User:Egm6321.f12.team1.cla/report7

=Problem 7.1 Finding the local maximum of a hypergeometric function=

Prove That:

 * Use Matlab to plot $$_2 F_1(5,-10;1;x)$$ near $$x=0$$ to display the local maximum (or minimum in this region)

Proof
To prove this equation, we will assume a relation, the proof of which has been left for future work.


 * $$n! \sum\limits_{k=0}^m \dbinom{m}{k} \dbinom{k+n}{n} (-x)^k = \sum\limits_{k=0}^m \dbinom{m}{k} n! \dbinom{k+n}{n} (-x)^k $$
 * $$= \sum\limits_{k=0}^m \dbinom{m}{k} \frac {(k+n)!}{k!} (-x)^k $$
 * $$= \sum\limits_{k=0}^m \dbinom{m}{k}(-1)^k \frac {\mathrm{d}^n}{\mathrm{d}x^n} x^{k+n} $$
 * $$=  \frac {\mathrm{d}^n}{\mathrm{d}x^n} \sum\limits_{k=0}^m (-1)^k\dbinom{m}{k} x^{k+n} $$
 * $$=  \frac {\mathrm{d}^n}{\mathrm{d}x^n} \left[\left( \sum\limits_{k=0}^m \dbinom{m}{k} (-x)^k \right) x^n \right]$$

Similarly

Using ($$), ($$) and ($$), we get


 * $$\frac {\frac {\mathrm{d}^n} {\mathrm{d}x^n} (1-x)^m x^n}{\frac {\mathrm{d}^m} {\mathrm{d}x^m} (1-x)^n x^m} = \frac {n !}{m !} \frac {(1-x)^m}{(1-x)^n}.\quad\blacksquare$$

Proof
The Jacobi polynomials are defined via the hypergeometric function as follows :

Setting $$\alpha = 0, \beta = a+b-1, n=-a$$ we get
 * $$_2F_1(a,b;1;\frac{1-z}{2}) = \frac{(-a)!}{(1)_{-a}} P_{-a}^{(0,a+b-1)}(z)$$

Since $$ (1)_{-a} = (-a)!$$
 * $$_2F_1(a,b;1;\frac{1-z}{2}) = P_{-a}^{(0,a+b-1)}(z)$$

Using Rodrigues' formula as
 * $$P_n^{(\alpha,\beta)} (z) = \frac{(-1)^n}{2^n n!} (1-z)^{-\alpha} (1+z)^{-\beta} \frac{d^n}{dz^n} \left\{ (1-z)^\alpha (1+z)^\beta (1 - z^2)^n \right\} $$

Setting $$\alpha = 0$$


 * $$\Rightarrow\,_2F_1(a,b;1;\frac{1-z}{2}) = \frac{(-1)^{-a}}{2^{-a} (-a)!} (1+z)^{-a-b+1} \frac{d^n}{dz^n} \left\{(1-z)^{-a} (1+z)^{a+b-1 - a } \right\} $$

Setting $$z=1-2x$$


 * $$\Rightarrow \frac{\mathrm{d}f}{\mathrm{d}x} = \frac{\mathrm{d}f}{\mathrm{d}z}\frac{\mathrm{d}z}{\mathrm{d}x}$$
 * $$\Rightarrow \frac{\mathrm{d}f}{\mathrm{d}x} = \frac{\mathrm{d}f}{\mathrm{d}z}\left(-\frac{1}{2}\right)$$

Also and

Substituting ($$), ($$), ($$), ($$) into ($$) and simplifying, we get

Similarly, setting $$\alpha = 0, \beta = -a-b+1, n=b-1, z=1-2x$$ we get


 * $$\Rightarrow \frac {_2F_1(a,b;1;x)} {_2F_1(-b+1,-a+1;1;x)} = (1-x)^{-a-b+1}  \frac {(b-1)!} {(-a)!} \times   \frac {(1-x)^{-a}} {(1-x)^{b-1}}  \times  \frac {\frac {\mathrm{d}^{-a}} {\mathrm{d}x^{-a}} \left[(1-x)^{b-1}x^{-a}\right]} {\frac {\mathrm{d}^{b-1}} {\mathrm{d}x^{b-1}} \left[(1-x)^{-a}x^{b-1}\right]}$$

Setting $$n=-a, m=b+1$$ in ($$)
 * $$\frac {\frac {\mathrm{d}^{-a}} {\mathrm{d}x^{-a}} \left[(1-x)^{b-1}x^{-a}\right]} {\frac {\mathrm{d}^{b-1}} {\mathrm{d}x^{b-1}} \left[(1-x)^{-a}x^{b-1}\right]} = \frac {(-a)!} {(b-1)!} \times \frac {(1-x)^{b-1}} {(1-x)^{-a}} $$
 * $$\Rightarrow \frac {(b-1)!} {(-a)!} \times  \frac {(1-x)^{-a}} {(1-x)^{b-1}}  \times  \frac {\frac {\mathrm{d}^{-a}} {\mathrm{d}x^{-a}} \left[(1-x)^{b-1}x^{-a}\right]} {\frac {\mathrm{d}^{b-1}} {\mathrm{d}x^{b-1}} \left[(1-x)^{-a}x^{b-1}\right]} = 1 $$
 * $$\Rightarrow \frac {_2F_1(a,b;1;x)} {_2F_1(-b+1,-a+1;1;x)} = (1-x)^{-a-b+1}$$
 * $$_2F_1(a,b;1;x) = (1-x)^{-a-b+1}\,_2F_1(-b+1,-a+1;1;x)\qquad \blacksquare$$

Solution:
Using ($$) and setting $$a=5, b=-10$$, we get
 * $$_2F_1(5,-10;1;x) = (1-x)^{-5+10+1}\,_2F_1(-4,11;1;x)$$
 * $$\Rightarrow\,_2F_1(5,-10;1;x) = (1-x)^6\,_2F_1(-4,11;1;x)$$

Since in the second term, $$a$$ is a negative integer, the summation will terminate after $$-(-4)+1=5$$ terms
 * $$\Rightarrow\,_2F_1(5,-10;1;x) = (1-x)^6\sum\limits_{k=0}^4 \frac{(-4)_k(11)_k}{(1)_k}\frac{x^k}{k!}$$

Since $$(1)_k = k!$$
 * $$\Rightarrow\,_2F_1(5,-10;1;x) = (1-x)^6\sum\limits_{k=0}^4 (-4)_k(11)_k \frac{x^k}{(k!)^2}$$
 * $$\Rightarrow\,_2F_1(5,-10;1;x) = (1-x)^6\left[ (-4)_0(11)_0 \frac{x^0}{(0!)^2} + (-4)_1(11)_1 \frac{x^1}{(1!)^2} + (-4)_2(11)_2 \frac{x^2}{(2!)^2} + (-4)_3(11)_3 \frac{x^3}{(3!)^2} + (-4)_4(11)_4 \frac{x^4}{(4!)^2}\right]$$
 * $$\Rightarrow\,_2F_1(5,-10;1;x) = (1-x)^6\left[ 1 + (-4) \times 11 x + (-4)\times(-3)\times 11 \times 12 \times\frac{x^2}{4} + (-4)\times(-3)\times(-2)\times 11 \times 12 \times 13\times \frac{x^3}{36}\right.$$
 * $$\left. + (-4)\times(-3)\times(-2)\times(-1)\times 11 \times 12 \times 13 \times 14 \times \frac{x^4}{(24)^2} \right]$$
 * $$\Rightarrow\,_2F_1(5,-10;1;x) = (1-x)^6 (1 - 44x + 396x^2 - 1144x^3 + 1001x^4)\qquad\blacksquare$$

Since $$_2F_1(5,-10;1;x)$$ is a polynomial with six of its roots are at $$x=1$$, the first derivative of $$_2F_1(5,-10;1;x)$$ will have at least 5 roots at $$x=1$$. Also since $$_2F_1(5,-10;1;x)$$ is of degree 10, the first derivative will have a degree of 9 and a maximum of 9 real roots. Since 5 of the roots are at the same point $$x=1$$, at max we can see 5 points where the first derivative will be 0. So $$_2F_1(5,-10;1;x)$$ can have a maximum of 5 optima combined. The following graph displays all of the 5 points of optima with the blue circles.

Pavel Bhowmik (talk) 00:11, 5 December 2012 (UTC)

=Problem 7.2* Use Variation of Parameters =

Given
And intial conditions,

Where,

Find
The total solution $$ y(x) $$ for ($$)

Solution
First let's find $$ y_H(t) $$ using ($$) and ($$)
 * $$y_H(x)=AP_1(x)+BQ_1(x)$$
 * $$P_1(x)=u_1(x), BQ_1(x)=u_2(x)$$

In order to solve for the homogeneous solution we need to assume a first homogeneous solution below which we get from King 2003,
 * $$P_1(x)=u_1(x)=cos(a_{0}t)$$

Now because of our initial conditions we can not assume that the $$ t_0 $$ is equal to zero and therefore have to add it to our first homogeneous solution.
 * $$u_1(x)=cos(a_{0}(t-t_{0}))$$

And the second homogeneous equation which we also get from King 2003 with the only difference being the addition of our initial conditions,
 * $$u_2(x)=sin(a_{0}(t-t_{0}))$$

Plugging these into ($$) we get,
 * $$y_H(x)=Acos(a_{0}(t-t_{0}))+Bsin(a_{0}(t-t_{0}))$$

Now using our first and second homogeneous solutions we can find the particular solution $$ y_p(t)$$ using ($$).
 * $$y_P(t)=\int^{t}\frac{W(s)}ds - \int^{t}\frac{W(s)}ds$$

where we know every function and variable in the particular solution formula except the Wronskian which we can find using the equation below and the first and second homogeneous solutions,
 * $$W=u_{1}u_{2}'-u_{2}u_{1}'$$
 * $$W=cos(a_{0}(t-t_{0})cos(a_{0}(t-t_{0}))-sin(a_{0}(t-t_{0}))(-sin(a_{0}(t-t_{0})))$$
 * $$W=cos^2(a_{0}(t-t_{0}))+sin^2(a_{0}(t-t_{0}))=a_0$$

Now we have all the functions we need to solve for the particular solution of ($$)
 * $$y_P(t)=\int^{t}\frac{W(s)}ds - \int^{t}\frac{W(s)}ds$$
 * $$y_P(t)=\int^{t}\frac{a_0}ds - \int^{t}\frac{a_0}ds$$

Now we can combine the two parts under one integral
 * $$y_P(t)=\frac{1}{a_0} \int^{t}[ - ]ds$$

Now we can the apply the trigonometric identity below to the particular solution for further simplification,
 * $$ sin(\alpha+-\beta)=sin(\alpha)cos(\beta)+-cos(\alpha)sin(\beta)$$

Using this identity on the particular solution we get,
 * $$y_P(t)=\frac{1}{a_0} \int^{t}[f(s)sin(a_{0}(t-s))]ds$$

Now that we have our total homogeneous solution and our particuar solution of ($$) we can plug these into ($$) to get the complete solution for ($$)
 * $$y(x)=Acos(a_{0}(t-t_{0}))+Bsin(a_{0}(t-t_{0}))+ \frac{1}{a_0} \int^{t}[f(s)sin(a_{0}(t-s))]ds$$

This compares nicely to the given solution for ($$) from the lecture notes,
 * $$y(x)=Acos(a_{0}(t-t_{0}))+Bsin(a_{0}(t-t_{0}))+ \frac{1}{a_0} \int^{t}[f(s)sin(a_{0}(t-s))]ds$$

With the one of the key differences being the use of the dummy variable $$ s $$ instead of $$ \tau $$ but since it is a dummy variable the letter is interchangeable. The second difference being the known values of $$ A $$ and $$ B $$ from the lecture notes.

=Problem 7.3* Finding of the Infinitesimal Length in Spherical Coordinates =

Given:
$$ds^2 = dx_i dx_i = \sum_{i=1}^3(dx_i)^2$$

Find:
Show that the infinitesimal length ds in (2) p. 38-6 can be written in spherical coordinates as

$$ds^2 = 1 \cdot dr^2 + r^2 d\theta^2 + r^2(\cos\theta)^2\, d \varphi^2$$

Solution
$$\mathbf{ds} = dx_i \mathbf{e}_i \Rightarrow \ ds^2= \mathbf{ds} \cdot \mathbf{ds} = (dx_i \mathbf{e}_i) \cdot (dx_j \mathbf{e}_j) = dx_i\, dx_j(\mathbf{e}_i \cdot \mathbf{e}_j)$$

$$\mathbf{e}_i \cdot \mathbf{e}_j = \delta_{ij}$$

$$\delta_{ij} := \begin{cases}1, & \text{for} i = j \\ 0, & \text{for} i \ne j \end{cases}$$

The form of the infinitesimal length ds is expressed in Cartesian coordinates.

$$ds^2 = dx_i dx_i = \sum_{i=1}^3(dx_i)^2$$

This can be rewritten in the following form, which is the key for derivation steps, which are rigorous with trigonometric identities.

$$ds^2 = dx_1^2 + dx_2^2 + dx_3^2$$

The spherical coordinates can be utilized. Simply identify the appropriate identities. Recall that we are using the "astronomical" coordinate system.

$$x = r\cos {\theta} \cos {\phi} = \xi_1 \cos{\xi_2}\cos{\xi_3}$$

$$y = r\cos {\theta} \sin {\phi} = \xi_1 \cos{\xi_2} \sin{\xi_3}$$

$$z = r\sin {\theta} = \xi_1 \sin{\xi_2}$$

$$ds^2 = (dr\cos{\theta} \cos{\phi} - (r\sin{\theta} \cos{\phi} d\theta + r\cos{\theta} \sin{\phi} d\phi))^2 +(dr\cos{\theta} \sin{\phi} - (r\sin{\theta} \sin{\phi} d\theta - r\cos{\theta} \cos{\phi} d\phi))^2 + (dr\sin{\theta} + r\cos{\theta} d\theta)^2$$

$$\alpha = (r\sin{\theta} \cos{\phi} d\theta + r\cos{\theta} \sin{\phi} d\phi)$$

$$\beta = (r\sin{\theta} \sin{\phi} d\theta - r\cos{\theta} \cos{\phi} d\phi)$$

The substitutions then give the shortened form, which allows derivation of the squared terms.

$$ds^2 = (dr\cos{\theta} \cos{\phi} - \alpha)^2 +(dr\cos{\theta} \sin{\phi} - \beta)^2 + (dr\sin{\theta} + r\cos{\theta} d\theta)^2$$

$$ds^2 = (dr\cos{\theta} cos{\phi})^2 - 2\alpha dr\cos{\theta} \cos{\phi} + \alpha^2 + (dr\cos{\theta} \sin{\phi})^2 - 2\beta dr\cos{\theta} \sin{\phi} + \beta^2$$

The algebraic manipulation is long, but proper addition and subtraction eliminate unnecessary terms. The key is to appply trigonometric identities and grouping with the identity of $$\sin^2{x} + \cos^2{x} = 1$$.

$$ds^2 = (dr^2\cos^2{\theta}(cos^2{\phi} + \sin^2{\phi}) - 2\alpha dr \cos{\theta} \cos{\phi} + \alpha^2 - 2\beta dr\cos{\theta} \sin{\phi} + \beta^2 + dr^2 \sin{\theta}^2 + 2rdr \sin{\theta} \cos{\theta}d\theta + r^2 \cos^2{\theta} d{\theta}^2$$

$$ds^2 = dr^2\cos^2{\theta} + dr^2\sin^2{theta} - 2\alpha dr\cos{\theta} \cos{\phi} + \alpha^2 - 2\beta dr\cos{\theta} \sin{\phi} + \beta^2 + 2rdr\sin{\theta} \cos{\theta}d\theta + r^2\cos^2{\theta} d{\theta}^2$$

$$ds^2 = dr^2 - 2\alpha dr\cos{\theta} \cos{\phi} + \alpha^2 - 2\beta dr\cos{\theta} \sin{\phi} + \beta^2 + 2rdr\sin{\theta} \cos{\theta}d\theta + r^2\cos^2{\theta} d{\theta}^2 $$

The next set of steps involves solutions utilizing our substitutions of $$\alpha$$ and $$\beta$$. Recall the "inner" and "outer" terms from the FOIL multiplications. Each of those terms must now be expanded and simplified.

$$A = - 2\alpha dr\cos{\theta} \cos{\phi}$$

$$A = - 2rdr(\sin{\theta}\cos{\theta}\cos^2{\phi}d\theta + \cos^2{\theta} \cos{\phi} \sin{\phi}d{\phi})$$

$$B = - 2\beta dr\cos{\theta} \sin{\phi}$$

$$B = - 2rdr(\sin{\theta} \cos{\theta} \sin^2{\phi} d\phi - \cos^2{\theta} \cos{\phi} \sin{\phi} d\phi)$$

$$A + B = -2rdr \sin{\theta}\cos{\theta}(\cos{\phi}^2 + \sin^2{\phi})d\theta = -2rdr \sin{\theta} \cos{\theta}d\theta$$

Recall equation 7.1.4. The terms A and B can be put back into the equation to get the simplification.

$$ds^2 = dr^2 + A + \alpha^2 + B + \beta^2 + 2rdr\sin{\theta} \cos{\theta}d\theta + r^2\cos^2{\theta} d{\theta}^2 $$

$$ds^2 = dr^2 - 2rdr \sin{\theta} \cos{\theta} d\theta + \alpha^2 + \beta^2 + 2rdr\sin{\theta} \cos{\theta}d\theta + r^2\cos^2{\theta} d{\theta}^2 $$

Next, the terms for $${\alpha}^2 + {\beta}^2$$ must be expanded.

$${\alpha}^2 = r^2 \sin^2{\theta} \cos^2{\phi} d{\theta}^2 + 2r^2 \sin{\theta} \cos{\theta} \cos{\phi} \sin{\phi} d\theta d\phi r^2 \cos^2{\theta} \sin^2{\phi} d{\phi}^2$$

$${\beta}^2 = r^2 \sin^2{\theta} \sin^2{\phi} d{\theta}^2 - 2r^2 \sin{\theta} \cos{\theta} \cos{\phi} \sin{\phi} d\theta d\phi r^2 \cos^2{\theta} \cos^2{\phi} d{\phi}^2$$

$${\alpha}^2 + {\beta}^2 = r^2\sin^2{\theta} d{\theta}^2 (\cos^2{\phi} + \sin^2{\phi}) + r^2 \cos^2{\theta}(\sin^2{\phi} + \cos^2{\phi})d{\phi}^2$$

$${\alpha}^2 + {\beta}^2 = r^2\sin^2{\theta} d{\theta}^2 + r^2 \cos^2{\theta} d{\phi}^2$$

Moving along, all inner and outer terms from the FOIL calculation are added and simplified below.

$$ds^2 = dr^2 - rdr \sin{\theta} \cos{\theta} d\theta + r^2 \cos^2{\theta} d{\theta}^2 + r^2 \sin^2{\theta} d{\theta}^2 + r^2 \cos^2{\theta} d{\phi}^2 + 2rdr \sin{\theta} \cos{\phi} d\theta - 2rdr \sin{\theta} \cos{\theta} d\theta$$

Finally, after addition and subtracting terms above, the solution is derived in the desired form to solve the problem.

$$ds^2 = dr^2 + r^2 d{\theta}^2 + r^2 \cos^2{\theta} d{\phi}^2$$

Each coordinate line has a tangent vector with a magnitude of h.

$$dr^2 = 1 * dr^2 = {h_1}^2{\xi_1}^2$$

$${h_1}^2 = 1$$ $${h_1} = 1$$

Similarly, the other h parameters can be found for each variable. $$r^2 d{\theta}^2 = {h_2}^2{\xi_2}^2$$

$${h_2}^2 = r^2$$

$${h_2} = r$$

$$r^2 \cos^2{\theta} d{\phi}^2 = {h_3}^2{\xi_3}^2$$

$${h_3}^2 = r^2 \cos^2{\theta}$$

$${h_3} = r\cos{\theta}$$

=Problem 7.4 Plot the Altitude =

Given:

 * $$z(t):=v_{y}(t)$$
 * $$n=2,3, a=k/m=2, b=mg=10$$
 * $$z(0)=v_{y}(0)=50$$

Find:
Plot $$z(t)$$ versus $$t$$ for each value of n. Also plot the altitude $$y(t)$$ vs time and find the time the projectile hits the ground

1. The case when n=2
For n=2 we can integrate ($$) by hand using the following integration rule.

Plugging in the parameter values given into ($$) and taking the derivative involves the following steps


 * $$\frac{1}{10}\int \frac{1}{\frac{z^2}{5}+1}du$$

To get this equation into the form of ($$) we can say $$u=\frac{z}{\sqrt{5}}$$ which would make $$du=\frac{1}{\sqrt{5}}dz$$. Substituting these values in gives us...


 * $$\frac{\sqrt{5}}{10}\int \frac{1}{u^2+1}du=\frac{\sqrt{5}}{10}\tan^{-1}\left(u\right)+c$$

We can now plug $$ u$$ back in.


 * $$-t=\frac{\sqrt{5}}{10}\int \frac{1}{u^2+1}du=\frac{\sqrt{5}}{10}\tan^{-1}\left(\frac{z}{\sqrt{5}}\right)+c$$

The same solution was found using Wolfram Alpha and can be found at this link

Solve for z:
 * $$z=\sqrt{5}tan\left(-2\sqrt{5}t+c2\sqrt{5}\right)$$

Once again we know that at $$t=0 z=50$$ so we can find the constant
 * $$c=-1.78e14$$

Since we cannot solve for z we will use Matlab to plot z(t) vs t

We know that at $$t=0; z=50$$ so we can solve for the constant.
 * $$c=\frac{\sqrt{5}}{10}\tan^{-1}(10\sqrt{5})=1.706$$

Therfore...
 * $$z=\sqrt{5}tan\left(-2\sqrt{5}t+1.526\right)$$

To find the altitude as a function time we can just take the integral of the velocity with respect to time
 * $$y(t)=\int\left[\sqrt{5}tan\left(-2\sqrt{5}t+1.526\right)\right]+c$$

Using matlab to solve symbolically we obtain...
 * $$y(t)=\frac{1}{2}\log\left(\cos\left(2\sqrt{5}t-\frac{763}{500}\right)\right)+c$$

Assuming $$y=0$$ when $$t=0$$ we can solve for c...
 * $$c=-\frac{1}{2}\log\left[\cos\left(\frac{-763}{500}\right)\right]$$

And...
 * $$y(t)=\frac{1}{2}\log\left(\cos\left(2\sqrt{5}t-\frac{763}{500}\right)\right)-\frac{1}{2}\log\left[\cos\left(\frac{-763}{500}\right)\right]$$

The particle returns to the ground at approximately 0.68 seconds

2. Solve using Matlab 'roots' function
For $$n=3$$, we have
 * $$\int\frac{dz}{2z^3+10}=\frac{1}{10}z\;_{2}F_{1}\left(1,\frac{1}{3} ; 1+\frac{1}{3} ; -\frac{z^3}{5} \right )+k$$

and


 * $$\Rightarrow -t = \frac{1}{10}z\;_{2}F_{1}\left(1,\frac{1}{3} ; \frac{4}{3} ; -2\frac{z^3}{10} \right )+k$$

At $$z=50, t=0$$
 * $$\Rightarrow k = -\frac{1}{10}50\;_{2}F_{1}\left(1,\frac{1}{3} ; \frac{4}{3} ; -\frac{50^3}{5} \right ) = -5\;_{2}F_{1}\left(1,\frac{1}{3} ; 1+\frac{1}{3} ; -\frac{50^3}{5} \right )$$
 * $$\Rightarrow t = 5\;_{2}F_{1}\left(1,\frac{1}{3} ; \frac{4}{3} ; -\frac{50^3}{5} \right ) - \frac{1}{10}z\;_{2}F_{1}\left(1,\frac{1}{3} ; 1+\frac{1}{3} ; -\frac{z^3}{5} \right )$$


 * $$\Rightarrow t = 5\;_{2}F_{1}\left(1,\frac{1}{3} ; \frac{4}{3} ; -\frac{50^3}{5} \right )- \frac{z}{10}\,\left(1-\frac{1\cdot \frac{1}{3}}{\frac{4}{3}}\frac{\frac{z^3}{5}}{1!}+ \frac{1\cdot2\cdot\frac{1}{3} \cdot\frac{4}{3}}{\frac{4}{3}\cdot \frac{7}{3}}\frac{(\frac{z^3}{5})^2}{2!}-\frac{1\cdot2\cdot3\cdot\frac{1}{3}\cdot\frac{4}{3}\cdot\frac{7}{3}}{\frac{4}{3}\cdot\frac{7}{3}\cdot\frac{10}{3}}\frac{(\frac{z^3}{5})^3}{3!}+\cdots\right)$$

Discarding higher order terms with decreasing coefficients
 * $$t \approx 5\;_{2}F_{1}\left(1,\frac{1}{3} ; \frac{4}{3} ;-\frac{50^3}{5} \right ) - 0.1z+0.005z^4-0.00057z^7+0.00008z^{10}$$

But unfortunately the roots obtained by the approximation, that is correct to a high extent, failed to yield a root, that is close to the initial value in the problem. In fact all of the roots were in pairs of complex conjugate. Which led to the idea that it is probably not a feasible way to calculate the solution.

3. Solve using Matlab 'hypergeom' function
For $$n=3$$, we have
 * $$\int\frac{dz}{2z^3+10}=\frac{1}{10}z\;_{2}F_{1}\left(1,\frac{1}{3} ; 1+\frac{1}{3} ; -\frac{z^3}{5} \right )+k$$

and


 * $$\Rightarrow -t = \frac{1}{10}z\;_{2}F_{1}\left(1,\frac{1}{3} ; 1+\frac{1}{3} ; -2\frac{z^3}{10} \right )+k$$

At $$z=50, t=0$$
 * $$\Rightarrow k = -\frac{1}{10}50\;_{2}F_{1}\left(1,\frac{1}{3} ; 1+\frac{1}{3} ; -\frac{50^3}{5} \right ) = -5\;_{2}F_{1}\left(1,\frac{1}{3} ; 1+\frac{1}{3} ; -\frac{50^3}{5} \right )$$
 * $$\Rightarrow t = 5\;_{2}F_{1}\left(1,\frac{1}{3} ; 1+\frac{1}{3} ; -\frac{50^3}{5} \right ) - \frac{1}{10}z\;_{2}F_{1}\left(1,\frac{1}{3} ; 1+\frac{1}{3} ; -\frac{z^3}{5} \right )$$

Using this information, we find $$z(t)$$, which we use to determine $$y(t)$$

Experimentally we found that the minimum value $$z$$ can attend using the relation ($$) is $$-1.71$$ (In reality, with these parameters, at this velocity, the gravitational acceleration and the deceleration due to the resistance counteract each other). So we limited the range of $$z$$ in the code to that extent only. Below the range, the hypergeometric function produces complex numbers.



n=2
For n = 2, the graphs obtained are presented below



n=3
For n = 3, the graphs obtained are presented below



n=2
For n = 2, the graph obtained is presented below



n=3
For n = 3, the graph obtained is presented below

Pavel Bhowmik (talk) 14:41, 5 December 2012 (UTC)

=Problem 7.5* Heat conduction on a cylinder =

Find

 * 1) $$ \{dx_i \}=\{dx_1,dx_2,dx_3 \} $$ 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) $$ ds^2=\sum_{i}(dx_i)^2= \sum_{k}(h_k)^2(d\xi_k)^2 $$ Identify $$ \{h_i\} $$ in terms of $$ \{\xi_i\} $$
 * 3) $$ \Delta u $$ in cylindrical coordinates
 * 4) Use separation of variables to find the separated equations and compare to the Bessel Equation
 * $$ (1-x^2)y''-2xy'+(x^2-v^2)y=0, v\in \mathbb{R} $$

Part 2
Plugging Equations ($$),($$)($$) into Equation ($$), we get

$$ ds^2= (dx_1)^2+(dx_2)^2+(dx_3)^2 $$
 * $$ = [(cos\xi_2)^2\,(d\xi_1)^2-2\,\xi_1\,cos\xi_2\,sin\xi_2\,d\xi_1\,d\xi_2+\xi_1^2\,(sin\xi_2)^2\,(d\xi_2)^2]+[(sin\xi_2)^2\,(d\xi_1)^2+2\,\xi_1\,cos\xi_2\,sin\xi_2\,d\xi_1\,d\xi_2+\xi_1^2\,(cos2\xi_2)^2\,(d\xi_2)^2]+(d\xi_3)^2 $$
 * $$ = \underbrace{1}_{(h_1)^2} \,(d\xi_1)^2+\underbrace{\xi_1^2 }_{(h_2)^2} \,(d\xi_2)^2+\underbrace{1}_{(h_3)^2} \,(d\xi_3)^2 $$

Therefore $$\{h_i\}=\{1,\xi_1,1\}$$

Part 3
Laplace operator in general curvilinear coordinates

From Part 2, we get$$ h_1\,h_2\,h_3=\xi_1 $$


 * $$i=1$$


 * $$i=2$$


 * $$i=3$$

Plugging Equations ($$)($$)($$)into Equation ($$) yeilds

$$ \Delta u= \frac{1}{\xi_1} \left[\frac{\partial}{\partial\xi_1}\left(\xi_1\,\frac{\partial u}{\partial \xi_1}\right)+\frac{1}{\xi_1}\,\frac{\partial^2 u}{\partial\xi_2^2}+\xi_1\,\frac{\partial^2 u}{\partial\xi_3^2}\right] $$
 * $$= \frac{1}{\xi_1}\,\frac{\partial}{\partial\xi_1}\left(\xi_1\,\frac{\partial u}{\partial \xi_1}\right)+ \frac{1}{\xi_1^2}\,\frac{\partial^2 u}{\partial\xi_2^2}+ \frac{\partial^2 u}{\partial\xi_3^2} $$

Part 4
Define $$ u$$ with separate variables

$$ \Delta u=\frac{Y Z}{\xi_1}\frac{d}{d\xi_1}\left(\xi_1\frac{dX}{d\xi_1}\right)+\frac{XZ}{\xi_1^2}\frac{d^2Y}{d\xi_2^2}+XY\frac{d^2Z}{d\xi_3^2} =0 $$

Divide by $$XYZ$$ yeilds $$ \Delta u=\underbrace {\frac{1}{X\xi_1}\frac{d}{d\xi_1}\left(\xi_1\frac{dX}{d\xi_1}\right) +\frac{1}{Y\xi_1^2}\frac{d^2Y}{d\xi_2^2}}_{\alpha(\xi_1,\xi_2)}+\underbrace {\frac{1}{Z}\frac{d^2Z}{d\xi_3^2}}_{\beta(\xi_3)}=0 $$

$$ \Rightarrow -\alpha(\xi_1,\xi_2)=\beta(\xi_3)=k_1 (constant) $$

$$ \frac{1}{Z}\frac{d^2Z}{d\xi_3^2}=k_1 $$

$$\frac{1}{X\xi_1}\frac{d}{d\xi_1}\left(\xi_1\frac{dX}{d\xi_1}\right) +\frac{1}{Y\xi_1^2}\frac{d^2Y}{d\xi_2^2}=-k_1$$

Multiply by $$\xi_1^2$$ yields

$$\underbrace{\frac{\xi_1}{X}\frac{d}{d\xi_1}\left(\xi_1\frac{dX}{d\xi_1}\right)+k_1\xi_1^2}_{\alpha_1(\xi_1)} +\underbrace{\frac{1}{Y}\frac{d^2Y}{d\xi_2^2}}_{\alpha_2(\xi_2)}=0$$

$$\Rightarrow -\alpha_1(\xi_1)=\alpha_2(\xi_2)=k_2 (constant)$$

$$\frac{1}{Y}\frac{d^2Y}{d\xi_2^2}=k_2$$

$$\frac{\xi_1}{X}\frac{d}{d\xi_1}\left(\xi_1\frac{dX}{d\xi_1}\right)+k_1\xi_1^2=-k_2$$

$$\Rightarrow \frac{\xi_1}{X}\left(\xi_1\frac{d^2X}{d\xi_1^2}+\frac{dX}{d\xi_1}\right)+k_1\xi_1^2+k_2=0$$

$$\Rightarrow \xi_1^2\frac{d^2X}{d\xi_1^2}+\xi_1\frac{dX}{d\xi_1}+(k_1\xi_1^2X+k_2X)=0$$

Here we get an L2-ODE-VC about $$\xi_1$$,replace $$\xi_1=x, X(\xi_1)=y(x)$$yields

$$x^2y''+xy'+(k_1x^2y+k_2y)=0$$

Let$$ k_1=-\frac{1}{2}, k_2=\frac{v^2}{2} $$ Equation ($$)becomes $$ -2x^2y''-2xy'+(x^2-v^2)y=0 $$

Comparing this equation with Bessel's Equation $$ (1-x^2)y''-2xy'+(x^2-v^2)y=0, v\in R $$

We find these two equations look similar in the structure except the coefficient of the highest order term.

=Problem 7.6* Find Spherical Coordinates =

Find

 * $$\Delta u$$ in spherical coordinates

Solution
Math/Physics convention for spherical coordinates is as follows


 * $$(\xi_{1},\xi_{2},\xi_{3})=(r,\theta,\phi)$$

From problem 7.3 we know the following expressions


 * $$h_{1}=1$$
 * $$h_{2}=r$$
 * $$h_{3}=r\cos \theta$$

According to Equation (2) 39-3 from the lecture notes


 * $$\frac{\partial}{\partial \xi_{1}}\left[\frac{h_{1}h_{2}h_{3}}{h_{1}^2}\frac{\partial u}{\partial \xi_{1}} \right ]=\frac{\partial}{\partial r}\left[\frac{r^2\cos \theta}{1^2}\frac{\partial u}{\partial r} \right ]$$

Therefore we know from the math/physics convention that the following expressions are true
 * $$\frac{\partial}{\partial \xi_{2}}\left[\frac{h_{1}h_{2}h_{3}}{h_{2}^2}\frac{\partial u}{\partial \xi_{2}} \right ]=\frac{\partial}{\partial \theta}\left[\frac{r^2\cos \theta}{r^2}\frac{\partial u}{\partial \theta} \right ]$$


 * $$\frac{\partial}{\partial \xi_{3}}\left[\frac{h_{1}h_{2}h_{3}}{h_{3}^2}\frac{\partial u}{\partial \xi_{3}} \right ]=\frac{\partial}{\partial \phi}\left[\frac{r^2\cos \theta}{(r\cos \theta)^2}\frac{\partial u}{\partial \phi} \right ]$$

Using these expressions and the math/physics convention we can find the Laplacian in spherical coordinates based on Equation 1


 * $$\Delta u=\frac{1}{r^2\cos \theta}\left[\frac{\partial}{\partial r}\left(\frac{r^2\cos \theta}{1^2}\frac{\partial u}{\partial r} \right )+\frac{\partial}{\partial \theta}\left(\frac{r^2\cos \theta}{r^2}\frac{\partial u}{\partial \theta} \right )+\frac{\partial}{\partial \phi}\left(\frac{r^2\cos \theta}{(r\cos \theta)^2}\frac{\partial u}{\partial \phi} \right )\right]$$

This can be simplified to


 * $$\Delta u=\frac{1}{r^2\cos \theta}\left[\frac{\partial}{\partial r}\left(r^2\cos \theta\frac{\partial u}{\partial r} \right )+\frac{\partial}{\partial \theta}\left(\cos \theta\frac{\partial u}{\partial \theta} \right )+\frac{\partial}{\partial \phi}\left(\frac{1}{\cos \theta}\frac{\partial u}{\partial \phi} \right )\right]$$

=Contribution Table=

= References =