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### Topics - Tristan Fraser

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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
##### 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.

3
##### 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?

4
##### APM346--Misc / Tutorial this week?
« on: April 04, 2018, 01:52:37 PM »
Is there a tutorial this week?

5
##### 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).

6
##### Quiz-5 / Quiz 5, T5102
« on: March 07, 2018, 06:28:31 PM »
Let $\a > 0$ , find the fourier transforms of $$(x^2 + a^2)^{-1}$$ and $$x(x^2 + a^2)^{-1}$$. HINT: Consider the fourier transform of $e^{-\beta|k|}$.

Starting with the hint, we can take:

$f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{ikx} e^{-\beta|k|} dk = \frac{1}{\sqrt{2\pi}}[ \int_{0}^{\infty} e^{-k(\beta - ix)} dk + \int_{-\infty}^{0} e^{k(\beta + ix)} dk]$

Integrating and evaluating at the bounds will leave us:

$\frac{1}{\sqrt{2\pi}} [\frac{1}{\beta + ix} + \frac{1}{\beta - ix} ] = \frac{2\beta}{x^2 + \beta^2} \times \frac{1}{\sqrt{2\pi}}$

Then, for problem 1:

let $\beta = a$

Then using the property that $\hat{f} = F$ and $\hat{F} = f$, we can note:

the fourier transform of the above is merely $\frac{1}{x^2 + a^2} \ times \frac{2a}{\sqrt{2\pi}}$. The above property then implies that some multiple of the function $e^{-a|k|}$ is our desired function for the fourier transform, specifically:

$g(k) = \frac{\sqrt{2\pi}}{2a} e^{-a|k|}$ 's fourier transform, $\hat{g(x)} = \frac{\sqrt{2\pi}}{2a} \ times \frac{2\beta}{x^2 + a^2} \times \frac{1}{\sqrt{2\pi}} = \frac{1}{x^2 + a^2}$ as desired.

For the second function's fourier transform: note that $g(x) = xf(x) \rightarrow \hat{g(k)} = i\hat{f'(k)}$ is a property, and that the second function is x times the previous problem's function.

Thus, take the derivative of $\hat{f(k)}$ i.e. $\frac{d}{dk} ( e^{-a|k|}) = \frac{-ak e^{-a|k|}}{|k|}$ So then:

the forward fourier transform is:

$\frac{-i\sqrt{\pi}k e^{-a|k|}}{\sqrt{2}|k|}$

7
##### APM346––Home Assignments / 5.2P, Question 3 approach
« on: March 06, 2018, 09:59:05 PM »
For 3a we are asked to take the Fourier transform of

$$(x^2 + a^2) ^ {-1}$$ For $a> 0$

i.e. to find $\tilde{f(k)} = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) e^{-ikx} f(x) dx = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) e^{-ikx} (x^2 + a^2) ^ {-1} dx$ and using Euler's identity, we can decompose the integral into cosines and sines (unlike in problems 2, where the trick seemed to lie doing the reverse) giving us:

$$\tilde{f(k)} = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) (coskx - isinkx) \frac{1}{x^2 + a^2} dx$$

The denominator is even, the first term is even , so we'll keep it (as it is even), the second term is odd, with an even denominator. Integrating over all space means that the integral of any odd function will be 0, leaving only the even function integral. Furthermore, since even functions are symmetric about x = 0, we can rewrite the integral:

$$\tilde{f(k)} = \frac{2}{\sqrt{2\pi}} \int_{0}^{\infty}\frac{coskx}{x^2 + a^2} dx$$

From here, I'm not quite sure how to approach this integral, as I'd normally try a trig substitution, but it seems like it would only complicate this...

Am I missing something obvious? What steps lack justification/ are incorrect?

NOTE: This question also applies to the approach in general for these problems in 3, as 3b,3c, and 3d are complications on this/add on more steps with integration by parts, so I'd also like to know if this even is the right treatment for this class of problems

8
##### APM346--Lectures / Stuck on Example 4 of 2.6 (modified)
« on: February 09, 2018, 10:19:46 PM »
Given:

$$u_{tt} - c^2 u_{xx} = 0$$(1)  for $0<x<\infty$
$$u |_{t=0} = g(x) , u_t | _{t=0} = h(x)$$(2)  for $0<x$, and additionally
$$(\alpha u + \beta u_t) |_{x=0} = q(t)$$(3) for $t> 0$

We are asked to evaluate and find the general solution for both regions $x > ct$ and $0<x<ct$

Given that the most general solution under the first condition is $$u(x,t) = \phi (x+ct) + \psi(x-ct)$$(I), we will focus on the latter case outlined, i.e. $x<ct$

We know from the definitions that:
$$\phi(x) = \frac{1}{2} g(x) + \frac{1}{2c} \int_{0}^{x} h(x')dx'$$(4) and
$$\psi(x) = \frac{1}{2} g(x) - \frac{1}{2c} \int_{0}^{x} h(x')dx'$$(5)

We can begin examining our boundary conditions. As usual, the particular issue is that $\psi(x-ct)$ is a problem in this region, as the values might be negative, while $\phi(x+ct)$ will not be

Jumping straight into (3), we apply it to (I), to get:

$$q(t) = \alpha ( \phi(ct) + \psi(-ct) ) + \beta c (\phi(ct) ' - \psi(-ct) ' )$$(6) and applying the relation $x = -ct$ to (6):
$$q(\frac{-x}{c}) = \alpha(\phi(-x) + \psi(x) ) + \beta c (\phi(-x)' - \psi(x)')$$(6')

From here, my steps get a bit more uncertain, where I take the (total) derivative of (4) to be :

$$\phi(x) ' = \frac{1}{2} g'(x) + \frac{1}{2c} (h(x) - h(0) ) + \int_{0}^{x} h'(x')dx'$$(7) using Leibnitz's Rule of Integration

plugging into (6') gives me:

$$q(\frac{-x}{c}) = \alpha( \frac{1}{2}g(-x) - \frac{1}{2c}\int_{0}^{-x} h(x')dx' + \psi(x)) + \beta c (\frac{1}{2} g'(-x) + \frac{1}{2c} (h(-x) - h(0) ) + \int_{0}^{-x} h'(x')dx - \psi(x)')$$(

NOTE:
I originally misread the example, as I meant to talk about example 4, but I have modified the Robin condition such that it involves $\alpha u$ and $\beta u_t$ as opposed to the original wording of the problem.

This leaves me two questions:
1. Is this the correct procedure to get the conclusion outlined in the example?
2. What steps do I need to take to get the conclusion outlined (an expression for $\psi$ only in terms of functions of $q,\psi', \phi'$)?
3. Is this procedure the correct one, when considering boundary conditions?

9
##### APM346--Lectures / Question from Monday's Lecture
« on: January 16, 2018, 10:04:18 PM »
When we were (re)introduced to the following Cauchy problem:
(i) $$u_{tt} - u_{xx} = 0$$
(ii) $$u|_{t =\frac{x^2}{2}} = x^3$$
(iii) $$u_{t} |_{t = \frac{x^2}{2}} = 2x$$

And we took the general solution of

(from i: 1) $$u(x,t) = \phi(x +t) + \psi (x-t)$$
and applying the conditions (ii,iii)
(from ii: 2) $$u(x, \frac{x^2}{2}) = \phi(x + \frac{x^2}{2}) + \psi(x - \frac{x^2}{2}) = x^3$$
(from iii : 3) $$u_{t} (x,\frac{x^2}{2}) = \phi'(x + \frac{x^2}{2}) - \psi'(x - \frac{x^2}{2}) = 2x$$
(2') $$3x^2 = (1+x) \phi'(x + \frac{x^2}{2}) + (1-x)\psi'(x - \frac{x^2}{2})$$

But from there we wrote that :
$$\phi'(x+ \frac{x^2}{2}) = \frac{\begin{vmatrix} 2x & -1 \\ 3x^2 & 1-x \end{vmatrix}}{\begin{vmatrix} 1 & -1 \\ 1+x & 1-x \end{vmatrix}} = \frac{2x+ x^2}{2} = x+ \frac{x^2}{2}$$

and similarly for $$\psi'(x - \frac{x^2}{2}) = -x + \frac{x^2}{2}$$

Where did these determinants come from?

10
##### APM346--Lectures / Classification of PDEs (Chapter Problems 1) clarification
« on: January 07, 2018, 08:08:17 PM »

From the textbook, we defined ( I might be paraphrasing)

Quote
"Equations of the form $Lu = f(X)$ where Lu is a linear partial differential expression with respect to an unknown function u is called a Linear Equation. The equation is homogenous if $f = 0$ and inhomogenous otherwise
and used the example of $Lu := a_{11}u_{xx} + 2a_{12}u_{xy} +a_{22}u_{yy} + a_{1}u_{x} + a_{2}u_{y} + au = f(x)$ (i) . We also know it is Non linear if it cannot be expressed in the form (i).

This leads me to my question as Problem 1 asks us to classify equations (1-10), if they are linear homogenous, inhomogenous or , non linear
and for two of them:
$u_{t} + uu_{x} = 0$ (2) and $u_{t} + uu_{x} + u = 0$ (4).

Paraphrasing the text:
Quote
If $Lu = f(x)$ , and both the coefficients and right hand side expression depend on lower order derivatives, then the equation is Quasilinear

So does this mean that eqns (2) and (4) are Quasilinear then? Is that wholly seperate from the options of Linear Homogenous, Linear Inhomogenous and Nonlinear, or are Quasilinear and Semilinear sub-catagories of those?

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