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Messages - Adam Gao

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Web Bonus Problems / Re: Week 13 -- BP2
« on: April 05, 2018, 03:07:00 PM »
New proof evaluating the limit at $x=0$:

$f'(\varphi) = -f(\varphi ') = -\int f(x) \varphi'(x)\,dx = -\int \mathrm{ln}|x| \varphi'(x)\,dx = -\int_{-\infty}^{\infty}\mathrm{ln}|x| \varphi'(x)\,dx$

Using limits to evaluate an improper integral

 $= -\lim_{\epsilon \to 0^+} \left[ \int_{-\infty}^{-\epsilon}\mathrm{ln}|x| \varphi'(x)\,dx + \int_{\epsilon}^{\infty}\mathrm{ln}|x| \varphi'(x)\,dx \right]$

Then by integration by parts, we have:

$\lim_{\epsilon \to 0^+} \int_{-\infty}^{-\epsilon}\mathrm{ln}|x| \varphi'(x)\,dx$

$ = \lim_{\epsilon \to 0^+} \left [\mathrm{ln}|-\epsilon| \varphi(-\epsilon) - \mathrm{ln}|-\infty|\varphi(-\infty)\right] - \lim_{\epsilon \to 0^+}\int_{-\infty}^{-\epsilon}x^{-1}\varphi(x)\,dx$


$\lim_{\epsilon \to 0^+} \int_{\epsilon}^{\infty}\mathrm{ln}|x| \varphi'(x)\,dx$

$ = \lim_{\epsilon \to 0^+} \left[ \mathrm{ln}|\infty| \varphi(\infty) - \mathrm{ln}|\epsilon|\varphi(\epsilon)\right] - \lim_{\epsilon \to 0^+}\int_{\epsilon}^{\infty}x^{-1}\varphi(x)\,dx$

By $\mathrm{ln}|x|$ and $\varphi(x)$ being even (is this correct?) we get:

$\lim_{\epsilon \to 0^+}\left [\mathrm{ln}|-\epsilon| \varphi(-\epsilon) - \mathrm{ln}|-\infty|\varphi(-\infty) + \mathrm{ln}|\infty| \varphi(\infty) - \mathrm{ln}|\epsilon|\varphi(\epsilon)\right] = 0$



$= -\lim_{\epsilon \to 0^+} \left[ \int_{-\infty}^{-\epsilon}\mathrm{ln}|x| \varphi'(x)\,dx + \int_{\epsilon}^{\infty}\mathrm{ln}|x| \varphi'(x)\,dx \right] = \lim_{\epsilon \to 0^+}\int_{-\infty}^{-\epsilon}x^{-1}\varphi(x)\,dx + \lim_{\epsilon \to 0^+}\int_{\epsilon}^{\infty}x^{-1}\varphi(x)\,dx = pv\int \mathrm{ln}|x| \varphi'(x)\,dx$

Web Bonus Problems / Re: Week 13 -- BP2
« on: April 05, 2018, 01:28:54 PM »
Wrong step in the correct direction meaning I shouldn't use limits but I should eventually integrate by parts?

Should I use local integration?

Web Bonus Problems / Re: Week 13 -- BP2
« on: April 04, 2018, 10:53:34 PM »
I had a sign error as well. I hope this solution makes more sense.

$f'(\varphi) = -f(\varphi ') = -\int f(x) \varphi'(x)\,dx = -\int \mathrm{ln}|x| \varphi'(x)\,dx = -\int_{-\infty}^{\infty}\mathrm{ln}|x| \varphi'(x)\,dx$

Using limits to evaluate an improper integral

 $= -\lim_{R \to \infty} \int_{-R}^{R}\mathrm{ln}|x| \varphi'(x)\,dx$

Then by integration by parts:

 $= -\lim_{R \to \infty}(\mathrm{ln}|R| \varphi(R) - \mathrm{ln}|-R| \varphi(-R)) + \lim_{R \to \infty}(\int_{-R}^{R}x^{-1}\varphi(x)\,dx) = 0 + \lim_{R \to \infty}\int_{-R}^{R}x^{-1}\varphi(x)\,dx = pv\int x^{-1}\varphi (x)\,dx$

The nonintegral limit term dissappeared because $ln|x|$ is even. I also had to assume $\varphi$ was even to make the proof work. I am not sure if that is true.

Web Bonus Problems / Re: Week 13 -- BP2
« on: April 04, 2018, 03:30:03 PM »
Sorry, I am still confused. Why do we assume $\varphi(0)\ne 0$?

Would I be starting in the right direction with this instead?

$f'(\varphi) = f(\varphi ') = \int f(x) \varphi'(x)\,dx = \int \mathrm{ln}|x| \varphi'(x)\,dx$

Web Bonus Problems / Re: Week 13 -- BP2
« on: April 04, 2018, 01:49:26 PM »
$f'(x) = x^{-1}$

$f'(\varphi) = \int f'(x) \varphi(x)\,dx = \int x^{-1}\varphi (x)\,dx = pv\int x^{-1}\varphi (x)\,dx$

Concern/Confusion: Since the derivative of log absolute value of x is same as the derivative of log of x, why do we use a principal value integral instead of a normal integral? Did I somehow abuse notation?

Web Bonus Problems / Re: Web bonus problem--Week 10-11
« on: March 24, 2018, 05:01:35 PM »
c. and d. Correction to Hamiltonian:

First express $\dot{q}$ in terms of generalized momenta $p$.

We have $p_i = \frac{\partial L}{\partial \dot q_i} =  m\dot q_{i} + eA(q)_{i}$


$\dot{q}_i = \frac{p_i}{m} - \frac{eA(q)_i}{m}$

For the Hamiltonian to really be a Hamiltonian, we must replace the $\dot{q}$ term:

$H = \frac{m}{2}{\dot{q}}^2 + V(q)$

Now replace $\dot{q}_i$:

$H = \frac{m}{2}\sum_{i=1}^{n}( \frac{p_i}{m} - \frac{eA(q)_i}{m})^2 + V(q)$

I suppose $A$ affects the Hamiltonian because it affects generalized momenta.


Find $\ddot{q}$ due to $A$.

The Euler-Lagrange equation is as follows:

$eA'(q) - V'(q) = m\ddot{q} + eA'(q)\dot q + \frac{\partial A}{\partial t}$

Solving for $\ddot{q}$:

$\ddot{q} = \frac{1}{m}(eA'(q) - V'(q) - A'(q)\dot{q} + \frac{\partial A}{\partial t})$

Web Bonus Problems / Re: Web bonus problem--Week 10-11
« on: March 24, 2018, 02:10:55 AM »
$L(\mathbf{q}, \dot{\mathbf{q}})=  \frac{m}{2}{\dot{\mathbf{q}}}^2 + e\mathbf{A}(\mathbf{q})\cdot \dot{\mathbf{q}} -V(\mathbf{q})$


Euler-lagrange equations are of the form

$\frac{\partial L}{\partial q} = \frac{d}{dt}\frac{\partial L}{\partial \dot q}$

Compute both sides:

$\frac{\partial L}{\partial q} = eA'(q)\cdot q - V'(q)$

$\frac{\partial L}{\partial\dot q} = m\dot q + eA(q)$

$\frac{d}{dt}\frac{\partial L}{\partial \dot q} = m\ddot{q} + eA'(q)\cdot\dot q + \frac{\partial A}{\partial t}$


$\frac{\partial L}{\partial q} = eA'(q)\cdot q - V'(q) = \frac{d}{dt}\frac{\partial L}{\partial \dot q} = m\ddot{q} + eA'(q)\cdot\dot q + \frac{\partial A}{\partial t}$


Generalized momenta in index form can be taken from the Euler Lagrange equation quantities:

$p_i = \frac{\partial L}{\partial \dot q_i} =  m\dot q_{i} + eA(q)_{i}$

Generalized forces take the following form:

$\frac{\partial L}{\partial q_{i}} = e\frac{\partial A}{\partial q_i} - \frac{\partial V}{\partial q_i}$


The Hamiltonian $H(q,p)$ is defined as $H := \sum_{k=1}^n
\frac{\partial L}{\partial \dot{q}_k} \dot{q}_k -L$

Plugging in the generalized momenta and Lagrangian we get:

$H =  \sum((m\dot q_{i} + eA(q)_{i})\dot{q}_i )- L = m\dot q^2 + eA(q)\cdot\dot{q} - \frac{m}{2}{\dot{q}}^2 - eA(q)\cdot \dot{q} + V(q) = \frac{m}{2}{\dot{q}}^2 + V(q)$


$A$ has no effect on the Hamiltonian (or in physics, energy). However, A does affect generalized momenta; and its positional derivative (derivative with respect to $q$) affects generalized forces.

Web Bonus Problems / Re: Web Bonus problem -- Reading Week
« on: February 17, 2018, 01:28:15 PM »
Part (a):

To check that $u = e^{\frac{1}{2}ax^2}$ satisfies $u'(x) = -ax u(x)$, differentiate $u$. I am not sure if we can just treat $u$ like an exponential or if we have to expand $u = e^{\frac{1}{2}\mathrm{Re}(a)x^2}e^{\frac{1}{2}\mathrm{Imaginary}(a)x^2}$ using euler's formula first.

Next, use $u'(x) = -ax u(x)$ to show $i\omega{u}(\omega) = -ia\hat{u}'(\omega)$.

First, use fourier transform property $g(x) = xf(x) \rightarrow \hat{g}(k) = i\hat{f}'(k)$.
Using $u'(x) = g(x)$ and $-au(x) = f(x)$, as well as the definition of a fourier transform, we get $\int_{\infty}^{\infty} u'(x)e^{-i\omega x} = -ia\hat{u}'(\omega)$.

Now use fourier transform property $g'(x) = f(x) \rightarrow \hat{g}(k) = ik\hat{f}(k)$. Using $-axu(x) = f(x)$ and $u'(x) = g'(x)$ we get $\int_{\infty}^{\infty} -axu(x)e^{-i\omega x} = i\omega \hat{u} (\omega)$. Replacing $-axu(x)$ with $u'(x)$ we get $\int_{\infty}^{\infty} u'(x)e^{-i\omega x} = i\omega \hat{u} (\omega)$. Apply this equality to the previously shown equation $\int_{\infty}^{\infty} u'(x)e^{-i\omega x} = -ia\hat{u}'(\omega)$ and get $i\omega \hat{u} (\omega) = -ia\hat{u} ' (\omega)$.

Finally, find $\hat{u}$. Consider that $i\omega \hat{u} (\omega) = -ia\hat{u} ' (\omega) \rightarrow  -\frac{\omega}{a} \hat{u} (\omega) = \hat{u} ' (\omega)$. This is an ODE with an exponential solution, $\hat{u} (\omega) = Ce^{-\frac{\omega ^2}{2a}}$.

Web Bonus Problems / Re: Web bonus problem--Week 3
« on: January 20, 2018, 10:32:30 AM »
$G(x') = \int g(x') dx'$
$g(x') = \frac{d}{dx'} G(x')$

Web Bonus Problems / Re: Web bonus problem--Week 3
« on: January 20, 2018, 12:18:29 AM »
The second part of the question asks us to prove (2), starting with the premises (1), (3), and (4).

First prove the hint:
v = \int_{0}^{t} u(x,t')\,dt'.
g(x') = \frac{d}{dx'}G(x');
We already defined:
u(x,t) = \frac{1}{2}(g(x+ct) + g(x-ct)
from (1) and
v(x,t) = \frac{1}{2c} \int_{x-ct}^{x+ct}g(x')\,dt'
from (3). Then:
\int_{0}^{t} u(x,t')dt' = \frac{1}{2} \Bigl(\int_{0}^{t} g(x + ct')dt' + \int_{0}^{t} g(x - ct')\,dt'\Bigr)\\
= \frac{1}{2} \Bigl(\frac{1}{c}G(x+ct) - \frac{1}{c}G(x) + \frac{1}{-c}G(x-ct) - \frac{1}{-c}G(x)\Bigr)\\
= \frac{1}{2c} \Bigl(G(x + ct) - G(x-ct)\Bigr) = \frac{1}{2c} \int_{x-ct}^{x+ct}g(x')dt' = v(x,t)
Now I have proved the hint and I will implement it to help prove $(2)$. From the hint: $v_t = u$.

Thus $v_{tt} = u_{t}$. We know from (4):

$v_{tt} = c^{2}v_{xx}$

Thus $u_{t} = c^{2}v_{xx}$

Thus $u_{tt} = c^{2}v_{xxt}$

Let us find $v_{xxt}$. (We hope to prove the first condition of $(2)$.)

Previously we have shown $v(x,t) = \frac{1}{2c}(G(x+ct) - G(x-ct))$

Thus $v_x = \frac{1}{2c}(G'(x+ct) - G'(x-ct)) = \frac{1}{2c}(g(x+ct) - g(x-ct))$

Thus $v_{xx} = \frac{1}{2c}(g'(x+ct) - g'(x-ct))$

Thus $v_{xxt} = \frac{1}{2}g''(x+ct) + g''(x-ct)$

Let us find $u_{xx}$ in hopes that it is equal to $v_{xxt}$. This will help complete our proof.

$u_{x} = \frac{1}{2}(g'(x+ct) + g'(x-ct)$

Thus $u_{xx} = \frac{1}{2}g''(x+ct) + g''(x-ct) = v_{xxt}$.

Thus $u_{tt} = c^{2}u_{xx}$.

So $u_{tt} - c^{2}u_{xx} = 0$, proving that the first condition of $(2)$ must follow from $(1)$, $(3)$, and $(4)$.

To prove that the rest of $(2)$ follows, we don't even need $(3)$ or $(4)$. Plug $t=0$ into $u(x,t)$ and its partial derivative $u_t$ to prove those conditions.

Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 12:12:41 PM »
The identity is impossible. In other words, there does not exist any $g(y)$ or $h(y)$ such that the equation $2x^2 + xg(y) + h(y) = 0 $ is true for all $x$ and $y$ on the domain of $u(x,y)$.


Denote $v(x,y) = 2x^2 + xg(y) + h(y)$. Say we find some value $x=x_0$ and $y=y_0$ such that $v(x_0, y_0) = 0$. If $g(y_0)$ is positive or 0, let $x_1 > x_0$. Now closely examine the expression $v(x_1, y_0)$ and compare to $v(x_0, y_0)$. $2x_{1}^2 > 2x_{0}^2$ and $x_{1}g(y_0) > x_{0}g(y_0)$. Then $v(x_1, y_0) \ne v(x_0, y_0) = 0$.

If $g(y_0)$ is negative or 0, let $x_2 < x_0$. We would similarly find $v(x_2, y_0) \ne v(x_0, y_0) = 0$.

Thus, there is always some set of points $x$ and $y$ where $v(x,y) \ne 0$.

How this applies to the problem (14):

The fact that the identity is never possible means that there does not exist a general solution for the constraints $u_{xx} = y^2$ and $u_{yy} = -x^2$.

Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 05, 2018, 09:49:04 PM »

I don't think you are making too many assumptions if your goal is to find one possible solution, but it is valid to be concerned if it there are leaps in logic for a general solution.

Your solutions (I) and (II) are interpreted to be the general solutions to uxx=0 and uyy=0, respectively. I believe your solution steps for ux and uy for respective constraints to be correct. However, I have modified the final steps for the general solutions:

(I) : u(x,y) =  xϕ(y)+ψ(y)+bx+C
(II) : u(x,y) = yα(x)+β(x)+cy+C

Note that if you take the derivative of (I) with respect to x, b gets absorbed into C. Analogous conditions apply to (II).

When constrained to variables x and y, C is a constant (and so are b and c).

Finally, note that the general solution which obeys both uxx=0 and uyy=0 is equal to both solution (I) and solution (II).

Thus, solve: xϕ(y)+ψ(y)+bx+C = yα(x)+β(x)+cy+C

Let's assume none of the terms are equal to 0. This will allow us to impose constraints that will deny trivial solutions. If any of ϕ(y), ψ(y), α(x), and β(x) are equal to 0, the general solution should still be valid.

Thus, xϕ(y) = yα(x).

Thus, ϕ(y)=ay and α(x)=ax. (This is pretty much what you said, except I imposed the extra constraint that the functions share the same coefficient). Check that xϕ(y) = yα(x) = axy.

So now we have axy + ψ(y) + bx + C = axy + β(x) + cy + C

Thus, ψ(y) = cy and  β(x) = bx.

Thus, the general solution is axy + bx + cy + C.

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