jkwiens.com: Answers

Answer: Heat Distribution in a Rod

Wednesday, 26 of March , 2008 at 12:55 pm

Question: A 1D rod of length L has an initial heat distribution of

$u(x,0) = - cos(\frac{8 \pi x}{L})$

If the rod has insulated ends ( $\frac{\partial u}{\partial x}(0,t) = \frac{\partial u}{\partial x}(L,t) = 0$ ) and obeys the heat equation

$\frac{\partial u(x,t)}{\partial t} = k \frac{\partial^2 u(x,t)}{\partial x^2}$ ,

What is the heat distribution of the rod as a function of time?

Answer:
First assume that the solution to the PDE

$\frac{\partial u(x,t)}{\partial t} = k \frac{\partial^2 u(x,t)}{\partial x^2}$ ,

has the form

$u(x,t) = \phi(x) G (t)$

Therefore, we can reduce the equation to the following:

$\frac{\partial u(x,t)}{\partial t} = k \frac{\partial^2 u(x,t)}{\partial x^2}$
$\phi(x) \frac{\partial G(t)}{\partial t} = k G(t) \frac{\partial^2 \phi(x)}{\partial x^2}$
$\frac{1}{k G(t)} \frac{\partial G(t)}{\partial t} =\frac{1}{\phi(x)} \frac{\partial^2 \phi(x)}{\partial x^2}$

Since this equation is true for all x and t, therefore both sides of the equation must equal a constant.

$\frac{1}{k G(t)} \frac{\partial G(t)}{\partial t} =\frac{1}{\phi(x)} \frac{\partial^2 \phi(x)}{\partial x^2} = - \lambda$

which can be written as two ODEs

$\frac{d G(t)}{dt} = - \lambda k G(t)$

$\frac{d^2 \phi(x)}{dx^2} = - \lambda \phi(x)$

Solving the first-order differential $\frac{d G(t)}{dt} = - \lambda k G(t)$ is trivial and can easily be shown to have the solution:

$G(t) = C e^{-\lambda k t}$

Next, we need to solve $\frac{d^2 \phi(x)}{dx^2} = - \lambda \phi(x)$

The second-order differential has 3 different cases ( $\lambda = 0$ , $\lambda > 0$ , $\lambda < 0$ ).

Case $\lambda = 0$
The ODE for this case would be

$\frac{d^2 \phi(x)}{dx^2} =0$

which has the solution

$\phi(x) = A_0 x + A_1$

If we apply the BC $\frac{d \phi}{dx}(0)=0$ , we get

$\frac{d \phi}{dx}(0) = A_0 = 0$

which implies

$\phi(x) = A_1$

Case $\lambda > 0$
The ODE for this case would be

$\frac{d^2 \phi(x)}{dx^2} = - \lambda \phi(x)$

which has the solution

$\phi(x) = C_1 cos(\sqrt{\lambda}x) + C_2 sin(\sqrt{\lambda}x)$

Next, we apply the BC $\frac{d \phi}{dx}(0)=0$ .

$\frac{d \phi}{dx}(0) = \sqrt{\lambda}(-C_1 sin(\sqrt{\lambda}0) + C_2 cos(\sqrt{\lambda}0))$

$\frac{d \phi}{dx}(0) = \sqrt{\lambda}C_2$

Since $\lambda > 0$ , this implies . Therefore,

$\phi(x) = C_1 cos(\sqrt{\lambda}x)$

If we apply the 2nd BC $\frac{d \phi}{dx}(L)=0$ , we find the following

$\frac{d \phi}{dx}(L) = -C_1 \sqrt{\lambda} sin(\sqrt{\lambda}L) = 0$

Therefore, for non-trivial solutions $\sqrt{\lambda}L = n \pi$ where $n=1,2,3 \ldots$ .

This means,

$\phi(x) = C_1 cos(\frac{n \pi x}{L})$

is a solution.

Case $\lambda < 0$
The ODE for this case would be

$\frac{d^2 \phi(x)}{dx^2} = s \phi(x)$ where $s = -\lambda$ and

which has the solution

$\phi(x) = B_1 e^{\sqrt{s}x} + B_2 e^{-\sqrt{s}x}$

If we apply the BC $\frac{d \phi}{dx}(0)=0$ ,we find

$\frac{d\phi}{dx}(0) = B_1 \sqrt{s} e^{\sqrt{s}0} - B_2 \sqrt{s} e^{-\sqrt{s}0} = 0$

$\sqrt{s}(B_1 - B_2) = 0$

Since , we know that which means

$\phi(x) = B_1 (e^{\sqrt{s}x} + e^{-\sqrt{s}x})$

is a solution.

If we apply the 2nd BC $\frac{d \phi}{dx}(L)=0$ , we find

$\frac{d\phi}{dx}(L) = B_1 \sqrt{s} (e^{\sqrt{s}L} - e^{-\sqrt{s}L}) = 0$

Since and $e^{\sqrt{s}L} \neq e^{-\sqrt{s}L}$ , we know that .

This means that there is no solution where $\lambda < 0$ .

Since the PDE will satify any linear combination of the above solutions, we find that

$u(x,t) = A_0 + \sum_{n=1}^{\infty} A_n cos(\frac{n \pi x}{L}) e^{-(\frac{n\pi}{L})^2 kt}$

If we apply the IC $u(x,0) = - cos(\frac{8 \pi x}{L})$ to the solution, we find that and every other is zero.

Therefore, the solution would be

$u(x,t) =-3 cos(\frac{8 \pi x}{L}) e^{-(\frac{8\pi}{L})^2 kt}$

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Category: Answers, Differential Equations

Answer: Solve Laplace’s Equation

Monday, 17 of March , 2008 at 9:10 pm

Question: Find the solutions to Laplace’s Equation:

$\frac{\partial^2 u(x,y)}{\partial x^2} + \frac{\partial^2 u(x,y)}{\partial y^2} = 0$

Answer:
First assume that the solution to the PDE

$\frac{\partial^2 u(x,y)}{\partial x^2} + \frac{\partial^2 u(x,y)}{\partial y^2} = 0$

has the form

$u(x,y) = \phi(x) \xi (y)$

Therefore, we can reduce the equation to the following:

$\xi (y) \frac{\partial^2 \phi(x)}{\partial x^2} + \phi(x) \frac{\partial^2 \xi (y)}{\partial y^2} = 0$
$\frac{1}{\phi (x)} \frac{\partial^2 \phi(x)}{\partial x^2} = \frac{-1}{\xi (y)}\frac{\partial^2 \xi (y)}{\partial y^2}$

Since this equation is true for all x and y, therefore both sides of the equation must equal a constant.

$\frac{1}{\phi (x)} \frac{\partial^2 \phi(x)}{\partial x^2} = \frac{-1}{\xi (y)}\frac{\partial^2 \xi (y)}{\partial y^2} = \lambda$

This implies that we need to solve two ODEs.

$\frac{\partial^2 \phi(x)}{\partial x^2} =\lambda\phi (x)\frac{\partial^2 \xi (y)}{\partial y^2} = -\lambda\xi (y)$

Since the solutions to the two ODEs will be very similar, I will solve the ODE

$\frac{\partial^2 \Psi}{\partial r^2} =\kappa \Psi(r)$

and apply the results to the two ODEs.

There are 3 cases which we need to solve ( $\kappa > 0$ , $\kappa = 0$ , and $\kappa < 0$ ).

Case $\kappa = 0$
The ODE for this case would be

$\frac{\partial^2 \Psi}{\partial r^2} =0$

which has the solution

$\Psi = C_1 r + C_2$

Case $\kappa < 0$ and $\kappa > 0$
The ODE

$\frac{\partial^2 \Psi}{\partial r^2} =\kappa \Psi(r)$

which will have the solution

$\Psi = B_1 e^{\sqrt{\kappa}r} + B_2 e^{-\sqrt{\kappa}r}$

Please note that $\sqrt{\kappa}$ will be an imaginary number when $\kappa < 0$ .

Therefore, if we apply the above solution, we can find the functions that solve the ODEs

$\frac{\partial^2 \phi(x)}{\partial x^2} =\lambda\phi (x)\frac{\partial^2 \xi (y)}{\partial y^2} = -\lambda\xi (y)$

which would be

$\phi (x)= A_1 e^{i\sqrt{\lambda}x} + A_2 e^{-i\sqrt{\lambda}x}$

$\xi (y)= B_1 e^{\sqrt{\lambda}y} + B_2 e^{-\sqrt{\lambda}y}$ when $\lambda < 0$

$\phi (x)= A_1 e^{\sqrt{\lambda}x} + A_2 e^{-\sqrt{\lambda}x}$

$\xi (y)= B_1 e^{i\sqrt{\lambda}y} + B_2 e^{-i\sqrt{\lambda}y}$ when $\lambda > 0$

$\phi (x)= C_1 x + C_2$

$\xi (y)= D_1 y + D_2$ when $\lambda = 0$

Therefore, the solution would have the form

$u(x,y) = \begin{cases}(A_1 e^{i\sqrt{\lambda}x} + A_2 e^{-i\sqrt{\lambda}x})(B_1 e^{\sqrt{\lambda}y} + B_2 e^{-\sqrt{\lambda}y}) & for~\lambda < 0 \\ (A_1 e^{\sqrt{\lambda}x} + A_2 e^{-\sqrt{\lambda}x})(B_1 e^{i\sqrt{\lambda}y} + B_2 e^{-i\sqrt{\lambda}y}) & for~\lambda > 0 \\ (C_1 x + C_2)(D_1 y + D_2) & for~\lambda = 0\end{cases}$

Any superposition of the above equation will satisfy Laplace’s equation. In order to reduce this solution more, we would need to be given Boundary and Initial Conditions.

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Category: Answers, Differential Equations

Answer: Falling Infinite Rope

Friday, 7 of March , 2008 at 8:21 pm

Question: An infinite rope with a linear density of $\lambda$ is placed on a frictionless table. If the end of the rope is placed at the end of the table and starts falling, what is the velocity of the rope as a function of distance?

Answer: According to Newton’s 2nd Law of Motion, we know that

$F_{net} = \frac{\partial p}{\partial t}$

If we create a force diagram, we can easily see that $F_{net} = mg$ .

where $m = \lambda x$ .

Therefore, we can create the equation of motion as follows:

$\frac{\partial p}{\partial t} = \frac{\partial (m \dot{x})}{\partial t} = \dot{x}\dot{m} + m\ddot{x} = mg$
$\dot{x}\frac{\partial (\lambda x)}{\partial t} + \lambda x \ddot{x} = \lambda x g$
$\lambda\dot{x}^2 + \lambda x \ddot{x} = \lambda x g$
$\dot{x}^2 + x \ddot{x} = x g$

In order to solve this differential equation, let $y = \frac{1}{2} \dot{x}^2$ .

This means:

$\dot{y} = \dot{x} \ddot{x} ~\Longrightarrow ~ \ddot{x} = \frac{\dot{y}}{\dot{x}} = \frac{\frac{\partial y}{\partial t}}{\frac{\partial x}{\partial t}} = \frac{\partial y}{\partial x}$

Therefore, the differential equation becomes

$2y + x \frac{dy}{dx} = gx ~\Longrightarrow ~ \frac{dy}{dx} + \frac{2}{x}y = g$

We can solve this using the integrating factor method. According to our differential equation, our integrating factor will be $e^{\int \frac{2 dx}{x}}= x^2$ . If we multiple our integrating factor to our ODE, we get

$x^2 \frac{dy}{dx} + 2xy = g x^2$
$\frac{d(x^2 y)}{dx} = gx^2$
$x^2 y = \int g x^2 dx = \frac{gx^2}{3} + C$
$\therefore y = \frac{g}{3}x + \frac{C}{x^2}$

However, since $y = \frac{1}{2} \dot{x}^2$ , we know that

$\frac{1}{2}\dot{x}^2 = \frac{g}{3}x + \frac{C}{x^2}$
$\dot{x} = \sqrt{\frac{2}{3}gx + \frac{2C}{x^2}}$

If we add the boundary condition, $\dot{x}(0) =0$ , the equation reduces to

$\dot{x} = \sqrt{\frac{2}{3}gx}$

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Category: Answers, Classical Mechanics

Answer: General Form of Sequence

Saturday, 1 of March , 2008 at 9:59 pm

Question: The solution to $x = \sqrt{10y+1}$ where $x,y \in \mathbb N$ yields the following sequence:

$8,~12,~36,~44,~84,~96,~152,~168,~240~\ldots$

Find the general form for this sequence.

Answer: When analyzing this sequence, the first thing I tried was comparing the difference between each number in the sequence.

Lets investigate this new sequence.

$4,~24,~8,~40,~12,~56,~16,~72~\ldots$

It appears the odd terms are related to each other and so are the even terms. Lets break the odd terms and even terms into two different sequences.

$odd~terms:~4,~8,~12,~16,\ldots$

$even~terms:~24,~40,~56,~72,\ldots$

As you can see, the odd terms has the form:

$\phi_n = 4n$

The even terms, likewise, can be seen to have the form:

$\phi_n = \phi_{n-1} + 16$ or

$\phi_n = 16n + 8$

If we merge these results, we can get pattern for the following sequence

$4,~24,~8,~40,~12,~56,~16,~72~\ldots$

which is

$\phi_n = \begin{cases}4(\frac{n+1}{2}) & odd~n \\ 16(\frac{n}{2}) +8 & even~n\end{cases}$

$\phi_n = \begin{cases}2(n+1) & odd~n \\ 8(n+1) & even~n\end{cases}$

We can then use this result to find the solution to

$8,~12,~36,~44,~84,~96,~152,~168,~240~\ldots$

which would be

$\psi_n = \psi_{n-1} + \phi_n$

$\psi_n= \psi_{n-1} + \begin{cases}2(n+1) & odd~n \\ 8(n+1) & even~n\end{cases}$

where $\psi_0 = 8$ .

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Category: Algebra, Answers

Answer: Infinite Potential Well

Saturday, 23 of February , 2008 at 10:43 am

Question: A quantum particle is inside a one-dimensional infinitely deep potential well.

Find the wavefunction that satisfies Schrödinger’s equation.

$i \hbar \frac{\partial}{\partial t} \Psi (\vec{r}, t)= \frac{-\hbar^2}{2m} \nabla^2 \Psi (\vec{r}, t) + V(\vec{r}) \Psi (\vec{r}, t)$

Answer:

First assume that $\Psi (\vec{r}, t)$ can be represented as $\Psi (\vec{r}, t) = \phi (\vec{r}) \Phi (t)$ . This means that Schrödinger’s equation can be manipulated as followed:

$i \hbar \frac{\partial}{\partial t} \Psi (\vec{r}, t)= \frac{-\hbar^2}{2m} \nabla^2 \Psi (\vec{r}, t) + V(\vec{r}) \Psi (\vec{r}, t)$
$i \hbar \phi (\vec{r}) \frac{\partial}{\partial t} \Phi (t)= \frac{-\hbar^2}{2m}\Phi (t) \nabla^2 \phi (\vec{r}) + V(\vec{r}) \phi (\vec{r}) \Phi (t)$
$\frac{i \hbar}{\Phi (t)} \frac{\partial}{\partial t}\Phi (t)= \frac{-\hbar^2}{2m \phi (\vec{r})} \nabla^2 \phi (\vec{r}) + V(\vec{r})$

Since the this equation holds for all $\vec{r}$ and , both sides of the equation must equal a constant.

$\frac{i \hbar}{\Phi (t)} \frac{\partial}{\partial t}\Phi (t)= \frac{-\hbar^2}{2m \phi (\vec{r})} \nabla^2 \phi (\vec{r}) + V(\vec{r}) = E$

Time Component
Next, we will solve for the time component of the above equation.

$\frac{i \hbar}{\Phi (t)} \frac{\partial}{\partial t}\Phi (t) = E$
$i \hbar \int \frac{d\Phi}{\Phi (t)} = E \int dt$
$i \hbar ln(\Phi (t)) = E t$
$\Phi (t) = e^{\frac{-iE}{\hbar}t}$

Time-Independent Component
The time-independent component of the Schrödinger equation would be:

$\frac{-\hbar^2}{2m} \nabla^2 \phi (\vec{r}) + \phi (\vec{r})V(\vec{r}) = E\phi (\vec{r})$

The solution to this problem can be easily solved for the case $x \le 0$ and $x \ge L$ . Since $V(x) = \infty$ , the only way to solve

$\frac{-\hbar^2}{2m} \frac{d^2}{dx^2} \phi (x) + \phi (x)V(x) = E\phi (x)$

is to have $\phi (x) = 0$ .

For the case when , we know that . Therefore, the time-independent equation will become:

$\frac{-\hbar^2}{2m} \frac{d^2}{dx^2} \phi (x) = E\phi (x)$

This equation should look rather familiar because it is the differential equation describing a simple harmonic oscillator. Therefore, the general solution to this differential equation would be:

$\phi (x) = A sin(\frac{\sqrt{2mE}}{\hbar}x) + B cos(\frac{\sqrt{2mE}}{\hbar}x)$

Therefore, the solution would be:

$\phi (x)=\begin{cases}0 & x \le 0~and~x \ge L \\A sin(\frac{\sqrt{2mE}}{\hbar}x) + B cos(\frac{\sqrt{2mE}}{\hbar}x)& 0<x<L\end{cases}$

If we add a requirement that $\phi (x)$ needs to be piecewise-smooth, we can add boundary conditions $\phi (0) = 0$ and $\phi (L) = 0$ to the problem:

$\phi (x) = A sin(\frac{\sqrt{2mE}}{\hbar}x) + B cos(\frac{\sqrt{2mE}}{\hbar}x)$

Therefore, when $\phi (0)$ , we can see that because

$\phi (x) = A sin(0) + B cos(0) = B = 0$

Now that we know , we can apply the second boundary condition $\phi (L)=0$ to the remaining portion of the equation.

$\phi (L) = A sin(\frac{\sqrt{2mE}}{\hbar}L) = 0$
$\therefore ~ \frac{\sqrt{2mE}}{\hbar}L = n \pi$ where $n \in \mathbb N$

This implies

$E = \frac{n^2 \pi^2 \hbar^2}{2mL^2}$

Therefore, we can see that the general solution can be reduced to:

$\phi (x)=\begin{cases}0 & x \le 0~and~x \ge L \\A sin(\frac{n\pi}{L}x) & 0<x<L\end{cases}$

The last step is to normalize this wave-function

$1 = \int_{-\infty}^{\infty} |\phi (x)|^2 dx$
$= |A|^2 \int_{0}^{L} sin^2(\frac{n\pi}{L}x) dx$
$= |A|^2 \frac{L}{n\pi}(-\frac{1}{2}cos(\frac{n\pi}{L}x)sin(\frac{n\pi}{L}x) + \frac{1}{2}\frac{n\pi}{L}x)\arrowvert_{0}^{L}$
$= |A|^2 \frac{L}{n\pi}(\frac{1}{2}n\pi) = |A|^2 \frac{L}{2}$

Therefore

$A = \sqrt{\frac{2}{L}}$

Therefore, the final solution would be

$\Psi (x, t)=\begin{cases}0 & x \le 0~and~x \ge L \\ \sqrt{\frac{2}{L}} sin(\frac{n\pi}{L}x)e^{\frac{-iE}{\hbar}t} & 0<x<L\end{cases}$
where $E = \frac{n^2 \pi^2 \hbar^2}{2mL^2}$

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Category: Answers, Quantum Mechanics