User:Egm6321.f09.Team06.sada/HW6

PROBLEM 8
The Laplace Equation in circular cylinder coordinates $$\displaystyle \Delta \psi = 0 $$

Obtain the separated equations for Laplace equation on circular cylinder coordinates, and identify the Bessel differential equation

$$\displaystyle x^2 y'' + x y' - (x^2 - \nu^2) y = 0 $$ Eq. 1

Solution
Assume the total solution of the Laplace equation is given by

$$\displaystyle f(\xi_1, \xi_2, \xi_3) = X_1(\xi_1) X_2(\xi_2) X_3(\xi_3) $$

Using Laplace operator on the function $$\displaystyle f$$, we can write Laplace equation

$$\displaystyle \Delta f = \frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left(\xi_1 \frac{\partial f}{\partial \xi_1} \right) + \frac{1}{\xi_1^2} \frac{\partial^2 f}{\partial \xi_2^2} + \frac{\partial^2 f}{\partial \xi_3^2} = 0 $$

$$\displaystyle X_2 X_3 \frac{1}{\xi_1} \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + X_1 X_3 \frac{1}{\xi_1^2} \frac{\partial^2 X_2}{\partial \xi_2^2} + X_1 X_2 \frac{\partial^2 X_3}{\partial \xi_3^2} = 0 $$

Divide by $$\displaystyle X_1 X_2 X_3$$ $$\displaystyle \frac{1}{\xi_1 X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} + \frac{1}{X_3} \frac{\partial^2 X_3}{\partial \xi_3^2} = 0 $$

The last term is only dependent on $$\displaystyle \xi_3$$, and is the only term dependent on $$\displaystyle \xi_3$$, hence it must be a constant. So we substitute $$\displaystyle \frac{1}{\xi_1 X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} - \beta^2 = 0 $$

where $$\displaystyle \beta^2 = -\frac{1}{X_3} \frac{\partial^2 X_3}{\partial \xi_3^2} $$

Now, multiply by $$\displaystyle \xi_1^2$$ $$\displaystyle \frac{\xi_1}{ X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} - \xi_1^2 \beta^2 = 0 $$

Now, the second term is only dependent on $$\displaystyle \xi_2$$, and it is the only term dependent on $$\displaystyle \xi_2$$. It must also be a constant. We can substitute $$\displaystyle \nu^2 = \frac{1}{X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} $$

So that $$\displaystyle \frac{\xi_1}{ X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \nu^2 - \xi_1^2 \beta^2 = 0 $$

Expand the first term $$\displaystyle \frac{\xi_1^2}{ X_1 } \frac{\partial^2 X_1}{\partial \xi_1^2} + \frac{\xi_1}{ X_1 } \frac{\partial X_1}{\partial \xi_1} + \nu^2 - \xi_1^2 \beta^2 = 0 $$

Multiply by $$\displaystyle X_1$$ $$\displaystyle \xi_1^2 \frac{\partial^2 X_1}{\partial \xi_1^2} + \xi_1 \frac{\partial X_1}{\partial \xi_1} - \left( \xi_1^2 \beta^2 - \nu^2 \right) X_1 = 0 $$

Now, substitute $$\displaystyle x = \xi_1 \beta$$ $$\displaystyle x^2 \frac{\partial^2 X_1}{\partial x^2} + x \frac{\partial X_1}{\partial x} - \left( x^2 - \nu^2 \right) X_1 = 0 $$

Then, with $$\displaystyle X_1 = y(x)$$, we get $$\displaystyle x^2 \frac{\partial^2 y}{\partial x^2} + x \frac{\partial y}{\partial x} - \left( x^2 - \nu^2 \right) y = 0 $$

or

$$\displaystyle x^2 y'' + x y' - \left( x^2 - \nu^2 \right) y = 0 $$

which is the same as Eq. 1, i.e. the Bessel differential equation

PROBLEM 9
Let $$\displaystyle f = \sum_i g_i $$

Show that,

a) if $$\displaystyle \{g_i\}$$ is odd, then $$\displaystyle f$$ is odd.

b) if $$\displaystyle \{g_i\}$$ is even, then $$\displaystyle f$$ is even.

Solution
The definition of an odd function is that $$\displaystyle f(x) = -f(-x)$$.

a) So, if every $$\displaystyle g_i$$ is odd, then

$$\displaystyle f(x) = \sum_i g_i(x) = \sum_i \left[-g_i(-x)\right] = -\sum_i g_i(-x) = -f(-x) $$

Thus the first case is proven.

The definition of an even function is that $$\displaystyle f(x) = f(-x)$$. So, if every $$\displaystyle g_i$$ is even, then

b) So, if every $$\displaystyle g_i$$ is even, then

$$\displaystyle f(x) = \sum_i g_i(x) = \sum_i g_i(-x) = f(-x) $$

And, thus the second case also hold true.

PROBLEM 10
Show that, for $$\displaystyle k = 0,1,2,...$$,

a) $$\displaystyle P_{2k}(x)$$ is even, and

b) $$\displaystyle P_{2k+1}(x)$$ is odd

Solution
We know that, the Legendre polynomials are given by

$$\displaystyle P_n(x) = \sum_{i=0}^{[n/2]} (-1)^i \frac{(2 n - 2 i)! x^{n - 2i}}{2^n i! (n-i)! (n-2i)!} $$  Eq.(1)

a) The definition of an even function is that $$\displaystyle f(x) = f(-x)$$. Substitute $$\displaystyle n=2k$$ in Eq.(1)

$$\displaystyle P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! x^{2k - 2i}}{2^{2k} i! (2k-1)! (2k-2i)!} $$

$$\displaystyle P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! \left[(x)^2\right]^{k - i}}{2^{2k} i! (2k-1)! (2k-2i)!} $$

Then, $$\displaystyle P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! \left[(-x)^2\right]^{k - i}}{2^{2k} i! (2k-1)! (2k-2i)!} = P_{2k}(-x) $$

Thus $$\displaystyle P_{2k}(x) $$ is even by definition.

b) The definition of an odd function is that $$\displaystyle f(x) = -f(-x)$$. Substitute $$\displaystyle n=2k+1$$ in Eq.(1)

$$\displaystyle P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! x^{2k + 1 - 2i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$

Now, the above equation can be written as

$$\displaystyle P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (x) \left[(x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$   Eq.(2)

$$\displaystyle P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (x) \left[(-x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$   Eq.(3)

Now Eq.(2) at $$\displaystyle -x$$ is

$$\displaystyle P_{2k+1}(-x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (-x) \left[(-x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$   Eq.(4)

Comparing Eqs. (3) and (4) shows that

$$\displaystyle P_{2k+1}(x) = -P_{2k+1}(-x) $$

Thus it is proven that $$\displaystyle P_{2k+1}(x) $$ is odd by definition.

The polynomial
 * {| style="width:100%" border="0" align="left"

q(x) = \sum_{i=0}^4 c_i x^i $$ $$ with the coefficients
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

c_0 = 3, \quad c_1 = 10, \quad c_2 = 15, \quad c_3 = -1, \quad c_4 = 5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

PROBLEM 11
The polynomial

$$\displaystyle q(x) = \sum_{i=0}^4 c_i x^i $$

with

$$\displaystyle c_0 = 3, \quad c_1 = 10, \quad c_2 = 15, \quad c_3 = -1, \quad c_4 = 5 $$

a) Find $$\displaystyle \{ a_i \}$$ S.T. $$\displaystyle q(x) = \sum_{i=0}^4 a_i P_i(x)$$, where $$\displaystyle P_i(x)$$ are Legendre polynomials.

b) Plot $$\displaystyle q = \sum_i c_i x^i = \sum_i a_i P_i(x) $$.

Solution
From Lecture 31-3 Eqs. 1,2,3,4&5


 * $$ \begin{align}

P_{0}\left(x\right) &= 1 \\ P_{1}\left(x\right) &= x \\ P_{2}\left(x\right) &= \frac{1}{2}\left(3x^{2}-1\right) \\ P_{3}\left(x\right) &= \frac{1}{2}\left(5x^{3}-3x\right) \\ P_{4}\left(x\right) &= \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} \end{align} $$


 * Plugging the Legendre polynomials into the expression containing $$\left\{a_{i}\right\}$$, then group the terms by power of $$x$$:


 * $$ \begin{align}

q\left(x\right) &= \sum_{i=0}^{4}\ a_{i} P_{i} \\ &= a_{0} + a_{1} x + a_{2} \frac{1}{2}\left(3x^{2}-1\right) + a_{3}\frac{1}{2}\left(5x^{3}-3x\right) + a_{4}\left(\frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8}\right) \\ &= \left(a_{0}-\frac{1}{2}a_{2}+\frac{3}{8}a_{4}\right) + x\left(a_{1}-3a_{3}\right) + x^{2}\left(\frac{3}{2}a_{2}-\frac{15}{4}a_{4}\right) + x^{3}\left(\frac{5}{2}a_{3}\right) + x^{4}\left(\frac{35}{8}a_{4}\right) \end{align} $$


 * Then, we can set each term of this equation equal to the corresponding term in the definition of $$q\left(x\right)$$,


 * $$ \begin{align}

\sum_{i=0}^{4}\ a_{i} &P_{i} = \sum_{i=0}^{4}\ c_{i} x^{i}& \\ \\ &\text{(1)  }a_{0} - \frac{1}{2}a_{2} + \frac{3}{8}a_{4} &= 3 \\ &\text{(2) }a_{1} - 3a_{3} &= 10 \\ &\text{(3) }\frac{3}{2}a_{2} - \frac{15}{4}a_{4} &= 15 \\ &\text{(4) }\frac{5}{2}a_{3} &= -1 \\ &\text{(5) }\frac{35}{8}a_{4} &= 5 \end{align} $$


 * The last two equations yield the values of $$a_{3}$$ and $$a_{4}$$ directly:


 * $$ \begin{align}

a_{3} &= \frac{-2}{5} \\ a_{4} &= \frac{8}{7} \end{align} $$


 * $$a_{2}$$ is then obtained from equation (3) by plugging in $$a_{4}$$:


 * $$ \begin{align}

\frac{3}{2}a_{2} &= 15 + \frac{15}{4}a_{4} \\ a_{2} &= 10 + \frac{10}{4}\cdot \frac{8}{7} \\ &= \frac{90}{7} \end{align} $$


 * Similarly, $$a_{1}$$ is obtained from equation (2) by plugging in $$a_{3}$$:


 * $$ \begin{align}

a_{1} &= 10 + 3a_{3} \\ &= 10 + \frac{-6}{5} \\ &= \frac{44}{5} \end{align} $$


 * Finally, $$a_{0}$$ is obtained by inserting $$a_{2}$$ and $$a_{4}$$ into equation (1):


 * $$ \begin{align}

a_{0} &= 3 + \frac{1}{2}a_{2} - \frac{3}{8}a_{4} \\ &= 3 + \frac{45}{7} - \frac{3}{7} \\ &= 3 + \frac{42}{7} \\ &= 9 \end{align} $$


 * In summary, the following expressions for $$\left\{a_{i}\right\}$$ were obtained:


 * $$ \begin{align}

a_{0} &= 9,

\quad a_{1} = \frac{44}{5}\\ \\ a_{2} &= \frac{90}{7},

\quad a_{3} = \frac{-2}{5}\\ \\ a_{4} &= \frac{8}{7} \end{align} $$