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Quiz-4 / Re: Quiz 4
« on: March 02, 2018, 02:39:27 PM »
Sorry, forgot the actual variable so
$$ T(t) = E\cos(\omega^2\sqrt{k}t) + F\sin(\omega^2\sqrt{k}t) $$
$$ T(t) = E\cos(\omega^2\sqrt{k}t) + F\sin(\omega^2\sqrt{k}t) $$
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Quiz-4 / Re: Quiz 4 Thursday
« on: March 01, 2018, 11:36:02 PM »
There is also $ T(t) $, which has the ODE $ T'' + k\omega^4T = 0 $
$$ \implies T = E\cos(\omega^2\sqrt{k}) + F\sin(\omega^2\sqrt{k}) $$
So the full solution for $ u(x, y) = X(x)T(t) $ with X given in Jingxuan's answer below
$$ \implies T = E\cos(\omega^2\sqrt{k}) + F\sin(\omega^2\sqrt{k}) $$
So the full solution for $ u(x, y) = X(x)T(t) $ with X given in Jingxuan's answer below
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Term Test 1 / Re: P3
« on: February 15, 2018, 11:09:40 PM »
Yeah you're right, looks like I forgot to write it with the chain rule
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Term Test 1 / Re: P3
« on: February 15, 2018, 09:51:06 PM »
General solution for $ x >-t $: $ u = \psi(x+2t) + \phi(x-2t) $
Given initial conditions, we have $ \psi(x) = \phi(x) = 0 $ so $$ u = 0, x > 2t $$
Using the boundary conditions for $ -t < x < 2t $, ie where $\phi(x), x<0 $
$$ \psi'(t) + \phi'(-3t) = \sin(t) $$
From previous solution $$ \psi(x) = 0 \implies \psi'(x) = 0 $$
So $$ \phi(t) = -\cos(\frac{-t}{3}) + constant, t <0 $$
$$ \phi(t) = -\cos(\frac{t}{3}) + constant, t<0 $$
$$ \phi(x - 2t) = -\cos(\frac{(x-2t)}{3}) + constant $$
For $\phi$ to be constant at x = 2t, ie the same value from both sides of the characteristic equation:
$$ 0 = -\cos(0) + c \implies c = 1 $$
So $ u = -\cos(\frac{(x-2t)}{3}) + 1 , -t <x<2t $
Given initial conditions, we have $ \psi(x) = \phi(x) = 0 $ so $$ u = 0, x > 2t $$
Using the boundary conditions for $ -t < x < 2t $, ie where $\phi(x), x<0 $
$$ \psi'(t) + \phi'(-3t) = \sin(t) $$
From previous solution $$ \psi(x) = 0 \implies \psi'(x) = 0 $$
So $$ \phi(t) = -\cos(\frac{-t}{3}) + constant, t <0 $$
$$ \phi(t) = -\cos(\frac{t}{3}) + constant, t<0 $$
$$ \phi(x - 2t) = -\cos(\frac{(x-2t)}{3}) + constant $$
For $\phi$ to be constant at x = 2t, ie the same value from both sides of the characteristic equation:
$$ 0 = -\cos(0) + c \implies c = 1 $$
So $ u = -\cos(\frac{(x-2t)}{3}) + 1 , -t <x<2t $
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Quiz-1 / Re: Thursday Session
« on: January 25, 2018, 10:15:48 PM »
The original integral is $$ \frac{dt}{1} = \frac{dx}{x^2 + 1} = 0 $$
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Web Bonus Problems / Re: Web bonus problem -- Week 4
« on: January 24, 2018, 05:54:17 PM »
Jingxuan: why should there be only continuity at one point? The entire derivative function should be continuous on its domain.
My thinking in equating the two functions is that the derivative with respect to t is given by the two functions, so to ensure that they both give the same value at a point x they should be equal everywhere (since they are already smooth themselves).
The arguments of the functions threw me off a bit in the beginning. Now I'm thinking it should be $r'(0) = h(x) = u_{t}, t=0$
My thinking in equating the two functions is that the derivative with respect to t is given by the two functions, so to ensure that they both give the same value at a point x they should be equal everywhere (since they are already smooth themselves).
The arguments of the functions threw me off a bit in the beginning. Now I'm thinking it should be $r'(0) = h(x) = u_{t}, t=0$
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Web Bonus Problems / Re: Web bonus problem -- Week 4
« on: January 24, 2018, 01:22:42 AM »
A $C^1$ function has continuous derivative so maybe a condition should be that $$ u_{t} = h(x) = g'(x)$$ Otherwise if $ h(x) \neq g'(x) \forall x $ then there could be points of discontuity
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Web Bonus Problems / Re: Web bonus problem -- Week 2
« on: January 15, 2018, 11:32:33 PM »
Simplifying further, we get $$ u = 2xt + C $$ but using the second initial condition implies C = 0 so the solution is $ u(x,t) = 2xt $ (thanks for going over this in class!)
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Web Bonus Problems / Re: Web bonus problem -- Week 2
« on: January 15, 2018, 12:06:43 AM »
Thank you! I'm pretty sure my integration was wrong to begin with - I was confused with the functions' arguments and what the constants of integration should be (like Jaisen said)
If we can't integrate B, maybe we can differentiate A and solve for the functions this way?
If we can't integrate B, maybe we can differentiate A and solve for the functions this way?
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Web Bonus Problems / Re: Web bonus problem -- Week 2
« on: January 14, 2018, 09:20:00 PM »
General solution for wave equation:
$$u(t,x)= \phi(x+t) + \psi(x-t)$$
For the first initial condition: $u|_{t=x^2/2}= x^3$
$$ x^3 = \phi(x+x^2/2) + \psi(x-x^2/2)\tag{A}$$
Second condition: $u_t|_{t=x^2/2}$
$$ 2x = \phi'(x+x^2/2) - \psi'(x-x^2/2)\tag{B}$$
$$ \implies x^2 = \phi(x+x^2/2) - \psi(x-x^2/2)\tag{C}$$
Combining the two equations,
$$ \phi(x+x^2/2) = (x^3 + x^2) /2 $$
$$ \psi(x+x^2/2) = (-x^2 + x^3) /2 $$
But now I'm not sure how to change the functions to get their argument to $x + t$ and $x - t$ to find $u$ in terms of x and t
$$u(t,x)= \phi(x+t) + \psi(x-t)$$
For the first initial condition: $u|_{t=x^2/2}= x^3$
$$ x^3 = \phi(x+x^2/2) + \psi(x-x^2/2)\tag{A}$$
Second condition: $u_t|_{t=x^2/2}$
$$ 2x = \phi'(x+x^2/2) - \psi'(x-x^2/2)\tag{B}$$
$$ \implies x^2 = \phi(x+x^2/2) - \psi(x-x^2/2)\tag{C}$$
Combining the two equations,
$$ \phi(x+x^2/2) = (x^3 + x^2) /2 $$
$$ \psi(x+x^2/2) = (-x^2 + x^3) /2 $$
But now I'm not sure how to change the functions to get their argument to $x + t$ and $x - t$ to find $u$ in terms of x and t
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Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 09, 2018, 12:36:43 AM »
For the necessity $u_{xxyy}=f_{yy}=u_{yyxx}=g_{xx}$, can we say that this is because mixed partial derivatives of the same type, ie same number of differentiations with respect to the same variables, are equal? (By Green's Theorem)
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