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Messages - Wanying Zhang

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Home Assignment 2 / 2.4 problem4
« on: February 27, 2019, 10:30:59 PM »
The problem is given:
$$u_{tt} - u_{xx} = (x^2 -1)e^{-\frac{x^2}{2}}$$
$$u(x,0) = -e^{-\frac{x^2}{2}}, u_t(x,0) = 0$$

I have already got the general solution as followed, but I have trouble solving the integral,
$$\int_{0}^{t} \int_{x-t+s}^{x+t-s} (y^2-1)e^{-\frac{x^2}{2}}dyds$$
and I tried $\Delta$ method, but it seems to make the equation more complex. Professor, could you please give a hint of solving this problem?

Home Assignment 3 / Problem2(17) even or odd?
« on: February 01, 2019, 10:40:32 AM »
Given the conditions:
$$g(x) = 0,   
h(x) =
       \text{1} & {|x| < 1}\\
       \text{0} & {|x| \geq 1} \\
I know since both $g, h$ are even functions, as a result, $u$ is even with respect to $x$. But I don't quite understand how to get the conclusion that $u$ is odd with respect to $t$?

Home Assignment 3 / Re: S2.3 problem2(17)(18)
« on: January 31, 2019, 08:47:07 AM »
Professor, may I ask how to derive all the five conditions mentioned above? Why it's not necessary to consider the condition like ${x+ct>1, x-ct<-1}$ or ${x-ct>1, x+ct,-1}$ something like this?

I have difficulties obtaining the general solution of the equation $u_{tt} - c^2u_{xx} = 0$. From the online textbook Section2.3, it mentions $v = u_t +cu_x$, and it gets the result $v_t - cv_x = 0$ by chain rule. But when I expand $v_t - cv_x = 0$, I get an extra term $x'(t)$ when applies chain rule to $v_t$ because I think $u_{t}$ in $v$ can be two separated into to parts which are $u_t(t)$ and $u_t(x(t))$. Then applying chain rule, it becomes $v_t = u_{tt} + x'(t) +c(u_{xt} + x'(t))$ where the term $x'(t)$ can not be cancelled. I wonder where is the problem of my thought.

Home Assignment 2 / Assignment 2.2 Problem3 (16)
« on: January 24, 2019, 12:25:14 AM »
Given the question $u_t + 3u_x -2u_y = xyu$. My steps are:
$$\frac{dt}{1} = \frac{dx}{3} = \frac{dy}{-2} = \frac{du}{xyu} \Rightarrow x=C_1+3t, y=C_2-2t$$
$$xydt = \frac{du}{u} \Rightarrow (C_1+3t)(C_2-2t)dt = \frac{du}{u}$$
$$\ln u = C_1C_2 t + \frac{t^2}{2} (3C_2 - 2C_1) - 2t^3 +\phi(C_1, C_2)$$
I know then I need to insert base $e$ into both sides to get $u$, but it would be complex. I wonder is there any simple way to solve this? Or do I have something wrong in above equations so that leads to the complex solution?

Home Assignment 2 / Re: Home Assignment 2 Problem 2(a)
« on: January 23, 2019, 05:13:27 PM »
I wonder why the integral curves are half-lines from (0, 0)? I agree that these integral curves are straight since the equation $C = \frac{x}{y}$, but it seems no given range for either $x$ and $y$. How is the characteristic of being half determined? Any reply would be appreciated.

Home Assignment 2 / Re: Concept: Where does the (-) come from
« on: January 22, 2019, 10:53:03 PM »
I guess that since $u$ is of $x$ and $x$ is of $t$, they are not independent and you need to use $\partial$ in your steps. Also, there should be chain rule in this case, so you cannot simply take integrals as in your second equation.

Home Assignment 2 / Re: problem 5 (23)
« on: January 20, 2019, 11:53:29 AM »
I tried to use the trigonometric identities to solve this problem and think I got the correct answer, which is $-\frac{1}{2} xy + (\frac{1}{2})(x^2 + y^2)arcsin(\frac{y}{\sqrt{x^2 + y^2}})$. But then I'm confused about "In one instance solution does not exist" at end of this problem because I obtain solutions for all 4 subproblems, except two of them with arbitrary function. I wonder which subproblem may not have solutions.

Home Assignment 2 / problem 5 (23)
« on: January 20, 2019, 12:53:01 AM »
I have trouble solving this problem: $yu_x - xu_y = x^2$. I know the characteristic equation is $\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{x^2}$ and then have $C = \frac{x^2}{2} + \frac{y^2}{2}$. Then the following should be the integration relative to $du$, but either $\frac{du}{dx}$ or $\frac{du}{dy}$ will contain not only one variable, like $\frac{du}{dx} = \frac{x^2}{y}$ contain both $x$ and $y$. I wonder if $x$ and $y$ are independent here. If not, should I rewrite the expression $C = \frac{x^2}{2} + \frac{y^2}{2}$ in order to get the expression of y in terms of x , and then applies it into the integration relative to $du$? Any reply would be appreciated.

Home Assignment 1 / Re: problem 4 (25)
« on: January 18, 2019, 08:04:40 PM »
I would start from the beginning.
$$\frac{u_{xy}}{u_x} = \frac{u_x u_y}{u_x} \Rightarrow \frac{u_{xy}}{u_x} = u_y$$
Integrate both sides,
$$\ln{u_x} = u + f(x)$$
$$u_x = e^{u+f(x)} = e^u \cdot g(x)$$
$$\frac{\partial u}{\partial x} = g(x)e^u$$
$$\frac{\partial u}{e^u} = g(x)\partial x$$
Integrate both sides,
$$-e^{-u} = G(x) + h(y)$$
$$u(x,y) = -\ln (-G(x) - h(y))$$
OK. V.I.

Home Assignment 2 / Secondary Textbook Chp1.2 Exercise 10
« on: January 18, 2019, 06:16:16 PM »
I have difficulties solving this problem: $u_x + u_y + u = e^{x+2y}$. I get $\frac{dy}{dx} = 1$ from the characteristic equation and have no ideas what should be followed. Any reply would be appreciated.

Home Assignment 1 / Re: Home Assignment 1
« on: January 16, 2019, 01:42:48 PM »
This is my new try: Follow the hint and I got,
$$\frac{u_{xy}}{u_x} = \frac{u_y}{u}$$
integrate both sides,
$$\ln (u_x) = \ln (u) + f(x)$$
where $f(x)$ is any function only of x. Then, we have,
$$u_x = g(x) \cdot u$$
where $g(x)$ is a function of $f(x)$ (i.e. $g(x) = e^{f(x)}$). Rewrite this equation,
$$\frac{\partial u}{\partial x} = g(x) \cdot u$$
Above is ok, but below is not: you should wright $dx$ and $du$ (because standalone $\partial u$ and $\partial x$ do not make sense), and you need to integrate rather than differentiate
$$\frac{\partial u}{u} = g(x) \partial x$$
$$-\frac{1}{u^2} = g'(x) + h(y)$$
where $h$ is any function that only of y. Therefore,
$$u = \frac{1}{\sqrt{-g'(x) - h(y)}}$$

Home Assignment 1 / Re: Home Assignment 1
« on: January 15, 2019, 12:02:54 AM »
I have a guess but also not quite sure if I'm in the right way. Integrate the equation you got
$$\int_ {}{}\frac{(uu_x)_y}{uu_x} dy = \int {}{}\frac{u_y}{u}dy \qquad\qquad\qquad\qquad \color{red}{\text{Error in the left part; the rest incorrect. V.I.}}$$
$$ \ln uu_x = \ln u + f(x)$$
where $f$ is any functions that is only of $x$,
$$uu_x = u \cdot g(x)$$
where $g$ is a function of $f$ mentioned above. Divide $u$ from both sides,
$$ u_x = g(x)$$
then integrate both sides,
$$ u = G(x) + h(y)$$
where $G'(x) = g(x)$, and $h$ is any function that is only of y.

Final Exam / Re: FE-P2
« on: December 18, 2018, 11:32:06 AM »
Here's the solution.

Quiz-7 / Re: TUT 0301
« on: December 01, 2018, 07:31:54 PM »
Sorry.. I think the following one is the correct solution. I had wrong arg computation above.
This one maybe the correct answer.

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