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### Messages - Ioana Nedelcu

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1
##### Quiz-4 / Re: Quiz 4
« on: March 02, 2018, 02:39:27 PM »
Sorry, forgot the actual variable so

$$T(t) = E\cos(\omega^2\sqrt{k}t) + F\sin(\omega^2\sqrt{k}t)$$

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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

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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

4
##### 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 $u = -\cos(\frac{(x-2t)}{3}) + 1 , -t <x<2t$

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##### APM346--Lectures / Re: Stuck on Example 3 of 2.6
« on: February 09, 2018, 10:36:47 PM »
Equations (6) and (6') can be written as ODEs with respect to $\psi(x)$:

$$q(t) = \alpha ( \phi(ct) + \psi(-ct) ) + \beta c (\phi(ct) ' - \psi(-ct) ' )$$
$$\alpha \psi(-ct) - \beta c (\psi(-ct))' = q(t) -\alpha \phi(ct) - \beta c (\phi(ct))'$$

and you can relate the above equation to the general $y + y' = p(t)$ equation from ODEs to solve.

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##### APM346--Lectures / Re: Section 2.6 Example 2 correction
« on: February 07, 2018, 05:03:26 PM »
Yeah it should be $\phi(-x)$ but in the next line they correctly included -x in the g(x) and h(x) functions so I don't think it should change any following equations

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##### APM346--Lectures / Re: Section 2.6 Example 1 correction
« on: February 06, 2018, 02:48:20 PM »
Yeah I think it means to say that you use the initial boundary condition to get $\phi(ct)+\psi(−ct)=p(t)$. Then you solve for $\psi(x), x <0$ and plug d'Alembert's solution for $\phi(-x)$ into the equation

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##### APM346––Home Assignments / Re: Boundary condition
« on: February 05, 2018, 11:33:01 AM »
Right so here we would use the boundary condition at x=0 to find $\psi(x-ct)$ where the argument is negative (0<x<ct).

So $u = \phi(x+ct) + \chi(\frac{t}{c}) - \phi(ct-x)$

Then I assume we use the initial conditions at t =0 to solve.

9
##### APM346––Home Assignments / Boundary condition
« on: February 05, 2018, 02:05:49 AM »
For section 2.6, question 1, how do we use the boundary condition at x = 0?

Solving using the initial conditions with d'Alembert's formula, the solution is $u = \phi(x + ct)$, for x>ct and 0<x<ct since it has a positive argument always

10
##### APM346––Home Assignments / Re: Section 2.3, Q 5
« on: February 02, 2018, 03:39:46 AM »
Do you mean that the numerator is also zero when the functions are equal? At r=0, u = 0/0 is indeterminate then.

Can't we just say it's continuous when r doesn't equal 0?

11
##### APM346--Lectures / Re: section 2.4 Goursat problem
« on: January 31, 2018, 10:57:35 AM »
What are $\xi$ and $/eta$? You'll see that simply plugging them in satisfies the condition

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##### APM346––Home Assignments / Section 2.3, Q 5
« on: January 31, 2018, 12:17:06 AM »
Since we found the solution for $v(r,t)$ and $u = v/r$, then the general solution will be

$$u = \frac{1}{2r}\Phi (r+ct)+\Phi (r-ct)+\frac{1}{2rc}\int_{r-ct}^{r+ct}\Psi (s) \,ds$$

Clearly, it is not continuous at r = 0 so do we only consider as r approaches 0?

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##### APM346––Home Assignments / Re: S2.3 P3
« on: January 31, 2018, 12:09:27 AM »
This was the web bonus question for week 2, see the solution there

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##### APM346––Home Assignments / Re: Goursat Problem
« on: January 31, 2018, 12:08:14 AM »
My take on it: it is a linear combination of the solution to the homogeneous equations given the boundary conditions (1) and the inhomogeneous solution (2)

(2) is simple, using the Duhamel integral or change of variables to the characteristic curves: $$\frac{-1}{4c^2} \int _{0}^{t} \int_{x-c(t-t')}^{x+c(t-t')} f(x',t')\,dx'dt'$$

(1) is a bit confusing because of the intervals: $$g(t) = \phi(2ct) + \psi(0), t>0$$ $$h(t) = \phi(0) + \psi(2ct), t<0$$ Do we use an odd extension/ method of reflection for the functions h and g so we can solve them?

15
##### 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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