If $\theta_t=\frac{1}{r}\frac{\partial}{\partial{r}}(r\theta_r)$, show that $\int_0^{\infty} \theta(r,t)rdr=\int_0^{\infty} \theta(r,0)rdr$? We have $r>0, t>0$ and we're given the conditions: $$r\theta_r \to 0  \text{ as } r\to \infty$$ and $$\theta(r,t)\leq K\in \mathbb{R}\text{ as }r\to 0$$
I tried taking integrals with respect to $r$ over the PDE. Rearranging the PDE and applying the integral:
$\int_0^{\infty}\big(r\theta_t-\frac{\partial}{\partial{r}}(r\theta_r)\big)dr=\int_0^{\infty}r\theta_tdr+\int_0^{\infty}\frac{\partial}{\partial{r}}(r\theta_r)dr=\int_0^{\infty}r\theta_tdr+r\theta_r |_{0}^{\infty}=\int_0^{\infty}r\theta_t(r,t)dr=0 $
Is this correct? From here I can't see how to proceed to get the result in the title?
 A: We can study the how the integral changes in time:
$$\frac{d}{dt}\int_{(0,\infty)}x\theta (x,t)dx\underbrace{=}_{\textrm{Leibniz}}\int_{(0,\infty)}x\frac{\partial}{\partial t}\theta (x,t)dx= \\ =\int_{(0,\infty)}\bigg[\frac{\partial}{\partial r}\bigg(r\frac{\partial \theta}{\partial r}\bigg)\bigg](x,t)dx=r\frac{\partial \theta}{\partial r}\bigg\vert^\infty_0=0$$
This implies that the integral never changes. Therefore
$$\int_{(0,\infty)}x\theta (x,0)dx=\int_{(0,\infty)}x\theta (x,t)dx \ \ \ \forall t \geq 0$$
A: $$\begin{align}
\phantom{=}&\frac{\mathrm{d}}{\mathrm{d}t}\int_0^\infty\theta r\,\mathrm{d}r\\
=&\int_0^\infty\partial_t\left[r\theta\right]\,\mathrm{d}r\\
=&\int_0^\infty r\theta_t\,\mathrm{d}r\\
=&\int_0^\infty\partial_r\left[r\theta_r\right]\,\mathrm{d}r\\
=&{\large{\left.r\theta_r\right|_{r=0}^\infty}}\\
=&0
\end{align}$$
This tells us that the $\int_0^\infty r\theta\,\mathrm{d}r$ is independent of $t$, and so it is a given that:
$$\int_0^\infty r\theta(r,t)\,\mathrm{d}r=\int_0^\infty r\theta(r,0)\,\mathrm{d}r\qquad\forall t\in\mathbb{R}^{+}_0$$
A: More generally, for your future work with the heat equation in a fixed domain $\Omega$ with $\partial u/\partial n = 0$ on $\partial \Omega$, the divergence theorem provides help:
\begin{align}
\frac{d}{dt}\int_\Omega\, u(x, t)\, dx &= 
\int_\Omega u_t(x, t)\, dx \\
&= \int_\Omega\, \Delta u (x, t)\, dx \\
&= \int_{\partial\Omega}\, \frac{\partial u}{\partial n}(x, t)\, dS(x) \\
&= 0.
\end{align}
