Hanqing Li Hanqing Li 李汉青


Gronwall inequalities turn integral or differential error estimates into explicit bounds. They are the basic mechanism for uniqueness, continuous dependence, global existence, and numerical convergence.

Integral Gronwall inequality

Theorem. Let \(u\ge0\) be continuous on \([t_0,T]\), let \(a\ge0\), and let \(b\ge0\) be integrable. If \[ u(t)\le a+\int_{t_0}^tb(s)u(s)\,\mathrm{d}s, \] then \[ u(t)\le a\exp\left(\int_{t_0}^tb(s)\,\mathrm{d}s\right). \]

Proof. Put \[ v(t)=a+\int_{t_0}^tb(s)u(s)\,\mathrm{d}s. \] Then \(u\le v\) and \[ v'(t)=b(t)u(t)\le b(t)v(t). \] Where \(v>0\), \((\log v)'\le b\). Integrate and use \(v(t_0)=a\). The case \(a=0\) follows by replacing \(a\) with \(\varepsilon\) and letting \(\varepsilon\downarrow0\).

Inhomogeneous form

Theorem. If \[ u(t)\le a(t)+\int_{t_0}^tb(s)u(s)\,\mathrm{d}s, \] where \(a\) is nonnegative and nondecreasing, then \[ u(t)\le a(t) \exp\left(\int_{t_0}^tb(s)\,\mathrm{d}s\right). \] More generally, variation of constants gives \[ u(t)\le a(t)+ \int_{t_0}^t a(s)b(s) \exp\left(\int_s^tb(r)\,\mathrm{d}r\right)\mathrm{d}s \] under suitable regularity.

Bellman--Gronwall kernel form

If \[ u(t)\le a(t)+\int_{t_0}^tk(t,s)u(s)\,\mathrm{d}s \] with a nonnegative Volterra kernel, repeated substitution produces a resolvent series. Fractional kernels \((t-s)^{\alpha-1}\) lead to Mittag--Leffler rather than exponential bounds. The ordinary exponential is special to a constant first-order accumulation rate.

Discrete Gronwall

Theorem. If \[ u_n\le a+\sum_{j=0}^{n-1}h_jb_ju_j, \qquad u_n,a,b_j,h_j\ge0, \] then \[ u_n\le a\prod_{j=0}^{n-1}(1+h_jb_j) \le a\exp\left(\sum_{j=0}^{n-1}h_jb_j\right). \]

Proof. Define \(v_n=a+\sum_{j<n}h_jb_ju_j\). Then \[ v_{n+1}\le(1+h_nb_n)v_n. \] Iterate and use \(1+x\le e^x\).

ODE stability estimate

If \(f\) is \(L\)-Lipschitz and \(x,y\) solve the same ODE with different initial conditions, then \[ \|x(t)-y(t)\|\le e^{L|t-t_0|}\|x_0-y_0\|. \] The bound can be sharpened using a one-sided Lipschitz constant \[ \langle f(x)-f(y),x-y\rangle \le\mu\|x-y\|^2, \] which yields \(e^{\mu t}\) and can reveal contraction even when \(L\) is large.