User:Egm6321.f12.team3/rephid7

=R7.1 Find local maximum and verify the validity of equation=

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem
1. Use MATLAB to plot $$F(5,-10;1;x)$$ near x=0 to show the local maximum (or maxima) in this region.

Solution
As will be shown in Part 2 of this problem, function $$F(5,-10;1;x)$$ has a singular point at x=1, thus we plot the hypergeometric function over the interval less than 1, that is, $$-0.05\leq x \leq 0.99$$.

The hypergeometric function $$F(5,-10;1;x)$$ and its derivative function are ploted using MATLAB code as follows,





We can conclude from the plot that, within interval $$-0.05\leq x \leq 0.5$$, derivative function of the hypergeometric function under consideration has 3 zero points, in which the zero point at $$x=0.23$$ is the one corresponding to the local maximum in this region, that is, $$F(5,-10;1;0.23)=0.1481$$.

By definition of the hypergeometric function,

We can express $$F(5,-10;1;x)$$ as

Where,

To check coefficient of each term of the power series of ($$), we have coefficients of the 9th, 10th, 11th, 12th terms of the series as follows

the 9th term:

the 10th term:

the 11th term:

the 12th term:

Obviously we can reach the conclusion that, because the factor $$ (-10)_{k}$$ in numerator equals to zero when $$ k\geqslant 11$$, the coefficients of the kth terms of the series representing hypergeometric function are all zero when k satisfies $$ k\geqslant 11$$, which means that the power series under consideration has 11 terms only.

Thus we have,

$$\displaystyle F(5,-10;1;x)= \sum^{10}_{k=0} (5)_k(-10)_k\frac{x^k}{k!k!} $$

To show the validity of equation $$F(5,-10;1;x) = (1-x)^6(1001x^4-1144x^3+396x^2-44x+1)$$, we should expand the RHS of this equation first, then compare the coefficients of each term of expanded RHS with their counterparts of LHS.

First expand the factor $$ (1-x)^6$$ on RHS, we have

Thus, the product of $$ (1-x)^6$$ and $$ (1001x^4-1144x^3+396x^2-44x+1)$$ can be expanded as a power series with 11 terms whose power of x increases from 0 to 10. The terms from power 0 to power 10 are,

Power 0: $$\displaystyle 1\times 1=1 $$

Power 1: $$\displaystyle -44x-6x=-50x $$

Power 2: $$\displaystyle 15 x^2+396 x^2+44x\times 6x=675 x^2 $$

Power 3: $$\displaystyle -20x^3-1144x^3-6x \times396x^2+44x \times 15x^2=-4200 x^3$$

Power 4: $$\displaystyle 1001x^4+15x^4+6x\times1144x^3+44x\times20x^3+396x^2\times 15x^2=14700 x^4 $$

Power 5: $$\displaystyle -6x^5-15x^4\times44x -20x^3\times396x^2-15x^2\times1144x^3-6x\times 1001x^4=-31752 x^5 $$

Power 6: $$\displaystyle x^6+6x^5\times44x+15x^4\times396x^2+20x^3\times1144x^3+15x^2\times 1001x^4=44100 x^6 $$

Power 7: $$\displaystyle - x^6\times44x-6x^5\times396x^2 -15x^4\times1144x^3 -20x^3\times1001x^4=-39600 x^7 $$

Power 8: $$\displaystyle x^6\times396x^2+6x^5\times1144x^3 +15x^4\times1001x^4 =22275 x^8 $$

Power 9: $$\displaystyle -x^6\times1144x^3 - 6x^5\times1001x^4 =-7150 x^9$$

Power 10: $$\displaystyle x^6\times1001x^4=1001 x^{10}$$

Then, compare the coefficients of each terms above with their counterparts of LHS. Clearly sum of the 11 terms listed above is the same as LHS of Eq. the validity of ($$) is now proved.

The alternative way to prove the validity of equation $$F(5,-10;1;x) = (1-x)^6(1001x^4-1144x^3+396x^2-44x+1)$$ is just subtract LHS from RHS of this equation using MotionGenesis code listed as follows,

MotionGenesis returned Which means that LHS equals to RHS of ($$), the validity of ($$) is now proved.

=R 7.2 Variation of Parameters=

Given
Linear Second Order ODE with Constant Coefficients :

$$\displaystyle y'' + a_0 = f(t), t \geq t_0$$

Initial conditions are : $$\displaystyle y(t_0), y'(t_0)$$

Problem
Use Variation of Parameters to show that

$$\displaystyle y(t) = y_0 cos(a_0(t-t_0)) + \frac {y'(t_0)}{a_0}sin(a_0(t-t_0)) + \frac{1}{a_0}\int_{t_0}^{t} f(\tau)sin(a_0(t-\tau))d\tau$$

Solution
Taking a trial solution for the homogeneous equation as

$$\displaystyle y= e^{rt}$$

$$\displaystyle y' = re^{rt}$$

$$\displaystyle y'' = r^2 e^{rt}$$

$$\displaystyle \therefore y'' + a_0y = r^2 e^{rt} + a_0 e^{rt} = 0 $$

$$\displaystyle \therefore r^2 + a_0 = 0$$

$$\displaystyle r = \pm i \sqrt a_0$$

Thus we have got both the homogeneous solutions instead of 1, and the homogeneous solution being

$$\displaystyle y_h = C_1 cos(\sqrt a_0 t) + C_2 sin (\sqrt a_0 t)$$

Comparing it to the homogeneous part of the solution to be shown, we can see that the terms above have $$\displaystyle \sqrt a_0$$ instead of $$\displaystyle a_0$$. We assume that there is a printing mistake as replace $$\displaystyle a_0$$ by $$\displaystyle a_0 ^2$$ in the original differential equation.

$$\displaystyle y_h = C_1 cos( a_0 t) + C_2 sin ( a_0 t)$$

$$\displaystyle y_h ' = a_0(- C_1 sin ( a_0 t) + C_2 cos( a_0 t)) $$

At $$\displaystyle t=t_0$$

$$\displaystyle y(t_0) = C_1 cos( a_0 t_0) + C_2 sin ( a_0 t_0)$$

$$\displaystyle \frac{y'(t_0)}{a_0} = - C_1 sin ( a_0 t) + C_2 cos( a_0 t) $$

Solving for $$\displaystyle C_1$$ and $$\displaystyle C_2$$, we get

$$ \displaystyle C_2=y(t_0)cos(a_0t_0)-\frac{y'(t_0)}{a_0}sin(a_0t_0) $$

$$ \displaystyle C_1=y(t_0)sin(a_0t_0)+\frac{y'(t_0)}{a_0}cos(a_0t_0) $$

Plugging it in the equation for homogeneous solutions, and simplifying using trigonometric formulas, we get,

$$ \displaystyle y_h= y(t_0)cos(a_0(t-t_0))+\frac{y'(t_0)}{a_0}sin(a_0(t-t_0)) $$

$$\displaystyle \therefore u_1 = y(t_0)cos(a_0(t-t_0))$$

$$\displaystyle \therefore u_2 = \frac{y'(t_0)}{a_0}sin(a_0(t-t_0))$$

From equation 4, Pg 34-4,

$$\displaystyle h(t) = u_1 ^2 exp\int(a_1).dt$$, but a_1 = 0 is our ODE

$$\displaystyle h(t) = y^2(t_0)cos^2(a_0(t-t_0))$$

From Equation 1, Pg 34-6,

$$\displaystyle y_p = u_1(t)\int \frac{1}{h(t)}\left[\int h(t)\frac{f(t)}{u_1(t)}dt \right]dt$$

Now, $$\displaystyle \frac{h(t)}{u_1(t)} = u_1(t)$$

$$\displaystyle y_p = u_1(t)\int \frac{1}{h(t)}\left[\int u_1(t)f(t)dt \right]dt$$

Even with the "simplification", the integral will be difficult to solve. We will thus use the formula given by King and others (R*6.7).

$$\displaystyle y_p = \int^{t} f(\tau) \left[ \frac{u_1(\tau)u_2(t) - u_1(t)u_2(\tau)}{u_1(\tau) u_2'(\tau) - u_2(\tau) u_1 '(\tau)}\right]d\tau$$

$$\displaystyle u_1 ' = - a_0 y(t_0)sin(a_0(t-t_0))$$

$$\displaystyle u_2 ' = y'(t_0)cos(a_0(t-t_0))$$

$$\displaystyle y_p = \int^{t} f(\tau) \left [ \frac{ y(t_0) y'(t_0) (cos (a_0(\tau-\tau_0))sin (a_0(t-t_0))-cos (a_0(t-t_0))sin (a_0(\tau-\tau_0))}{y(t_0) y'(t_0) a_0 (cos ^2 (a_0(\tau-\tau_0) + sin ^2 (a_0(\tau-\tau_0))} \right] d \tau $$

$$\displaystyle y_p = \frac{1}{a_0}\int^{t} f(\tau) (cos (a_0(\tau-\tau_0))sin (a_0(t-t_0))-cos (a_0(t-t_0))sin (a_0(\tau-\tau_0)) )d \tau $$

$$\displaystyle y_p = \frac{1}{a_0}\int^{t} f(\tau) (sin (a_0(t-t_0))-(a_0(\tau-\tau_0)) )d \tau $$

$$\displaystyle y_p = \frac{1}{a_0}\int^{t} f(\tau) (sin (a_0(t-\tau)))d \tau $$

$$ y(t) = y_h + y_p = y(t_0)cos(a_0(t-t_0))+\frac{y'(t_0)}{a_0}sin(a_0(t-t_0)) + \frac{1}{a_0}\int^{t} f(\tau) (sin (a_0(t-\tau)))d \tau$$

This is equivalent to the expression that needs to be shown.

= R*7.3 Infinitesimal Length ds in terms of spherical coordinates =

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem
Show that the infinitesimal length $$ \displaystyle ds$$ in 3-D space can be written in spherical coordinates as

Given
From lecture notes ,

Cartesian Coordinates $$ \displaystyle (x,y,z)$$ and in Spherical Coordinates $$ \displaystyle (r,\theta,\phi)$$

$$ \displaystyle x_1 = x = r\cos {\theta} \cos {\phi} = \xi_1 \cos{\xi_2}\cos{\xi_3}$$

$$ \displaystyle x_2 = y = r\cos {\theta} \sin {\phi} = \xi_1 \cos{\xi_2} \sin{\xi_3}$$

$$ \displaystyle x_3 = z = r\sin {\theta} = \xi_1 \sin{\xi_2}$$

Infinitesimal length in Cartesian coordinates can be expressed as:

The Laplace operator in spherical coordinates is given by

Solution
We know,

For $$ \displaystyle i=1,2,3$$

Using the following equations,

we get,

$$ \displaystyle ds^2=(cos\theta)^2 dr^2-r sin\theta cos\theta dr d\theta-r cos\theta sin\theta d\theta dr+r^2(sin\theta)^2 (d\theta)^2 r^2(cos\theta)^2 (d\varphi)^2+(sin\theta)^2 dr^2+r cos\theta sin\theta dr d\theta+r sin\theta cos\theta d\theta dr+r^2(cos\theta)^2 (d\theta)^2$$

where,

$$ \displaystyle (h_1)^2=1$$

$$ \displaystyle dr^2=(d\xi_1)^2$$

$$ \displaystyle (h_2)^2=r^2$$

$$ \displaystyle d\theta^2=(d\xi_2)^2$$

$$ \displaystyle (h_3)^2=r^2(cos\theta)^2$$

$$ \displaystyle d\varphi^2=(d\xi_3)^2$$

Now, from the above results, we can deduce,

Now in ($$),

for $$ \displaystyle i=1, $$

for $$ \displaystyle i=2, $$

for $$ \displaystyle i=3, $$

Plugging the above results in ($$), we get,

This proves ($$).

=R7.4 Determine the time at which a projectile returns to the ground =

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem
The objective is to find expressions for the velocity and altitude of a projectile, which has an initial vertical velocity. After determining functions for the velocity and altitude, plot these functions and find the time at which the projectile returns to the ground.

Given
where $$v_y(t)$$ is the vertical velocity of the projectile.

Case I : n=2
Substituting the values of n,a and b in ($$), we get:

$$ \int \frac{dz}{2z^2+10} = \frac{1}{10}z\ _2F_1 \left(1,\frac{1}{2};1+ \frac{1}{2};-2\frac{z^2}{10}\right) $$

$$ = \frac{1}{10}z\ _2F_1 \left(1,\frac{1}{2}; \frac{3}{2};\frac{-2z^2}{10}\right) $$

Or we can directly integrate the given expression,

$$ \int \frac{dz}{2z^2+10} = \frac {1}{10} \int \frac{dz}{\frac{z^2}{5} + 1} $$

Let $$ u = \frac {z}{\sqrt 5}, \therefore u^2 = \frac{z^2}{5}, dz = \sqrt 5 du $$

$$ \int_{z(0)=50}^{z(t)} \frac{dz}{2z^2+10} = \frac {\sqrt 5}{10} \int \frac{du}{u^2 + 1} $$

$$ \int_{z(0)=50}^{z(t)} \frac{dz}{2z^2+10} = \frac {1}{2 \sqrt 5} tan^{-1}(u) + k $$

$$ \int_{z(0)=50}^{z(t)} \frac{dz}{2z^2+10} = \frac {1}{2 \sqrt 5} tan^{-1}(\frac {z}{\sqrt 5}) + k = -t$$

We have initial condition as at t = 0, z = 50

Putting it in the above integral get $$k = -\frac {1}{2 \sqrt 5} tan^{-1}(\frac {50}{\sqrt 5})$$

Thus we have,

$$\frac {1}{2 \sqrt 5} tan^{-1}(\frac {z}{\sqrt 5}) = \frac {1}{2 \sqrt 5} tan^{-1}(\frac {50}{\sqrt 5})- t $$

$$tan^{-1}(\frac {z}{\sqrt 5}) = tan^{-1}(\frac {50}{\sqrt 5})- 2 \sqrt 5 t $$

$$\frac {z}{\sqrt 5} = tan(tan^{-1}(\frac {50}{\sqrt 5})- 2 \sqrt 5 t)$$

Now,

$$y(t) = \int z(t) = \sqrt 5 \int tan(tan^{-1}(\frac {50}{\sqrt 5})- 2 \sqrt 5 t)$$

Integrating using Wolfram Alpha we get,

$$y(t) = \frac{1}{2\sqrt 5} log(cos(2 \sqrt 5 t - tan^{-1}(\frac {50}{\sqrt 5}))$$

The plot for y(t) and z(t) vs t are given below.

From the plot, we see that the projectile hits the ground at t = 0.6825 sec. It may noted that the Plot of Altitude y(t) has been translated to (0,0) for convenience of calculation of Time t when the projectile reaches the ground.

The MATLAB Code for the above plot is as follows:

Case II : n=3
=R*7.5 Heat Conduction in a Cylinder=

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given
$$\displaystyle x= r cos\theta = \xi_1 cos (\xi_2)$$

$$\displaystyle y = r sin\theta = \xi_1 sin (\xi_2)$$

$$\displaystyle z= \xi_3$$

Find

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


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


 * 3.  Find $$\displaystyle \Delta u$$ in cylindrical co-ordinates.


 * 4.  Use separation of variables to find separated equation and compare to the Bessel Equation.

Part 1
$$\displaystyle x_1 = x = r cos\theta = \xi_1 cos (\xi_2)$$

Using the chain rule, we get

Similarly,

Part 2
$$\displaystyle ds^2 = \sum_{i} (dx_i)^2 = \sum_{k} (h_k)^2 (d\xi_k)^2$$

$$\displaystyle (dx_1)^2 = \left(cos (\xi_2) d\xi_1 - \xi_1 sin (\xi_2) d\xi_2\right)^2$$

$$\displaystyle (dx_2)^2 = \left(sin (\xi_2) d\xi_1 + \xi_1 cos (\xi_2) d\xi_2\right)^2$$

Adding ($$),($$), and ($$)

$$\displaystyle ds^2 = cos^2 (\xi_2) (d\xi_1)^2 + (\xi_1)^2 sin^2 (\xi_2) (d\xi_2)^2 - 2 \xi_1 cos (\xi_2) sin (\xi_2) d\xi_1 d\xi_2 + sin^2 (\xi_2) (d\xi_1)^2 + (\xi_1)^2 cos^2 (\xi_2) (d\xi_2)^2 + 2 \xi_1 sin (\xi_2) cos (\xi_2) d\xi_1 d\xi_2 + (d\xi_3)^2$$

$$\displaystyle ds^2 = \left(cos^2 (\xi_2) + sin^2 (\xi_2)\right) (d\xi_1)^2 + (\xi_1)^2 \left(sin^2 (\xi_2) + cos^2 (\xi_2)\right) (d\xi_2)^2 + (d\xi_3)^2$$

$$\displaystyle ds^2 = (d\xi_1)^2 + (\xi_1)^2 (d\xi_2)^2 + (d\xi_3)^2$$

Part 3
$$\displaystyle \Delta u =\frac{1}{h_{1}h_{2}h_{3}}\sum^{3}_{i=1}\frac{\partial}{\partial \xi_{i}}\left[\frac{h_{1}h_{2}h_{3}}{(h_{i}^{2})}\frac{\partial u}{\partial \xi_{i}}\right] $$

$$\displaystyle h_{1}h_{2}h_{3} = \xi_1 = r$$

For $$ \displaystyle i=1, $$

for $$ \displaystyle i=2, $$

for $$ \displaystyle i=3, $$

Putting it all together,we get

Part 4
Let $$\displaystyle \Delta u = R(r)\Theta(\theta)Z(z) = 0$$

Substituting in (EquationNote 7.5.13)

$$\displaystyle \Delta u = \frac{\Theta Z}{r} \frac{\partial}{\partial r}\left[{r} \frac{\partial R}{\partial r}\right] + \frac{RZ}{r^2} \frac{\partial}{\partial \theta}\left[\frac{\partial \Theta}{\partial \theta}\right] + {R\Theta} \left[\frac{{\partial}^2 Z}{\partial z^2}\right]$$ $$\displaystyle $$

Dividing through by $$\displaystyle R \Theta Z$$ and multiplying by $$\displaystyle r^2$$

$$\displaystyle \Delta u = \frac{r}{R} \frac{\partial}{\partial r}\left[{r} \frac{\partial R}{\partial r}\right] + \frac{1}{\Theta} \frac{\partial}{\partial \theta}\left[\frac{\partial \Theta}{\partial \theta}\right] + \frac{r^2}{Z} \left[\frac{{\partial}^2 Z}{\partial z^2}\right]$$

Now, the functions of $$\displaystyle Z$$ and $$\displaystyle \Theta$$ are constants in terms of $$\displaystyle r$$

Taking

$$\displaystyle \frac{1}{\Theta} \frac{\partial}{\partial \theta}\left[\frac{\partial \Theta}{\partial \theta}\right] = k_1 $$ and

$$\displaystyle \frac{r^2}{Z} \left[\frac{{\partial}^2 Z}{\partial z^2}\right] = r^2 k_2$$

$$\displaystyle \frac{r}{R} \frac{\partial}{\partial r}\left[{r} \frac{\partial R}{\partial r}\right] = \frac{r}{R} \left[ \frac{\partial R}{\partial r} + {r} \frac{{\partial}^2 R}{\partial r^2}\right] = \frac{r}{R} \frac{\partial R}{\partial r} + \frac{r^2}{R} \frac{{\partial}^2 R}{\partial r^2}$$

Substituting for $$\displaystyle \Delta u$$

$$\displaystyle \Delta u = \frac{r}{R} \frac{\partial R}{\partial r} + \frac{r^2}{R} \frac{{\partial}^2 R}{\partial r^2} + r^2 k_2 + k_1 = 0$$

Multiplying by R, and dividing by $$\displaystyle k_2$$ and putting $$\displaystyle \frac{k_1}{k_2} = -\nu ^2$$

$$\displaystyle \Delta u = \frac{r^2}{k_2} \frac{{\partial}^2 R}{\partial r^2} + \frac{r}{k_2} \frac{\partial R}{\partial r} + (r^2 - \nu ^2)R = 0$$

This, when compared to the Bessel Equation

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

Shows us that the two equations are different only in their coefficients for $$\displaystyle y' '$$ and $$\displaystyle \frac{{\partial}^2 R}{\partial r^2}$$

$$\displaystyle$$

= R*7.6 Laplace Operator in Spherical Coordinates using Maths/Physics convention =

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem
Show that the infinitesimal length $$ \displaystyle ds$$ in 3-D space can be written in spherical coordinates as

$$ \displaystyle ds^2 = 1.dr^2 + r^2 d{\theta}^2 + r^2 \cos^2{\theta} d{\phi}^2$$

Also, find the Laplace Operator in Spherical Coordinates using the Maths/Physics Convention

Given
From lecture notes ,

Cartesian Coordinates $$ \displaystyle (x,y,z)$$ in Spherical Coordinates $$ \displaystyle (r,\theta,\phi)$$ and Maths/Physics Convention $$ \displaystyle (r,\bar \theta,\phi)$$

$$ \displaystyle x_1 = x = \xi_1 \cos{\xi_2}\cos{\xi_3} = r\sin{\bar \theta} \cos {\phi}$$

$$ \displaystyle x_2 = y = \xi_1 \cos{\xi_2} \sin{\xi_3} = r\sin {\bar \theta} \sin {\phi}$$

$$ \displaystyle x_3 = z = \xi_1 \sin{\xi_2} = r\cos {\bar \theta}$$

where $$ \displaystyle \bar \theta = \frac{\pi}{2}-\theta$$

Infinitesimal length in Cartesian coordinates can be expressed as:

The Laplace operator in spherical coordinates is:

Solution
For ease let, $$\displaystyle (\xi_{1},\xi_{2},\xi_{3})=(r,\bar\theta,\phi)$$

Taking the partial derivatives,

Taking the squares of ($$), ($$) and ($$),


 * $$\displaystyle (dx_{1})^{2} = (\sin\xi_{2}\cos\xi_{3}d\xi_{1})^{2} + (\xi_{1}\cos\xi_{2}\cos\xi_{3}d\xi_{2})^{2} + (\xi_{1}\sin\xi_{2}\sin\xi_{3}d\xi_{3})^{2} + 2\xi_{1}\sin\xi_{2}\cos\xi_{2}\cos^{2}\xi_{3}d\xi_{1}d\xi_{2} - 2\xi_{1}^{2}\cos\xi_{2}\sin\xi_{2}\cos\xi_{3}\sin\xi_{3}d\xi_{2}d\xi_{3}-2\xi_{1}\sin^{2}\xi_{2}\cos\xi_{3}\sin\xi_{3}d\xi_{1}d\xi_{3}$$


 * $$\displaystyle (dx_{2})^{2} = (\sin\xi_{2}\sin\xi_{3}d\xi_{1})^{2} + (\xi_{1}\cos\xi_{2}\sin\xi_{3}d\xi_{2})^{2} + (\xi_{1}\sin\xi_{2}\cos\xi_{3}d\xi_{3})^{2} + 2\xi_{1}\cos\xi_{2}\sin\xi_{2}\sin^{2}\xi_{3}d\xi_{1}d\xi_{2} + 2\xi_{1}^{2}\sin\xi_{2}\cos\xi_{2}\sin\xi_{3}\cos\xi_{3}d\xi_{2}d\xi_{3}+2\xi_{1}\sin^{2}\xi_{2}\sin\xi_{3}\cos\xi_{3}d\xi_{1}d\xi_{2}$$


 * $$\displaystyle (dx_{3})^{2} = (\cos\xi_{2}d\xi_{1})^{2} + (\xi_{1}\sin\xi_{2}d\xi_{2})^{2}-2\xi_{1}\sin\xi_{2}\cos\xi_{2}d\xi_{1}d\xi_{2} $$

Substituting the squares of the partial derivatives of x into ($$) and simplifying,

Further simplification, yields:

From ($$), we can identify $$\displaystyle h_{1}=1,\ h_{2}=\xi_{1},\ h_{3}=\xi_{1}\sin\xi_{2} $$.

Substituting this in ($$), the laplace operator in spherical co-ordinates with a math/physics convention can be expressed as:

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