Our Lecturer gaves us the following problem:
$u_{tt}=u_{rr}+\frac{u_r}{r}$ , $0<r<a,0<t$ (1)
with Initial Conditions (IC): $u(r,0)=f(r)$, $u_t(r,0)=0$
and Bounadry Conditions (BC): $u(a,t)=0$
The above equations are in polar coordinates and come when someone examines the phainomenon of a vibrating drum. Actually the real equation (instead of (1)) is: $u_{tt}=u_{rr}+\frac{u_r}{r}+\frac{u_{φφ}}{r^2}$ (2)
One of my collegues said that from these IC and BC and equation (1) we can deduce that the problem given is axessymmetric, i.e it does not depend on the angle $φ$...
So I have tried to prove his words writing:
From (1) and (2) we have: $u_{φφ}=0$
or
$u(r,φ,t)=F(r,t)φ+G(r,t)$
but I could not get anything significant step using the IC and BC...
Is my Collegue right?
Please take in mind that I do not need the solution of the problem of the membrane, I just want to know if it is axissymmetric or not. Thanks.
Yes, the solution is axially symmetric. What you essentially need to show is under given IC and BC, the solution to the PDE is unique.
Let $u_1(r,\phi,t)$ be a solution of PDE under given IC and BC. Since IC and BC is invariant under rotation, for any constant $K$, the function $u_2(r,\phi,t) \stackrel{def}{=} u_1(r,\phi+K,t)$ is also a solution of the problem satisfying same IC and BC.
Let $v = u_1 - u_2$. By construction, it satisfies
$$\begin{array}{rcll} \text{PDE} &:& v_{tt} = \nabla \cdot (\nabla v)\\ \text{IC} &:& v(r,\phi,0) = v_t(r,\phi,0) = 0,& \forall r \le a, \forall \phi\\ \text{BC} &:& v(a,\phi,t) = 0,& \forall r \ge 0, \forall \phi \end{array}$$
Consider following $t$-dependent integral,
$$Q(t) = \int_{B(0,a)} (v_t^2 + |\nabla v|^2) dxdy$$ It is clear $Q(0) = 0$. With help of a 2-d version of divergence theorem, we find $$\begin{align} Q'(t) &= 2 \int_{B(0,a)} (v_t v_{tt} + (\nabla v_t)\cdot (\nabla v)) dxdy\\ &= 2 \int_{B(0,a)} (v_t \nabla \cdot (\nabla v) +(\nabla v_t)\cdot (\nabla v)) dxdy\\ &= 2 \int_{B(0,a)} \nabla \cdot ( v_t \nabla v) dxdy\\ &= 2 \int_{r = a} (v_t \nabla v) \cdot \hat{r} ad\phi \end{align}$$ Since $v_t = 0$ at $r = a$, last integral evaluates to $0$.
This means $Q'(t) = 0$ for $t \ge 0$. This in turn implies $Q(t) = Q(0) = 0$ for $t \ge 0$.
As a result, $v = 0$ and hence $u_2 = u_1$ for $t \ge 0$.
Since this is true for all $K$, the solution $u_1$ of PDE is axially symmetric.