmatrix} \bar{x} \\ \bar{y} \end{Bmatrix}



\mathbf{V}^T

\begin{Bmatrix} x \\ y \end{Bmatrix}

$$



\Rightarrow

\begin{Bmatrix} x \\ y \end{Bmatrix}

= \mathbf{V}

\begin{Bmatrix} \bar{x} \\ \bar{y} \end{Bmatrix}

$$


 * where $$\mathbf{V} (\mathbf{V}^T) = \mathbf{I}$$

$$ \therefore

\left \lfloor \bar{x} \; \bar{y} \right \rfloor

\mathbf{\Lambda}

\begin{Bmatrix} \bar{x} \\ \bar{y} \end{Bmatrix}

+

\left \lfloor \bar{d} \; \bar{e} \right \rfloor

\begin{Bmatrix} \bar{x} \\ \bar{y} \end{Bmatrix}

+ f

= 0

$$

HW: Prove $$\mathbf{\Lambda}= \mathbf { V^T A V}$$ --Egm6322.s09.lapetina 00:42, 24 April 2009 (UTC)

Prove:

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

where:

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

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

and

$$\mathbf{VV^{T}}=\mathbf{I}$$

If we premultiply the statement:

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

by $$\mathbf{V^{T}}$$, we are left with:

$$\mathbf{ V^{T} V \Lambda V^T}=\mathbf{V^{T} A}$$, and if we post-multiply this by

$$\mathbf{V}$$, we are left with:

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

Substituting $$\mathbf{V^{-1}}$$ for $$\mathbf{V}$$, we're left with:

$$\mathbf{ V^{-1} V \Lambda V^{-1} V}=\mathbf{V^{T} A V}$$.

By definition, $$\mathbf{V^{-1} V}=\mathbf{VV^{-1}}=\mathbf{I}$$, and

$$\mathbf{ I \Lambda I}=\mathbf{V^{T} A V}$$, equivalent to:

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

Note that $$\mathbf{\Lambda} $$ is a diagonal matrix

$$

\begin{bmatrix} \lambda_1     & \cdots & 0      \\ \vdots & \ddots & \vdots \\ 0     & \cdots & \lambda_n \end{bmatrix}

$$

OR

$$

\mathbf{\Lambda} =

\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}

\Rightarrow

\left[ \lambda_1 \bar{x}^2 + \lambda_2 \bar{y}^2 \right]

+\bar{d} \bar{x} + \bar{e}\bar{y}

+ f = 0

$$

Consider the cases where:

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

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

HW: Show case#1 and case#2

Egm6322.s09.bit.sahin 16:44, 24 April 2009 (UTC)

We know that Report 6. $$\lambda^2-\lambda(a+c)+(ac-b^2)=0$$

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)

Hyperbolic Case
Complete the details to show how we arrive at:

$$\xi ^2- \eta^2= \pm 1$$, the canonical form of hyperbolae,

from the equation:

$$\lambda_1 (\bar\bar x)^2+ \lambda_2 (\bar\bar x)^2=g$$,

where

$$g=- \left [ \lambda_1 r^2 + \lambda_2 s^2 -\bar d r -\bar e s +f \right ]$$

and

$$\lambda_1 \lambda_2 < 0$$. We will assume $$\lambda_1 > 0$$, $$ \lambda_2 < 0$$ and $$g>0$$.

Elliptic Case
--Egm6322.s09.xyz 16:46, 24 April 2009 (UTC)

Cases $$\lambda_1 \lambda_2 >0$$ and $$\lambda_1 \lambda_2 <0$$,

$$\lambda_1 \lambda_2 >0$$is the case of ellipse

$$\lambda_1 \lambda_2 <0$$is the case of hyperbola

In both of these two cases:$$\lambda_1 \lambda_2 \neq0$$

Geometric interpretation:



Eigenvalue problem $$\Leftrightarrow$$ Rotation of $$(x,y)$$to$$(\overline {x},\overline{y})$$which are parallel to principal axes of conics. Orthogonal of $$\underline {V}$$ $$(\underline {v} \ \underline {v}^T$$ $$=\underline {I})$$

Next Step: Get rid of linear terms in

$$\left \lfloor \overline {x} \ \overline {y} \right \rfloor$$ $$\underline {\Lambda}$$ $$	\binom{\overline {x}}{\overline {y}}   $$ + $$\left \lfloor \overline {d} \ \overline {e} \right \rfloor$$ $$	\binom{\overline {x}}{\overline {y}}   $$ +$$f=0$$ Equation(1)

Translation of Coordinate System.

Define the new coordinate system $$(\overline{\overline {x}} \ \overline{\overline {y}}) $$as:

$$	\binom{\overline{\overline {x}}}{\overline{\overline {y}}}$$ = $$\binom {\overline {x}+r}{\overline {y}+s}$$

(r,s)are unknowns.It's the "center" of ellipse or hyperbola.

$$\binom {\overline {x}}{\overline {y}}$$ = $$	\binom{\overline{\overline {x}}-r}{\overline{\overline {y}}-s}$$

Plugging it into Equation (1), we obtain:

$$\lambda_1 (\overline{\overline{x}}-r)^2+\lambda_2 (\overline{\overline{y}}-s)^2$$ $$+\overline {d} (\overline{\overline{x}}-r)+\overline {e} (\overline{\overline{y}}-s) +f =0$$

--EGM6322.S09.TIAN 21:36, 23 April 2009 (UTC)

HW:Find Out r and s Expanding

$$\lambda_1 (\overline{\overline{x}}-r)^2+\lambda_2 (\overline{\overline{y}}-s)^2$$ $$+\overline {d} (\overline{\overline{x}}-r)+\overline {e} (\overline{\overline{y}}-s) +f =0$$

We obtain:

$$\lambda_1 \overline {\overline {x}}^2 + \lambda_2 \overline {\overline {y}}^2$$ $$+(\overline {d}- 2 \lambda_1 r )\overline {\overline {x}}$$ $$+(\overline {e}- 2 \lambda_2 s )\overline {\overline {y}}$$ $$+(\lambda_1 r^2 + \lambda_2 s^2 - \overline {d} r -\overline {e} s +f)=0$$

Because we need to get rid of linerar terms,

$$ \overline {d} - 2 \lambda_1 r =0$$

$$ \overline {e} - 2 \lambda_2 s =0$$,

Finally we have:

$$r= \overline {d} / 2 \lambda_1$$

$$s= \overline {e} / 2 \lambda_2$$

--EGM6322.S09.TIAN 21:36, 23 April 2009 (UTC)

HW:Find Out g $$\lambda_1 (\overline{\overline{x}}-r)^2+\lambda_2 (\overline{\overline{y}}-s)^2=g$$

Comparing it to

$$\lambda_1 \overline {\overline {x}}^2 + \lambda_2 \overline {\overline {y}}^2$$ $$+(\lambda_1 r^2 + \lambda_2 s^2 - \overline {d} r -\overline {e} s +f)=0$$

we get,

$$g=-(\lambda_1 r^2 + \lambda_2 s^2 - \overline {d} r -\overline {e} s +f)$$

--EGM6322.S09.TIAN 21:36, 23 April 2009 (UTC)

=The Geometric Interpretation of Classification= =The Canonical Form of Conics=

Egm6322.s09.Three.nav 13:49, 24 April 2009 (UTC)

The modified form of the Equation of Conics is derived at the end of the section on The EigenValue Problem. This standard form on further simplification reduces to

$$\lambda_{1}\bar{\bar{x}}^2+\lambda_{2}\bar{\bar{y}}^2+f-\frac{d^2}{4\lambda_{1}}-\frac{e^2}{4\lambda_{2}}=0$$ ---(1)

Define $$ g= -(f-\frac{d^2}{4\lambda_{1}}-\frac{e^2}{4\lambda_{2}})$$. Then (1) reduces to the form

$$\lambda_{1}\bar{\bar{x}}^2+\lambda_{2}\bar{\bar{y}}^2= g$$--(2)

Here $$\lambda_{1}$$ and $$\lambda_{2}$$ denote the eigenvalues obtained by solving the eigenvalue problem for the original equation of conics: $$Ax^2+Bxy+Cy^2+\bar{d}x+\bar{e}y+\bar{f}=0$$

Equation (2) can be further reduced to equations representing the canonical forms for an ellipse, hyperbola and parabola.

Case (i):Ellipse: $$\lambda_{1}$$ and $$\lambda_{2}$$ >0 $$\Rightarrow$$ g>0

The canonical form representing the ellipse then reduces to: $$\xi^2+\eta^2=1$$ --(3)

Case (ii): Hyperbola: $$\lambda_{1}$$ and $$\lambda_{2}$$ <0 $$\Rightarrow g\neq 0 $$

The canonical form representing the hyperbola takes two different forms depending on the sign of 'g'.

Case ii(a): g>0 $$\to \xi^2-\eta^2=+1$$ ---(4a) (This is also true for $$\lambda_{1} < 0$$,$$\lambda_{2} > 0$$ and g>0)

Case ii(b): g<0 $$\to \xi^2-\eta^2=-1$$ ---(4b) (This is also true for $$\lambda_{1} < 0$$,$$\lambda_{2} > 0$$ and g<0) Case ii(c) $$\bar{\xi}\bar{\eta}=\pm 1$$--(4c)

This is a different representation of the canonical form for hyperbolas. It is derived from either of the equations (4a) or (4b)

Note that $$[\xi,\eta]$$ represent a different coordinate system using which (2) has been transformed into the various canonical forms shown. The derivation of forms(3),(4a) and (4b) are shown in separate sections. The derivation of form (4c) from (4a)or (4b) is shown below.

Define a new coordinate system $$[\bar{\xi}, \bar{\eta}]$$ in terms of the old coordinate system $$[\xi, \eta]$$ such that $$ \begin{Bmatrix}\bar{\xi} \\ \bar{\eta}\end{Bmatrix}= \begin{Bmatrix}\xi-\eta \\ \xi+\eta\end{Bmatrix}$$

$$ \Rightarrow \begin{Bmatrix}\bar{\xi}\\ \bar{\eta}\end{Bmatrix}= \begin{bmatrix}\ \ 1 \ -1 \\ -1 \ -1\end{bmatrix}\begin{Bmatrix}\xi-\eta \\ \xi+\eta\end{Bmatrix}$$

$$\Rightarrow \begin{Bmatrix}\xi \\ \eta \end{Bmatrix}= \begin{bmatrix}\ \ 1 \ -1 \\ -1 \ -1\end{bmatrix}^{-1}\begin{Bmatrix} \bar{\xi} \\ \bar{\eta} \end{Bmatrix}= \frac {1}{2}\begin{bmatrix}\ \ 1 \ -1 \\ -1 \ -1\end{bmatrix}\begin{Bmatrix}\bar{\xi} \\ \bar{\eta} \end{Bmatrix}$$

Eqn.(4a) or (4b) can be re-written in the form: $$\begin{Bmatrix}\xi \ \eta \end{Bmatrix}\begin{bmatrix}1 \ \ \ \ 0 \\ 0 \ -1 \end{bmatrix}\begin{Bmatrix}\xi \\ \eta \end{Bmatrix}= \pm 1$$--(5)

Rewriting (5) in the new coordinate system $$[\bar{\xi}, \bar{\eta}]$$, we get:

$$ \begin{Bmatrix}\bar{\xi} \ \bar{\eta} \end{Bmatrix}\frac{1}{2}\begin{bmatrix} \ \ 1 \  -1 \\ -1 \ -1 \end{bmatrix}\begin{bmatrix}1 \ \ \ \  0 \\ 0 \ -1 \end{bmatrix}\frac{1}{2}\begin{bmatrix} \ \ 1 \   -1 \\ -1 \ -1 \end{bmatrix}\begin{Bmatrix}\bar{\xi} \\ \bar{\eta} \end{Bmatrix}= \pm 1$$

$$ \Rightarrow \frac{1}{4} \begin{Bmatrix}\bar{\xi} \ \bar{\eta} \end{Bmatrix}\begin{bmatrix} \ \ 1 \  -1 \\ -1 \ -1 \end{bmatrix}\begin{bmatrix}1 \ \ \ \  0 \\ 0 \ -1 \end{bmatrix}\begin{bmatrix} \ \ 1 \   -1 \\ -1 \ -1 \end{bmatrix}\begin{Bmatrix}\bar{\xi} \\ \bar{\eta} \end{Bmatrix}= \pm 1$$

$$\Rightarrow \frac{1}{4}\begin{Bmatrix}\bar{\xi} \ \bar{\eta} \end{Bmatrix}\begin{bmatrix}\ \ \ 0 \ -2 \\ -2 \ \ \ \ 0 \end{bmatrix}\begin{Bmatrix}\bar{\xi} \\ \bar{\eta} \end{Bmatrix}= \pm 1$$

$$\Rightarrow \bar{\xi}\bar{\eta}=\pm 1$$ -(4c)

Thus we see that by a rotation of coordinates, form (4a) or (4b), which is the standard canonical form representing a hyperbola reduces to form (4c). This simplified form is used mostly in algebraic calculations and even in high-school geometry lessons.

Canonical Form of a Parabola
Egm6322.s09.Three.nav 13:49, 24 April 2009 (UTC)

Given the general,modified form of a PDE: $$\lambda_{1}\bar{x}^2+\lambda_{2}\bar{y}^2+d\bar{x}+e\bar{y}+f=0$$--(1)

Derive the canonical form representing a parabola from (1).

The canonical form representing a parabola is given by the following equation

$$\bar{\xi}^{2}- \bar{\eta} = 0$$ --(2)

where $$\bar{\xi}$$ and $$\bar{\eta}$$ represent coordinates in a different coordinate system.

So how do we derive (2) from (1)? We apply rotation once and translation of coordinates twice to $$\bar{x}$$ and $$\bar{y}$$.

Step(1)- First Translation

Define $$\begin{bmatrix}\bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}= \begin{bmatrix}\bar{x}+r \\ \bar{y}\end{bmatrix}$$

Substituting $$\bar{\bar{x}}$$ and $$\bar{\bar{y}}$$ in (1), we get

$$\lambda_{1}(\bar{\bar{x}}-r)^2+\lambda_{2}\bar{\bar{y}}^2+d(\bar{\bar{x}}-r)+e\bar{\bar{y}}+f=0$$

$$\Rightarrow \lambda_{1}\bar{\bar{x}}^{2}+\lambda_{2}\bar{\bar{y}}^2+\bar{\bar{x}}(d-2\lambda_{1}r)+e\bar{\bar{y}}+\lambda_{1}r^{2}+f=0$$--(3)

For equation (3) to resemble (2), we need the terms: $$\bar{\bar{y}}^2, \bar{\bar{x}} \ and \ \lambda_{1}r^{2}+f$$ to vanish.

Note that for a parabola, the equation of conics: $$Ax^2+Bxy+Cy^2+\bar{d}x+\bar{y}+\bar{f}=0$$, $$B^2-4AC=0$$ i.e only one eigenvalue exists. Thus assume $$\lambda_{2}$$= 0.

Also in order for the $$\bar{\bar{x}}$$ term to vanish, we need its coefficient to be equal to 0 $$\Rightarrow d- 2\lambda_{1}r=0$$

$$\Rightarrow r= \frac{d}{2\lambda_{1}r}$$

Substituting $$\lambda_{2}$$ and r in (3), the equation reduces to $$\lambda_{1}\bar{\bar{x}}^{2}+e\bar{\bar{y}}+\frac{d}{4\lambda_{1}^{2}}+f=0$$-(4)

Denote g= $$\frac{d}{4\lambda_{1}^{2}}+f$$. Thus (4) reduces to

$$\Rightarrow \lambda_{1}\bar{\bar{x}}^{2}+e\bar{\bar{y}}+g=0$$-(5)

Step(2)- Rotation

Eqn.(4) can be rewritten as $$\left \lfloor \bar{\bar{x}} \ \bar{\bar{y}} \right \rfloor\begin{Bmatrix} \lambda_{1} \ 0 \\ 0 \ \ 0 \end{Bmatrix}\begin{bmatrix}\bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}+\left \lfloor 0 \ e \right \rfloor\begin{bmatrix}\bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}+g=0$$ (6)

Now apply rotation of coordinates by defining a new coordinate system $$(\xi, \eta)$$ where $$\begin{bmatrix} \xi \\ \eta \end{bmatrix}= \mathbf{J_{\beta}}\begin{bmatrix}\bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}$$

where $$\mathbf{J_{\beta}}= \begin{bmatrix}\frac{1}{\sqrt{\lambda_{1}}} \ \ \ 0 \\ \ \ \ 0 \ \ \ \frac{1}{e}\end{bmatrix}$$

Due to $$\mathbf{J_{\beta}}$$ being an orthogonal matrix, $$\mathbf{J_{\beta}}^{-1}= \mathbf{J_{\beta}}^{T}= \mathbf{J_{\beta}}$$

$$\Rightarrow \begin{bmatrix}\bar{\bar{x}}\\\bar{\bar{y}} \end{bmatrix}= \mathbf{J_{\beta}}^{-1}\begin{bmatrix}\xi\\\eta \end{bmatrix}= \mathbf{J_{\beta}}\begin{bmatrix}\xi\\\eta \end{bmatrix}$$

Substituting this rotation of coordinates in (6), we get

$$\begin{bmatrix}\xi\ \eta \end{bmatrix} \mathbf{J_{\beta}} \begin{Bmatrix}\lambda_{1} \ 0 \\ 0 \ \ 0 \end{Bmatrix} \mathbf{J_{\beta}}\begin{bmatrix}\xi\\\eta \end{bmatrix}+\left \lfloor 0 \ e \right \rfloor \mathbf{J_{\beta}}\begin{bmatrix}\xi\\\eta \end{bmatrix}+g=0$$

$$\Rightarrow \begin{bmatrix}\xi\ \eta \end{bmatrix} \begin{bmatrix}\frac{1}{\sqrt{\lambda_{1}}} \ \ \ 0 \\ \ \ \ 0 \ \ \ \frac{1}{e}\end{bmatrix} \begin{Bmatrix}\lambda_{1} \ 0 \\ 0 \ \ 0 \end{Bmatrix} \begin{bmatrix}\frac{1}{\sqrt{\lambda_{1}}} \ \ \ 0 \\ \ \ \ 0 \ \ \ \frac{1}{e}\end{bmatrix} \begin{bmatrix}\xi\\\eta \end{bmatrix}+\left \lfloor 0 \ e \right \rfloor \begin{bmatrix}\frac{1}{\sqrt{\lambda_{1}}} \ \ \ 0 \\ \ \ \ 0 \ \ \ \frac{1}{e}\end{bmatrix} \begin{bmatrix}\xi\\\eta \end{bmatrix}+g=0$$

Simplifying the above equation further, we obtain

$$\xi^2- \eta+g=0$$ ---(7)

Step(3)- Second Translation

Now define $$\begin{bmatrix}\bar{\xi}\\ \bar{\eta}\end{bmatrix}= \begin{bmatrix}\xi \\ \eta-g \end{bmatrix}$$

This is a valid translation because $$\lambda_{1}$$ which is a function of the coefficients A,C (as defined in the equation of conics) is a constant value. Hence g which is a function of $$\lambda_{1}$$ and f is also aconstant. Thus all we are doing in the second translation is translate the $$\bar{\bar{y}}$$ by a constant. Hence making this substitution in (7) gives us the final canonical form (2)

$$\bar{\xi}^{2}- \bar{\eta} = 0$$

Egm6322.s09.Three.nav 03:39, 24 April 2009 (UTC)

Canonical Form of a Hyperbola
HW:Deriving the Canonical Form for a Hyperbola

--Egm6322.s09.lapetina 01:18, 24 April 2009 (UTC)

Complete the details to show how we arrive at:

$$\xi ^2- \eta^2= \pm 1$$, the canonical form of hyperbolae, where $$\xi$$ and $$\eta$$ represent axes in coordinate system that coincide with the principle axes of the ellipse.

We begin with the equation:

$$\lambda_1 (\bar\bar x)^2+ \lambda_2 (\bar\bar x)^2=g$$,

where

$$g=- \left [ \lambda_1 r^2 + \lambda_2 s^2 -\bar d r -\bar e s +f \right ]$$.

This has been derived in other sections of the report here.

We will assume that $$\lambda_1 \lambda_2 < 0$$, $$\lambda_1 > 0$$, $$ \lambda_2 < 0$$ and $$g>0$$.

The eigenvalue equation can be re-expressed in matrix form as:

$$

\left \lfloor \bar{\bar{x}} \ \bar{\bar{y}} \right \rfloor

\begin{bmatrix} \lambda_{1} \ 0 \\ 0 \ \lambda_2 \end{bmatrix}

\begin{bmatrix}

\bar{\bar{x}}\\ \bar{\bar{y}}

\end{bmatrix}

=g $$

Here we want to apply a rotation transformation of the form:

$$\begin{Bmatrix}

\xi \\ \eta

\end{Bmatrix}

=

\mathbf{J_\beta}

\begin{Bmatrix}

\bar{\bar{x}}\\ \bar{\bar{y}}

\end{Bmatrix}$$.

We need to determine the values of

$$\mathbf{J_\beta}$$.

For a hyperbola of this type:

$$

\mathbf{J_\beta}=

\begin{bmatrix} \frac{1}{d_x} & 0 \\ 0 & \frac{1}{d_y} \end{bmatrix}

$$

where $$d_x = \frac{\sqrt{g}}{\sqrt{|\lambda_1|}}$$

and

$$d_y = \frac{\sqrt{g}}{\sqrt{|\lambda_2|}}$$

The inverse of this Jacobian matrix is :

$$

\mathbf{J_\beta}=

\begin{bmatrix} d_x & 0 \\ 0 & d_y \end{bmatrix}

$$,

and

$$ \begin{Bmatrix}

\bar{\bar{x}}\\ \bar{\bar{y}}

\end{Bmatrix}

= \begin{bmatrix} d_x & 0 \\ 0 & d_y \end{bmatrix}

\begin{Bmatrix}

\xi \\ \eta

\end{Bmatrix} $$

and

$$ \left \lfloor \bar{\bar{x}} \ \bar{\bar{y}} \right \rfloor

=

\left \lfloor \xi \ \eta \right \rfloor

\begin{bmatrix} d_x & 0 \\ 0 & d_y \end{bmatrix}$$.

Substituting this directly into the eigenvalue equation yields:

$$

\left \lfloor \xi \ \eta \right \rfloor

\begin{bmatrix} d_x & 0 \\ 0 & d_y \end{bmatrix}

\begin{bmatrix} \lambda_{1} \ 0 \\ 0 \ \lambda_2 \end{bmatrix}

\begin{bmatrix} d_x & 0 \\ 0 & d_y \end{bmatrix}

\begin{Bmatrix}

\xi \\ \eta

\end{Bmatrix}

=g $$

which simplifies to a polynomial form of:

$$d_x^2 \lambda_1 (\xi)^2- d_y^2 \lambda_2 (\eta)^2=g$$.

The sign in the middle is negative because $$\lambda_2<0$$.

This in turn simplifies to:

$$(\frac{\sqrt{g}}{\sqrt{|\lambda_1|}})^2 \lambda_1 (\xi)^2- (\frac{\sqrt{g}}{\sqrt{|\lambda_2|}})^2 \lambda_2 (\eta)^2=g$$.

We only divide the terms on the left by the square of the root of the absolute values of $$\lambda_1$$ and $$\lambda_2$$, preserving their signs. Because g is a function of constants, as described in the above reference, it retains its sign.

The result is the canonical form:

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

for the case where $$\lambda_1>0$$, $$\lambda_2<0$$, and $$g>0$$. To arrive at the form:

$$\xi ^2- \eta^2= -1$$ for the case where $$\lambda_1>0$$, $$\lambda_2<0$$, and $$g<0$$ is trivial. Since we divide by $$g$$ by its absolute value, it becomes a $$-1$$ using the same Jacobian.

However, for the case where:

$$\lambda_1<0$$, $$\lambda_2>0$$, and $$g<0$$, to arrive at:

$$\xi ^2- \eta^2= 1$$, we need to use different values of $$d_x$$ and $$d_y$$:

$$d_x = \frac{i \sqrt{g}}{ \sqrt{|\lambda_1|}}$$

and

$$d_y = \frac{i \sqrt{g}}{\sqrt{|\lambda_2|}}$$.

with a Jacobian of the form

$$

\mathbf{J_\beta}=

\begin{bmatrix} 0 & d_y \\ d_x & 0 \end{bmatrix}

$$.

The inverse of this is:

$$

\mathbf{J_\beta^{-1}}=

\begin{bmatrix} 0 & \frac{1}{d_y} \\ \frac{1}{d_x} & 0 \end{bmatrix}

$$

By substitution, the matrix operator form of the conic then becomes:

$$

\left \lfloor \xi \ \eta \right \rfloor

\begin{bmatrix} 0 & \frac{1}{d_y} \\ \frac{1}{d_x} & 0 \end{bmatrix}

\begin{bmatrix} \lambda_{1} \ 0 \\ 0 \ \lambda_2 \end{bmatrix}

\begin{bmatrix} 0 & \frac{1}{d_y} \\ \frac{1}{d_x} & 0 \end{bmatrix}

\begin{Bmatrix}

\xi \\ \eta

\end{Bmatrix}

=g $$

This leaves us with the polynomial equation:

$$ -1 \left [ \lambda_2 \frac{\xi^2}{d_x^2}+ \lambda_1 \frac{\eta^2}{d_y^2} \right ]=g$$.

Upon substitution, this becomes

$$(\frac{\sqrt{g}}{\sqrt{|\lambda_2|}})^2 \lambda_2 (\xi)^2- (\frac{\sqrt{g}}{\sqrt{|\lambda_1|}})^2 \lambda_1 (\eta)^2=-g$$.

Constant values cancel, leaving:

$$\xi ^2- \eta^2= 1$$ for the case where:

$$\lambda_1<0$$, $$\lambda_2>0$$, and $$g<0$$.

For the case where

$$\lambda_1<0$$, $$\lambda_2>0$$, and $$g>0$$,

to arrive at the form $$\xi ^2- \eta^2= -1$$,

we can use the same Jacobian, because the sign of $$g$$ will change the sign on the right side.

Canonical Form of an Ellipse
Egm6322.s09.Three.ge 20:30, 23 April 2009 (UTC)

HW:Deriving the Canonical Form for an Ellipse

Given the general,modified form of a PDE: $$\lambda_{1}\bar{x}^2+\lambda_{2}\bar{y}^2+d\bar{x}+e\bar{y}+f=0 \qquad \ldots (1)$$

Derive the canonical form representing an ellipse from (1).

The canonical form representing an ellipse is given by the following equation

$$\xi^{2}+ \eta^2 = 1 \qquad \ldots (2)$$

where $$\xi$$ and $$\eta$$ represent axes in coordinate system that coincide with the principle axes of the ellipse.

In order to show equation (1) in the form of equation (2)the coordinate system $$(\bar{x},\bar{y})$$ must be translated and then rotated.

Step(1)- Translation

Define $$\begin{bmatrix} \bar{\bar{x}}\\ \bar{\bar{y}}\end{bmatrix}= \begin{bmatrix}\bar{x}+r \\ \bar{y}+s\end{bmatrix}$$

Substituting $$\bar{\bar{x}}$$ and $$\bar{\bar{y}}$$ in (1) yields:

$$\lambda_{1}(\bar{\bar{x}}-r)^2+\lambda_{2}(\bar{\bar{y}}-s)^2+d(\bar{\bar{x}}-r)+e(\bar{\bar{y}}-s)+f=0$$

Expanding the squared terms gives

$$\begin{matrix} \Rightarrow \lambda_{1}\bar{\bar{x}}^{2}+\lambda_{2}\bar{\bar{y}}^2+\bar{\bar{x}}(d-2\lambda_{1}r)+\bar{\bar{y}}(e-2\lambda_{2}s)+ f=0 \\ \therefore \\ r= \frac{d}{2\lambda_{1}} \qquad s= \frac{e}{2\lambda_{2}} \end{matrix}\qquad \ldots (3)$$

Where the values for $$r$$ and $$s$$ are set so that equation (3) resembles (2). Specifically, $$r$$ and $$s$$ are found by solving the following equations.

$$\begin{matrix}(d-2\lambda_{1}r)=0 \\ \\ (e-2\lambda_{2}s)=0 \end{matrix}$$

Substituting $$r$$ and $$s$$ into equation (3), the equation reduces to:

$$\begin{matrix}\lambda_{1}\bar{\bar{x}}^2+\lambda_{2}\bar{\bar{y}}^2-\frac{d}{4\lambda_{1}^{2}}-\frac{e}{4\lambda_{2}^{2}}+f=0 \\ \textrm{Let}\quad g=\left[-\frac{d}{4\lambda_{1}^{2}}-\frac{e}{4\lambda_{2}^{2}}+f \right ] \\ \end{matrix}$$

Thus,

$$ \lambda_{1}\bar{\bar{x}}^2+\lambda_{2}\bar{\bar{y}}^2=g \quad\ldots (4)$$

Step(2)- Rotation

Equation (4) can be rewritten as $$\left \lfloor \bar{\bar{x}} \ \bar{\bar{y}} \right \rfloor

\begin{bmatrix} \lambda_{1} & 0 \\ 0 & \lambda_{2} \\ \end{bmatrix}

\begin{Bmatrix} \bar{\bar{x}}\\ \bar{\bar{y}} \end{Bmatrix}=g \quad \ldots (5)$$

Now one applies rotation of coordinates by defining to arrive at the new coordinate system $$(\xi, \eta)$$.

Define the rotation as:

$$\begin{bmatrix} \xi \\ \eta \end{bmatrix}=

\mathbf{J_{\beta}} \begin{bmatrix} \bar{\bar{x}}\\ \bar{\bar{y}} \end{bmatrix}$$

where $$\mathbf{J_{\beta}}= \begin{bmatrix} \sqrt{\frac{g}{\lambda_{1}}} & 0 \\ 0 & \sqrt{\frac{g}{\lambda_{2}}} \end{bmatrix}$$

Since $$\mathbf{J_{\beta}}$$ being an orthogonal matrix, (i.e. $$\mathbf{J_{\beta}}^{-1}= \mathbf{J_{\beta}}^{T}= \mathbf{J_{\beta}}$$) it can be shown that:

$$\begin{bmatrix} \bar{\bar{x}}\\ \bar{\bar{y}} \end{bmatrix}=

\mathbf{J_{\beta}}^{-1} \begin{bmatrix}\xi \\ \eta \end{bmatrix}=

\mathbf{J_{\beta}} \begin{bmatrix}\xi \\ \eta \end{bmatrix}$$

Substituting this rotation of coordinates into equation (5) results in:

$$\left \lfloor \xi \quad \eta \right \rfloor

\mathbf{J_{\beta}} \begin{bmatrix} \lambda_{1} & 0 \\ 0 & \lambda_{2} \end{bmatrix}

\mathbf{J_{\beta}} \begin{Bmatrix} \xi \\ \eta \end{Bmatrix} =g$$

Plugging in the value of $$\mathbf{J_{\beta}}$$ yields the following expression.

$$\left\lfloor \xi \quad \eta \right\rfloor

\begin{bmatrix} \sqrt{\frac{g}{\lambda_{1}}} & 0 \\ 0 & \sqrt{\frac{g}{\lambda_{2}}} \end{bmatrix}

\begin{bmatrix} \lambda_{1} & 0 \\ 0 & \lambda_{2} \end{bmatrix}

\begin{bmatrix} \sqrt{\frac{g}{\lambda_{1}}} & 0 \\ 0 & \sqrt{\frac{g}{\lambda_{2}}} \end{bmatrix}

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

=g$$

Simplifying the above equation further, we obtain

$$g\xi^{2}+ g\eta^{2} = g$$

Which clearly reduces to:

$$\xi^{2}+ \eta^{2} = 1$$

 For the ellipse case, $$\displaystyle \lambda_1 \lambda_2 > 0$$. Assume that $$\displaystyle \lambda_1 > 0$$ and $$\displaystyle \lambda_2 > 0$$, so that $$\displaystyle \lambda_1 \lambda_2 > 0$$, it follows from Eq.(4) in the collapsible box that $$\displaystyle g > 0$$, and thus $$\displaystyle \sqrt(g)$$ is a real number. For the case in which $$\displaystyle \lambda_1 < 0$$ and $$\displaystyle \lambda_2 < 0$$, a similar argument can be made with $$\displaystyle - \lambda_1 > 0$$, $$\displaystyle - \lambda_2 > 0$$, and $$\displaystyle - g > 0$$. Egm6322.s09 10:09, 3 May 2009 (UTC)

=The Relationship to Transformations of Variables=

Homework: Get the Canonical form for 2nd order linear Elliptic PDE Egm6322.s09.bit.la 19:58, 24 April 2009 (UTC) 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:

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

 For PDEs, it is better to put the function $$\displaystyle u$$ at the end of the matrix operator, and even though $$\displaystyle \mathbf J_B$$ is diagonal, it is still better to specify the transpose in the general formula for transformation of coordinates, i.e., $$\displaystyle \left \lfloor \partial_{\xi} \partial_{\eta} \right \rfloor \mathbf J_B \begin{bmatrix} \lambda_1 & 0 \\  0 & \lambda_2\end{bmatrix} \mathbf J_B^T \left\{ \begin{matrix} \partial_{\xi} \\ \partial_{\eta}\end{matrix} \right\} (u) = g $$. Egm6322.s09 10:09, 3 May 2009 (UTC)

Homework: Get the Canonical form for 2nd order linear Hyperbolic PDE For hyperbolic PDE: $$\lambda_1 u_{xx}-\lambda_2 u_{yy}=g$$

$$\begin{Bmatrix} \partial_{x} \\ \partial_{y} \end{Bmatrix} =J_B \begin{Bmatrix} \partial_{\xi} \\ \partial_{\eta} \end{Bmatrix}$$  ... (1)

The Jacobian matrix is:

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

so the hyperbolic PDE can be written as:

$$\left \lfloor \partial_x \partial_y \right \rfloor \begin{bmatrix} u \lambda_1 & 0 \\ 0 & -u \lambda_2 \end{bmatrix} \begin{bmatrix} \partial_x \\ \partial_y \end{bmatrix}=g$$

substituting (1) into the above equation gives:

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

So the canonical form for this hyperbolic PDE is

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

Egm6322.s09.three.liu 16:21, 24 April 2009 (UTC)

HW:Getting the Canonical Form for Parabola PDEs Now we have $$\lambda_1 u_{xx} + e u_y =h$$ (1)

and we want $$u_{\xi \xi} - u_{\eta} =g$$ (2)

Rewriting (1) in matrix form, we have:

$$\left \lfloor \partial {x} \ \partial {y} \right \rfloor$$ $$\begin{bmatrix} \lambda_1     &  0      \\ 0             & 0 \end{bmatrix} $$ $$\begin{bmatrix} \partial_x u           \\ \partial_y u            \end{bmatrix} $$ $$+ \left \lfloor 0 \ e \right \rfloor$$ $$\begin{bmatrix} \partial_x u           \\ \partial_y u            \end{bmatrix}=g $$(3)

We have already derived that,

$$\begin{bmatrix} \partial_x            \\ \partial_y \end{bmatrix} $$ = $$J_{\beta} \begin{bmatrix} \partial_{\xi}            \\ \partial_{\eta} \end{bmatrix} $$ (4)

$$J_{\beta}= \begin{bmatrix} 1/ \sqrt{\lambda_1}     &  0      \\ 0             & 0 \end{bmatrix} $$(5)

Plugging (4) and (5) into (3), we obtain:

$$\left \lfloor \partial {\xi} \ \partial {\eta} \right \rfloor J_{\beta}^T$$ $$\begin{bmatrix} \lambda_1     &  0      \\ 0             & 0 \end{bmatrix} J_{\beta}$$ $$\begin{bmatrix} \partial_{\xi} u           \\ \partial_{\eta} u            \end{bmatrix} $$ $$+ \left \lfloor 0 \ e \right \rfloor J_{\beta}$$ $$\begin{bmatrix} \partial_{\xi} u           \\ \partial_{\eta} u            \end{bmatrix}=g $$

$$\left \lfloor \partial {\xi} \ \partial {\eta} \right \rfloor \begin{bmatrix} 1/ \sqrt{\lambda_1}     &  0      \\ 0             & 0 \end{bmatrix}$$ $$\begin{bmatrix} \lambda_1     &  0      \\ 0             & 0 \end{bmatrix} \begin{bmatrix} 1/ \sqrt{\lambda_1}     &  0      \\ 0             & 0 \end{bmatrix}$$ $$\begin{bmatrix} \partial_{\xi} u           \\ \partial_{\eta} u            \end{bmatrix} $$ $$+ \left \lfloor 0 \ e \right \rfloor J_{\beta}$$ $$\begin{bmatrix} \partial_{\xi} u           \\ \partial_{\eta} u            \end{bmatrix}=g $$

The above equation is:

$$u_{\xi \xi} - u_{\eta} =g$$ , which is exactly we want.

--EGM6322.S09.TIAN 21:40, 23 April 2009 (UTC)

Homework: General classification of 2nd order linear PDEs to 2nd order nonlinear PDEs 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 equation without knowing the specific solution.The type of the equation will be changing with the specific solution at the same point (x,y)

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

Egm6322.s09.bit.gk 19:43, 24 April 2009 (UTC)

 The above collapsible box described the linearization of a nonlinear PDE with respect to the second order derivatives to do the classification. On the other hand, note also a PDE like $$u_{\xi \xi }+u_{\eta \eta }=g$$ is also a nonlinear PDE if $$\displaystyle g(\xi, \eta, u, u_\xi, u_\eta)$$ is nonlinear with respect to its arguments. Egm6322.s09 10:09, 3 May 2009 (UTC)

Homework: Verify (2) in Solution of (1). Derivation of d'Alembert solution for 1-D wave equation 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

$$\frac{\partial u}{\partial \xi}={F}'$$

$$\Rightarrow \frac{\partial^2 u}{\partial \xi\partial \eta}=0$$

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

Hence the solution.

Egm6322.s09.bit.gk 20:11, 24 April 2009 (UTC)

Homework: Finding $$\bar{a}$$, $$\bar{b}$$, $$\bar{c}$$

Egm6322.s09.bit.sahin 16:52, 24 April 2009 (UTC)

--Egm6322.s09.lapetina 18:01, 24 April 2009 (UTC)

$$\mathbf{J}=\begin{bmatrix} \phi _{x} &\phi _{y} \\ \psi _{x}&\psi_{y} \end{bmatrix}$$, $$\mathbf{J^{T}}=\begin{bmatrix} \phi _{x} &\psi _{x} \\ \phi _{y}&\psi_{y} \end{bmatrix}$$ 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 matrices into the expression

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

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

Therefore,

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

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

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

The sign of $$det{\underline A}$$ would not change under only change of coordinates, including nonlinear change of coordinates.

Therefore, the classification of 2nd order PDEs includes the canonical forms and the invariance of $$det{\underline A}$$.

=Summary : Differences Between conics and PDE's= Egm6322.s09.bit.la 17:31, 24 April 2009 (UTC)

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

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

 Hyperbolic PDEs 

Egm6322.s09.bit.sahin 17:38, 24 April 2009 (UTC)

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

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

Egm6322.s09.bit.gk 19:17, 24 April 2009 (UTC)

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

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 derivation of d'Alembert solution from Form 2 of the hyperbolic canonical form is missing. Egm6322.s09 10:09, 3 May 2009 (UTC)

= References =

=Signatures= --Egm6322.s09.xyz 21:38, 23 April 2009 (UTC)

--EGM6322.S09.TIAN 21:40, 23 April 2009 (UTC)

Egm6322.s09.Three.nav 03:44, 24 April 2009 (UTC)

--Egm6322.s09.lapetina 14:44, 24 April 2009 (UTC)

Egm6322.s09.Three.ge 15:26, 24 April 2009 (UTC)

Egm6322.s09.three.liu 16:21, 24 April 2009 (UTC)

Egm6322.s09.bit.la 17:31, 24 April 2009 (UTC)

Egm6322.s09.bit.sahin 17:39, 24 April 2009 (UTC)

Egm6322.s09.bit.gk 20:14, 24 April 2009 (UTC)