User:Egm6322.s09.bit.la/r1

=Exercises=

Exercise 1
Given the data:

$$\mathbf{i}=m\mathbf{i}+n\mathbf{j}=2\mathbf{i}+1\mathbf{j}$$

$$\mathbf{j}=p\mathbf{i}+q\mathbf{j}=1\mathbf{i}+5\mathbf{j}$$

$$(x_\mathbf{0},y_\mathbf{0})=(3,4) $$ pag 15-1

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

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

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

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

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

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

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

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

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

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

Find $$\frac{d y^2}{d x}$$

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

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

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

$$tan(y)=\frac{sin(y)}{cos(y)}=1+tan^2y=1+\frac{sin^2y}{cos^2y}$$

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

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