16

**Home Assignment 2 / Re: problem 5 (23)**

« **on:**January 27, 2019, 07:07:51 PM »

There is NO root. You need to parametrize before integration

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

16

There is NO root. You need to parametrize before integration

17

If series has an infinite radius of convergence then it converges on the whole plane. If the radius of convergence is finite ...

However, even if the radius of convergence is infinite, it does not answer to many questions. F.e. from the decomposition of $e^z$ one cannot derive that $e^x$ fast rapidly as $\mathbb{R}\ni x\to +\infty$ and rapidly decays as $\mathbb{R}\ni x\to -\infty$.

However, even if the radius of convergence is infinite, it does not answer to many questions. F.e. from the decomposition of $e^z$ one cannot derive that $e^x$ fast rapidly as $\mathbb{R}\ni x\to +\infty$ and rapidly decays as $\mathbb{R}\ni x\to -\infty$.

18

Quote

Given the question $u_t + 3u_x -2u_y = xyu$. My steps are:Since $C_1=x-3t$, $C_2=y+2t$ we plug

$$\frac{dt}{1} = \frac{dx}{3} = \frac{dy}{-2} = \frac{du}{xyu} \Rightarrow x=C_1+3t, y=C_2-2t$$

$$xydt = \frac{du}{u} \Rightarrow (C_1+3t)(C_2-2t)dt = \frac{du}{u}$$

$$\ln u = C_1C_2 t + \frac{t^2}{2} (3C_2 - 2C_1) - 2t^3 +\phi(C_1, C_2)$$

$$

u= f(x-3t, y+2t) \exp \bigl( (x-3t)(y+2t)t +\frac{t^2}{2}(3y+6t-2x+6t) -2t^3\bigr)

$$

with an arbitrary function $f.,.)$ of two variables ($f=e^\phi$).

19

What you really need to do is to replace $C_1,C_2$ by their expression through $x,y,t$

20

Please all these lines tend to $(0,0)$ in one as $s\to -\infty$ ($\frac{dx}{ds}=x$, $\frac{dy}{ds}=y$) but never reach it. Please read again ODE, Chapter 7.

21

Yes, I do mean that equation. When I try it myself however, (lines 2 to 3 of my initial post) the math comes out that dx = -tdt.You need to read Section 2.1 of textbook to understand that equation $au_t+bu_x=f$ requires equation of integral curves $\frac{dt}{a}=\frac{dx}{b}=\frac{du}{f}$. What is on your post are incomprehensibly written senseless manipulations.

22

Do you mean $u_t + tu_x=0$? Then $\frac{dt}{1} = \frac{dx}{t}=\frac{du}{0}$. The first equation implies $$dx=tdt \implies x=\frac{1}{2}t^2+C\implies x-\frac{1}{2}t^2=C.$$

23

I obtain solutions for all 4 subproblems, except two of them with arbitrary function.No, your solution is incorrect because it is not a continuous single-valued function. I gave you a hint: what is the natural parameter along integral curves?

24

Sure, $x$ and $y$ are not independent along integral curves. To proceed you need to parametrize the integral curve. Think: what is the best way to parametrize it?

25

Please learn how to post math properly Also, asking for help, copy the problem.

26

Since integral curves are rays (straight half-lines) from $(0,0)$ the solution in the plane wit the punched out origin is any function, constant along these rays, in particular $u=f(y/x)$.

But if we want solution in the whole plane, $u$ must be continuous at $(0,0)$ and since all rays intersect there $u$ is just a constant.

But if we want solution in the whole plane, $u$ must be continuous at $(0,0)$ and since all rays intersect there $u$ is just a constant.

27

Read Section 2.1, Subsection "semilinear equations"

28

Junjing, you are right but I am not sure if anyone but me would be able to read your solution

29

You draw just a bit more than one period and say "$\pi$-periodic"

30

Physical interpretation is useful. However Quizzes, Tests and Exam neither test nor require any knowledge of Physics. Like in ODE or Calculus I, II classes.