APM346-2018S > Final Exam

FE-P6

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Victor Ivrii:
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,\\
&0 &&r\ge \pi,\end{aligned}\right.\qquad \label{6-2}
\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
\begin{equation}
r  u_{rr}+2 u_r= (r u)_{rr}.
\label{6-3}
\end{equation}

Andrew Hardy:
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

Jingxuan Zhang:
Andrew,

Heed your trig! Beside there should be cases.

George Lu:
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 $u(r,0)=0$

Andrew Hardy:
That's what I had originally before JX confused me.

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