on Grönwall's inequality

Overview

This post explores various extensions and applications of Grönwall’s inequality, moving from the classic integral form to operator-theoretic perspectives and nonlinear transport problems.

🏷️ Introduction

We start with a simple application of Grönwall’s inequality.

Grönwall's inequality

If the following relation holds:

$$0\le f(x)\le C{\int }_0^{x}f(s)ds,x\in [0,A],$$

then $f(x)=0$.

The traditional way is letting $h(x)=e^{-Cx}{\int }_0^{x}f(s)ds$, and utilize the relation:

$$h^{\prime }(x)=e^{-Cx}\left(f(x)-C{\int }_0^{x}f(s)ds\right)\ge 0.$$

Here, we attempt with an alternative way. We define the integral operator $\mathcal{A}:L_{+}^{\infty }[0,A]\mapsto L_{+}^{\infty }[0,A]$ by the following:

$$\mathcal{A}f:=C{\int }_0^{x}f(s)ds.$$

Then $\mathcal{A}$ is a positive and compact operator. According to the Krein-Rutman theorem, if its spectral radius $r(\mathcal{A})$ is a positive eigenvalue, there must exist a corresponding eigenfunction $\varphi (x)>0$ strictly positive. However, the definition of $\mathcal{A}$ implies that $\mathcal{A}\varphi (0)=0$, which gives a contradiction. Therefore, the spectral radius $r(\mathcal{A})=0$.

Observe that $0\le f(x)\le \mathcal{A}f(x)$ implies $0\le f(x)\le {\mathcal{A}}^{n}f(x)$ for all $n\in \mathbb{N}$. Gelfand’s formula implies $\Vert {\mathcal{A}}^k{\Vert }_{op}\to 0$ as $k\to \infty$, hence $f(x)=0$ by taking the limit.

🏷️ Extension

If there is another operator $\mathcal{B}$ that commutes with $\mathcal{A}$, then the spectral radius $r(\mathcal{A}+\mathcal{B})$ can be estimated. We consider the abstract problem as follows.

Extension of Grönwall's inequality

Let $\mathcal{B}$ be a linear positive operator with $r(B)<1$ that commutes with $\mathcal{A}$, and satisfy:

$$0\le f(x)\le \mathcal{A}f+\mathcal{B}f,$$

then $f(x)=0$.

The commutativity implies an estimate $r(\mathcal{A}+\mathcal{B})\le r(\mathcal{A})+r(\mathcal{B})<1$. Then the same argument holds since $(\mathcal{A}+\mathcal{B}{)}^k\to 0$ as $k\to \infty$.

🏷️ Background of Transport Equation

The transport equation describes the dynamics of radiative particles interacting with the environment. For a homogeneous medium, the governing equation is:

$$\begin{array}{rl}v\cdot \nabla u(x,v)+{\sigma }_{a}u& ={\sigma }_s(\mathcal{K}u-u)\text{ in }D\times {\mathbb{S}}^{d-1},\\ u{|}_{{\Gamma }_{-}}& =h(x,v),\end{array}$$

where $\mathcal{K}$ is the scattering operator and $h(x,v)$ is the source on the incoming boundary ${\Gamma }_{-}$:

$${\Gamma }_{-}=\left\{(x,v)\in \partial D\times {\mathbb{S}}^{d-1}\mid v\cdot n(x)<0\right\}.$$

🏷️ Cone-beam Source

Definition: Cone-beam source

If $h(x,v)$ satisfies:

• $h$ is non-negative;
• There exists a set $V\subset {\mathbb{S}}^{d-1}$ such that $\mathrm{supp}h\subset \partial D\times V\cap {\Gamma }_{-}$.

Then $h$ is called a cone-beam source.

In the special case ${\sigma }_s=0$, the solution can be solved directly:

$$u(x,v)=h(x-{\tau }_{-}(x,v)v,v)E(x,x-{\tau }_{-}(x,v)v),$$

where $E(x,x-sv)=\exp \left(-{\int }_0^s{\sigma }_a(x-tv)dt\right)$.

🏷️ Nonlinearity

In practical scenarios, ${\sigma }_a$ may depend on the solution’s flux, leading to a nonlinear equation:

$$u(x,v)=h(x-{\tau }_{-}(x,v)v,v)\exp \left(-{\int }_0^{{\tau }_{-}(x,v)}{\sigma }_a(x-tv,⟨u⟩)dt\right),$$

where $⟨u⟩={\int }_{{\mathbb{S}}^{d-1}}u(x,v)dv$. We assume Lipschitz continuity:

Lipschitz Assumption

$$|{\sigma }_a(x,⟨u⟩)-{\sigma }_a(x,⟨w⟩)|\le L|⟨u⟩(x)-⟨w⟩(x)|$$

Let $f:=⟨u-w⟩$. Integrating over ${\mathbb{S}}^{d-1}$ and using the inequality $|e^{-a}-e^{-b}|\le e^{-c}|a-b|$ for $a,b\ge c\ge 0$, we find:

$$|f(x)|\le \overline{h}{\int }_V{\int }_0^{{\tau }_{-}(x,v)}|{\sigma }_a(x-tv,⟨u⟩)-{\sigma }_a(x-tv,⟨w⟩)|dtdv\le C{\int }_V{\int }_0^{{\tau }_{-}(x,v)}|f(x-tv)|dtdv,$$

where $\overline{h}={\sup }_{{\Gamma }_{-}}h$. This is an analog of the classic Grönwall inequality, forcing $f(x)=0$.

🏷️ Isotropic Scattering

With constant scattering ${\sigma }_s>0$, and ${\sigma }_t:={\sigma }_a+{\sigma }_s$, the solution satisfies:

$$\begin{array}{rl}u(x,v)=& h(x-{\tau }_{-}(x,v)v,v)\exp \left(-{\int }_0^{{\tau }_{-}(x,v)}{\sigma }_t(x-tv,⟨u⟩)dt\right)\\ & +{\int }_0^{{\tau }_{-}(x,v)}\exp \left(-{\int }_0^s{\sigma }_t(x-tv,⟨u⟩)dt\right){\sigma }_s⟨u⟩(x-sv)ds.\end{array}$$

Let $f:=⟨u-w⟩$. Integrating over ${\mathbb{S}}^{d-1}$ yields:

$$\begin{array}{rl}|f(x)|& \le C{\int }_V{\int }_0^{{\tau }_{-}(x,v)}|f(x-tv)|dtdv+\frac{1}{{\nu }_{d-1}}{\int }_D\frac{e^{-{\sigma }_s|x-y|}}{|x-y{|}^{d-1}}{\sigma }_s|f(y)|dy\\ & +{\int }_{{\mathbb{S}}^{d-1}}{\int }_0^{{\tau }_{-}(x,v)}{\int }_0^s{e}^{-s{\sigma }_s}L|f(x-tv)|{\sigma }_s⟨u⟩(x-sv)dtdsdv\\ & \le C{\int }_V{\int }_0^{{\tau }_{-}(x,v)}|f(x-tv)|dtdv+\frac{1+L\ell \overline{h}}{{\nu }_{d-1}}{\int }_D\frac{e^{-{\sigma }_s|x-y|}}{|x-y{|}^{d-1}}{\sigma }_s|f(y)|dy,\end{array}$$

where $\ell =\mathrm{diam}(D)$.

Commutativity Lemma

Let $\mathcal{A}$ and $\mathcal{S}$ be operators from $L^2(D)$ to $L^2(D)$:

$$\mathcal{A}f:={\int }_V{\int }_0^{{\tau }_{-}(x,v)}|f(x-tv)|dtdv,Sf:={\int }_D\frac{e^{-{\sigma }_s|x-y|}}{|x-y{|}^{d-1}}f(y)dy.$$

Then $\mathcal{A}$ commutes with $\mathcal{S}$.

This implies the relation:

$$|f(x)|\le C\mathcal{A}|f|+\frac{(1+L\ell \overline{h})}{{\nu }_{d-1}}{\sigma }_s\mathcal{S}|f|.$$

Uniqueness Theorem

If $(1+L\ell \overline{h})(1-e^{-{\sigma }_s\ell })<1$, then the cone-beam source permits a unique solution.

🏷️ Notes

• This uniqueness result serves as an exercise utilizing the operator-theoretic version of Grönwall’s inequality.
• Commutativity is a powerful tool for estimating spectral radii of operator sums (Zima, 1993).

🔗 See Also

• on Moser iteration --- Moser’s bootstrap uses a discrete “growth tracking” logic similar to Grönwall, applied to the scale of integrability exponents rather than time.

📚 References

🐻  Zima, M. 1993. A theorem on the spectral radius of the sum of two operators and its application. Bulletin of the Australian Mathematical Society 48(3), 427–434.