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I'm reading a paper studying the following IVP of a density $\rho: \mathbb{R}^{n} \times \mathbb{R}^{\geq 0} \rightarrow \mathbb{R}$

$\rho_t + \nabla \cdot (\rho v) =0$

$\rho(\alpha,0) \equiv \rho_0 (\alpha)$

where $v = k \ast \rho$ and $k: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is a radially symmetric, compactly supported smooth function (the convolution is performed component wise).

It states that its easy to show that given radially symmetric initial conditions, the solution will remain radially symmetric. Of course, I have not found this so easy.

user61387
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1 Answers1

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The standard approach is to show that if $\rho$ is a solution and $T:\mathbb R^n\to\mathbb R^n$ is an orthogonal linear transformation, then $\rho\circ T$ is also a solution. The symmetry follows from the fact that the solution for given initial condition $\rho_0$ is unique: since $\rho_0\circ T=\rho_0$, we have $\rho\circ T=\rho$.

So, all we do it plug $\tilde \rho = \rho\circ T$ into the equation and check that everything commutes with the composition operator. First, $\tilde \rho_t=\rho_t\circ T$. Then $$ \tilde v(x) = \int k(y)\tilde \rho(x-y)\,dy=\int k(y) \rho(Tx-Ty)\,dy = \int k(\tilde y) \rho(Tx-\tilde y)\,d\tilde y = v(Tx) $$ where I changed the variable $\tilde y=Ty$ and took advantage of orthogonality $d\tilde y=dy$ and radial symmetry $k(Ty)=k(y)$. Finally, $\nabla \cdot((\rho v)\circ T) = (\nabla \cdot(\rho v))\circ T$ because divergence is cool like that. Everything checks out.

  • Actually, I realized I made a mistake. The convolution kernel is of the form $k(r) \vec{x}$. Luckily, your method works perfectly with only a slight modification. Much obliged! – user61387 Feb 09 '13 at 07:32