User:Egm6322.s09.bit.sahin/r6

Cauchy Equation

The so-called Cauchy equation or Euler equation

$$x^{2}y''+axy'+by=0$$, (a,b constant)

can also be purely algebraic manipulations. By substituting

$$y=x^{m}$$

and its derivatives into the first equation we find

$$x^{2}m\left ( m-1 \right )x^{m-2}+axmx^{m-1}+bx^{m}=0$$.

By omitting the common power $$x^{m}$$, which is not zero when $$x\neq 0$$, we obtain the auxiliary equation

$$m^{2}+\left ( a-1 \right )m+b=0$$.

If the roots $$m_{1}$$ and $$m_{2}$$ of this equation are different, then the functions

$$y_{1}\left ( x \right )=x^{m_{1}}$$ and $$y_{2}\left ( x \right )=x^{m_{2}}$$

constitute a fundamental system of solutions for all $$x$$ for which these functions are defined. The corresponding general solution is

$$y=c_{1}x^{m_{1}}+c_{2}x^{m_{2}}$$

Gradient in Polar Coordinates
Gradient in cartesian coordinates is

$$grad T=\frac{\partial T}{\partial x}\mathbf{i}+\frac{\partial T}{\partial y}\mathbf{j}$$

Rectangular coordinates and polar coordinates are related as follows

$$x=rcos\theta $$, $$y=rsin\theta $$

and additionally

$$r^{2}=x^{2}+y^{2}$$, $$\theta =tan^{-1}\left ( y/x \right )$$

two polar basis vectors expressed in terms of the rectangular basis vectors as following

$$\mathbf{e_{r}}=cos\theta \mathbf{i}+sin\theta \mathbf{j}$$,

$$\mathbf{e_{\theta }}=-sin\theta \mathbf{i}+cos\theta \mathbf{j}$$

or

$$\mathbf{i}=cos\theta \mathbf{e_{r}}-sin\theta \mathbf{e_{\theta }}$$,

$$\mathbf{j}=sin\theta \mathbf{e_{r}}+cos\theta \mathbf{e_{\theta }}$$

By using the chain rule, it can be written that

$$\frac{\partial T}{\partial x}\mathbf{i}=\left (\frac{\partial T}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial T}{\partial \theta}\frac{\partial \theta}{\partial x} \right )\left ( cos\theta \mathbf{e_{r}}-sin \theta \mathbf{e_{\theta}} \right )$$

$$\frac{\partial T}{\partial x}\mathbf{i}=\left (\frac{\partial T}{\partial r} cos \theta+\frac{\partial T}{\partial \theta} \frac{-sin \theta}{r} \right )\left ( cos\theta \mathbf{e_{r}}-sin \theta \mathbf{e_{\theta}} \right )$$

$$\frac{\partial T}{\partial x}\mathbf{i}=\frac{\partial T}{\partial r}cos^{2}\theta\mathbf{e_{r}}-\frac{\partial T}{\partial r}sin\theta cos\theta \mathbf{e_{\theta }}-\frac{\partial T}{\partial \theta }\frac{1}{r}sin\theta cos\theta \mathbf{e_{r}}+\frac{\partial T}{\partial \theta }\frac{1}{r}sin^{2} \theta \mathbf{e_{\theta}}$$

Similarly

$$\frac{\partial T}{\partial y}\mathbf{j}=\left (\frac{\partial T}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial T}{\partial \theta}\frac{\partial \theta}{\partial y} \right )\left ( sin\theta \mathbf{e_{r}}+cos \theta \mathbf{e_{\theta}} \right )$$

$$\frac{\partial T}{\partial y}\mathbf{j}=\left (\frac{\partial T}{\partial r}sin \theta+ \frac{\partial T}{\partial \theta}\frac{cos \theta}{r} \right )\left ( sin\theta \mathbf{e_{r}}+cos \theta \mathbf{e_{\theta}} \right )$$

$$\frac{\partial T}{\partial y}\mathbf{j}=\frac{\partial T}{\partial r}sin^{2}\theta\mathbf{e_{r}}+\frac{\partial T}{\partial r}sin\theta cos\theta \mathbf{e_{\theta }}+\frac{\partial T}{\partial \theta }\frac{1}{r}sin\theta cos\theta \mathbf{e_{r}}+\frac{\partial T}{\partial \theta }\frac{1}{r}cos^{2} \theta \mathbf{e_{\theta}}$$

Eventually,

$$\frac{\partial T}{\partial x} \mathbf{i}+\frac{\partial T}{\partial y}\mathbf{j}=\frac{\partial T}{\partial r}\mathbf{e_{r}}cos^{2} \theta+\frac{\partial T}{\partial \theta}\frac{sin^{2} \theta \mathbf{e_{\theta}}}{r}+\frac {\partial T}{\partial r}sin^{2} \theta \mathbf{e_{r}}+\frac{\partial T}{\partial \theta}\frac{cos^{2} \theta \mathbf{e_{\theta}}}{r}$$

$$\frac{\partial T}{\partial x} \mathbf{i}+\frac{\partial T}{\partial y}\mathbf{j}=\frac{\partial T}{\partial r}\mathbf{e_{r}}+\frac{1}{r}\frac{\partial T}{\partial \theta}\mathbf{e_{\theta}}$$