User:Egm6321.f12.team01.Zhang

Problem Statement


Cylindrical coordinates ,which are shown in the above figure, are expressed as follows:

1) Find $$ \{dx_i \}=\{dx_1,dx_2,dx_3 \} $$
 * in terms of
 * $$ \{\xi_j \}=\{\xi_1,\xi_2,\xi_3 \} $$ and  $$ \{d\xi_k \}=\{d\xi_1,d\xi_2,d\xi_3 \} $$

2) Find $$ ds^2=\sum_{i}(dx_i)^2= \sum_{k}(h_k)^2(d\xi_k)^2 $$. Identify $$ \{h_i\} $$ in terms of $$ \{\xi_i\} $$

3) Find $$ \Delta u $$ in cylindrical coordinates

4) Use separation of variables to find the separated equations and compare to the Bessel equ.(1)P.27-1

Given
Laplace equation:


 * $$ \Delta u= 0 $$

Laplace operator in general curvilinear coordinates:

Bessel L2-ODE-VC:

Nomenclature

 * $$ (\xi_1,\xi_2,\xi_3) $$: General curvilinear coordinates
 * $$ f'(x)= \frac{df(x)}{dx} $$

Part 1) Find {dxi}
Expand $$ \{dx_i \} $$ by using product rule:

Part 2) Find (ds)^2 and Identify {hi}

 * $$ ds^2= (dx_1)^2+(dx_2)^2+(dx_3)^2 $$
 * $$ ds^2= (cos\xi_2\,d\xi_1-\xi_1\,sin\xi_2\,d\xi_2)^2+(sin\xi_2\,d\xi_1+\xi_1\,cos\xi_2\,d\xi_2)^2+(d\xi_3)^2 $$
 * $$ ds^2= cos^2\xi_2\,(d\xi_1)^2-2\,\xi_1\,sin\xi_2\,cos\xi_2\,d\xi_1\,d\xi_2+\xi_1^2\,sin^2\xi_2\,(d\xi_2)^2+sin^2\xi_1\,(d\xi_1)^2+2\,\xi_1\,sin\xi_2\,cos\xi_2\,d\xi_1\,d\xi_2+\xi_1^2\,cos^2\xi_2\,(d\xi_2)^2+(d\xi_3)^2 $$

Therefore, we have


 * (5.10)
 * }

Part 3) Find Laplace Operator in Cylindrical Coordinates
Eq.(5.9) gives

When $$ i=1 $$,

When $$ i=2 $$,

When $$ i=3 $$,

Pluging the results derived from eq.(5.11) to eq.(5.14) into eq.(5.4) yields


 * $$ \Delta u= \frac{1}{h_1h_2h_3}\sum_{i=1}^{3}\frac{\partial}{\partial\xi_i}[\frac{h_1h_2h_3}{(h_i)^2}\frac{\partial u}{\partial\xi_i}] $$
 * $$ \Delta u= \frac{1}{\xi_1} [\frac{\partial}{\partial\xi_1}(\xi_1\,\frac{\partial u}{\partial \xi_1})+\frac{1}{\xi_1}\,\frac{\partial^2 u}{\partial\xi_2^2}+\xi_1\,\frac{\partial^2 u}{\partial\xi_3^2}] $$


 * (5.15)
 * }

Part 4) Separation of Variables and Comparison with Bessel Equation
To solve Laplace Equation $$ \Delta u=0 $$, we need to employ the method of Separation of Variables introduced on lecture notes P.39-5. Assume that

Plug eq.(5.16) into eq.(5.15) and expand:

Divide eq.(5.17) by $$ R(\xi_1)\Theta(\xi_2)Z(\xi_3) $$:

Since the third term of eq.(5.18) is neither a function of $$ \xi_1 $$ nor $$ \xi_2 $$, we can denote it as a constant with respect to the first two terms.

Let $$ k_1= \frac{1}{Z}\frac{d^2Z}{d\xi_3^2} $$, and multiply eq.(5.18) with $$ \xi_1^2 $$, then we have

Now the second term of eq.(5.19) is not a function of $$ \xi_1 $$, we can further denote it as a constant with respect to the first and third terms.

Let $$ k_2= \frac{1}{\Theta}\frac{d^2\Theta}{d\xi_2^2} $$, and expand eq.(5.19) using product rule:

Replace $$ \xi_1=x, R=y $$. Eq.(5.20) becomes

Which could be further simplified into:

Divide eq.(5.22) by $$ k_1 $$:

Recall Bessel's Equation:

Let $$ k_1=-\frac{1}{2}, k_2=\frac{v^2}{2} $$,eq(5.23) becomes


 * (5.24)
 * }

One can tell that the coeffcients of $$ y' $$ and $$ y $$ in eq.(5.24) correspond to those in Bessel's Equation.

However, the coefficient of $$ y' $$, which is $$ -2x^2 $$, is different from $$ 1-x^2 $$ in Bessel's Equation.

Given:
Taylor series at x=0 (ala Maclaurin series)

Derive:

 * Bulleted list item


 * Bulleted list item

Solution:
Let $$f(x)=\frac{1}{1-x}$$

Take derivatives of f(x) we got

$$f'(x)=\frac{1}{(1-x)^2}$$

$$f''(x)=\frac{2}{(1-x)^3}$$

$$f'''(x)=\frac{3*2}{(1-x)^4}$$

$$.....$$

$$f^{(n)}(x)=\frac{n!}{(1-x)^{(n+1)}}$$

Plugging into Taylor series at Equation ($$) yeilds


 * $$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}}{n!}x^n $$

$$\Rightarrow f(x)=\sum_{n=0}^{\infty}\frac{n!}{n!(1-x)^{(n+1)}}|_{x=0}x^n $$

Let $$f_1(x)=\frac{1}{x}$$

Take derivatives of $$f(x)$$ we got

$$f_1'(x)=-\frac{1}{x^2}$$

$$f_1''(x)=+\frac{2}{x^3}$$

$$f_1'''(x)=-\frac{3*2}{x^4}$$

$$\,\,\,\,.....$$

$$f_1^{(n)}(x)=(-1)^{n}\frac{n!}{x^{(n+1)}}$$

Plugging into Taylor series at Equation ($$) yeilds

$$f_1(x)=\frac{1}{x}=\sum_{n=0}^{\infty}\left.(-1)^{n}\frac{n!}{n!x^{(n+1)}}\right| _{x=0}x^n $$


 * $$\Rightarrow \frac{1}{x}=\sum_{n=0}^{\infty}(-1)^{n}x^n$$


 * $$\Rightarrow \frac{1}{x}=1-x+x^2-x^3...$$

Given:
$$(x-1)y^{''}-xy^'+y=0$$ First homogeneous solution: $$ \displaystyle u_1(x)=e^x$$ Trail solution: $$ \displaystyle  y=e^{rx}, r=\mbox{constant} $$ Characteristic equation: $$ \displaystyle (x-1)r^2-xr+1 = 0$$ WA: $$r_1 = 1 $$ $$r_2(x) = \frac{1}{x-1}$$

Explain:
Explain why $$\displaystyle r_2(x)$$ is not a valid root, i.e., $$\displaystyle u_2(x) = e^{xr_{2}(x)}$$ is not a valid solution.