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### Messages - Zarak Mahmud

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31
##### Home Assignment 3 / Re: problem 3
« on: October 10, 2012, 11:04:48 PM »
Hahah, yeah I think I would have had to spend about 3 hours typing that all up.

32
##### Home Assignment 3 / Re: Problem 2
« on: October 10, 2012, 09:34:11 PM »
Part (a):

$$\begin{equation*} u(x,t) = U(x - vt, t)\\ Let \xi = x - vt, \eta = t \\ u_x = U_x = \frac{\partial U}{\partial \xi}\frac{\partial \xi}{\partial x} + \frac{\partial U}{\partial \eta}\frac{\partial \eta}{\partial x} = U_x\\ u_{xx} = U_{\xi\xi}\\ u_t = U_t = \frac{\partial U}{\partial \xi}\frac{\partial \xi}{\partial t} + \frac{\partial U}{\partial \eta}\frac{\partial \eta}{\partial t} = -vU_{\xi} + U_{\eta}\\ -vU_{\xi} + U_{\eta} + vU_{\xi} = kU_{\xi\xi} \\ U_{\eta} = kU_{\xi\xi}\\ U(\xi,\eta) = \int_{-\infty}^{\infty}G(\xi,y,\eta)g(y)dy\\ u(x,t) = \int_{-\infty}^{\infty}G(x - vt,y,t)g(y)dy \\ = \int_{-\infty}^{\infty} \frac{1}{2 \sqrt{k \pi t}} e^{\frac{-(x-vt-y)^2}{4kt}}g(y)dy \end{equation*}$$

Part (b):

Let $$\begin{equation*} u(x,t) = e^{\alpha t + \beta x}V(x,t)\\ u_t = \alpha e^{\alpha t + \beta x}V + e^{\alpha t + \beta x}V_t\\ u_x = \beta e^{\alpha t + \beta x} + e^{\alpha t + \beta x}V_t\\ u_{xx} = \beta^2 e^{\alpha t + \beta x}V + 2\beta e^{\alpha t + \beta x}V_x + e^{\alpha t + \beta x}V_{xx}\\ \end{equation*}$$

Now plug these into the given heat equation.

$$\begin{equation*} \alpha V + V_t + v\beta V + vV_x = k\beta^2 V + 2k\beta V_x + kV_{xx} \end{equation*}$$
set $\alpha + v\beta = k\beta^2$ and $v = 2k\beta$ then,

$$\begin{equation*} \beta = \frac{v}{2k}, \alpha = \frac{-v^2}{4k} \\ u(x,t) = e^{\alpha t + \beta x} \int_{0}^{\infty} G_D(x,y,t)g(y)dy\end{equation*} \\$$
This works because if $u(0,t) = 0,$ then $V(0,t) = 0.$

33
##### Home Assignment 3 / Re: Problem 1
« on: October 10, 2012, 09:31:36 PM »
(a)

G(x,y,t) = \frac{1}{2 \sqrt{k \pi t}} e^{\frac{-(x-y)^2}{4kt}}\

\begin{equation*}
u(x,t) = \int_{-\infty}^{\infty} G(x,y,t)g(y)dy \\
= \int_{-\infty}^{0} G(x,y,t)g(y)dy + \int_{0}^{\infty} G(x,y,t)g(y)dy \\
\end{equation*}
Let $-y = z$,
$$= \int_{\infty}^{0} G(x,-z,t)g(-z)(-dz) + \int_{0}^{\infty} G(x,y,t)g(y)dy$$

We have Dirichlet boundary condition so g(z) is odd;$g(-z) = -g(z).$

\begin{equation*}
= -\int_{\infty}^{0} G(x,-z,t)g(z)dz + \int_{0}^{\infty} G(x,y,t)g(y)dy\\
= -\int_{\infty}^{0} G(x,-y,t)g(y)dy + \int_{0}^{\infty} G(x,y,t)g(y)dy\\
= \int_{0}^{\infty}\big[G(x,y,t)g(y) - G(x,-y,t)g(y)\big]dy\\
= \int_{0}^{\infty} \frac{1}{2 \sqrt{k \pi t}} \left[ e^{\frac{-(x-y)^2}{4kt}} - e^{\frac{-(x+y)^2}{4kt}} \right] g(y)\,dy.\\
\end{equation*}

(b)
Neumann boundary conditions: $g(y)$ is even $\implies g(x) = g(-x)$
Making the same substitution as in (a), we have:

\begin{equation*}
u(x,t) = \int_{0}^{\infty} \frac{1}{2 \sqrt{k \pi t}} \left[ e^{\frac{-(x-y)^2}{4kt}} + e^{\frac{-(x+y)^2}{4kt}} \right] g(y)\,dy.
\end{equation*}

34
##### Home Assignment 1 / Re: WTH?
« on: October 01, 2012, 10:34:20 PM »
BTW, has anyone any idea how much time it takes to prepare 1 hour of lecture notes or 1 home assignment? 2-3 hours :-)

I started using vim recently in an effort to type TeX commands faster. But it has a steep learning curve

35
##### Home Assignment 2 / Problem 2
« on: October 01, 2012, 09:01:03 PM »
Part (a): Make the variable change and use product rule to find $u_{rr}$, $u_r$, and $u_{tt}$.
$$\begin{equation*} u = vr \\ \frac{\partial u}{\partial t} = \frac{1}{r}\frac{\partial v}{\partial t} \\ \frac{\partial^2 u}{\partial t^2} = \frac{1}{r} \frac{\partial^2 v}{\partial t^2} \\ u_{tt} = \frac{1}{r}v_{tt} \\ \frac{\partial u}{\partial r} = \frac{\partial}{\partial r}\frac{v}{r} = \frac{1}{r}\frac{\partial v}{\partial r} - \frac{v}{r^2}\\ \frac{\partial^2 u}{\partial r^2} = \frac{\partial}{\partial r}\frac{\partial u}{\partial v} \\ = \frac{\partial}{\partial r} \left(\frac{1}{r}\frac{\partial v}{\partial r} - \frac{v}{r^2} \right) \\ = -\frac{1}{r^2}\frac{\partial v}{\partial r} + \frac{1}{r}\frac{\partial^2 v}{\partial r^2} - \frac{1}{r^2} + \frac{2v}{r^3} \\ u_{rr} + \frac{2}{r}u_r = \frac{v_{rr}}{r} \\ u_{tt} = \frac{v_{tt}}{r} \\ \frac{v_{tt}}{r} = c^2\frac{v_{rr}}{r} \\ v_{tt} = c^2v_{rr} \end{equation*}$$

Part (b):
$$\begin{equation*} v_{tt} = c^2v_{rr} \\ u(r, t) = \frac{f(r + ct) + g(r - ct)}{r} \\ \end{equation*}$$

Part (c): Take initial conditions with change of variable and then plug into D'Alambert formula.
$$\begin{equation*} \phi (r) = v(r, 0) = ru(r, 0) = r \Phi (r) \\ \psi (r) = v_t(r, 0) = ru_t(r, 0) = r \Psi (r) \\ u(r, t) = \frac{1}{2r} \left[(r+ct) \Phi (r+ct) + (r - ct) \Phi (r - ct) \right] + \frac{1}{2cr} \int_{r-ct}^{r+ct}s \Psi (s) ds \\ \end{equation*}$$

Part (d): For function to be continuous at $r = 0$, we must have
$$\begin{equation*} \lim\limits_{r \to 0} \left[(r+ct) \Phi (r+ct) + (r - ct) \Phi (r - ct) \right] = 0 \\ ct \lim\limits_{r \to 0} \left[ \Phi (r+ct) - \Phi (r - ct) \right] = 0 \\ \Phi (ct) - \Phi (-ct) = 0 \\ \implies \Phi (ct) = \Phi(-ct) \end{equation*}$$
Thus, $\Phi$ is an even function. Similarly, $\Psi$ must be odd since the integral of an odd function between symmetric bounds, i.e., $[-ct, ct]$ is equal to $0$.

36
##### Home Assignment 2 / Re: Problem 3
« on: October 01, 2012, 09:00:43 PM »
Part (a):

$$\begin{equation*} x > 2t > 0 \\ u(x, t) = \frac{1}{4} \int_{x-2t}^{x+2t} dx' \\ = \frac{1}{4}(x + 2t - x + 2t) \\ = t. \\ \end{equation*}$$

$$\begin{equation*} 0<x<2t \\ u(x, t) = \frac{1}{4} \int_{2t - x}^{2t + x}dx' \\ = \frac{1}{4}(2t + x - 2t + x) \\ = \frac{x}{2}. \\ \end{equation*}$$

Part (b):
For $x > 2t > 0$ we obtain $u(x,t) = t$ as in (a).
$$\begin{equation*} 0<x<2t \\ u(x, t) = \frac{1}{4} \int_{0}^{2t - x} dx' + \frac{1}{4} \int_{0}{2t + x} dx' \\ = \frac{1}{4}(2t - x + 2t + x) \\ = t. \\ \end{equation*}$$

Part (c):
$$\begin{equation*} x > 2t > 0 \\ u(x, t) = \frac{1}{4} \int_{x-2t}^{x+2t} x' dx' \\ = \frac{1}{8}[(x+2t)^2 - (x-2t)^2] \\ = xt. \\ \end{equation*}$$

$$\begin{equation*} 0 < x < 2t \\ u(x,t) = \frac{1}{4} \int_{2t - x}^{2t+x} x'dx' \\ = \frac{1}{8}[(2t+x)^2 - (2t - x)^2] \\ = xt. \\ \end{equation*}$$

Part (d):
$x > 2t > 0$ as in (c): $u(x,t) = xt$
$$\begin{equation*} 0 < x < 2t \\ u(x,t) = \frac{1}{4} \int_{0}^{2t-x}x'dx' + \frac{1}{4} \int_{0}^{2t-x} x'dx' \\ = \frac{1}{8}[(2t-x)^2 + (2t + x)^2] \\ = t^2 + \frac{x^2}{4}. \\ \end{equation*}$$

37
##### Home Assignment 2 / Re: Problem4
« on: October 01, 2012, 08:59:59 PM »
Part (a):

$$\begin{equation*} \rho \frac{\partial^2u}{\partial t^2} - T \frac{\partial^2u}{\partial x^2} = 0 \\ \frac{1}{c^2} \frac{\partial^2u}{\partial t^2} - \frac{\partial^2u}{\partial x^2} = 0 \\ c = \sqrt{\frac{T}{\rho}} = 1 \\ \end{equation*}$$

$$\frac{\partial e}{\partial t} = \frac{1}{2}\left(u_t \frac{\partial u_t}{\partial t} + \frac{\partial u_t}{\partial t} u_t + \frac{\partial u_x}{\partial t}u_x + u_x\frac{\partial u_x}{\partial t}\right) \\ =u_t \frac{\partial u_t}{\partial t} + u_x\frac{\partial u_x}{\partial t} \\ =u_tu_{tt} + u_xu_{xt}$$
$$\frac{\partial p}{\partial x} = \frac{\partial}{\partial x}\left(u_tu_x \right) \\ =\frac{\partial u_t}{\partial x}u_x + u_t\frac{\partial u_x}{\partial x} \\ = u_{tx}u_x + u_tu_{xx} \\$$
$$\frac{\partial e}{\partial t} = \frac{\partial p}{\partial t} \\ u_tu_{tt} + u_xu_{xt} = u_{tx}u_x + u_tu_{xx} \\ u_tu_{tt} = u_tu_{xx} \\ u_t(u_{tt} - u_{xx}) = 0 \\$$
$u_{tt} - u_{xx} = 0$ because this is precisely the wave equation with $c = 1$.

$$\frac{\partial p}{\partial t} = \frac{\partial}{\partial t}\left(u_tu_x \right) \\ =u_{tt}u_{xt} + u_tu_{xt} \\$$
$$\frac{\partial e}{\partial x} = \frac{1}{2}\frac{\partial}{\partial x}(u_t^2 + u_x^2) \\ \frac{1}{2}(u_{tx}u_t + u_tu_{tx} + u_{xx}u_x + u_xu_{xx}) \\ = u_{tx}u_t + u_{xx}u_x \\$$
$$\frac{\partial p}{\partial t} = \frac{\partial e}{\partial x} \\ u_{tt}u_x + u_tu_{xt} = u_{tx}u_t + u_{xx}u_x \\ u_x(u_{tt} - u_{xx}) = 0 \\$$

Part (b):
$$\frac{\partial^2 e}{\partial t^2} = \frac{\partial}{\partial t}\frac{\partial e}{\partial t} = \frac{\partial}{\partial t}\frac{\partial p}{\partial x} = \frac{\partial}{\partial x}\frac{\partial p}{\partial t} = \frac{\partial}{\partial x}\frac{\partial e}{\partial x} = \frac{\partial^2 e}{\partial x^2} \\ e_{tt} - e_{xx} = 0 \\$$
$$\frac{\partial^2 p}{\partial t^2} = \frac{\partial}{\partial t}\frac{\partial p}{\partial t} = \frac{\partial}{\partial t}\frac{\partial e}{\partial x} = \frac{\partial}{\partial x}\frac{\partial e}{\partial t} = \frac{\partial}{\partial x}\frac{\partial p}{\partial x} = \frac{\partial^2 p}{\partial x^2} \\ p_{tt} - p_{xx} = 0 \\$$

38
##### Home Assignment 2 / Re: Problem4
« on: October 01, 2012, 01:20:04 PM »
hi y'all, a quick question,
would it be appropriate to assume x & t are independent variables in this question? as like, they are presumably not correlated in any function of each other.

Also, what does it mean by rho = T = 1 ? (is T the the kinetic energy or something?)

When modeling a physical string, $T$ represents the tension force and $\rho$ is the mass density. I think they are just asking us to consider $c = 1$ for this problem, since $c = \sqrt{\frac{T}{\rho}}$.

39
##### Misc Math / Re: Classification criteria for PDEs
« on: September 30, 2012, 10:35:56 PM »
Thanks for the very detailed post.

So, for example, for second order linear PDEs in two variables and real constant coefficients, the classification depends only on the coefficients of the second derivatives?

If we have something like
$$$$3u_{xx} + 7u_{xy} + 2u_{yy} = 0$$$$

Here $A = 3$, $B = 7$ and $C = 2$, and since $B^2 - 4AC = 25 > 0$, the PDE is hyperbolic. Is that correct?

40
##### Misc Math / Classification criteria for PDEs
« on: September 29, 2012, 05:26:38 PM »
I read somewhere (and I think it was mentioned in class) that all linear PDEs can be categorized into either parabolic, hyperbolic, or elliptic types according to: $B^2 - 4AC$. For example, if we have
$$$$u_{t} = u_{xx}$$$$
How do we determine what the values of $A$, $B$ and $C$ are?
And does this only apply to second order and smaller PDEs?

41
##### Misc Math / 1-D Wave equation derivation
« on: September 28, 2012, 01:27:42 PM »
Starting with

$$$$\frac{\partial}{\partial x} \left[ T(x,t) \sin{\theta (x,t)} \right] = \rho (x) u_{tt}$$$$

where $\rho$ is the density and $T(x,t)$ is the tension force, we made the assumption that the vibrations are small, which gave us a linearized wave equation. I can see why some of the other assumptions (i.e. full flexibility, and no horizontal tension component) make sense, but I don't think I understand the insight behind this one.

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