# $L^1$ estimate of the heat equation

The classical heat equation read $$\partial_tu-\Delta u=0$$ in $$\Bbb R^3$$. If $$u$$ is sufficintly smooth, then we can multiply both sides by $$|u|^{p-2}u$$ to get the $$L^p$$ estimate, where $$1. That is, $$\frac{1}{p}\frac{d}{dt}||u||_{L^p}^p+c(p)||\nabla |u|^\frac{p}{2}||_{L^2}^2=0.$$

What about the $$L^1$$ case? I have not seen this in a book or paper.

Maybe we can use the heat kernel to do such one. But can we just do $$L^1$$ estimate on the solution. In fact, when considering axiysmmetric Navier-Stokes equations, the swirl component of the voriticity divides by $$r$$: $$\Omega=\omega/r$$ satisfies $$\partial_t\Omega+(u^r\partial_r+u^z\partial_z)\Omega-(\Delta+\frac{2}{r}\partial_r)\Omega=f$$ for some $$f$$. In this case, how can we do $$L^1$$ estimate?

• Welcome to MathOverflow! Could you please add a few more details to your question, such as: (i) Are you interested in the heat equation on the whole space $\mathbb{R}^n$ or an a domain $\Omega \subseteq \mathbb{R^n}$? (ii) In case that you are interested in the heat equation on a domain, do you assume the domain to be bounded? And what are the boundary conditions you impose? (iii) What exactly do you mean by an $L^1$-estimate''? Simply an (optimal) upper bound for the $L^1$-norm of the solution $u$? Jun 22 '19 at 9:02
• Thank you for the clarifications! If the initial value of the heat equation fulfils $u(0) \ge 0$, then $u(t) \ge 0$ for all times $t \ge 0$ (if you assume the solution to be in $L^1$ for all times $t$). Hence, by integrating against the constant function $1$ you get that $\frac{d}{dt} \|u(t)\|_1 = \langle 1, \frac{d}{dt} u(t) \rangle = \langle 1, \Delta u(t) \rangle = 0$, which shows that $\|u(t)\|_1 = \|u(0)\|_1$ for all times if the initial value is non-negative. Jun 22 '19 at 12:49
• Singular integrals (such that the Leray projection on vector fields with null divergence) are bounded on $L^p$ for $1<p<\infty$ but not on $L^1$ nor on $L^\infty$. Mar 26 '21 at 19:44