User:Egm6322.s09.mafia/HW3

 My second wave of comments is color coded in green. Egm6322.s09 00:31, 2 April 2009 (UTC)

 See my comments below. Egm6936.s09 02:16, 21 February 2009 (UTC)

After you made a correction for a section with a comment box, you want to put a comment in that same comment box on what you did. Egm6936.s09 11:44, 26 February 2009 (UTC)

 NOTE: This page is being posted at roughly 19 UTC because of problems observed on the Wikiversity Domain. Anticipate revisions prior to 22 UTC, as well as possibly afterwards, if all edits are not possible due to the failures of the Wikiversity domain. --Egm6322.s09.lapetina 19:12, 20 February 2009 (UTC)

A few formatting and citation edits were made. No further edits are anticipated.

--Egm6322.s09.lapetina 20:11, 20 February 2009 (UTC)

Many edits have been made throughout. Editors were instructed to mention actions within comment boxes as indicated.

--Egm6322.s09.lapetina 21:05, 6 March 2009 (UTC)

=Transformations of Coordinates=

Linear Transformation of Coordinates
--Egm6322.s09.xyz 16:13, 4 April 2009 (UTC)

Example of Jacobian for Linear Transformation
Example: An example of a linear transformation is rotation in 2D. This can be expressed as:

$$

\begin{Bmatrix} \bar{x_1} \\ \bar{x_2} \end{Bmatrix}

=

\bar{R}

\begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix}

$$

The transpose of the Jacobian matrix for this example is given as follows:

$$

\bar{J}^T

=

\begin{bmatrix} \frac{\partial \bar{x_1}} {\partial {x_1}} & \frac{\partial \bar{x_2}} {\partial {x_1}}\\ \frac{\partial \bar{x_1}} {\partial {x_2}} & \frac{\partial \bar{x_2}} {\partial {x_2}} \end{bmatrix}

=

\begin{bmatrix} R_{11} & R_{21}\\ R_{12} & R_{22} \end{bmatrix}

=

\bar{R}^T

$$

Nonlinear Transformation of Coordinates
--Egm6322.s09.xyz 16:14, 4 April 2009 (UTC)

Expression of general form in matrix notation using the Jacobian Matrix
Recall the general form of a 2nd order PDE that is linear in all orders expressed in matrix notation (Eqn. 2, pg. 8-1):

$$

\left \lfloor \partial_x \; \partial_y \right \rfloor \begin{bmatrix} a & b\\ b & c \end{bmatrix}

\begin{Bmatrix} \partial_x u \\ \partial_y u \end{Bmatrix} +

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

\begin{Bmatrix} \partial_x u \\ \partial_y u \end{Bmatrix}

+ fu + g = 0

$$

Let $$\bar{A}$$ be the following coefficient matrix:

$$ \bar{A} =

\begin{bmatrix} a & b\\ b & c \end{bmatrix} $$

Recall (Eqn. 1, pg. 12-3):

$$ \begin{Bmatrix} \partial_x \\ \partial_y \end{Bmatrix}

\left( u \right)

=

\begin{bmatrix} \frac{\partial \bar{x}}{\partial x} & \frac{\partial \bar{y}}{\partial x}\\ \frac{\partial \bar{x}}{\partial y} & \frac{\partial \bar{y}}{\partial y} \end{bmatrix}

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

\left( \bar u \right)

=

{\bar J}^T

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

\left( \bar u \right)

$$

The general form in matrix notation of this non-linear transformation using the Jacobian matrix is given as:

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

\bar{J}

\left (

\bar{A} \bar{J}^T

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

\right )

\left( \bar{u} \right)

+

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

\bar{J}^T

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

\left( \bar{u} \right)

+ f\left( \bar{u} \right)

+ g = 0

$$

note: $$\bar{A}, d, e, f,$$ and $$g$$ are all functions of $$\bar{x}$$ and $$\bar{y}$$

Expressions of Transformations of Coordinates
--Egm6322.s09.lapetina 17:48, 24 April 2009 (UTC)

Recall the Jacobian Matrix for $$n$$ variables:

$$\bar x_{i}=\bar x_{i}(x_1,...,x_n)$$

$$\bar J=\begin{bmatrix} \partial {\bar x_i}     \\ \partial {\bar x_j} \end{bmatrix} $$.

In two dimensions this can be expressed as:

$$\bar J=\begin{bmatrix} J_{11} & J_{12} \\ J_{21} & J_{22} \end{bmatrix}=\begin{bmatrix} \phi_{x} & \phi_{y} \\ \psi_{x} & \psi_{y} \end{bmatrix}$$

Or, as Lapidus and Pinder show:

$$u_{xx}=u_{\bar x \bar x}(J_{11})^2+2 u_{\bar x \bar y} J_{11} J_{21}+u_{\bar y \bar y}^2$$

To derive Pinder's equations in matrix operator form, we use equation(1) from page 12-3,

$$\begin{Bmatrix} \partial_x \\ \partial_y \end{Bmatrix} \left \lfloor \partial_x \; \partial_y \right \rfloor =\begin{bmatrix} \partial_{xx}     & \partial_{xy}      \\ \partial_{yx}     & \partial_{yy} \end{bmatrix}=\bar J^T \begin{Bmatrix} \partial_{\bar x} \\ \partial_{\bar y} \end{Bmatrix} \left \lfloor \partial_x \; \partial_y \right \rfloor \bar J $$

Note:

$$\left \lfloor \partial_x \; \partial_y \right \rfloor= \left \lfloor \partial_{\bar x} \; \partial_{\bar y} \right \rfloor \bar J$$

$$(\bar A \bar B)^T={\bar B}^T{\bar A}^T$$

$$\alpha={\bar J}^T \begin{bmatrix} \partial_{\bar x \bar x}     & \partial_{\bar x \bar y}      \\ \partial_{\bar y \bar x}     & \partial_{\bar y \bar y}     \end{bmatrix} \bar J $$ To find $$\alpha$$, we must complete some matrix multiplication

$$ \alpha = \begin{bmatrix} J_{11}     & J_{21}      \\ J_{12}     & J_{22} \end{bmatrix} \begin{bmatrix} \partial_{\bar x \bar x}     & \partial_{\bar x \bar y}      \\ \partial_{\bar y \bar x}     & \partial_{\bar y \bar y}  \end{bmatrix} \begin{bmatrix} J_{11}     & J_{12}      \\ J_{21}     & J_{22} \end{bmatrix} = \begin{bmatrix} J_{11}     & J_{21}      \\ J_{12}     & J_{22} \end{bmatrix}

\begin{bmatrix} \partial_{\bar x \bar x} J_{11} +\partial_{\bar x \bar y} J_{21} & \partial_{\bar x \bar x} J_{12} +\partial_{\bar x \bar y} J_{22}    \\ \partial_{\bar y \bar x} J_{11} +\partial_{\bar y \bar y} J_{21}  & \partial_{\bar y \bar x} J_{12} +\partial_{\bar y \bar y} J_{22} \end{bmatrix}

$$

$$ \alpha=

\begin{bmatrix}

\partial_{\bar x \bar x} (J_{11})^2 +2 \partial_{\bar x \bar y} J_{21} J_{11} + \partial_{\bar y \bar y } {J_{21}}^2 & \partial_{\bar x \bar x} J_{11} J_{21} + \partial_{\bar x \bar y} J_{22} J_{11} + \partial_{\bar x \bar y} J_{12} J_{21} + \partial_{\bar y \bar y} J_{22} J_{21} \\

\partial_{\bar x \bar x} J_{11} J_{21} + \partial_{\bar x \bar y} J_{21} J_{12} + \partial_{\bar x \bar y} J_{11} J_{22} + \partial_{\bar y \bar y} J_{21} J_{22} & \partial_{\bar x \bar x} (J_{12})^2 +2 \partial_{\bar x \bar y} J_{12} J_{22} + \partial_{\bar y \bar y } {J_{22}}^2

\end{bmatrix}

$$

$${\alpha}_{11}=(\partial_{\bar x \bar x}(J_{11})^2)+\partial_{\bar x \bar y}J_{11}J_{21}+\partial_{\bar y \bar x}J_{11}J_{21}+\partial_{\bar y \bar x}J_{11}J_{21}+\partial_{\bar y \bar y}(J_{21})^2 $$

Egm6322.s09.three.liu 16:46, 24 April 2009 (UTC)

Assume $$\partial_{\bar x \bar y}=\partial_{\bar y \bar x}$$

With conversions, it can be shown that our expression is equivalent to that of Lapidus and Pinder: $$ \partial_{xx}=\partial_{\bar x \bar x}(J_{11})^2+\partial_{\bar x \bar y}J_{11}J_{21}+\partial_{\bar y \bar y}(J_{21})^2=\partial_{\xi \xi}(\phi_{x})^2+2\partial_{\xi \eta}\phi_{x}\psi_{x}+\partial_{\eta \eta}(\psi_{x})^2$$

This proof completed in one of the homework assignments, found here.

 It is best to report the assignment right here, instead of elsewhere in the report, since that would create some confusion while reading the report. At least you should create in internal link to help the readers jump to the part of the report where the proof was done; see further below. To reedit and resubmit. Egm6322.s09 13:04, 23 February 2009 (UTC)

Link added as requested. --Egm6322.s09.lapetina 20:23, 6 March 2009 (UTC) Homework: Transform the Laplace Equation into polar coordinates, and compare to published results

Egm6322.s09.Three.nav 13:29, 24 April 2009 (UTC)

The Laplace's Equation is often encountered in electrostatics, heat and mass transfer theory and in other areas of mechanics and physics. Simply worded, the Laplace's Equation states that div(grad u)=0.

It has the following mathematical form:

In Cartesian coordinates (x,y)$$: \ \ u_{xx}+u_{yy}= 0 $$

In Polar coordinates (r,$$\theta$$)$$: \ \ \ \ \ u_{rr}+\frac{u_{r}}{r}+\frac{u_{\theta\theta} }{r^{2}}=0 $$

where x= rcos($$\theta$$) and y= rsin($$\theta$$)

$$ Recognize\ that\ u_{xx}+u_{yy}= \begin{bmatrix} \partial _{x}\ \partial_{y} \end{bmatrix} \begin{bmatrix} \partial _{x} \\ \partial_{y}\end{bmatrix} \left \{u \right \} =\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}JJ^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix}\left \{u \right \}$$

where $$ J= \begin{bmatrix} \frac {\partial r}{\partial x} \ \frac {\partial r}{\partial y}\\ \frac {\partial \theta }{\partial x} \ \frac{\partial \theta}{\partial y} \end{bmatrix}$$

Denote cos($$\theta$$)=C and sin($$\theta$$)=S

$$ \Rightarrow J= \begin{bmatrix} \ C \ S \\\frac{-S}{r} \ \frac{C}{r} \end{bmatrix}$$

Differentiating by parts

$$\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}JJ^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix}\left \{u \right \}= \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\left \langle JJ^{T}\right \rangle\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}+\left \langle \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix} J\right \rangle J^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}$$---(A)

where the terms in the angled brackets are kept fixed throughout differentiation.

Evaluating the RHS of equation (A) term by term '''First term= $$\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\left \langle JJ^{T}\right \rangle\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \} $$'''(1)

$$ JJ_{T}=\begin{bmatrix} \ C \ S \\ \frac{-S}{r} \ \frac{C}{r} \end{bmatrix}\begin{bmatrix} \ C \ -S/r \\ S \ \ C/r \end{bmatrix}= \begin{bmatrix} 1 \  0 \\  0 \ \frac{1}{r^{2}} \end{bmatrix}$$

$$\Rightarrow \ (1)= \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\left \langle \begin{bmatrix} 1 \ \ \ 0 \\ \ 0 \ \frac{1}{r^{2}} \end{bmatrix} \right \rangle\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}$$

$$\Rightarrow\ (1)=\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\begin{bmatrix} \partial_{r} \\ \frac{\partial_{\theta}}{r^2}\end{bmatrix}\left \{u \right \}$$

$$\Rightarrow\ (1)= \frac {\partial^2{u}}{\partial{r^2}}+\frac{1}{r^2}\frac{\partial^2{u}}{\partial{\theta^2}} $$ Second term= $$\left \langle \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix} J\right \rangle J^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}'''$$-(2)

By definition of the Jacobian, note that the term in the angled brackets is the same as $$\begin{bmatrix} \partial_{x} \ \partial_{y} \end{bmatrix}$$

$$\Rightarrow\ (2)= \left (\begin{bmatrix} \partial_{x} \ \partial_{y} \end{bmatrix}J^{T} \right ) \begin{Bmatrix} \partial_{r}\\ \partial_{\theta} \end{Bmatrix} \left(u\right )$$

Evaluating the term in angled brackets for equation(2)

$$\left \langle \begin{bmatrix} \partial_{x} \ \partial_{y} \end{bmatrix}J^{T} \right \rangle= \begin{bmatrix} \left(\partial_{x}C+\partial_{y}S \right) \ \left( -\partial_{x}\frac {S}{r}+\partial_{y}\frac{C}{r} \right) \end{bmatrix}$$

$$= \begin{bmatrix} \frac{1}{r} \ 0 \end{bmatrix}$$

$$\Rightarrow\ (2)= \begin{bmatrix} \frac{1}{r} \ 0 \end{bmatrix}\begin{bmatrix} \partial_{r} \\ \partial_{\theta} \end{bmatrix} \left \{u \right \} $$

$$\Rightarrow\ (2)= \frac{1}{r} \partial_{r}u$$

Finally putting terms (1) and (2) together into equation (A)

$$u_{xx}+u_{yy}= \begin{bmatrix} \partial _{x}\ \partial_{y} \end{bmatrix} \begin{bmatrix} \partial _{x} \\ \partial_{y}\end{bmatrix} \left \{u \right \} =\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}JJ^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix}\left \{u \right \}$$

$$\Rightarrow\ (A)= \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}JJ^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix}\left \{u \right \}= \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\left \langle JJ^{T}\right \rangle\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}+\left \langle \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix} J\right \rangle J^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}$$

$$\Rightarrow\ u_{xx}+u_{yy}= \frac {\partial^2{u}}{\partial{r^2}}+\frac{1}{r^2}\frac{\partial^2{u}}{\partial{\theta^2}}+\frac{1}{r}\frac {\partial{u}}{\partial{r}}$$

Hence we have proved, using transformation of coordinates, that:

$$ u_{xx}+u_{yy} =\ u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^2}u_{\theta\theta}$$

Comparing with published results, we see that our derived result matches perfectly.

The Jacobian for Linear and Affine Maps
Egm6322.s09.Three.ge 17:32, 24 April 2009 (UTC)

The Jacobian matrix is a function of $$x$$ and $$y$$ in general.

$$\underline J \left ( x,y \right )$$ in general is not a constant matrix.

However, $$\underline J $$ is a constant matrix for linear and affine maps. In summary,

$$\begin{Bmatrix} \underline{x}\\ \underline{y} \end{Bmatrix} = \underline{E} \begin{Bmatrix} x\\ y \end{Bmatrix} + \underline{F}$$

Where:

$$\underline{J}=\underline{E}$$

An Example of Affine Mapping
Egm6322.s09.bit.sahin 16:23, 24 April 2009 (UTC); Egm6322.s09.Three.ge 17:32, 24 April 2009 (UTC)

Example of affine map.



$$\begin{matrix} \widehat{\underline{i}}=m \widehat{i}+n \widehat{j}\\ \widehat{\underline{j}}=p \widehat{i}+q \widehat{j} \end{matrix}$$

To get the transformation matrix, note that:

$$\overrightarrow{OP}=\overrightarrow{O\overline{O}}+\overrightarrow{\overline{O}\mathbf{P}}$$

Also note:

$$\overrightarrow{O\mathbf{P}}=x{\mathbf{i}}+y{\mathbf{j}}$$

$$\overrightarrow{O\overline{O}}=x_{o}\mathbf{i}+y_{o}\mathbf{j}$$

$$\overrightarrow{O\mathbf{P}}=\bar{x}{\mathbf{i}}+\bar{y}{\mathbf{j}}=\bar{x}(m\mathbf{i}+n\mathbf{j})+\bar{y}(p\mathbf{i}+q\mathbf{j})$$

$$=(m\bar{x}+p\bar{y})\mathbf{i}+(n\bar{x}+q\bar{y})\mathbf{j}$$

where:

$$x=m\bar{x}+p\bar{y}+x_0$$

$$y=n\bar{x}+q\bar{y}+x_0$$

Using the matrix form:

$$\begin{Bmatrix}x\\y\end{Bmatrix}= \begin{bmatrix}m & p\\n & q\end{bmatrix}\begin{Bmatrix}\bar{x}\\\bar{y}\end{Bmatrix}+ \begin{Bmatrix}x_0\\y_0\end{Bmatrix}$$

then,

$$x=x_{0}+\overline{x}m+\overline{y}p$$

$$y=y_{0}+\overline{x}n+\overline{y}q$$

from the equation system $$\overline{x}$$ and $$\overline{y}$$ are obtained as

$$\overline{x}=\frac{q}{\left (mq-np \right )}x-\frac{p}{\left (mq-np \right )}y-\frac{q}{\left (mq-np \right )}x_{0}+\frac{p}{\left (mq-np \right )}y_{0}$$

$$\overline{y}=\frac{m}{\left (mq-np \right )}y-\frac{n}{\left (mq-np \right )}x+\frac{n}{\left (mq-np \right )}x_{0}-\frac{m}{\left (mq-np \right )}y_{0}$$

Recall:

$$\begin{Bmatrix} \overline{x}\\ \overline{y} \end{Bmatrix} = \underline{E} \begin{Bmatrix} x\\ y \end{Bmatrix} + \underline{F}$$

$$\begin{Bmatrix} \overline{x}\\ \overline{y}

\end{Bmatrix} =\begin{bmatrix} e_{11} &e_{12} \\ e_{21} &e_{22} \end{bmatrix}\begin{Bmatrix} x\\ y

\end{Bmatrix}+\begin{Bmatrix} f_{11}\\ f_{21}

\end{Bmatrix}$$

therefore,

$$\overline{x}=e_{11}x+e_{12}y+f_{11}$$

$$\overline{y}=e_{21}x+e_{22}y+f_{21}$$

Hence the coefficient are,

$$e_{11}=\frac{q}{\left (mq-np \right )}$$

$$e_{12}=\frac{-p}{\left (mq-np \right )}$$

$$f_{11}=\frac{py_{0}-qx_{0}}{\left (mq-np \right )}$$

$$e_{21}=\frac{m}{\left (mq-np \right )}$$

$$e_{22}=\frac{-n}{\left (mq-np \right )}$$

$$f_{21}=\frac{nx_{0}-my_{0}}{\left (mq-np \right )}$$

Eventually we obtain $$\underline{E}$$ and $$\underline{F}$$ as following

$$\underline{E}=\begin{bmatrix} \frac{q}{\left (mq-np \right )}&\frac{-p}{\left (mq-np \right )} \\ \frac{m}{\left (mq-np \right )}&\frac{-n}{\left (mq-np \right )} \end{bmatrix}$$

$$\underline{F}=\begin{Bmatrix} \frac{py_{0}-qx_{0}}{\left (mq-np \right )}\\ \frac{nx_{0}-my_{0}}{\left (mq-np \right )} \end{Bmatrix}$$

Exercise 1 below shows this with values for $$n,m,p,q x_0$$ and $$y_0$$.

 Not clear. To redo and resubmit. Re-explain and redo in class if still have problems. Egm6936.s09 02:20, 21 February 2009 (UTC)

Changes needed were made. Egm6322.s09.bit.sahin 15:52, 10 April 2009 (UTC)

Homework Problems
Homework: Find the Jacobian Matrix for the transformation from Cartesian to Polar Coordinate

One example of nonlinear coordinate transformation is from Cartesian Coordinates to Polar coordinates:

e.g.

Cartesian coordinates can be expressed as:$$\left ( x,y \right )$$

while polar coordinates are expressed as:$$(r,\theta)$$.

The relationship between the two coordinates is:

$$x=r\cos\theta=\bar{x}\cos\bar{y}$$

$$y=r\sin\theta=\bar{x}\sin\bar{y}$$

To find J: $$ J=\begin{bmatrix} \frac{\partial {\bar x_i}}   {\partial {x_j}} \end{bmatrix} $$ = $$\begin{bmatrix} J_{11}     & J_{12}     \\ J_{21}     & J_{22} \end{bmatrix}$$

$$x=rcos\Theta =\bar{x_1}cos\bar{x_2}=x_1(\bar{x_1}, \bar{x_2})$$

$$y=rsin\Theta =\bar{x_1}sin\bar{x_2}=x_2(\bar{x_1}, \bar{x_2})$$

$$\bar{x_1}=r=(x^2+y^2)^{1/2}=\bar{x_1}(x_1, x_2)$$

$$\bar{x_2}=\Theta =tan^{-1}(\frac{y}{x})=\bar{x_2}(x_1, x_2)$$

$$J_{11}=\frac{\partial \bar{x}_1}{\partial x}=\frac{\partial r}{\partial x}=(x^2+y^2)^{-1/2}x$$

$$J_{12}=\frac{\partial \bar{x}_2}{\partial r}=\frac{\partial r}{\partial x}=(x^2+y^2)^{-1/2}y$$

$$J_{21}=\frac{\partial \bar{x}_2}{\partial x_1}=\frac{\partial \Theta}{\partial x}=\frac{\partial \tan^{-1}(s)}{\partial s}\frac{\partial s}{\partial x}$$

where: $$s:=\frac{y}{x}$$

It becomes clear that one must take the derivative of $$tan^{-1}(\frac {y} {x})$$.

Recall: $$y=tan^{-1}x\Rightarrow x=tan(y)=\frac{sin(y)}{cos(y)}$$

Find $$\frac{d y}{d x}$$

$$\frac{d y}{d x}=\frac{1}{\frac{d x}{d y}}$$

$$\frac{d x}{d y}=\frac{(cosy)^2-siny(-siny)}{cos^2y}=\frac{1}{cos^2y}=1+tan^2y=1+x^2$$

Therefore,

$$\frac{d y}{d x}=cos^2y$$

$$tan(y)=\frac{sin(y)}{cos(y)}$$

Expressing J in terms of $$\begin{pmatrix}r,&\Theta \end{pmatrix} $$

$$J_{11}=\frac{x} {r}=\frac{rcos\Theta }{r}=cos\Theta$$

$$J_{12}=\frac{y}{\left (x^{2}+y^{2} \right )^{1/2}}=\frac{rsin\theta }{r}=sin\theta $$

$$J_{21}=\frac{1}{1+\left (\frac{y}{x} \right )^{2}}\left (-\frac{y}{x^{2}} \right )=\frac{-y}{x^{2}+y^{2}}=\frac{-rsin\theta }{r^{2}}=-\frac{sin\theta }{r}$$

$$J_{22}=\frac{\partial \overline{x}_{2}}{\partial x_{2}}=\frac{\partial \theta }{\partial y}$$

Remembering that $$\theta =tan^{-1}\left (\frac{y}{x} \right )$$, If we define $$s=y/x$$ then

$$\frac{\partial \theta }{\partial y}=\frac{\partial }{\partial s}tan^{-1}(s)\frac{\partial s}{\partial y}=\frac{1}{1+\left (y/x \right )^{2}}\left (\frac{1}{x} \right )=\frac{x}{x^{2}+y^{2}}=\frac{rcos\theta }{r^{2}}=\frac{cos\theta }{r}$$

As a result, we obtain Jacobian matrix as

$$J= \begin{bmatrix} J_{11}     & J_{12}     \\ J_{21}     & J_{22} \end{bmatrix} = \begin{bmatrix} cos\theta & sin\theta \\ -\frac{sin\theta }{r}&\frac{cos\theta }{r} \end{bmatrix}=\frac{1}{r}\begin{bmatrix} rcos\theta &rsin\theta  \\ -sin\theta &cos\theta \end{bmatrix}$$

 There is an error in the derivative of $$\displaystyle y = \tan^{-1} x$$ above. Egm6322.s09 17:11, 23 February 2009 (UTC)

 Put your reply inside the above comment box, and sign with 4-tilde command. There is still one more "misprint"; see further below for a hint about the same misprint. Egm6322.s09 10:04, 2 April 2009 (UTC)

 This part has been changed. Egm6322.s09.three.liu 14:33, 10 April 2009 (UTC)

Homework: Find Complete Expression for Eqn. 1.2.6, 1.2.7, & 1.2.8 in Lapidus & Pinder [1982] --Egm6322.s09.xyz 16:19, 4 April 2009 (UTC)

Compare the expansion of Equation 1 on page 12-3 to LP (Lapidus & Pinder, 1982), found on page 14 of their book. Translate their notation into ours.

Obtain the first partial derivative under transform of coordinates: Eq(1) in P12-3. (using chain rule)

In LP, P14: Obtain second partial derivative (using chain rule), translate it into new coordinate (ours).

Where ours: $$(\bar x_1, \bar x_2)=(\bar x, \bar y)$$

$$\bar {x_i}=\bar{x_i} (x_1,x_2) $$    (i=1,2,...)

But for Lapidus and Pinder: $$\xi=\phi(x,y)$$

$$\eta=\psi(x,y)$$

LP (1.2.5 ab): First partial derivative

LP (1.2.6-8): Second partial derivative

Obtain the complete form:

$$u_{xx}=u_{\xi\xi}(\phi_{x})^2+2 u_{\xi \eta} \phi_{x} \psi_{x}+u_{\eta \eta}+(\psi_{x})^2+...$$

Recall that a 2nd order PDE with two (2) independent variables has the following form (see L&P, eqn. 1.2.1):

$$a \left( \cdot \right) u_{xx} + 2b \left( \cdot \right) u_{xy} + c \left( \cdot \right) u_{yy} + d \left( \cdot \right) u_x + e \left( \cdot \right) u_y + f \left( \cdot \right) u + g \left( \cdot \right) = 0$$

where $$\xi = \phi \left(x,y \right)$$ and $$\eta = \psi \left(x,y \right)$$

Chain rule is used to develop the following expressions for the first and second order derivatives with respect to x,y:

$$\blacktriangleright u_x = u_{\xi}\phi_x + u_{\eta}\psi_x$$

$$\blacktriangleright u_y = u_{\xi}\phi_y + u_{\eta}\psi_y$$

$$\blacktriangleright u_{xx} = u_{\xi\xi}\phi_x^2 + 2u_{\xi\eta}\phi_x\psi_x + u_{\eta\eta}\psi_x^2 + u_{\xi}\phi_{xx} + u_{\eta}\psi_{xx}$$ (Eqn. 1.2.6)

$$\blacktriangleright u_{xy} = u_{\xi\xi}\phi_x\phi_y + u_{\xi\eta}\left( \phi_x\psi_y + \phi_y\psi_x \right) + u_{\eta\eta}\psi_x\psi_y + u_{\xi}\phi_{xy} + u_{\eta}\psi_{xy}$$ (Eqn. 1.2.7)

$$\blacktriangleright u_{yy} = u_{\xi\xi}\phi_y^2 + 2u_{\xi\eta}\phi_y\psi_y + u_{\eta\eta}\psi_y^2 + u_{\xi}\phi_{yy} + u_{\eta}\psi_{yy}$$ (Eqn. 1.2.8)

Solution:

The complete form of the equations 1.2.6 to 1.2.8 from Lapidus and Pinder:

Let,

$$  \xi =\phi (x,y)$$

and

$$ \eta =\psi (x,y) $$

$$\therefore $$ by chain rule we have

$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial x}$$

and

$$\frac{\partial u}{\partial y}=\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y}$$

EXPRESSION FOR $$\frac{\partial^2 u}{\partial x^2}$$

We Know that

$$\frac{\partial^2 u}{\partial x^2}=\frac{\partial }{\partial x}\left (\frac{\partial u}{\partial x} \right )$$ $$ \therefore$$ from chain rule we have

fex $$ \frac{\partial^2 u}{\partial x^2}=\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial x} \right )\frac{\partial \phi}{\partial x} + \frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial x} \right )\frac{\partial \psi}{\partial x}$$

$$\frac{\partial^2 u}{\partial x^2}=\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial x} \right )\frac{\partial \phi}{\partial x} + \frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial x} \right )\frac{\partial \psi}{\partial x} $$

Let,

$$\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial x} \right )\frac{\partial \phi}{\partial x} =I$$

and

$$\frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial x} \right )\frac{\partial \psi}{\partial x}=II

$$

Differentiating $$I$$ We have

$$I=\frac{\partial }{\partial \xi}\left(\frac{\partial u}{\partial \xi}\right)\left(\frac{\partial \phi}{\partial x}\right)^2 + \frac{\partial }{\partial \xi}\left(\frac{\partial \phi}{\partial x}\right )\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x}$$

$$ + \frac{\partial }{\partial \xi}\left(\frac{\partial u}{\partial \eta}\right)\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left(\frac{\partial \psi}{\partial x}\right)\frac{\partial \phi}{\partial x} $$

and

$$II=\frac{\partial }{\partial \eta}\left(\frac{\partial u}{\partial \xi}\right)\frac{\partial \phi}{\partial x}\frac{\partial \psi}{\partial x} + \frac{\partial }{\partial \eta}\left(\frac{\partial \phi}{\partial x}\right )\frac{\partial u}{\partial \xi}\frac{\partial \psi}{\partial x} $$

$$+\frac{\partial }{\partial \eta}\left(\frac{\partial u}{\partial \eta}\right)\left(\frac{\partial \psi}{\partial x}\right)^2 + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left(\frac{\partial \psi}{\partial x}\right)\frac{\partial \psi}{\partial x}

$$

$$\Rightarrow I=\frac{\partial^2 u}{\partial \xi^2}\left(\frac{\partial \phi}{\partial x}\right)^2 + \frac{\partial }{\partial \xi}\left(\frac{\partial \phi}{\partial x}\right )\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x} + \frac{\partial^2 u}{\partial \xi{\partial \eta}}\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left(\frac{\partial \psi}{\partial x}\right)\frac{\partial \phi}{\partial x}

$$

$$\Rightarrow II=\frac{\partial^2 u}{\partial \xi{\partial \eta}}\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x} + \frac{\partial }{\partial \eta}\left(\frac{\partial \phi}{\partial x}\right )\frac{\partial u}{\partial \xi}\frac{\partial \psi}{\partial x} + \frac{\partial^2 u}{\partial \eta^2}\left(\frac{\partial \psi}{\partial x}\right)^2 + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left(\frac{\partial \psi}{\partial x}\right)\frac{\partial \psi}{\partial x} $$

The complete expansion of $$u_{xx}=$$

$$ I+II=\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2  u}{\partial \xi^2}\left(\frac{\partial \phi}{\partial x}\right)^2+2\frac{\partial^2 u}{\partial \xi{\partial \eta}}\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x}+\frac{\partial^2 u}{\partial \eta^2}\left(\frac{\partial \psi}{\partial x}\right)^2$$

$$+\frac{\partial }{\partial \xi}\left(\frac{\partial \phi}{\partial x}\right )\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x} +\frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left(\frac{\partial \psi}{\partial x}\right)\frac{\partial \phi}{\partial x}+\frac{\partial }{\partial \eta}\left(\frac{\partial \phi}{\partial x}\right )\frac{\partial u}{\partial \xi}\frac{\partial \psi}{\partial x}+\frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left(\frac{\partial \psi}{\partial x}\right)\frac{\partial \psi}{\partial x}$$

EXPRESSION FOR $$\frac{\partial^2 u}{\partial y^2}$$

Similarly,

$$\frac{\partial^2 u}{\partial y^2}=\frac{\partial }{\partial y}\left (\frac{\partial u}{\partial y} \right )$$ $$ \therefore$$ from chain rule we have

$$ \frac{\partial^2 u}{\partial y^2}=\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial y} \right )\frac{\partial \phi}{\partial y} + \frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial y} \right )\frac{\partial \psi}{\partial y}$$

$$\frac{\partial^2 u}{\partial y^2}=\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \phi}{\partial y} + \frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \psi}{\partial y} $$

Let,

$$\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \phi}{\partial y} =I$$

and

$$\frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \psi}{\partial y}=II

$$

Differentiating $$I$$ We have

$$I=\frac{\partial }{\partial \xi}\left(\frac{\partial u}{\partial \xi}\right)\left(\frac{\partial \phi}{\partial y}\right)^2 + \frac{\partial }{\partial \xi}\left(\frac{\partial \phi}{\partial y}\right )\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y}$$

$$ + \frac{\partial }{\partial \xi}\left(\frac{\partial u}{\partial \eta}\right)\frac{\partial \psi}{\partial y}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left(\frac{\partial \psi}{\partial y}\right)\frac{\partial \phi}{\partial y} $$

and

$$II=\frac{\partial }{\partial \eta}\left(\frac{\partial u}{\partial \xi}\right)\frac{\partial \phi}{\partial y}\frac{\partial \psi}{\partial y} + \frac{\partial }{\partial \eta}\left(\frac{\partial \phi}{\partial y}\right )\frac{\partial u}{\partial \xi}\frac{\partial \psi}{\partial y} $$

$$+\frac{\partial }{\partial \eta}\left(\frac{\partial u}{\partial \eta}\right)\left(\frac{\partial \psi}{\partial y}\right)^2 + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left(\frac{\partial \psi}{\partial y}\right)\frac{\partial \psi}{\partial y}

$$

$$\Rightarrow I=\frac{\partial^2 u}{\partial \xi^2}\left(\frac{\partial \phi}{\partial y}\right)^2 + \frac{\partial }{\partial \xi}\left(\frac{\partial \phi}{\partial y}\right )\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial^2 u}{\partial \xi{\partial \eta}}\frac{\partial \psi}{\partial y}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left(\frac{\partial \psi}{\partial y}\right)\frac{\partial \phi}{\partial y}

$$

$$\Rightarrow II=\frac{\partial^2 u}{\partial \xi{\partial \eta}}\frac{\partial \psi}{\partial y}\frac{\partial \phi}{\partial y} + \frac{\partial }{\partial \eta}\left(\frac{\partial \phi}{\partial y}\right )\frac{\partial u}{\partial \xi}\frac{\partial \psi}{\partial y} + \frac{\partial^2 u}{\partial \eta^2}\left(\frac{\partial \psi}{\partial y}\right)^2 + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left(\frac{\partial \psi}{\partial y}\right)\frac{\partial \psi}{\partial y} $$

The complete eypansion of $$u_{yy}=$$

$$ I+II=\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2  u}{\partial \xi^2}\left(\frac{\partial \phi}{\partial y}\right)^2+2\frac{\partial^2 u}{\partial \xi{\partial \eta}}\frac{\partial \psi}{\partial y}\frac{\partial \phi}{\partial y}+\frac{\partial^2 u}{\partial \eta^2}\left(\frac{\partial \psi}{\partial y}\right)^2$$

$$+\frac{\partial }{\partial \xi}\left(\frac{\partial \phi}{\partial y}\right )\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} +\frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left(\frac{\partial \psi}{\partial y}\right)\frac{\partial \phi}{\partial y}+\frac{\partial }{\partial \eta}\left(\frac{\partial \phi}{\partial y}\right )\frac{\partial u}{\partial \xi}\frac{\partial \psi}{\partial y}+\frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left(\frac{\partial \psi}{\partial y}\right)\frac{\partial \psi}{\partial y}$$

EXPRESSION FOR $$\frac{\partial^2 u}{\partial x\partial y}$$

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

$$\Rightarrow \frac{\partial^2 u}{\partial x\partial y}=\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \phi}{\partial x} + \frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \psi}{\partial x}$$

$$I=\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \phi}{\partial x}$$

$$II=\frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \psi}{\partial x}$$

$$\Rightarrow I=\frac{\partial^2 u }{\partial \xi^2}\frac{\partial \phi}{\partial x}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \xi}\frac{\partial }{\partial \xi}\left (\frac{\partial \phi}{\partial y} \right )\frac{\partial \phi}{\partial x} + \frac{\partial^2 u}{\partial \xi\partial \eta}\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left (\frac{\partial \psi}{\partial y} \right )\frac{\partial \phi}{\partial x} $$

$$II=\frac{\partial^2 u}{\partial \xi\partial \eta}\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \xi}\frac{\partial }{\partial \eta}\left (\frac{\partial \phi}{\partial y} \right )\frac{\partial \psi}{\partial x} + \frac{\partial^2 u }{\partial \eta^2}\frac{\partial \psi}{\partial y}\frac{\partial \psi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left (\frac{\partial \psi}{\partial y} \right )\frac{\partial \psi}{\partial x} $$

$$\Rightarrow \frac{\partial^2 u}{\partial x\partial y}=I+II=\frac{\partial^2 u }{\partial \xi^2}\frac{\partial \phi}{\partial x}\frac{\partial \phi}{\partial y}+ \frac{\partial^2 u}{\partial \xi\partial \eta}\left(\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x}+\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x}\right)+\frac{\partial^2 u }{\partial \eta^2}\frac{\partial \psi}{\partial y}\frac{\partial \psi}{\partial x}$$

$$+\frac{\partial u}{\partial \xi}\frac{\partial }{\partial \xi}\left (\frac{\partial \phi}{\partial y} \right )\frac{\partial \phi}{\partial x}+ \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left (\frac{\partial \psi}{\partial y} \right )\frac{\partial \phi}{\partial x}+ \frac{\partial u}{\partial \xi}\frac{\partial }{\partial \eta}\left (\frac{\partial \phi}{\partial y} \right )\frac{\partial \psi}{\partial x}+ \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left (\frac{\partial \psi}{\partial y} \right )\frac{\partial \psi}{\partial x}$$

The matrix form of the above expressions

We Have

$$\begin{bmatrix} \frac{\partial }{\partial x} \\ \frac{\partial }{\partial y} \end{bmatrix} =J^T\begin{bmatrix} \frac{\partial }{\partial \overline x} \\ \frac{\partial }{\partial \overline y} \end{bmatrix}$$

Let this be 1

and

$$\begin{bmatrix} \frac{\partial }{\partial x} \\ \frac{\partial }{\partial y} \end{bmatrix}

\begin{bmatrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial y} \end{bmatrix}= \begin{bmatrix} \partial_{xx} &\partial_{xy} \\ \partial_{yx} &\partial_{yy} \end{bmatrix}$$

Let this be 2

Also

$$\because \left (AB \right )^T=B^TA^T$$

$$\begin{bmatrix} \partial_x & \partial_y \end{bmatrix}= \begin{bmatrix} \partial_ \bar x & \partial_\bar y \end{bmatrix}J$$

Let this be 3

Applying equation 2 and 3 in equation 1 ,we have

$$\begin{bmatrix} \partial_x \\ \partial_y \end{bmatrix} \begin{bmatrix} \partial_x & \partial_y \end{bmatrix} = J^T \begin{bmatrix} \partial_\bar x \\ \partial_\bar y \end{bmatrix} \begin{bmatrix} \partial_\bar x & \partial_\bar y \end{bmatrix} J$$

$$\Rightarrow \begin{bmatrix} \partial_{xx} &\partial_{xy} \\ \partial_{yx} &\partial_{yy} \end{bmatrix} =\begin{bmatrix} J_{11} & J_{21}\\ J_{12} & J_{22} \end{bmatrix} \begin{bmatrix} \partial_{\bar x\bar x} &\partial_{\bar x\bar y} \\ \partial_{\bar y\bar x} &\partial_{\bar y\bar y} \end{bmatrix} \begin{bmatrix} J_{11} &J_{12} \\ J_{11}& J_{22} \end{bmatrix}$$

$$\Rightarrow \begin{bmatrix} \partial_{xx} &\partial_{xy} \\ \partial_{yx} &\partial_{yy} \end{bmatrix}$$

$$= \begin{bmatrix}

\partial_{\bar x \bar x} (J_{11})^2 +2 \partial_{\bar x \bar y} J_{21} J_{11} + \partial_{\bar y \bar y } {J_{21}}^2 & \partial_{\bar x \bar x} J_{11} J_{21} + \partial_{\bar x \bar y} J_{22} J_{11} + \partial_{\bar x \bar y} J_{12} J_{21} + \partial_{\bar y \bar y} J_{22} J_{21} \\

\partial_{\bar x \bar x} J_{11} J_{21} + \partial_{\bar x \bar y} J_{21} J_{12} + \partial_{\bar x \bar y} J_{11} J_{22} + \partial_{\bar y \bar y} J_{21} J_{22} & \partial_{\bar x \bar x} (J_{12})^2 +2 \partial_{\bar x \bar y} J_{12} J_{22} + \partial_{\bar y \bar y } {J_{22}}^2

\end{bmatrix}$$

$$\therefore u_{xx}=\partial_{\bar x \bar x}(J_{11})^2+\partial_{\bar x \bar y}J_{11}J_{21}+\partial_{\bar y \bar y}(J_{21})^2=\partial_{\xi \xi}(\phi_{x})^2+2\partial_{\xi \eta}\phi_{x}\psi_{x}+\partial_{\eta \eta}(\psi_{x})^2$$

$$u_{xy}=\partial_{\bar x \bar x} J_{11} J_{21} + \partial_{\bar x \bar y} J_{22} J_{11} + \partial_{\bar x \bar y} J_{12} J_{21} + \partial_{\bar y \bar y} J_{22} J_{21} =u_{\xi\xi}\phi_x\phi_y + u_{\xi\eta}\left( \phi_x\psi_y + \phi_y\psi_x \right) + u_{\eta\eta}\psi_x\psi_y $$

$$u_{yy}=\partial_{\bar x \bar x} (J_{12})^2 +2 \partial_{\bar x \bar y} J_{12} J_{22} + \partial_{\bar y \bar y } {J_{22}}^2 =u_{\xi\xi}\phi_y^2 + 2u_{\xi\eta}\phi_y\psi_y + u_{\eta\eta}\psi_y^2 + $$

Egm6322.s09.bit.gk 20:32, 24 April 2009 (UTC)

 The proof by the matrix-operator method was incomplete (wrong); review the lecture notes. Egm6322.s09 13:04, 23 February 2009 (UTC)

 As mentioned in class, there are missing first-order terms in the proof by the matrix-operator method as compared to using chain-rule done further above. Needs correction. Egm6322.s09 16:30, 2 April 2009 (UTC)

Pierre-Simon Laplace (1749-1827)

Egm6322.s09.Three.nav 13:32, 24 April 2009 (UTC)

Pierre-Simon Laplace was born on 23 March, 1747 in Beaumont-en-Auge, Normandy, France. At the age of 16, while in university studying theology, he discovered a natural talent and a deep love for mathematics. He then set sail for Paris to become a student under the guidance of D'alembert. At the tender age of 19, he began produced a remarkable stream of Mathematical papers. His research was as diverse as it was trendsetting. In 1773, at the age of 24, he was elected an adjoint in the Académie des Sciences (French Academy of Sciences). It was here that he maintained many superior posts and was widely considered one of the Academy's best mathematicians.

The Laplace Equation is first mentioned in Laplace's most important work, Traité de Mécanique Céleste. Though we now name this equation after Laplace, it was actually known well before the time of Laplace. Laplace died on the morning of March 5, 1827. The Academy, seldom known for cancelling any of its meetings, did so on that day as a mark of respect for him.

Laplace

Homework: Transform the Laplace Equation into polar coordinates, and compare to published results

The Laplace's Equation is often encountered in electrostatics, heat and mass transfer theory and in other areas of mechanics and physics. Simply worded, the Laplace's Equation states that div(grad u)=0.

It has the following mathematical form:

In Cartesian coordinates (x,y)$$: \ \ u_{xx}+u_{yy}= 0 $$

In Polar coordinates (r,$$\theta$$)$$: \ \ \ \ \ u_{rr}+\frac{u_{r}}{r}+\frac{u_{\theta\theta} }{r^{2}}=0 $$

where x= rcos($$\theta$$) and y= rsin($$\theta$$)

$$ Recognize\ that\ u_{xx}+u_{yy}= \begin{bmatrix} \partial _{x}\ \partial_{y} \end{bmatrix} \begin{bmatrix} \partial _{x} \\ \partial_{y}\end{bmatrix} \left \{u \right \} =\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}JJ^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix}\left \{u \right \}$$

where $$ J= \begin{bmatrix} \frac {\partial r}{\partial x} \ \frac {\partial r}{\partial y}\\ \frac {\partial \theta }{\partial x} \ \frac{\partial \theta}{\partial y} \end{bmatrix}$$

Denote cos($$\theta$$)=C and sin($$\theta$$)=S

$$ \Rightarrow J= \begin{bmatrix} \ C \ S \\\frac{-S}{r} \ \frac{C}{r} \end{bmatrix}$$

Differentiating by parts

$$\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}JJ^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix}\left \{u \right \}= \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\left \langle JJ^{T}\right \rangle\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}+\left \langle \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix} J\right \rangle J^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}$$---(A)

where the terms in the angled brackets are kept fixed throughout differentiation.

Evaluating the RHS of equation (A) term by term '''First term= $$\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\left \langle JJ^{T}\right \rangle\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \} $$'''(1)

$$ JJ_{T}=\begin{bmatrix} \ C \ S \\ \frac{-S}{r} \ \frac{C}{r} \end{bmatrix}\begin{bmatrix} \ C \ -S/r \\ S \ \ C/r \end{bmatrix}= \begin{bmatrix} 1 \  0 \\  0 \ \frac{1}{r^{2}} \end{bmatrix}$$

$$\Rightarrow \ (1)= \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\left \langle \begin{bmatrix} 1 \ \ \ 0 \\ \ 0 \ \frac{1}{r^{2}} \end{bmatrix} \right \rangle\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}$$

$$\Rightarrow\ (1)=\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\begin{bmatrix} \partial_{r} \\ \frac{\partial_{\theta}}{r^2}\end{bmatrix}\left \{u \right \}$$

$$\Rightarrow\ (1)= \frac {\partial^2{u}}{\partial{r^2}}+\frac{1}{r^2}\frac{\partial^2{u}}{\partial{\theta^2}} $$ Second term= $$\left \langle \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix} J\right \rangle J^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}'''$$-(2)

By definition of the Jacobian, note that the term in the angled brackets is the same as $$\begin{bmatrix} \partial_{x} \ \partial_{y} \end{bmatrix}$$

$$\Rightarrow\ (2)= \left \lfloor \partial_{x} \ \partial_{y} \right \rfloor J^{T} \begin{Bmatrix} \partial_{r}\\ \partial_{\theta} \end{Bmatrix} \left(u\right)$$

Evaluating the term in angled brackets for equation(2)

$$\left \lfloor \partial_{x} \ \partial_{y} \right \rfloor J^{T} \begin{Bmatrix} \partial_{r}\\ \partial_{\theta} \end{Bmatrix} \left(u\right)= \begin{bmatrix} \left(\partial_{x}C+\partial_{y}S \right) \ \left (-\partial_{x}\frac {S}{r}+\partial_{y}\frac{C}{r} \right) \end{bmatrix}$$

$$= \begin{bmatrix} \frac{1}{r} \ 0 \end{bmatrix}$$

$$\Rightarrow\ (2)= \begin{bmatrix} \frac{1}{r} \ 0 \end{bmatrix}\begin{bmatrix} \partial_{r} \\ \partial_{\theta} \end{bmatrix} \left \{u \right \} $$

$$\Rightarrow\ (2)= \frac{1}{r} \partial_{r}u$$

Finally putting terms (1) and (2) together into equation (A)

$$u_{xx}+u_{yy}= \begin{bmatrix} \partial _{x}\ \partial_{y} \end{bmatrix} \begin{bmatrix} \partial _{x} \\ \partial_{y}\end{bmatrix} \left \{u \right \} =\begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}JJ^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix}\left \{u \right \}$$

$$\Rightarrow\ (A)= \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}JJ^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix}\left \{u \right \}= \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix}\left \langle JJ^{T}\right \rangle\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}+\left \langle \begin{bmatrix} \partial_{r}\ \partial_{\theta} \end{bmatrix} J\right \rangle J^{T}\begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left \{u \right \}$$

$$\Rightarrow\ u_{xx}+u_{yy}= \frac {\partial^2{u}}{\partial{r^2}}+\frac{1}{r^2}\frac{\partial^2{u}}{\partial{\theta^2}}+\frac{1}{r}\frac {\partial{u}}{\partial{r}}$$

Hence we have proved, using transformation of coordinates, that:

$$ u_{xx}+u_{yy} =\ u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^2}u_{\theta\theta}$$

Comparing with published results, we see that our derived result matches perfectly.

 If the terms in the angle brackets are kept fixed, then why do you differentiate the following:

$$\left \langle \begin{bmatrix} \partial_{x} \ \partial_{y} \end{bmatrix}J^{T} \right \rangle= \begin{bmatrix} \partial_{x}C+\partial_{y}S \ -\partial_{x}\frac {S}{r}+\partial_{y}\frac{C}{r} \end{bmatrix}$$

which should be better written as follows:

\displaystyle \left\lfloor \partial_{x} \ \partial_{y} \right\rfloor \mathbf J ^{T} =  \left\lfloor \left(	 \partial_{x}C+\partial_{y}S     \right) \       \left(         -\partial_{x}\frac {S}{r}+\partial_{y}\frac{C}{r}      \right) \right\rfloor $$ In the expression

$$\Rightarrow\ (2)= \left \langle \begin{bmatrix} \partial_{x} \ \partial_{y} \end{bmatrix}J^{T} \right \rangle \begin{bmatrix} \partial_{r}\\ \partial_{\theta} \end{bmatrix} \left\{u\right \}$$

just omit the angle brackets, and write as follows instead

\displaystyle \Rightarrow \    (2)   =   \left(      \left\lfloor	 \partial_{x} \ \partial_{y}      \right\rfloor      \mathbf J ^{T}   \right) \left\{ \begin{array}{l} \partial_{r} \\	 \partial_{\theta} \end{array} \right\} \left( u \right) $$ Then it makes sense to carry out the differentiation in

\displaystyle \left(     \left\lfloor	 \partial_{x} \ \partial_{y}      \right\rfloor      \mathbf J ^{T}   \right) $$

The necessary corrections have been made. By saying "the terms in the angled brackets are kept fixed", we mean to say that these terms are not subject to differentiation.

Egm6322.s09.Three.nav 19:00, 6 March 2009 (UTC)Egm6322.s09.Three.nav Egm6322.s09 17:51, 23 February 2009 (UTC)

 Move your reply to below my signature, then delete this comment box. Egm6322.s09 00:31, 2 April 2009 (UTC)

Rotation of Coordinates


Egm6322.s09.Three.nav 13:31, 24 April 2009 (UTC)

What is the form of the Jacobian when implementing a rotation of coordinates?

Consider one coordinate system, x-y (cartesian) and another coordinate system, $$\bar{x}-\bar{y}$$, obtained by rotating x-y clockwise by angle $$\theta$$ (as shown in the figure)

Assuming $$\widehat{\underline{i}}$$ and $$\widehat{\underline{j}}$$ are the unit vectors in the new coordinate system, $$\hat{i}$$ and $$\hat{j}$$ are the unit vectors in the old coordinate system

Then

$$ \widehat{\underline{i}}= icos\left(\theta \right)+ jsin\left(\theta \right)$$

$$ \widehat{\underline{j}}= -isin\left(\theta \right)+ jcos\left(\theta \right)$$

Comparing the above vector equations with those given in the example of Affine Mapping  considered previously,

$$\begin{matrix} \widehat{\underline{i}}=m \widehat{i}+n \widehat{j}\\ \widehat{\underline{j}}=p \widehat{i}+q \widehat{j} \end{matrix}$$

We see that m= cos$$\theta$$, n= sin$$\theta$$, p= -sin$$\theta$$, q= cos$$\theta$$

Hence the Jacobian ( J ) is given by

$$\underline{J}= \underline{E}= \begin{bmatrix} \frac{q}{\left (mq-np \right )}&\frac{-p}{\left (mq-np \right )} \\ \frac{m}{\left (mq-np \right )}&\frac{-n}{\left (mq-np \right )} \end{bmatrix}$$

$$\because$$mq-np= $$ \left(cos\theta \times cos\theta \right)$$ - $$\left(sin\theta \times -sin\theta \right)$$ = $$ cos^{2}\theta+ sin^{2}\theta$$ = 1

$$ \therefore \underline{J}= \underline{E}= \frac {1} {mq-np}\begin{bmatrix} \ \ q \ \ -p \\ -n \ \ \ m \end{bmatrix} = \frac {1} {1}\begin{bmatrix} cos\theta \ \ \ sin\theta \\ -sin\theta \ \ cos\theta \end{bmatrix} = \begin{bmatrix} cos\theta \ \ \ sin\theta \\ -sin\theta \ \ cos\theta \end{bmatrix}$$

 The above problem could be done simply and elegantly by the more general method described above in an example of affine mapping. Egm6322.s09 17:11, 23 February 2009 (UTC) Do not exactly understand how to apply the affine mapping to this example. How do you determine m,n,p,q? Egm6322.s09.Three.nav 19:22, 6 March 2009 (UTC)Egm6322.s09.Three.nav

 Add a figure to better explain the problem. The simpler method had been explained in class on several occasions. Ask me if not clear. Egm6322.s09 10:04, 2 April 2009 (UTC)

Simpler method has been implemented. Egm6322.s09.Three.nav 15:17, 10 April 2009 (UTC)

Example, Rotation and Translation
 The above section title does not make sense in view of the content below. The goal of what is done below is to obtain the expression of the second derivatives. Egm6322.s09 00:31, 2 April 2009 (UTC)

$$\alpha=\begin{Bmatrix} \partial_{x}\\ \partial_{y} \end{Bmatrix} \left \lfloor \partial_{x} \ \partial_{y} \right \rfloor$$

$$\begin{Bmatrix} \partial_{x}\\ \partial_{y} \end{Bmatrix}\left ( \lfloor \partial_{\overline{x}} \; \partial_{\overline{y}} \right \rfloor  \underline{J}(x,y))$$

If J is not a constant, take the product rule of the above function.

$$\alpha = \underline{J}^{T}\begin{Bmatrix} \partial_{\underline{x}}\\ \partial_{\underline{y}} \end{Bmatrix} \left \lfloor \partial_{\overline{x}} \; \partial_{\overline{y}} \right \rfloor \underline{J}+ \begin{Bmatrix} \partial_{x}\\ \partial_{y} \end{Bmatrix} \left \lfloor \partial_{\overline{x}} \; \partial_{\overline{y}} \right \rfloor \underline{J}$$

Let

$$ \ \beta=\underline{J}^{T}\begin{Bmatrix} \partial_{\underline{x}}\\ \partial_{\underline{y}} \end{Bmatrix} \left \lfloor \partial_{\overline{x}} \; \partial_{\overline{y}} \right \rfloor \underline{J}$$ and $$ \gamma= \begin{Bmatrix} \partial_{x}\\ \partial_{y} \end{Bmatrix} \left \lfloor \partial_{\overline{x}} \; \partial_{\overline{y}} \right \rfloor \underline{J}$$

In $$\beta$$ $$ \underline J $$ is treated as a constant while in $$\gamma$$ the operator $$\left \lfloor \partial_{\overline{x}} \;  \partial_{\overline{y}} \right \rfloor$$ is treated as a constant. Thus $$\beta$$ is 2nd order on the unknown function u, and $$\gamma$$ is a 1st order on u. This results in a non-zero $$\gamma$$ matrix of:

$$\gamma= \begin{bmatrix} \partial_{\overline{x}} \phi_{xx}+\partial_{\overline{y}} \psi_{xx} & \partial_{\overline{x}} \phi_{yx}+\partial_{\overline{y}} \psi_{yx}\\ \partial_{\overline{x}} \phi_{xy}+\partial_{\overline{y}} \psi_{xy} & \partial_{\overline{x}} \phi_{yy}+\partial_{\overline{y}} \psi_{yy}\end{bmatrix}$$

If J is a constant then the $$\gamma$$ matrix is equal to zero.

Homework: The Complete Form Of $$\gamma$$ --EGM6322.S09.TIAN 17:37, 24 April 2009 (UTC)

$$\gamma =$$ $$\begin{bmatrix} ( \partial_{\bar{x}}\phi_{xx} + \partial_{\bar{y}})\psi_{xx} & ( \partial_{\bar{x}}\phi_{xy} + \partial_{\bar{y}})\psi_{xy}\\ ( \partial_{\bar{x}}\phi_{xy} + \partial_{\bar{y}})\psi_{xy} & ( \partial_{\bar{x}}\phi_{xy} + \partial_{\bar{y}}\psi_{xy}) \end{bmatrix} $$

 Why was the above collapsible box needed? The expression for $$\displaystyle \gamma$$ is almost the same as the same expression for $$\displaystyle \gamma$$ above the collapsible box. Egm6322.s09 00:31, 2 April 2009 (UTC)

=Exercises =

Exercise 1
--Egm6322.s09.lapetina 01:59, 17 April 2009 (UTC)

Given $${\overline {\mathbf {i}}} = m \mathbf{i}+ n \mathbf{j} =2 \mathbf{i} + 1 \mathbf{j}$$ and $$\overline{\mathbf {j}} = p \mathbf{i}+ q \mathbf{j} =1 \mathbf{i} + 5 \mathbf{j}$$

and that

$$\left ( x_0, y_0 \right )= \left ( 3, 4 \right )$$, map $$\left ( \bar x,\bar y \right ) $$.

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #006600; background-color: #CCFFFF; text-align: left;"> Stick to the notation for tensors (underline = boldface) used in lectures, i.e., $$  \displaystyle \overline \mathbf i  = m  \mathbf i   + n  \mathbf j $$ and $$  \displaystyle \overline \mathbf j  = p  \mathbf i   + q  \mathbf j $$ Egm6322.s09 00:31, 2 April 2009 (UTC)

Edit made a second time.

--Egm6322.s09.lapetina 20:57, 10 April 2009 (UTC)

This answer will take the form:

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

=

\mathbf E

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

+

\mathbf F $$

This is an affine map, because 0 does not map to 0. In affine maps, $$\bar E$$ is equivalent to the Jacobian, where

$$ \mathbf J= \begin{bmatrix} \frac {\partial \bar x}{\partial x} & \frac {\partial \bar x}{\partial y}\\ \frac {\partial \bar y}{\partial x} & \frac {\partial \bar y}{\partial y} \end{bmatrix} $$

Therefore:

$$ \mathbf J= \frac{1}{mq-np}

\begin{bmatrix} q & -p\\ -n & m \end{bmatrix}

= \frac {1}{9} \begin{bmatrix} 5 & -1\\ -1 & 2 \end{bmatrix}

$$

and

$$\mathbf F =- \mathbf J

\begin{Bmatrix} x_0 \\ y_0 \end{Bmatrix} $$

Therefore:

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

= \mathbf J

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

-

\begin{Bmatrix} x_0 \\ y_0 \end{Bmatrix}

\right ]

$$

or

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

=

\begin{bmatrix} \frac{5}{9} & \frac{-1}{9}\\ \frac{-1}{9} & \frac{2}{9} \end{bmatrix}

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

-

\begin{Bmatrix} 3 \\ 4 \end{Bmatrix}

\right ]

$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Why do you want to use the above form? There is a simple and elegant way to do this problem. Egm6322.s09 17:11, 23 February 2009 (UTC)

More elegant way included. Old form is still included.

--Egm6322.s09.lapetina 14:05, 26 February 2009 (UTC)

Edited once again

--Egm6322.s09.lapetina 04:59, 27 March 2009 (UTC)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Identify the matrix $$\displaystyle \bar J$$, i.e., give its numerical value. Egm6322.s09 17:11, 23 February 2009 (UTC)

$$J$$ and $${\bar J}^{-1}$$ restated below. --Egm6322.s09.lapetina 14:05, 26 February 2009 (UTC)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> It is better to keep this exercise directly after the section on an example of affine mapping. above. Egm6936.s09 11:42, 26 February 2009 (UTC)

Link added. --Egm6322.s09.lapetina 20:39, 6 March 2009 (UTC)

Exercise 2
2 Dimensional Laplacian in polar coordinates:

$$\displaystyle {\rm div}( {\rm grad} \, u )=u_{xx}+u_{yy}=:\alpha$$

pag 13-1 Eq.(1) 2nd order linear PDE in the new coordinate $$(\bar{x}, \bar{y})$$ under non-linear coordinate transformation

$$\alpha = \left \lfloor \partial_{\bar{x}} \; \partial_{\bar{y}} \right \rfloor \partial_{\bar{x}}\partial_{\bar{y}}$$ $$\mathbf{J} \mathbf{1}\mathbf{J^T}\begin{Bmatrix}\partial_{\bar{x}}\\\partial_{\bar{y}}\end{Bmatrix}$$

where $$\mathbf{1}$$ is the identity matrix

Find the J pag 14-2

$$x=rcos\theta =\bar{x_1}cos\bar{x_2}=x_1(\bar{x_1}, \bar{x_2})$$

$$y=rsin\theta =\bar{x_1}sin\bar{x_2}=x_2(\bar{x_1}, \bar{x_2})$$

$$\bar{x_1}=r=(x^2+y^2)^{1/2}=\bar{x_1}(x_1, x_2)$$

$$\bar{x_2}=\theta =tan^{-1}(\frac{y}{x})=\bar{x_2}(x_1, x_2)$$

$$J_{11}=\frac{\partial \bar{x}_1}{\partial x}=\frac{\partial r}{\partial x}=(x^2+y^2)^{-1/2}x$$

$$J_{12}=\frac{\partial \bar{x}_2}{\partial r}=\frac{\partial r}{\partial x}=(x^2+y^2)^{-1/2}y$$

$$J_{21}=\frac{\partial \bar{x}_2}{\partial x_1}=\frac{\partial \theta}{\partial x}=\frac{\partial \tan^{-1}(s)}{\partial s}\frac{\partial s}{\partial x}$$

where: $$s:=\frac{y}{x}$$

Recall: $$y=tan^{-1}x\Rightarrow x=tan(y)=\frac{sin(y)}{cos(y)}$$

Find $$\frac{d y}{d x}$$

$$\frac{d y}{d x}=\frac{1}{\frac{d x}{d y}}$$

$$\frac{d x}{d y}=\frac{(cosy)^2-siny(-siny)}{cos^2y}=\frac{1}{cos^2y}=1+tan^2y=1+x^2$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Same error as above. Also, keep these related topics together in one place to ease the reading; one of the goals in reediting report R3. Egm6322.s09 17:11, 23 February 2009 (UTC)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> This part has been changed. Egm6322.s09.three.liu 14:33, 10 April 2009 (UTC)

$$\frac{d y}{d x}=cos^2y$$

$$tan(y)=\frac{sin(y)}{cos(y)}$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #006600; background-color: #CCFFFF; text-align: left;"> Same error as above. Egm6322.s09 00:31, 2 April 2009 (UTC)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #006600; background-color: #CCFFFF; text-align: left;"> This part has been changed. Egm6322.s09.three.liu 14:33, 10 April 2009 (UTC)

Expressing J in terms of $$\begin{pmatrix}r,\theta \end{pmatrix} $$

Egm6322.s09.bit.sahin 16:29, 24 April 2009 (UTC)

$$J_{11}=\frac{x} {r}=\frac{rcos\theta }{r}=cos\theta$$

$$J_{12}=\frac{y}{\left (x^{2}+y^{2} \right )^{1/2}}=\frac{rsin\theta }{r}=sin\theta $$

$$J_{21}=\frac{1}{1+\left (\frac{y}{x} \right )^{2}}\left (-\frac{y}{x^{2}} \right )=\frac{-y}{x^{2}+y^{2}}=\frac{-rsin\theta }{r^{2}}=-\frac{sin\theta }{r}$$

$$J_{22}=\frac{\partial \overline{x}_{2}}{\partial x_{2}}=\frac{\partial \theta }{\partial y}$$

Remembering that $$\theta =tan^{-1}\left (\frac{y}{x} \right )$$, If we define $$s=y/x$$ then

$$\frac{\partial \theta }{\partial y}=\frac{\partial }{\partial s}tan^{-1}(s)\frac{\partial s}{\partial y}=\frac{1}{1+\left (y/x \right )^{2}}\left (\frac{1}{x} \right )=\frac{x}{x^{2}+y^{2}}=\frac{rcos\theta }{r^{2}}=\frac{cos\theta }{r}$$

As a result, we obtain Jacobian matrix as

$$\underline{J}=\begin{bmatrix} cos\theta & sin\theta \\ -\frac{sin\theta }{r}&\frac{cos\theta }{r} \end{bmatrix}=\frac{1}{r}\begin{bmatrix} rcos\theta &rsin\theta  \\ -sin\theta &cos\theta \end{bmatrix}$$

In addition to this method, a simpler method can be used to determine the Jacobian matrix directly in polar coordinates.

$$x_{i}=x_{i}\left (\overline{x}_{1},\overline{x}_{2} \right )=x_{i}\left (r,\theta \right )$$

Instead of inversing the equation, differentiating $$\frac{\partial x_{i}}{\partial \overline{x}_{j}}$$ is simpler (e.g., $$\frac{\partial x}{\partial r},\frac{\partial x}{\partial \theta})$$.

$$\underline{A}=\begin{bmatrix} \frac{\partial x_{i}}{\partial \overline{x}_{j}} \end{bmatrix} =\begin{bmatrix} cos\theta &-rsin\theta  \\ sin\theta & rcos\theta \end{bmatrix}$$

Here $$\underline{A}=\underline{J}^{-1}$$. Therefore it can be written that

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Revision of lecture notes - Explanation of $$\displaystyle \mathbf{A} = \mathbf{J}^{-1}$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #006600; background-color: #CCFFFF; text-align: left;"> Sign your comment. The above explanation was not complete; see me after class. Egm6322.s09 00:31, 2 April 2009 (UTC)

Using:

$$\mathbf{A}=\mathbf{J}^{-1}=\frac{\partial x_{i}}{\partial \overline{x}_{j}}=\begin{bmatrix} m & p\\ n & q\end{bmatrix}$$

With the determinant:

$$\frac{1}{det\mathbf{J}^{-1}}\begin{bmatrix}q & -p\\ -n & m\end{bmatrix}\Rightarrow det\mathbf{J}^{-1}=mq-np$$

It is possible to see that:

$$\begin{Bmatrix}\bar{x}\\\bar{y}\end{Bmatrix}=\mathbf{J}\begin{bmatrix} \begin{Bmatrix}x\\y\end{Bmatrix}-\begin{Bmatrix}x_0\\y_0\end{Bmatrix} \end{bmatrix}$$

$$\mathbf{E}=\mathbf{J},\mathbf{F}=-\mathbf{J}\begin{Bmatrix} x_0\\y_0\end{Bmatrix}$$ from pag. 15-1

$$\underline{J} \underline{A}=\underline{I}=\left [\delta _{ij} \right ]=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Wrong notation $$\displaystyle \partial_{ij}$$; review lecture notes. Also the explanation for $$\displaystyle \underline{A} = \underline{J}^{-1}$$ is incomplete; review lecture notes. Egm6322.s09 17:11, 23 February 2009 (UTC)

$$\underline{J} $$ can easily be determined.

$$\underline{J}= \underline{A}^{-1}=\left (\underline{J}^{-1} \right )^{-1}=\begin{bmatrix} cos\theta & -rsin\theta \\ sin\theta & rcos\theta \end{bmatrix}^{-1}$$

$$\underline{A}^{-1}=\frac{1}{det\underline{A}}\begin{bmatrix} rcos\theta & rsin\theta \\ -sin\theta & cos\theta \end{bmatrix}$$

Eventually,

$$\underline{J}=\frac{1}{r}\begin{bmatrix} rcos\theta & rsin\theta \\ -sin\theta & cos\theta \end{bmatrix}$$

which is identical with the previous result.

=Complete Form Of Matrix=

--EGM6322.S09.TIAN 17:38, 24 April 2009 (UTC)

$$\alpha = \beta + \gamma$$

$$\beta$$ = $$\left \lfloor \partial_{\bar{x}} \; \partial_{\bar{y}} \right \rfloor $$ $$\mathbf{J} \mathbf{J^T}\begin{Bmatrix}\partial {\bar{x}}\\\partial {\bar{y}}\end{Bmatrix}$$

$$= \left \lfloor \partial_r \; \partial_{\theta} \right \rfloor $$ $$\frac {1} {r^2} $$ $$\begin{bmatrix} rc & rs\\ -s & c \end{bmatrix} $$ $$\begin{bmatrix} rc & -s\\ rs & c \end{bmatrix} $$ $$\begin{Bmatrix}\partial_r \\\ \partial_{\theta}\end{Bmatrix}$$

$$= \left \lfloor \partial_r \ \partial_{\theta} \right \rfloor $$ $$\begin{Bmatrix}\partial_r \\\ \partial_{\theta}\end{Bmatrix}$$

$$= \partial_{rr} +$$ $$\frac {1} {r^2} $$ $$\partial_{\theta \theta} $$

Note: $$\left \lfloor \partial_{\bar{x}} \ \partial_{\bar{y}} \right \rfloor $$$$\mathbf{J} \mathbf{J^T}$$ hold fixed, no further differentiate.

Three Methods To Figure Out $$\gamma$$
--EGM6322.S09.TIAN 17:39, 24 April 2009 (UTC)

Method One:

The Jacobian Matrix is:

$$J_{11}=\frac{x} {r}=\frac{rcos\Theta }{r}=cos\Theta$$

$$J_{12}=\frac{y}{\left (x^{2}+y^{2} \right )^{1/2}}=\frac{rsin\theta }{r}=sin\theta $$

$$J_{21}=\frac{1}{1+\left (\frac{y}{x} \right )^{2}}\left (-\frac{y}{x^{2}} \right )=\frac{-y}{x^{2}+y^{2}}=\frac{-rsin\theta }{r^{2}}=-\frac{sin\theta }{r}$$

$$J_{22}=\frac{\partial \overline{x}_{2}}{\partial x_{2}}=\frac{\partial \theta }{\partial y}$$

$$ \underline J=[J_{ij}]=[\frac {\partial {\bar {x_i}}}{\partial {x_j}}] =\begin{bmatrix} c & s     \\ -\frac {s}{r} & \frac{c}{r} \end{bmatrix} =\frac{1}{r}\begin{bmatrix} rc & rs     \\ -s & c \end{bmatrix} $$

$$\gamma =$$ $$= \left \lfloor \partial_r \ \partial_{\theta} \right \rfloor $$ $$\mathbf{J} \mathbf{J^T}$$ $$\begin{Bmatrix}\partial r \\\ \partial {\theta}\end{Bmatrix}$$

Method Two:

Get $$\underline {J}$$ directly in $$(r,\theta)$$ (polar coordinates)

$$x_{i}=x_{i}\left (\overline{x}_{1},\overline{x}_{2} \right )=x_{i}\left (r,\theta \right )$$

$$\begin{bmatrix} \frac{\partial x_{i}}{\partial \overline{x}_{j}} \end{bmatrix} =\begin{bmatrix} cos\theta &-rsin\theta  \\ sin\theta & rcos\theta \end{bmatrix}$$

$$\underline{A}=\begin{bmatrix} \frac{\partial x_{i}}{\partial \overline{x}_{j}} \end{bmatrix} =\begin{bmatrix} cos\theta &-rsin\theta  \\ sin\theta & rcos\theta \end{bmatrix}$$

$$\underline{A}=\underline{J}^{-1}=\frac{\partial x_{i}}{\partial \overline{x}_{j}} =\begin{bmatrix} a & b\\ c & d \end{bmatrix}$$

Therefore,

$$\frac{1}{det\mathbf{J}^{-1}}\begin{bmatrix}d & -b\\ -c & a\end{bmatrix}\Rightarrow det\mathbf{J}^{-1}=1$$

$$\underline{J}=\frac{1}{r}\begin{bmatrix} rcos\theta & rsin\theta \\ -sin\theta & cos\theta \end{bmatrix}=\frac{1}{r}\begin{bmatrix} rc & rs     \\ -s & c \end{bmatrix}$$

Therefore,

$$\gamma= \left \lfloor \partial_x \ \partial_y \right \rfloor $$ $$\mathbf{J} (x,y)$$ $$\begin{Bmatrix}\partial r \\\ \partial {\theta}\end{Bmatrix}$$

Method Three:

re-expect result in $$( r, $$ $$\theta$$ )

$$= \left \lfloor \partial_x \ \partial_y \right \rfloor $$ $$\mathbf{J} (r,\theta)$$ $$\begin{Bmatrix}\partial_r \\\ \partial_{\theta}\end{Bmatrix}$$

expressing result in $$( r, $$ $$\theta$$ )

Let us use Method Three:

$$= \left \lfloor \partial_x \ \partial_y \right \rfloor $$ $$\mathbf{J^T}$$ = $$= \left \lfloor \partial_x \ \partial_y \right \rfloor $$ $$\begin{bmatrix} c & -\frac {s}{r}\\ s & \frac {c}{s} \end{bmatrix} $$ = $$= \left \lfloor ( \partial_x c + \partial_y s) \;  (-\partial_x (\frac {s}{c})) + \partial_y (\frac {c}{r}) \right \rfloor $$

$$\partial_x c =$$ $$\frac{\partial}{\partial x}$$ $$cos {\theta}(x,y)$$ = $$\frac{\partial}{\partial r}$$ $$( cos {\theta})$$ $$\frac{\partial r}{\partial x}$$ + $$\frac{\partial}{\partial {\theta}}$$ $$( cos {\theta})$$ $$\frac{\partial {\theta}}{\partial x}$$

$$\partial_x f(r,{\theta})$$ = ($$\frac{\partial}{\partial r} \; f$$) $$\frac{\partial r}{\partial x} $$ + $$\frac{\partial f}{\partial {\theta}} $$ $$\frac{\partial {\theta}}{\partial x} $$

$$\frac{\partial {\theta}}{\partial x} $$ = $$\frac{\partial}{\partial x} $$ $$tan^{-1}$$ $$( \frac{y}{x} )$$ = $$J_{21} (r,{\theta})$$ = $$\frac{\partial {\bar{x_2}}}{\partial {x_1}} $$

$$J_{21}=$$ $$-\frac {sin{\theta}} {r}$$

$$\partial_x cos {\theta} $$

= $$(-sin{\theta})$$ $$(-\frac {s}{r})$$ = $$\frac {s^2}{r}$$

$$\partial_y sin {\theta}$$ = $$\frac {c^2}{r}$$

$$\Rightarrow$$ $$\partial_x c + \partial_y s$$ = $$\frac {1}{r}$$

$$\partial_x (\frac {s}{r})=$$ $$\frac{\partial} {\partial {r}} $$ $$(\frac {s}{r})$$ $$\frac{\partial r }{\partial x} +$$ $$\frac{\partial} {\partial {\theta}} $$ $$(\frac {s}{r})$$ $$\frac{\partial {\theta} }{\partial x} $$

= $$-\frac {2sc} {r^2}$$

Similarly, $$\partial_y (\frac {c}{r})=$$ $$\frac{\partial} {\partial {r}} $$ $$(\frac {c}{r})$$ $$\frac{\partial r }{\partial y} +$$ $$\frac{\partial} {\partial {\theta}} $$ $$(\frac {c}{r})$$ $$\frac{\partial {\theta} }{\partial y} $$

= $$-\frac {2sc} {r^2}$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> But you did not do Method 1 and Method 2. Also, it is not clear to me why you need to box "The complete form of $$\gamma$$" below, and the expression of $$\gamma$$ is wrong. The expression for $$\gamma$$ here is not the same expression for $$\gamma$$ in the section on Example, Rotation and Translation. You could have completed this topic, even though I only explained in words, without writing down. I did write down the end of the complete solution at the beginning of the following lecture. Egm6322.s09 17:11, 23 February 2009 (UTC)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #006600; background-color: #CCFFFF; text-align: left;"> You still haven't completed Method 1 and Method 2; there were also a lot of "misprints" in the above text around Method 1 and Method 2; clean up.

You deleted the collapsible box "The complete form of $$\gamma$$" but did not reply in the comment box above. Egm6322.s09 00:31, 2 April 2009 (UTC)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #006600; background-color: #CCFFFF; text-align: left;"> This part has been changed. Egm6322.s09.three.liu 15:20, 10 April 2009 (UTC)

= References =

= Signatures = Egm6322.s09.three.liu 19:02, 25 February 2009 (UTC)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> You need to sign with the "four tilde" command so you have the time-and-date stamp, just like everybody else; see below. Egm6322.s09 17:11, 23 February 2009 (UTC)

Egm6322.s09.Three.nav 20:37, 19 February 2009 (UTC)Egm6322.s09.three.nav

--Egm6322.s09.xyz 17:52, 18 February 2009 (UTC)

Egm6322.s09.bit.la 16:36, 20 February 2009 (UTC)

Egm6322.s09.bit.sahin 18:26, 20 February 2009 (UTC)

--Egm6322.s09.lapetina 19:12, 20 February 2009 (UTC)

Egm6322.s09.Three.ge 19:23, 20 February 2009 (UTC)

Egm6322.s09.bit.gk 20:05, 20 February 2009 (UTC)

--EGM6322.S09.TIAN 21:23, 20 February 2009 (UTC)