Author Topic: Section 2.3, Q 5  (Read 1157 times)

Ioana Nedelcu

  • Full Member
  • ***
  • Posts: 29
  • Karma: 3
    • View Profile
Section 2.3, Q 5
« on: January 31, 2018, 12:17:06 AM »
Since we found the solution for $ v(r,t) $ and $ u = v/r $, then the general solution will be

 $$ u = \frac{1}{2r}\Phi (r+ct)+\Phi (r-ct)+\frac{1}{2rc}\int_{r-ct}^{r+ct}\Psi (s) \,ds $$

Clearly, it is not continuous at r = 0 so do we only consider as r approaches 0?

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2487
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Section 2.3, Q 5
« Reply #1 on: January 31, 2018, 11:08:29 AM »
You went too far: don't solve IVP, just write down the general solution $u(r,t)$ of this equation and then decide when it is continuous.

Talking about future this would give you a spherical wave in the whole space, while singular at the origin solutions correspond to the spherical waves with sources or sinks there

Jingxuan Zhang

  • Elder Member
  • *****
  • Posts: 106
  • Karma: 20
    • View Profile
Re: Section 2.3, Q 5
« Reply #2 on: February 01, 2018, 08:21:24 AM »
$$u=\frac{1}{r}(f(r+ct)+g(r-ct))$$
If $f(x)=-g(-x)$ then the solution will be continuous at zero.

Ioana Nedelcu

  • Full Member
  • ***
  • Posts: 29
  • Karma: 3
    • View Profile
Re: Section 2.3, Q 5
« Reply #3 on: February 02, 2018, 03:39:46 AM »
Do you mean that the numerator is also zero when the functions are equal? At r=0, u = 0/0 is indeterminate then.

Can't we just say it's continuous when r doesn't equal 0?

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2487
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Section 2.3, Q 5
« Reply #4 on: February 02, 2018, 07:05:21 AM »
If there is a limit of $u(r,t)$ when $r\to 0$, we can always define $u(0,t)$ so that it will be continuous and, as long we deal with ordinary functions rather than distribution, we always fix functions this way. We call this a removable singularity.

For ordinary functions not only single points but the sets of measure zero (see Real Analysis class, in our class we will talk about this much later) do not matter.

For distributions it is not so, as they could be supported at single points.

Now, to finish this problem just write down $u$, plugging correct $g=-f$

Jingxuan Zhang

  • Elder Member
  • *****
  • Posts: 106
  • Karma: 20
    • View Profile
Re: Section 2.3, Q 5
« Reply #5 on: February 02, 2018, 06:58:56 PM »
So if I understand your hint right, professor, should I get:
$$\left\{
\begin{align*}
u(r,t)&=\frac{1}{r}(f(ct+r)-f(ct-r)) & r\neq0 \\
u(0,t)&=2f'(ct)
\end{align*}
\right.
$$

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2487
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Section 2.3, Q 5
« Reply #6 on: February 02, 2018, 07:16:15 PM »
Indeed