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APM346-2015F => APM346--Home Assignments => HA3 => Topic started by: Victor Ivrii on September 28, 2015, 01:04:56 PM

Title: HA4-P4
Post by: Victor Ivrii on September 28, 2015, 01:04:56 PM
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.3.P.html#problem-2.3.P.4 (http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.3.P.html#problem-2.3.P.4)
Title: Re: HA4-P4
Post by: Zaihao Zhou 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.
Title: Re: HA4-P4
Post by: Yumeng Wang on October 04, 2015, 03:15:05 PM
For (d). The following is my answer.

Given u is continuous when r=0, which means limrâ†’0 u(r,t) exists.
Then limrâ†’0 [f(r+ct) + g(r-ct)] = 0. Because otherwise, limrâ†’0 [f(r+ct) + g(r-ct)] â‰  0 implies limrâ†’0 u(r,t) tends to be infinity.

So f(ct) + g(-ct) = 0
f(ct) = - g(-ct)
-f(x) = g(-x)
So g(r-ct) = -f(ct-r). As a result, u = r-1 [f(r+ct) - f(ct-r)]
Title: Re: HA4-P4
Post by: Victor Ivrii on October 04, 2015, 04:24:35 PM
You did it collectively

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