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Messages - Tristan Fraser

Pages: [1] 2
1
APM346--Misc / Cap on bonus from karma?
« on: April 13, 2018, 10:39:08 AM »
Dr Ivrii,

I was wondering if there is any cap to the number of bonus points that can be gained from karma on the forum?

As I have reached 10 karma, and while I really wanted to try some of the bonus problems and contribute the discussion, I don't want to rob others of the opportunity of gaining bonus points if I've already reached my cap, and they haven't.

2
Final Exam / Re: FE-P1
« on: April 11, 2018, 08:13:38 PM »
@Jingxuan: Thank you!

The adjusted solution gives us;

$(e^{-x} \psi(x)) = 4e^{x} + C $

and then:

$\psi(x) =  4e^{2x} + Ce^{x}$

Now, this gives us:

$u(x,t) =  4e^{2(x-4t)} + Ce^{x-4t} + e^{-2(x+4t)}$

Then checking continuity gives us (x=4t):

$u(x,t) =  4+ C +e^{-2(8t)} = 3+ e^{-16t}$ implying that $C= -1$.

$$u(x,t) =  4e^{2(x-4t)} - e^{x-4t} + e^{-2(x+4t)}\qquad 0<x<4t$$

3
Final Exam / Re: FE-P7
« on: April 11, 2018, 06:34:09 PM »
Taking partial fourier transform gives us:   
 $\hat{u}_{xx}-k^2\hat{u}=0$ giving us an ODE, which then can be solved: $\hat{u}=A(k)e^{-|k|x}+B(k)e^{|k|y}$
   
The second term is discarded since we want a finite solution, otherwise it would violate boundary condition 3.

So now we take $\hat{u}_{x} = -|k|A(k)e^{-|k|x} = 0 =  \hat{h}$

What is $\hat{h}$

We know that $h(y) = -g'(y)$ and $g(y) = \frac{2}{y^2 + 1}$ with fourier transform of $e^{-|k|}$. We then use a property of the fourier transform, where $\hat{g} =  ik\hat{f}$ implies = $g = f'(x)$
Thus:

$\hat{h} =  -ik\hat{g}$

Thus
$ -|k|A(k) = -ik e^{-|k|}$, rearranging gives us:

$A(k) = \frac{ike^{-|k|}}{|k|} $

And $u(x,k) = \frac{ike^{-|k|}}{|k|}e^{-|k|x}  $

We can put this in a fourier integral, and integrate to get u(x,y):

$u(x,y) = \int_{-\infty}^{\infty} \frac{ike^{-|k|}}{|k|}e^{-|k|x} e^{iky} dk$
Split into two integrals, to simplify the process:

$u(x,y) = \int_{0}^{\infty} i e^{-k(1+x - iy)} + \int_{-\infty}^{0} -ie^{k(1+x+iy)} = \frac{(i)((1+x - iy) - (1+x+iy))}{(1+x)^2 + y^2} = \frac{2y}{(1+x)^2 + y^2}$

4
Final Exam / Re: FE-P1
« on: April 11, 2018, 05:55:16 PM »
General solution of

$u =  \phi (x+4t) + \psi (x-4t)$
$u_x =  \phi'(x+4t) + \psi'(x-4t)$
$u_t = 4(\phi' (x+4t) - \psi' (x-4t) $
For $x>4t$:

$4e^{-2x} =  \phi(x) + \psi(x)$
$4e^{-2x} = \phi'(x) - \psi'(x)$
integrate and rearrange to get:

$\phi(x) =  e^{-2x}$ and $\psi(x) = 3e^{-2x}$

$$u = e^{-2(x+4t)} + 3e^{-2(x-4t)} \qquad x>4t$$

For the case of $0<x<4t$:
$u_x =  \phi'(x+4t) + \psi'(x-4t)$ and $u$ are used in BC:

$\phi'(4t) + \psi'(-4t) - \phi(4t) -\psi(-4t) = e^{-8t}$

Then let $x = -4t$ and move the $\psi(x)$ to one side:

$\psi'(x) - \psi(x) = e^{2x} - \phi'(-x) + \phi(-x) $
We know two things: first the reason we're looking at $\psi$ instead of $\phi$ is that it has a chance of being negative, while the other does not, and thus, the second thing is that we can also plug in our previously used functions for $\phi$. The other nice thing is that you can rearrange to ODE:

$(e^{-x} \psi(x))' = e^{x} + (-2e^{x} + e^{x})$ integrate both sides:

$e^{-x} \psi(x) = C$
Thus:
$\psi(x) =  Ce^{x}$

$u =  e^{-2(x+4t)} + Ce^{x-4t}$

Then check continuity, plug in for $x = 4t$ for both u(x,t), and make sure they match up:

$u = e^{-16t} + C =  e^{-16t} + 3$

Therefore $C= 3$, ensuring continuity.

EDIT: There is a chance I might have misplaced a minus sign, please let me know if I have made such an error, as I will aim to correct it promptly.

5
Final Exam / Re: FE-P3
« on: April 11, 2018, 05:35:17 PM »
We let $u = X(x)T(t)$
then plug in:

$T''X - X''T + 4XT = 0 $

The boundary conditions imply:

$X(0) =  X(\pi) = 0 $

and

$u(x,0) = f(x)$
$u_t (x,0) = x^2 - \pi x$

Using same strategy as we always do with a separation of variables, divide by u, and move the $-4$ over to the other side.


$\frac{T''}{T}  - \frac{X''}{X} = -4$

Since all three terms are constants, set $\frac{X''}{X} =  -\lambda$ and $\frac{T''}{T} = -\lambda - 4$

Solving these ODEs:

$X(x) =  Acos\sqrt{\lambda x} + B\sin \sqrt{\lambda x}$
and plugging in the conditions for X will imply that $A = 0$
$X(\pi) = B\sin \sqrt{\lambda \pi} = 0$

and in order to have an eigenfunction with a nontrivial solution, $\sqrt{\lambda} \pi = \pi n$ where n is an integer. Therefore eigenfunction of $n^2$.

Finally, for T: (I dropped a minus sign on the exam, see my comment at the bottom)

$T = C\sin\sqrt{\lambda + 4} t + D\cos \sqrt{\lambda + 4}t $ becomes

$T =  C\sin\sqrt{n^2 + 4} t + D\cos \sqrt{n^2 + 4}t$

Then our general solution ought to have the form

$u(x,t) =  \sum_{n=1}^{\infty}(C_n\sin\sqrt{n^2 + 4} t + D_n\cos \sqrt{n^2 +4}t)\sin(nx)$

Apply the BC of $u(x,0) = 0$

$0 =  \sum_{n=1}^{\infty}(C_n\sin\sqrt{n^2 + 4} (0) + D_n\cos \sqrt{n^2 + 4}(0)\sin(nx)$

Second term's coefficient has to be 0 satisfy this BC:

$u_t(x,t) = \sum_{n=1}^{\infty}(\sqrt{n^2+4}(-C_n\cos\sqrt{n^2 + 4}(t) )\sin(nx)$

Then for the last condition:

$x^2 - \pi x = \sum_{n=1}^{\infty}(\sqrt{n^2+4}(-C_n )\sin(nx)$

Now to find the coefficient:

$C_n = \frac{-2}{\pi \sqrt{n^2+4}}\int_{0}^{\pi}\sin(nx)(x^2 - \pi x) dx$

Split this into two integrals,

$ \frac{-2}{\pi \sqrt{n^2+4}}(\int_{0}^{\pi}x^2\sin(nx - \int_{0}^{\pi} \sin(nx)\pi x)$

Integrate by parts to get

$ = (\frac{-2}{\pi \sqrt{n^2+4}})(\frac{-\pi^2 \cos \pi}{n} - \frac{2}{n^3}(1-\cos(n\pi)) + \frac{\pi^2 \cos \pi n}{n} =   \frac{4}{(\pi \sqrt{n^2+4}) n^3}(1-\cos(n\pi))$

Final series:

$u = \sum_{n=1}^{\infty} \frac{8}{(\pi \sqrt{n^2+4}) n^3})\sin(nx)(\cos(\sqrt{n^2+4})t)$

for odd n.

EDIT: After careful revision, I realized I must have dropped a minus sign on the final. The correct solution should be involving (n^2+4) not (n^2-4), and the above steps have been corrected to reflect that:

@Jingxuan and @George, thank you for pointing out my mistake!




6
APM346––Home Assignments / Re: Question 9 from 4.2P
« on: April 10, 2018, 10:47:02 PM »
So we basically take:

$ X'(l) = -A \sqrt{ \lambda}\sin \sqrt{ \lambda} l $
$ T'(t) = c \sqrt{ \lambda} [D \cos(c \sqrt{ \lambda}t) -C \sin(c \sqrt{ \lambda}t) $

Applying the tricky boundary condition:

$ A \sqrt{ \lambda} \sin \sqrt{\lambda}l  = i \alpha (c \sqrt{ \lambda} [D\cos(c \sqrt{ \lambda}t) -C\sin(c \sqrt{ \lambda}t)) $

Now here's where I might have done something very foolish;

$ A \sqrt{\lambda}\sin\sqrt{\lambda}l  = i\alpha (T'(t)) $ and integrated both sides:

$ A \sqrt{\lambda}\sin(\sqrt{\lambda}l) t  + F=  i\alpha T(t) $

Then I plugged it into *

$ \frac{T''}{c^2{\frac{A\sqrt{\lambda}\sin(\sqrt{\lambda}l) t + F}{i\alpha}}} = \frac{X''}{X} = - \lambda $

I'm afraid I'm a little lost. Perhaps another hint might help?



EDIT: I'm not sure why the LaTeX is producing this output, it looked fine in the editor I was working in (sharelatex)...

For clarity, refer to the screenshot:


7
APM346––Home Assignments / Question 9 from 4.2P
« on: April 08, 2018, 06:39:24 PM »
The problem states:

Quote
Consider wave equation with the Neumann boundary condition on the left and weird b.c. on the right:
$u_{tt}−c^2 u_{xx}=0 , \ \ u_x(0,t)=0, \ (u_x+i\alpha u_t)(l,t)=0, \ \ \ 0<x<l$
with $\alpha$∈ℝ.

Separate variables;
Find weird eigenvalue problem for ODE;
Solve this problem;
Find simple solution u(x,t)=X(x)T(t).
Hint. You may assume that all eigenvalues are real (which is the case).

I used the usual separation of variables: $u = X(x)T(t)$ and made

$\frac{T''}{c^2 T} = \frac{X''}{X} = - \lambda $, *

 to then get
$$X(x) = A\cos(\sqrt{\lambda}x) + B\sin(\sqrt{\lambda}x).$$
Resolving the "easy" boundary condition gives us

$X(x)  = A\cos(\sqrt{\lambda}x) $ OK

While for T:

T  = $C\cos(\frac{\sqrt{\lambda}}{c}t) + D\sin(\frac{\sqrt{\lambda}}{c}t)$

And then taking the "weird" boundary condition gives me something like: $\sqrt{\lambda}\tan(\sqrt{\lambda}l) = i\alpha\frac{T'}{T}$

But I wasn't quite sure where to go from there in terms of solving the problem. One approach that I think brings me closer (hopefully) to the solution is:

Note that $(\frac{1}{c^2}\frac{T'}{T})'$ = $\frac{1}{c^2} (\frac{T''}{T}  -  (\frac{T'}{T})^2)$

Then knowing that the first term on the RHS is constant ($-\lambda$), we can rewrite the expression wrt to $\frac{T'}{T}$ as follows:

$\frac{T'}{T} = c\sqrt{\frac{T''}{c^2 T} - \frac{1}{c^2}(\frac{T'}{T})'} = c \sqrt{-\lambda - (\frac{T'}{c^2T})'}$

which can be incorporated into the weird BC above :

$\frac{X'}{X}(l) = \alpha c \sqrt{\lambda + (\frac{T'}{c^2T})'}$

and then:

$\sqrt{\lambda} \tan(\sqrt{\lambda l}) = \alpha c \sqrt{\lambda + (\frac{T'}{c^2T})'} $

That being said, I'm worried that I've gone off an a tangent and have failed to simplify the problem. Any hints would be appreciated.

8
APM346--Lectures / Appendix on past finals
« on: April 07, 2018, 05:46:20 PM »
I've been looking through the past finals, and I was wondering if we would get an appendix of useful formulae, much like the one from 2015F, with useful Fourier properties and integrals for their coefficients, or no?

9
APM346--Misc / Tutorial this week?
« on: April 04, 2018, 01:52:37 PM »
Is there a tutorial this week?

10
Quiz-B / Re: Quiz-B P2
« on: April 04, 2018, 12:08:01 PM »
Here's my initial attempt for the induction:

We start with n =  1 case:

$< \partial f, \phi (x)> = <-f, \partial\phi (x)> $ for the general function $\phi(x)$. We can integrate by parts to get from LHS to get to RHS. This also holds for $\phi(x) = \delta(x)$. Then, for the n = 2 case, we can redefine a new function:

$<\partial^2 f, \phi(x)> = <-F, \partial^2\phi(x)>$ for the general function $\phi(x)$. We could also just redefine the functions in such a manner that we get the expression:

$<\partial F, \phi(x)> = <-f,\partial \phi(x)>$, it also holds for $\delta(x)$

Then generalizing to order  n = k, we can show it holds. Where F is the primitive for f. We can always reduce it to just a first order derivation instead of kth order.

11
Web Bonus Problems / Re: Web bonus problem--Week 10-11
« on: March 25, 2018, 09:12:44 PM »
Given a Lagrangian of $$ L(q,\dot{q})  = \frac{m\dot{\vec{q}}^2}{2} + eA( \vec {q})\dot{\vec{q}} - V(q) $$ and a Hamiltonian of $$ H (q,p) =  \frac{m}{2} \sum_{i =1}^{n} (\frac{p_i - eA(q)_i)}{m})^2  + V(q) $$

To find $ \ddot{q}$:

Use the Euler Lagrange equations: $ \frac{\partial L}{\partial \vec{q}} -  \frac{d}{dt} \frac{\partial L}{\partial \dot{\vec{q}}} = 0 $, we get:

$m\dot{q_j} + e\frac{\partial A}{\partial q_j} - m\ddot{q_j} - e (\frac{\partial A_j}{\partial q_i} \dot{q_j} - e \frac{\partial A_j}{\partial t})  = 0$ then $$ m\ddot{q_j} = m\dot{q_j} + e(\frac{\partial A_j}{\partial q_j}  - \frac{\partial A_j}{\partial q_i} \dot{q_j} - \frac{\partial A_j}{\partial t} ) $$

$$  \ddot{q_j}  =  \frac{1}{m} (m\dot{q_j} + e(\frac{\partial A_j}{\partial q_j} - \frac{\partial A_j}{\partial q_i} \dot{q_j} - \frac{\partial A_j}{\partial t} )) $$

Alternatively, one can use Hamilton's equations of motion to find $\dot{q}, \ddot{q}$ ,

$$ \frac{\partial H}{\partial p}  = \dot{q}  =  - \frac{2p}{2m} + \frac{eA}{2m} $$, where taking the time derivative gives us:

$$ \ddot{q}  = - \frac{\dot{p}}{m} + \frac{e}{2m} \frac{\partial A }{\partial q} \dot{q} $$

12
Term Test 2 / Re: TT2--P2N
« on: March 23, 2018, 08:38:42 PM »
As usual, let's setup our partial fourier transform, shifting from x basis to k basis, i.e.

$$ \hat{f}(k) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-ikx} f(x) dx  $$

Thus equations (1,2,3) become

$$ \hat{u_{yy}} - k^2 \hat{u} = 0 , \hat{u(k,0)} = 0 , (\hat{u_y}(k,1) + \alpha \hat{u}(k,1)) = \hat{g}(k)$$

Now, the fourier transform g(x) into $\hat{g}(k) $ is simply:

$\hat{g}(k) = \frac{1}{2\pi} \int_{-\infty}^{infty} e^{-ikx} g(x) dx  = \frac{1}{2\pi} \int_{-1}^{1} e^{-ikx}  dx = \frac{1}{2\pi} \frac{-e^{-ikx}}{ik} |_{-1}^{1} = \frac{1}{2\pi} \frac{e^{ik} - e^{-ik}}{ik} =  \frac{1}{\pi k} sin(k)$

The solution to the differential equation from the fourier transform of (1) is:

$\hat{u} =  A(k)e^{|k|y} +B(k)e^{-|k|y} $ ,  applying (2), we need: $-B(k) = A(k).$

$\hat{u} =  A(k)e^{|k|y} -A(k)e^{-|k|y}  = A(k)\sinh(|k|y)$

Finally, applying the condition (3):

$(A(k)|k| \cosh(|k|) + \alpha A(k)\\sinh(|k|) = \frac{sin(k)}{\pi k}$

So that: $A(k) = \frac{sin(k)}{\pi k} (|k|\cosh|k| + \alpha \sinh|k|)^{-1} $

Thus, our solution in x basis is given as:

$f(x) = \int_{-\infty}^{\infty} \hat{f(k)} e^{ikx} dk  = \int_{-\infty}^{\infty} (\frac{sin(k)}{\pi k} (|k|\cosh|k| + \alpha \sinh|k|)^{-1})\sinh(|k|y)e^{ikx}dk $


13
Term Test 2 / Re: TT2--P3N
« on: March 23, 2018, 08:09:03 PM »
We start by taking the $ u = X(x)Y(y) $, then plugging in gives us:

$$ X'' Y + Y'' X  = -\lambda XY $$

$$ X(0)Y(y) = X'(a)Y(y) = 0  \ \ and \ \ X(x)Y(0) = X(x)Y(b)' = 0 $$

Dividing both of these expressions by $XY$ gives us

$$\frac{X''}{X}  + \frac{Y''}{Y} = -\lambda $$ 

$$ X(0) = X'(a) = 0  \ \ and \ \ Y(0) = Y(b)' = 0 $$

Now we know that both $\frac{Y''}{Y} $ and $\frac{X''}{X} $ are independent of each other, i.e. they should be equivalent to some constant. Introduce constants $\lambda_1, \lambda_2$ such that $ \lambda = \lambda_1 + \lambda_2$, thus  $\frac{Y''}{Y} = -\lambda_{1} $ and $\frac{X''}{X} = -\lambda_{2} $

Then, we can examine the different cases of $\lambda_{1,2}$.

i) If both $\lambda_{1} = 0 = \lambda_{2} $:

We get a simplified eigenvalue problem of:
 $$ X'' = 0 , Y'' = 0 $$

Meaning that:

$$ X = A_0 x + B_0 , Y = C_0y + D_0  $$

Running it through the boundary conditions, we can easily show that: $ B_0 = 0 , D_0 = 0 , A_0 = 0 , C_0 = 0 $
I.e. this leads to a trivial solution of the eigenvalue problem.

For $\lambda_{1}, \lambda_{2} >0 $

We will get eigenvalue problem of $$X'' + \lambda_2 X = 0 , Y'' + \lambda_1  Y = 0$$

This results in:

$$X(x) = Acos\sqrt{\lambda_2}x + Bsin\sqrt{\lambda_2}x$$
$$Y(y) = Ccos\sqrt{\lambda_1}y + Dsin\sqrt{\lambda_1}y$$

Apply the boundary conditions, and we get:
$ A = 0 , C= 0, 0 =  \sqrt{\lambda_2}Bcos\sqrt{\lambda_2}a, 0 = \sqrt{\lambda_1}Dcos\sqrt{\lambda_1}b$

We're in search of nontrivial solutions, which can be attained if $\sqrt{\lambda_{1,2}}b,a =  \frac{\pi(2n+1)}{2} $, thus we have eigenvalues and eigenfunctions of:

$$ \lambda_1 =  (\frac{\pi(2m+1)}{2b})^2 , Y_{m} = sin(\frac{\pi(2m+1)}{2b})y , \lambda_2 = (\frac{\pi(2n+1)}{2a})^2, X_{n} =  sin(\frac{\pi(2n+1)}{2a}x) $$

For the case of $\lambda_1 , \lambda_2 < 0 $ we solve the eigenvalue problem of:

$$X''  - \lambda_2 X = 0 , Y'' - \lambda_1 Y = 0 $$, which gives us, in turn:

$$X(x) =  Ae^{\sqrt{\lambda_2} x} + Be^{-\sqrt{\lambda_2} x} , Y(y) = Ce^{\sqrt{\lambda_1} y} + De^{-\sqrt{\lambda_1} y} $$

Apply the boundary conditions to get: $ A+ B = 0, C+D = 0$ , $ 0 =  \sqrt{\lambda_2} A (e^{\sqrt{\lambda_2}a} +e^{-\sqrt{\lambda_2}a}) = 2A\sqrt{\lambda_2}\cosh(\sqrt{\lambda_2}a) $

and $ 0 = \sqrt{\lambda_1}C(e^{\sqrt{\lambda_1}b} +e^{-\sqrt{\lambda_1}b})  = 2C\sqrt{\lambda_1}\cosh(\sqrt{\lambda_1}b) $

But since the $\cosh$ function never reaches 0, we can't have a nontrivial solution. Therefore there only exists a trivial solution in this case.

Note: updated solution to reflect feedback





14
Term Test 2 / Re: TT2--P5
« on: March 23, 2018, 07:40:11 PM »
We first note that $ \cos^2 x  = (\frac{(e^{ix} + e^{-ix})}{2})^2  = \frac{2 + e^{2ix} + e^{-2ix}}{4} $

From there, we take the fourier transform:  $$\hat{f(}k) = \frac{1}{2\pi} \int_{-\infty }^{\infty } e^{-ikx} e^{-|x|} ( \frac{2 + e^{2ix} + e^{-2ix}}{4}) = \frac{1}{2\pi }\left ( \int_{0}^{\infty } \frac{2e^{-x(1+ki)} + e^{-x(1+ki - 2i)} + e^{-x(1+ki + 2i))}}{4} dx +\int_{-\infty }^{0} \frac{2e^{x(1- ki)} + e^{x(1- ki - 2i) }+ e^{x(1- ki + 2i))}}{4} dx\right )$$

We integrate that thing, and note that by evaluating at the bounds of $0,\pm \infty $, only the terms evaluated at 0 can be kept, netting us:

$$\hat{f(}k) =  \frac{1}{2\pi}( \frac{1}{2(1+ki)} + \frac{1}{4(1+ki - 2i)} + \frac{1}{4(1+ki+2i)} + \frac{1}{2(1-ki)} + \frac{1}{4(1-ki-2i)} + \frac{1}{4(1-ki+2i)}) $$

While this expression may be ugly, what is useful about it is the fact that these are fractions of complex numbers and their conjugates (noteably first and fourth terms, 2nd and 6th, 3rd and 5th) so we'd get the following fraction

$$\hat{f(}k) =  \frac{1}{2\pi}( \frac{2}{4(1+ k^2)} + \frac{2}{16(1+ (k-2)^2)} + \frac{2}{16(1+ (k+2)^2)} ) $$

Thus, we can plug this function into the IFT:

$$ f(x) =  \int_{-\infty}^{\infty} \hat{f}(k) e^{ikx} dk  = \int_{-\infty}^{\infty} \frac{1}{4\pi}( \frac{1}{(1+ k^2)} + \frac{1}{4(1+ (k-2)^2)} + \frac{1}{4(1+ (k+2)^2)}) e^{ikx} dk  $$

Leaving the function as a Fourier integral.

15
APM346--Lectures / Question 1 from TT2, 2015S
« on: March 21, 2018, 08:17:23 PM »
Quote
Consider the eigenvalue problem
$$x^2 X″+2xX′+\lambda X=0,\ \  x \ \ \epsilon (\frac{2}{3},\frac{3}{2}), \ \  X′(\frac{2}{3})=0; \ \ \ X′(32)=0 \ \ \ \ \ (0)$$

Assume $ \lambda \geq 0 $. Find all the eigenvalues and the corresponding eigenfunctions.
 
Hint: as (1) is Euler equation, look   for elementary  solutions  in the form $x^m$).

I wrote the same trick and got the same characteristic of

$$ k(k-1) + 2k + \lambda = 0 \ \ \ \ \ (1) $$

What I do not understand is why, and how we were able to make the substitution of $ t = ln (\frac{3x}{2}) $ to arrive at   $$ \ddot{X} + \dot{X} + \lambda X =  0 \ \ \ \ \  (2) $$

Since my solution instead relied on examining the cases of $\lambda \geq 0 $, but even after plugging in $ x = \frac{3}{2} e^{t} $ I do not see how we would get to the above eigenvalue problem (2).

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