### Show Posts

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.

### Messages - Zaihao Zhou

Pages: 1 [2]
16
##### HA4 / Re: HA4-P3
« on: October 15, 2015, 11:32:58 AM »
Sorry I'm confused about what form of boundary conditions should take.

Why do we need here condition Ux=vt but not Ux=0? If x = vt means "location of the wave is its velocity times t" then why don't we use Ux=ct boundary condition for cases without v mentioned? Shouldn't c be the velocity? If we have v, what is c?

Could someone or professor please explain? Thanks!

17
##### Textbook errors / 2.6 errors
« on: October 15, 2015, 11:15:42 AM »
In 2.6 Example 1 equation $$\psi(x) = p(-x/c) - \phi(x)$$ is not correct, the correct one should be $$\psi(x) = p(-x/c) - \phi(-x)$$. OK

Also in the end of this example, statement " this formula defines u(x,t) as 0 < x < ct solution is given by (4)" does not flow. I think you meant "this formula defines u(x,t) as 0 < x < ct, recall for x>ct solution is given by (4)". OK

Also before example 5, equation (31)' seems should have been minus sign before the first integral, not a plus. also for the first term shouldn't it be 1/2(g(ct+x) minus  g(ct-x)) instead of plus?

NO, this is even continuation (for "primed") equations

For 1D wave equation on the finite interval, below the graph, section (b), ...(f.e...Ux|x=a =...) should be (f.e. Ux|x=b = ...).

below this line ABB'B'' does not seem right, also AA'C'B'B should be AA'CB'B I think. and statement " u is defined in the intersection of..." should be " u is defined in the union of ..."

No, intersection: as $u$ is defined where both $\phi(.)$ and $\psi(.)$ are

18
##### HA4 / Re: HA4-P4
« on: October 15, 2015, 11:05:34 AM »
Hi for a) and b) I got exact opposite signs in the middle. Did you use the formula given in the textbook? I think there in 2.6 Example 1 equation $$\psi(x) = p(-x/c) - \phi(x)$$ is not correct, the correct one should be $$\psi(x) = p(-x/c) - \phi(-x)$$
I think this is the cause of your problem.

19
##### HA3 / Re: HA3-P5
« on: October 08, 2015, 08:57:34 AM »
Sorry my fault. Lost one of negative signs in the process throughout. Thank you for correcting me.
Confirmed all, you guys are correct

20
##### HA3 / Re: HA3-P5
« on: October 06, 2015, 06:08:14 PM »
The general solution is
$$u(x,t) = \frac{1}{2c}\int_0^t{\int_{x-c(t-t')}^{x+c(t-t')} f(x',t')dx'dt'}$$
For problem: $$u_{tt} - c^2u_{xx} = f(x,t) \quad and \quad g(x)=0,\ h(x)=0$$
To solve problems in P1 (4) - (7), substitute the f(x,t) to corresponding equations.

Below are my results, not sure if they are right.
a)
$$u(x,t) = \frac{1}{2\alpha^2c^2}[sin(\alpha x + \alpha ct) - sin(\alpha x - \alpha ct)]$$

b):  $$u(x,t) = \frac{sin(\alpha x)}{\beta^2 - \alpha^2c^2}[sin(\beta t) - sin(\alpha ct)]$$

c):  $$u(x,t) = \frac{1}{2c^2}[F(x+ct)-F(x-ct)]$$

d): stuck at  $$u(x,t) = \frac{1}{2c}\int^t_0[F''(x+ct-ct')-F''(x-ct+ct')]t'dt'$$

Kinda feel we can still do the integration but don't know how after this.

21
##### HA3 / Re: HA3-P3
« on: October 03, 2015, 11:02:35 PM »
For (b), I am thinking (c1 + c2)2 - 4c1c2 = (c1 - c2)2 which is always larger or equal to zero. And plug in expression(11),  it equals to (c1 - c2)2.
Given A2 always larger than 0 (ignore case of equal 0) so B2- 4AC always larger or equal to 0. We dont need to emphasize on this condition.
Therefore, is it possible the answer is A â‰  0 ?? I am not sure of this.
We can't even write c1 and c2 if the relationship between A, B, C is unknown. The very fact that you have written both implicitly require 4B2- 4AC >= 0, that's why you concluded it always greater or equal to zero.

22
##### HA3 / Re: HA3-P3
« on: October 03, 2015, 06:42:45 PM »
Just wondering, is there a typo in the original problem? The equation given is:
Au_{tt} + 2Bu_{tx} + Cu_{tt}
But I think the last term should be with respect to x:
Au_{tt} + 2Bu_{tx} + Cu_{xx}

Yes it has to be a typo or the whole question does not make sense.

23
##### HA3 / Re: HA4-P4
« on: October 03, 2015, 02:48:39 PM »
a)
$$u_r = \frac{v_rr-v}{r^2}$$
$$u_{rr} = \frac{v_{rr}r^2-2v_rr+2v}{r^3}$$
$$u_{tt} = \frac{v_{tt}}{r}$$

Thus, the original equation can be written as $$\frac{v_{tt}}{r} = c^2 [\frac{v_{rr}r^2-2v_rr+2v}{r^3} + \frac{2}{r}\frac{v_rr-v}{r^2}]$$
Then we get $$v_{tt} = c^2v_{rr}$$
b) $$u = \frac{1}{r}[f(r+ct)+g(r-ct)]$$
c)$$v_{t=0} = ru_{t=0} = \phi(r)$$, thus $$\phi(r) = r\Phi(r)$$
similarly, $$\psi(r) = r\Psi(r)$$
Substitute, we get$$u = \frac{1}{2r}[(r+ct)\Phi(r+ct)+(r-ct)\Phi(r-ct)]+\frac{1}{2cr}\int^{r+ct}_{r-ct}s\Psi(s)ds$$

d) not sure I understand the question. I tried to use L'Hopital rule to determine the limit of u, this seems to be well defined as long as $$\Phi'(x)$$ is well defined. However u(0,t) is infinite. That means $$\Phi'(x)$$ should be infinite? Not sure if I'm doing the right thing.

24
##### HA3 / Re: HA3-P3
« on: October 03, 2015, 01:45:04 PM »
Separate the operators and u:
$$(A\partial_{t}^{2} + 2B\partial_{t}\partial_{x} + C\partial_{x}^{2})u = 0$$
Provided $$4B^{2} - 4AC >= 0$$ we have roots r1, r2 s.t. $$\partial_{t} = r_1\partial_x, \partial_t = r_2\partial_x$$
if equal, then r1 = r2.

then we can transform the original equation to form: $$(\partial_{t} - r_1\partial_x)(\partial_t - r_2\partial_x)u = 0$$
we can solve this equation in the way taught in class, namely substitute the partial part and u to v and w then get u. The answer is: $$u = f(x+r_1t) + g(x+r_2t)$$
compare to the form given by the question, we have $$r_1 = -c_1, r_2 = -c_2$$ are the roots.
a). so they satisfy $$c_1 + c_2 = \frac{2B}{A}, c_1c_2 = \frac{C}{A}$$.
b). to have $$c_1 < c_2$$ we need $$4B^2 - 4AC = B^2 - AC > 0$$

25
##### HA3 / Re: HA3-P2
« on: October 03, 2015, 01:31:48 PM »
Not sure if I'm right. Since involve graphs I uploaded a picture. Skipped 4 and 6 since I don't see the point and whether I am doing the right thing. Please correct me if I'm wrong.

26
##### HA3 / Re: HA3-P1
« on: October 03, 2015, 01:16:22 PM »
General solution: $$u = \phi(x+ct) + \psi(x-ct)$$ for general form $$u_{tt} - c^{2}u_{xx} = 0$$
All of the equations in this question satisfy this general form with c=1, 2, 3, 1/2, 3 (possible mistake of the question, should put 9 in front of the first term, if so c = 1/3).
plug the corresponding c into the general solution we have the answer for each equation.

27
##### Web Bonus = Sept / Re: Web bonus problem : Week 3 (#5)
« on: September 29, 2015, 12:04:08 AM »
Part A

u_{xy} = u_{x}u_{y}u^{-1}

Rearrange the function, we have: $$uu_{xy} = u_{x}u_{y}$$
Divide both sides by $$uu_{x}$$, we have $$\frac{u_{xy}}{u_{x}} = \frac{u_{y}}{u}$$

which is $$(lnu_{x})_{y} = (lnu)_{y}$$.
Integrate and do some arrangements we have $$lnu = \int{e^{f(x)}dx}+g(y)$$
Thus $$u = e^{\int{e^{f(x)}dx}}e^{g(y)}$$

Part B arrange we have $$(lnu_{x})_{y} = u_{y}$$
Integrate and arrange we have $$-e^{-u} = \int{e^{f(x)}dx}+g(y)$$
thus we have $$u = -ln[-(\int{e^{f(x)}dx}+g(y))]$$

Part C

Pages: 1 [2]