# Toronto Math Forum

## APM346-2018S => APM346--Tests => Final Exam => Topic started by: Victor Ivrii on April 11, 2018, 02:46:28 PM

Title: FE-P6
Post by: Victor Ivrii on April 11, 2018, 02:46:28 PM
Solve as $t>0$
\begin{align}
&u_{tt}-\Delta u  =0, \label{6-1}\\
&u(x,y,z,0)=0,
&&u_t(x,y,z,0)=
\left\{\begin{aligned} &r^{-1}\sin(r) &&r:=\sqrt{x^2+y^2+z^2}<\pi,\\
\end{align}
and solve by a separation of variables.

Hint. Use spherical coordinates, observe that solution must be spherically symmetric: $u=u(r,t)$ (explain why).

Also, use equality

r  u_{rr}+2 u_r= (r u)_{rr}.
\label{6-3}
Title: Re: FE-P6
Post by: Andrew Hardy on April 11, 2018, 07:07:54 PM
The solution is spherically symmetric because the question has that symmetry.
Then using the hint I reduce the Laplacian to $$(ru)" = v"$$ then do separation of variables on v and I get the solution to the eigenvalue problem of v is that  $R = \sin(r)$ because the eigenvalue must be one and the cos term has a singularity at r = 0
So now my general solution is that $$u = \frac{\sin(r)}{r} (A\cos(t) + Bsin(t))$$ To satisfy the boundary conditions, A = 0 and
$$u = \frac{\sin(r)}{r} \sin(t) \text{ for } r< \theta$$ B = 1.
For the other boundary condition, the only conceivable option is that u is defined as the trivial solution outside of this domain, B = 0
Title: Re: FE-P6
Post by: Jingxuan Zhang on April 11, 2018, 07:09:45 PM
Andrew,

Heed your trig! Beside there should be cases.
Title: Re: FE-P6
Post by: George Lu on April 12, 2018, 01:17:19 PM
I have the same answer as Andrew, except without the constant B, and I have a $\sin(t)$ term instead, due to the fact that $u(r,0)=0$
Title: Re: FE-P6
Post by: Andrew Hardy on April 12, 2018, 03:06:16 PM
That's what I had originally before JX confused me.
Title: Re: FE-P6
Post by: Victor Ivrii on April 14, 2018, 05:23:18 AM
Unfortunately, I wrote "by separation of variables" and almost everybody got confused. If it was more advanced class, I would shrug "So what? You should think rather than follow wrong advices, even from Professor, Supervisor, ..." but it would be too cruel and unfair here.

So, I decided for those who got completely confused, to give mark "-" (effectively 0). So I am taking the sum of all problems except P5, and multiply by 7/6. I also take the sum of all problems. And finally I take the maximum of these two numbers.

Still, correct solution is pending.
Title: Re: FE-P6
Post by: Jingxuan Zhang on April 14, 2018, 07:14:10 AM
Alternatively, professor, you can consider it a bonus, which is rewarding for those who has worked out one of the previous year's final where this situation happened in almost exactly the same manner.

Observe indeed $u$ must be spherical symmetric as is the boundary. Let $v=ru$, then \eqref{6-1}-\eqref{6-2}become, once identity \eqref{6-3} is known,
\begin{align} &v_{tt}-v_{rr}  =0, \label{6-1'}\\ &v|_{t=0}=0, &&v_t|_{t=0}= \left\{\begin{aligned} &\sin(r) &&r<\pi,\\ &0 &&r\ge \pi,\end{aligned}\right.\qquad \label{6-2'} \end{align}
which is easily solved with a combined use of even continuation and D'Alembert's:
v=\left\{\begin{aligned}&0&&r>t+\pi, \\ &\sin r\sin t&&0<r<-t+\pi, \\ &\frac{\cos(r-t)+1}{2}&&|t-\pi|<r<t+\pi,\\&0&&0<r<t-\pi.\end{aligned}\right.\qquad\label{6-4}
So then
u=\left\{\begin{aligned}&0&&r>t+\pi, \\ &\frac{\sin r\sin t}{r}&&0<r<-t+\pi, \\ &\frac{\cos(r-t)+1}{2r}&&|t-\pi|<r<t+\pi,\\&0&&0<r<t-\pi.\end{aligned}\right.\qquad\label{6-5}
Title: Re: FE-P6
Post by: Victor Ivrii on April 14, 2018, 01:19:45 PM
.
Title: Re: FE-P6
Post by: Andrew Hardy on April 14, 2018, 01:46:20 PM
If this is still open for karma, I will go into more detail missing from the solution?

Title: Re: FE-P6
Post by: Victor Ivrii on April 14, 2018, 02:25:56 PM
There is nothing missing.