User:Egm6321.f12.team4/Report7

Problem R7.1 – Plotting Hypergeometric Function
sec64-9b

Statement
Use matlab to plot F(5, 10; 1; x) near x = 0 to display the local maximum (or maxima) in this region.

Show that

Solution
(a) Matlab Code 

(b) Matlab Result 

(c) Graph 

Author and References

 * Solved and Typed by -- Seong Hyeon Hong
 * Reviewed by -- Pushkar Mishra

Statement
Consider the L2-ODE-CC:

Initial conditions:

Use variation of parameters to show that:

Solution
In order to obtain at least the 1st homogeneous solution, we use trial solution:

and substitute it into homogeneous version of (7.2.1):

Here the exponential term $$e^{\lambda t}$$ can be canceled and then we derive the characteristic equation:

The two characteristic roots are

It is obvious that we have obtained both the 1st and the 2nd homogeneous solution, because both of the characteristic roots are constant. We then choose $$ y_{1} = e^{i \sqrt{a_0}t}$$ as the 1st homogeneous solution and use variation of parameters by assuming:

By substituting it into (7.2.1):

Assume Z:

By IFM, we assume $$h=h(t)$$, then

Then

Then U can be found to be:

Thus y(x) can be obtained:

Substituting expression for h, (7.2.13) into the above expression:

The first term turns out to be in a form of Fourier transformation where the function to be transformed is $$e^{-i\sqrt{a_0}s}f(s)$$. And through the convolution theorem， it can be derived as:

Thus the solution can be written as:

If we only consider real part of the solution, the expression (7.2.21) can be reduced to the form:

To determine the two constants, we plug the solution into two initial conditions:

We can then solve $$C_1',C_2$$ in terms of $$y(t_0),y'(t_0)$$ :

By substituting these two coefficients back into the expression (7.2.22) and reorganization, we can derive the final form of the solution:

(7.2.26) is the required expression and we note that it should be $$\sqrt{a_0}$$ instead of $${a_0}$$ in the solution. This is also reasonable in the perspective of unit, because in a spring-mass system y(t) has a unit of length, $$\sqrt{a_0}$$ has a unit of frequency, and y'(t) has a unit of velocity. Then $$\frac{y'(t)}{\sqrt{a_0}}$$ will have a unit of displacement, too.

Author and References

 * Solved and Typed by -- Rui Che
 * Reviewed by -- Kaitlin Harris

Problem R*7.3 – Infinitesimal Length in Spherical Coordinates
sec39-1

Statement
Show that the infinitesimal length

can be written in spherical coordinates as follows.

Also, show that the Laplace operator in general curvilinear coordinates,

can be written in spherical coordinates as follows,

Solution
The Eq.(2)sec38-6 can be expanded as below,

and it is given that the Cartesian coordinates, the general curvilinear coordinates and the spherical coordinates have the following relations,

According to the definition of total derivative we can say that,

Therefore if we apply the above equation to find $$dx_1$$, $$dx_2$$ and $$dx_3$$ we get,

Now we square $$dx_1$$, $$dx_2$$ and $$dx_3$$ and add them together to get $$ds^2$$

The result above shows that,

Now if we apply into the Eq(2)sec39-2,

i=1

'i=2'

'i=3'

Therefore now we have the following equation,

Author and References

 * Solved and Typed by -- Seong Hyeon Hong
 * Reviewed by -- Pushkar Mishra

Problem R7.4 – Determine the Path of a Particle Moving with Air Resistance
sec63-8 sec64-9b

Statement
Consider the integral with particular values,


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$$ \int_{z(0)=50}^{z(t)}\frac{dz}{az^n+b}= -\int_0^t dt = -t $$     (7.10.1)
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$$ a=k/m=2, b=g=10, z(t):=v_y(t) $$
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Initial vertical velocity: $$ z(0)=v_y(0)=50 $$ For each value of n, find the vertical velocity z(t) for different values of time (t), plot this function. Also find and plot the altitude y(t) vs time t, then find the time the projectile returns to the ground.
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If the explicit function cannot be found, use a numerical method to find z(t) for each given value of t.

(1) n=2

(2) n=3. Use the matlab command "roots" to find the appropriate root z for each given time t, verify with WA. Plot z versus t. Find y(t) by integrating z(t) using the trapezoidal rule.

(3) n=3 Use the matlab function "hypergom" to find the time t for each given value of z in the interval [-10,50]. Plot t versus z. Find y(t) by integrating z(t) using the trapezoidal rule. Compare with part 2

(4) Verify the results in Part 1 and 2 using matlab in 2 steps (a) use the command ode45 to integrate and obtain z(t), (b) use the trapezoidal rule to integrate z(t) to obtain y(t)

(5) The eq. of motion can be written as a system 1st-order ODE's to be integrated using matlab ode45:

Solution
Consider the general equation of a particle with air resistance,


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$$ m.\frac{\mathrm{d} v_y}{\mathrm{d} t}= -k.v_y^{n}- mg $$ (7.4.1)
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where $$ z(t):= v_y(t)$$, there (6.10.1) can also be written as,


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$$ mz^{'}+kz^{n}= -mg $$     (7.4.2)
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Rearranging the terms,


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$$ m.\frac{\mathrm{d} z}{\mathrm{d} t}= -(k.z^n+mg) $$     (7.4.3)
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$$ \frac{m.dz}{k.z^n+mg}= -dt $$     (7.4.4)
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$$ \int \frac{m.dz}{k.z^n+mg}= - \int dt $$ (7.4.5)
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$$ \int \frac{dz}{\frac{k}{m}z^n+g}= - \int dt $$ (7.4.6)
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Part # 1 - The value of n=2, $$ a = k/m = 2 $$ and $$ g=10 $$

(i) To calculate

Solving this equation by hand,


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$$ \int \frac{dz}{2.z^2+10} = - \int dt $$ (7.4.7)
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$$ \frac{1}{10} \int \frac{dz}{\frac{z^2}{5}+1} $$     (7.4.8)
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Consider $$ u =\frac{z}{\sqrt{5}} $$ and hence $$ du= \frac{dz}{\sqrt{5}}$$


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$$ \frac{1}{10} \int \frac{\sqrt {5}du}{u^2+1} $$     (7.4.9)
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$$ \frac{1}{2 \sqrt {5}} \int \frac{1}{u^2+1} $$     (7.4.10)
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$$ \frac{1}{2 \sqrt {5}} tan^{-1}(u) + constant(k1) $$     (7.4.11)
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$$ \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}}) + k1 = -t + k2 $$ (7.4.12)
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$$ \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}})-\frac{1}{2 \sqrt {5}} tan^{-1}(\frac{50}{\sqrt{5}}) = -t $$     (7.4.13)
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$$ t= 0.341247 - \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}}) $$     (7.4.14)
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WolframAlpha Calculation -> http://www.wolframalpha.com/input/?i={1%2F%282*5^0.5%29}{tan+inverse+%2850%2F5^0.5%29}

or to interpret z in terms of t


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$$ z = \sqrt{5}tan(1.52610 - 2\sqrt{5}t) $$     (7.4.15)
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Matlab code to plot z(t) against time t:

Figure: Part 1 / velocity z(t) vs. time

Figure: Part 1 / Vertical displacement y(t) vs. time

Part # 2 - n=3, Use of mathlab command "roots"

Matlab code to plot z(t) against time t:

We try our best following the step to get the solution. However, we found out it is not possible to get the explicit root even though we expand the equation in 100 items. So, we stop to solve the next problem.

Figure: Part 3 / Vertical displacement y(t) vs. time

Part # 3 - n=3, Use the matlab function "hypergom" in interval [-10,50], Plot t vs z, Use trapezoidal rule to calculate y(t)

Matlab code to plot z(t) vs t, and y(t) vs t

Figure: Part 3 / velocity z(t) vs. time

Figure: Part 3 / Vertical displacement y(t) vs. time

'''Part # 4 - Use mathlab command "ode45" to integrate and obtain z(t). Use trapezoidal rule to obtain y(t)'''

Figure: Part 4 / velocity z(t) vs. time, n=2

Figure: Part 4 / Vertical displacement y(t) vs. time, n=2

Figure: Part 4 / velocity z(t) vs. time, n=3

Figure: Part 4 / Vertical displacement y(t) vs. time, n=3

Part # 5 - Integrate a system 1st-order ODE's using matlab ode45

Author and References

 * Solved and Typed by -- Jinchao Lu and Jason Alphonso
 * Reviewed by -- Jinchao Lu and Jason Alphonso

Problem R*7.5 – Heat conduction on a cylinder
sec40-5

Statement
Given,

$$ x= rcos\theta = \xi_1cos\xi_2 $$

$$ y= rsin\theta = \xi_1sin\xi_2$$

$$ z= \xi_3 $$

1) Express, the infinitesimal change in length, in Cartesian coordinates, in terms of Cylindrical coordinates

2) Find, $$ ds^2 = \sum_i (dx)^2 = \sum_k (h_k)^2(d\xi_k)^2) $$, the length in cylindrical coordinates

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

4) Use separation of variable to find the separated equations and compare it to the Bessel Equation.

Solution
1.) From the problem statement,


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$$ x= \xi_1 cos \xi_2 $$ (7.5.1)
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Differentiating the above,


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$$ dx= d\xi_1 cos \xi_2 - \xi_1 sin \xi_2 d\xi_2$$ (7.5.2)
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Similarly for $$ dy $$ & $$ dz $$ ,


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$$ dy= d\xi_1 sin \xi_2 + \xi_1 cos \xi_2 d\xi_2$$ (7.5.3)
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$$ dz = d\xi_3 $$ (7.5.4)
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2.) Squaring Eq. (7.5.2), (7.5.3) & (7.5.4) and adding ,


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$$ ds^2 = dx^2 + dy^2 + dz^2 $$ (7.5.5)
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$$ ds^2 = (d\xi_1 cos \xi_2 - \xi_1 sin \xi_2 d\xi_2)^2 + (d\xi_1 sin \xi_2 + \xi_1 cos \xi_2 d\xi_2)^2 + d\xi_3^2 $$ (7.5.6)
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$$ ds^2 = ((d\xi_1 cos \xi_2)^2 + (\xi_1 sin \xi_2 d\xi_2)^2 -2( d\xi_1 cos \xi_2\xi_1 sin \xi_2 d\xi_2)+(d\xi_1 sin \xi_2)^2 + (\xi_1 cos \xi_2 d\xi_2)^2 + 2(d\xi_1 sin \xi_2\xi_1 cos \xi_2 d\xi_2)+ d\xi_3^2 $$     (7.5.7)
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Cancelling terms we get,


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$$ ds^2 = (d\xi_1)^2 + (\xi_1d\xi_2)^2 + (d\xi_3)^2 $$ (7.5.8)
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Comparing Eq. (7.5.8) with,


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$$ ds^2 = (h_1)^2(d\xi_1)^2 + (h_2)^2(d\xi_2)^2 + (h_3)^2(d\xi_3)^2 $$ (7.5.9)
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$$ h_1 = 1 $$ (7.5.10)
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$$ h_2 = \xi_1 $$ (7.5.11)
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$$ h_3 = 1 $$ (7.5.12)
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3.) We know that the Laplace equation in general curvilinear coordinates is,


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$$ \Delta u = \frac{1}{h_1 h_2 h_3} \sum^3_{i=1} \frac{\partial}{\partial \xi_i} \left [\frac{h_1 h_2 h_3}{(h_i)^2}\frac{\partial u}{\partial \xi_i} \right] $$ (7.5.13)
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Using the Eq. (7.5.10), (7.5.11), (7.5.12),


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$$ \Delta u = \frac{1}{\xi_1}[\frac {\partial}{\partial \xi_1}(\xi_1 \frac{\partial u}{\partial \xi_1})+ \frac {\partial}{\partial \xi_2}(\frac{1}{\xi_1}\frac{\partial u}{\partial \xi_2})+ \frac {\partial}{\partial \xi_3}(\xi_1 \frac{\partial u}{\partial \xi_3})] $$ (7.5.13)
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which can be rewritten as,


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$$ \Delta u = \frac{1}{\xi_1}\frac {\partial}{\partial \xi_1}(\xi_1 \frac{\partial u}{\partial \xi_1})+ \frac{1}{\xi_1^2}\frac {\partial}{\partial \xi_2}(\frac{\partial u}{\partial \xi_2})+ \frac{\partial^2 u}{\partial \xi_3^2}$$ (7.5.14)
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which is the Laplace equation in cylindrical coordinates.

Putting $$ \xi_1 = r $$, $$ \xi_2 = \theta $$ , $$ \xi_3 = z $$


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$$ \Delta u = \frac{1}{r}\frac {\partial}{\partial r}(r \frac{\partial u}{\partial r})+ \frac{1}{r^2}(\frac{\partial^2 u}{\partial \theta^2})+ \frac{\partial^2 u}{\partial z^2}$$ (7.5.15)
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$$ \Delta u = \frac{1}{r}\frac {\partial}{\partial r}(r \frac{\partial u}{\partial r})+ \frac{1}{r^2}(\frac{\partial^2 u}{\partial \theta^2})+ \frac{\partial^2 u}{\partial z^2} =0$$ (7.5.16)
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We start by assuming,


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$$ u (R, \Theta,Z) = R(r) \, \Theta (\theta) \, Z(z)$$ (7.5.17)
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Putting this in Eq. (7.5.16),


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$$ \frac{\Theta Z}{r} \frac {\partial}{\partial r}(r \frac{\partial R}{\partial r})+ \frac{R Z}{r^2}(\frac{\partial^2 \Theta}{\partial \theta^2}) + R \Theta \frac{\partial^2 Z}{\partial z^2}=0$$ (7.5.18)
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Dividing by $$ R \Theta Z $$ and multiplying by $$ r^2 $$,


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$$ \frac{r}{R} \frac {\partial}{\partial r}(r \frac{\partial R}{\partial r})+ \frac{1}{\Theta}(\frac{\partial^2 \Theta}{\partial \theta^2}) + \frac{r^2}{Z} \frac{\partial^2 Z}{\partial z^2}=0 $$
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which can be rewritten as ,


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$$ \frac{r}{R} \frac {\partial}{\partial r}(r \frac{\partial R}{\partial r}) + \frac{r^2}{Z} \frac{\partial^2 Z}{\partial z^2}= - \frac{1}{\Theta}(\frac{\partial^2 \Theta}{\partial \theta^2}) $$ (7.5.19)
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Since L.H.S is a function of r and z, and RHS is a function of \Theta only, for them to be equal, both should be equal to a constant k.


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$$ \frac{\partial^2 \Theta}{\partial \theta^2} = - k \Theta $$ (7.5.20)
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$$ \frac{r}{R} \frac {\partial}{\partial r}(r \frac{\partial R}{\partial r}) + \frac{r^2}{Z} \frac{\partial^2 Z}{\partial z^2} = k $$ (7.5.21)
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Dividing Eq. (7.5.21) by $$ r^2 $$ and rearranging,


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$$ \frac{1}{rR} \frac {\partial}{\partial r}(r \frac{\partial R}{\partial r})- \frac{k}{r^2} = - \frac{1}{Z} \frac{\partial^2 Z}{\partial z^2} = c $$ (7.5.22)
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$$ - \frac{1}{Z} \frac{\partial^2 Z}{\partial z^2} = c $$ (7.5.23)
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$$ \frac{1}{rR} \frac {\partial}{\partial r}(r \frac{\partial R}{\partial r})- \frac{k}{r^2} = c $$ (7.5.24)
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Eq. (7.5.24) can be written as,


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$$ r \frac {\partial}{\partial r}(r \frac{\partial R}{\partial r})+ (cr^2 - k)R =0 $$ (7.5.25)
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Differentiating by parts,


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$$ r^2 \frac{\partial^2 R}{\partial r^2} + r \frac{\partial R}{\partial r}+ (cr^2 - k)R =0 $$ (7.5.26)
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which is same as the Bessel Function. The other separated equations are, Eq. (7.5.20) and Eq. (7.5.23)

Author and References

 * Solved and Typed by -- Pushkar Mishra
 * Reviewed by -- Seong Hyeon Hong

Problem R*7.6 – Determining Laplacian of u in Spherical Coordinates Using Math/Physics Convention
sec40-6

Statement
The Laplacian of u, as shown in Problem R*7.3, is written in spherical coordinates as


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$$ \Delta u = \frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{\partial u}{\partial r})+\frac{1}{r^2(cos\theta)^2}\frac{\partial ^2u}{\partial \phi ^2}+\frac{1}{r^2 cos\theta} \frac{\partial }{\partial \theta}(cos\theta \frac{\partial u}{\partial \theta}) $$     (7.6.1)
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In the above equation, $$ \theta $$ is defined from the $$ r\theta $$ -plane up to a point, P, in the $$ r, \theta, \phi $$ coordinate system. In other words, it is the latitudinal definition, and it ranges from $$ \theta \in [\frac{-\pi }{2}, \frac{\pi }{2}] $$. For this problem, determine the Laplacian of u in spherical coordinates according to the mathematical and physics convention for $$ \theta $$.

Solution
The mathematical and physics convention for $$ \theta $$ defines $$ \bar{\theta} $$ from the positive $$ \phi $$ axis to the point, P, in the $$ r, \theta, \phi $$ coordinate system. In other words, it is the angle measured from the "North Pole," and it ranges from $$ \bar{\theta} \in [0, \pi ] $$. Thus one can see that the relation for transforming $$ \theta $$ to $$ \bar{\theta } $$ is


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$$ \bar{\theta } := \frac{\pi }{2} - \theta $$     (7.6.2)
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Plugging Equation 7.6.2 into Equation 7.6.1 for $$ \theta $$ yields


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$$ \Delta u = \frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{\partial u}{\partial r})+\frac{1}{r^2(cos(\frac{\pi }{2} - \theta))^2}\frac{\partial ^2u}{\partial \phi ^2}+\frac{1}{r^2 cos(\frac{\pi }{2} - \theta)} \frac{\partial }{\partial (\frac{\pi }{2} - \theta)}(cos(\frac{\pi }{2} - \theta) \frac{\partial u}{\partial (\frac{\pi }{2} - \theta)}) $$     (7.6.3)
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Using the trigonometric identity


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$$ cos(\alpha + \beta ) = cos\alpha cos\beta - sin\alpha sin\beta $$     (7.6.4)
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one can see that


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$$ cos(\frac{\pi }{2} + (-\theta ) ) = \underbrace{cos(\frac{\pi }{2})}_{0} cos(-\theta )- \underbrace{sin(\frac{\pi }{2})}_{1} \underbrace{sin(-\theta )}_{-sin\theta} = sin \theta $$     (7.6.5)
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Using Equation 7.6.5, Equation 7.6.2 becomes
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$$ \Delta u = \frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{\partial u}{\partial r})+\frac{1}{r^2(sin\theta )^2}\frac{\partial ^2u}{\partial \phi ^2}+\frac{1}{r^2 sin\theta } \frac{\partial }{\partial \theta}(sin\theta \frac{\partial u}{\partial \theta}) $$     (7.6.6)
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Thus Equation 7.6.6 is the Laplacian of u in spherical coordinates in the mathematical and physics convention.

Author and References

 * Solved and typed by -- Kaitlin Harris and Rui Che
 * Reviewed by -- Rui Che

Team Contributions

 * Seong Hyeon Hong (SH) Solved all problems.
 * Jason Alphonso (JA) Solved all problems.
 * Pushkar Mishra (PM) Solved all problems.
 * Jinchao Lu (JL) Solved all problems.
 * Che Rui (CH) Solved all problems.
 * Kaitlin Harris (KH) Solved all problems.