Gronwall's Inequality

Theorem 1 (Gronwall's Inequality): Let $r$ be a nonnegative, continuous, real-valued function on the interval $J = [a, b]$ and let $\delta, k \geq 0$ such that $\displaystyle{r(t) \leq \delta + \int_a^t kr(s) \: ds}$ for all $t \in J$. Then $\displaystyle{r(t) \leq \delta e^{k(t-a)}}$.

• Proof: Let $r$ be a nonnegative, continuous, real-valued function on $J = [a, b]$ with the property that there exists a $\delta, k \geq 0$ such that:

(1)

\begin{align} \quad r(t) \leq \delta + \int_a^t kr(s) \: ds \end{align}

• Define a new function $R$ on $J$ by:

(2)

\begin{align} \quad R(t) = \delta + \int_a^t kr(s) \: ds \end{align}

• Then $r(t) \leq R(t)$. Since $k > 0$ we have that $kr(t) \leq kR(t)$. So $kr(t) - R'(t) \leq kR(t) - R'(t)$. Multiplying both sides of this inequality by $-1$ yields and using the Fundamental theorem of Calculus to compute $R'(t)$ yields:

(3)

\begin{align} \quad R'(t) - kR(t) \leq R'(t) - kr(t) = \frac{d}{dt} \left ( \delta + \int_a^t kr(s) \: ds \right ) - kr(t) = kr(t) - kr(t) = 0 \end{align}

• Therefore:

(4)

\begin{align} \quad R'(t) - kR(t) \leq 0 \end{align}

• Now multiply both sides of this inequality by $\displaystyle{e^{k(a - t)} > 0}$ to get:

(5)

\begin{align} \quad e^{k(a - t)}R'(t) - e^{k(a - t)}kR(t) \leq 0 \quad (*) \end{align}

• Now consider the function $e^{k(a - t)}R(t)$. We compute the derivative of this function with respect to $t$ using the product rule for differentiation to get:

(6)

\begin{align} \quad \left [ e^{k(a-t)} R(t) \right ]' &= e^{k(a - t)}R'(t) -ke^{k(a - t)} R(t) \\ &= e^{k(a - t)} R'(t) - e^{k(a - t)}kR(t) \quad (**) \end{align}

• Comparing $(*)$ and $(**)$ we conclude that:

(7)

\begin{align} \quad \left [ e^{k(a - t)} R(t) \right ]' \leq 0 \end{align}

• We integrate both sides of the inequality above from $a$ to $t$ and use the Fundamental theorem of Calculus yet again to get:

(8)

\begin{align} \quad \int_a^t \left [e^{k(a - s)} R(s) \right ]' \: ds & \leq 0 \\ e^{k(a - t)} R(t) - e^{k(a - a)} R(a) & \leq 0 \\ e^{k(a - t)} R(t) - R(a) & \leq 0 \end{align}

• Since $\displaystyle{R(t) = \delta + \int_a^t kr(s) \: ds}$ we have that $R(a) = \delta$, and so:

(9)

\begin{align} \quad e^{k(a - t)} R(t) - \delta & \leq 0 \\ \quad e^{k(a - t)} R(t) & \leq \delta \\ \quad R(t) \leq & \delta e^{k(t - a)} \end{align}

• Since $r(t) \leq R(t)$ we conclude that for all $t \in J$:

(10)

\begin{align} \quad r(t) \leq \delta e^{k(t - a)} \quad \blacksquare \end{align}