Gronwall’s Inequality in SDE Estimates

Why Gronwall’s Inequality Appears in SDE Estimates

An estimate for an SDE solution often controls its size by an integral of its earlier size. The quantity to be bounded is then not isolated: a bound on its value at time $t$ involves values at every $s\le t$. Gronwall’s inequality converts this kind of recursive integral bound into an explicit exponential bound. The constants matter: in particular, an additive constant in an estimate must be retained when applying the inequality.

We first prove the integral inequality in the form used below. All functions in its statement are deterministic. Later, the nonnegative function will be an expectation such as $\mathbb{E}|X_t|^2$, and the inequality will hold for every $t$ in a fixed finite interval.

Theorem (Integral Gronwall inequality): Let $T>0$, let $a,b\ge0$, and let $u:[0,T]\to[0,\infty)$ be continuous. If $$ u(t)\le a+b\int_0^t u(s)\,ds,\qquad 0\le t\le T, $$ then $u(t)\le ae^{bt}$ for every $t\in[0,T]$.

Proof. Define $U(t)=a+b\int_0^t u(s)\,ds$. The assumed inequality gives $0\le u(t)\le U(t)$; in particular, $U\ge0$. Continuity of $u$ implies that $U$ is differentiable and $U'(t)=bu(t)\le bU(t)$. Therefore \[ \frac{d}{dt}\big(e^{-bt}U(t)\big)=e^{-bt}\big(U'(t)-bU(t)\big)\le0. \] Thus $e^{-bt}U(t)\le U(0)=a$, so $U(t)\le ae^{bt}$. Since $u(t)\le U(t)$, the desired bound follows. This argument includes $a=0$: then $0\le U(t)\le0$, so $u(t)=0$. It also includes $b=0$, when the original inequality itself says $u(t)\le a$. $\square$

A Version with a Forcing Term

Some estimates contain a known, time-dependent contribution in addition to the term involving $u$. The following form is useful when that contribution is integrable. Here $f$ is a nonnegative Lebesgue-integrable function on $[0,T]$; the inequality is required at every $t$, and $u$ is continuous and nonnegative.

Corollary: If $$ u(t)\le a+\int_0^t f(s)\,ds+b\int_0^t u(s)\,ds, $$ where $a,b\ge0$, then $$ u(t)\le e^{bt}\left(a+\int_0^t f(s)\,ds\right),\qquad 0\le t\le T. $$

Proof. Put $F(t)=a+\int_0^t f(s)\,ds$ and $v(t)=F(t)+b\int_0^t u(s)\,ds$. The assumed inequality gives $u\le v$. The function $v$ is absolutely continuous, and for almost every $t$, \[ v'(t)=f(t)+bu(t)\le f(t)+bv(t). \] Multiplying by $e^{-bt}$ and integrating from $0$ to $t$ gives \[ v(t)\le e^{bt}\left(a+\int_0^t e^{-bs}f(s)\,ds\right) \le e^{bt}\left(a+\int_0^t f(s)\,ds\right), \] where the last inequality uses $b,s\ge0$ and $f(s)\ge0$. Since $u\le v$, this proves the claim. If $a=0$ and $f=0$ almost everywhere, the same argument gives $u=0$; no positivity assumption on $a$ is needed. $\square$

Worked Example: A Bound with Constant Coefficients

Suppose $u:[0,1]\to[0,\infty)$ is continuous and $u(t)\le2+3\int_0^t u(s)\,ds$. Find a bound on the whole interval, and check that it is attained by a function satisfying equality.

Here $a=2$ and $b=3$, so the theorem gives $u(t)\le2e^{3t}$ for $0\le t\le1$. To check attainment, set $u_*(t)=2e^{3t}$. Direct integration gives \[ 2+3\int_0^t 2e^{3s}\,ds =2+3\cdot\frac{2}{3}(e^{3t}-1) =2e^{3t} =u_*(t). \] Thus $u_*$ satisfies the integral inequality with equality at every time. At $t=0$, both sides are $2$; at $t=1$, the bound is $2e^3$, which is also $u_*(1)$. This check confirms both the constants and the exponential rate.

Worked Example: A Time-Dependent Forcing Term

Suppose $u$ is continuous and nonnegative on $[0,1]$, and $u(t)\le2+\int_0^t s\,ds+\int_0^t u(s)\,ds$. Apply the forcing-term bound, and check it against the function for which this inequality is an equality.

Take $a=2$, $b=1$, and $f(s)=s$. Since $\int_0^t s\,ds=t^2/2$, the corollary gives \[ u(t)\le e^t\left(2+\frac{t^2}{2}\right). \] For a check, equality in the integral inequality corresponds to $u(0)=2$ and $u'(t)=t+u(t)$. The function $u_*(t)=3e^t-t-1$ has $u_*(0)=2$ and \[ u_*'(t)=3e^t-1=t+(3e^t-t-1)=t+u_*(t). \] Integrating this differential identity from $0$ to $t$ verifies $u_*(t)=2+\int_0^t s\,ds+\int_0^t u_*(s)\,ds$. It is nonnegative on $[0,1]$, since $u_*(0)=2$ and $u_*'(t)=3e^t-1>0$ there. At $t=1$, the stated bound is $5e/2$, while $u_*(1)=3e-2$; their difference is $2-e/2>0$, since $e<3<4$. The forcing-term estimate therefore bounds this equality case as claimed.

Applying Gronwall to an SDE Moment Estimate

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Let $X$ be an adapted continuous solution on $[0,T]$ of \[ X_t=\xi+\int_0^t\mu(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\,dB_s. \] Assume $\xi$ is $\mathcal{F}_0$-measurable with $\mathbb{E}|\xi|^2<\infty$, and assume $\mathbb{E}\sup_{0\le s\le T}|X_s|^2<\infty$. The stochastic integral here is an Itô integral: its integrand $H_s=\sigma(s,X_s)$ is progressively measurable and belongs to $\mathcal{L}^2$, the class satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. This integrability follows from the growth condition below and the assumed bound on $\mathbb{E}\sup_{s\le T}|X_s|^2$. The drift integral is an ordinary time integral.

Assume the uniform linear growth condition \[ |\mu(t,x)|^2+|\sigma(t,x)|^2\le K^2(1+|x|^2) \] for some $K\ge0$, all $t\in[0,T]$, and all $x\in\mathbb{R}$. A global Lipschitz condition, when used for existence or uniqueness, has the form \[ |\mu(t,x)-\mu(t,y)|^2+|\sigma(t,x)-\sigma(t,y)|^2\le L^2|x-y|^2. \] The moment estimate below uses linear growth, not the Lipschitz condition. It assumes a solution is already given and makes no claim about strong versus weak existence or uniqueness.

Proposition: Under these assumptions, for every $0\le t\le T$, $$ \mathbb{E}|X_t|^2\le \bigl(1+3\mathbb{E}|\xi|^2\bigr) \exp\!\bigl(3K^2(T+1)t\bigr)-1. $$

Proof. Fix $t\in[0,T]$, and write $D_t=\int_0^t\mu(s,X_s)\,ds$ and $M_t=\int_0^t\sigma(s,X_s)\,dB_s$. The integrability assumptions make these quantities well-defined. By Cauchy–Schwarz for the time integral and the Itô isometry for the stochastic integral, \[ \mathbb{E}|D_t|^2\le t\int_0^t\mathbb{E}|\mu(s,X_s)|^2\,ds \le T\int_0^t\mathbb{E}|\mu(s,X_s)|^2\,ds, \qquad \mathbb{E}|M_t|^2=\int_0^t\mathbb{E}|\sigma(s,X_s)|^2\,ds. \] For real numbers $x,y,z$, Cauchy–Schwarz gives $|x+y+z|^2\le3(|x|^2+|y|^2+|z|^2)$. Apply this to $X_t=\xi+D_t+M_t$, take expectations, and use the two bounds above. If $m(t)=\mathbb{E}|X_t|^2$, the growth condition yields \[ m(t)\le3\mathbb{E}|\xi|^2+ 3(T+1)K^2\int_0^t\mathbb{E}(1+|X_s|^2)\,ds =3\mathbb{E}|\xi|^2+C\int_0^t(1+m(s))\,ds, \] where $C=3K^2(T+1)$. Now set $u(t)=1+m(t)$. Adding $1$ to both sides gives \[ u(t)\le1+3\mathbb{E}|\xi|^2+C\int_0^t u(s)\,ds. \] The function $u$ is continuous: $X$ has continuous paths and the assumed square-integrable supremum provides an integrable bound for $|X_t|^2$, so dominated convergence applies as $t$ varies. The integral Gronwall inequality, with $a=1+3\mathbb{E}|\xi|^2$ and $b=C$, gives $u(t)\le(1+3\mathbb{E}|\xi|^2)e^{Ct}$. Subtracting $1$ proves the stated result. If $K=0$, then $C=0$, and the same argument gives $u(t)\le1+3\mathbb{E}|\xi|^2$, as the formula states. $\square$

Worked Example: Bounding the Second Moment of a Brownian SDE

Consider $dX_t=dB_t$, $X_0=1$, on $[0,1]$. Apply the proposition with a valid growth constant, then check the estimate against the exact second moment.

Here $\mu(t,x)=0$ and $\sigma(t,x)=1$, so $|\mu(t,x)|^2+|\sigma(t,x)|^2=1\le1+|x|^2$; thus $K=1$ is valid. Also $\mathbb{E}|\xi|^2=1$. The proposition gives, for $0\le t\le1$, \[ \mathbb{E}|X_t|^2\le(1+3\cdot1)e^{3\cdot1^2(1+1)t}-1 =4e^{6t}-1. \] In particular, at $t=1$ the bound is $4e^6-1$. For an independent check, the SDE itself gives $X_t=1+B_t$. Since $B_t$ has mean zero and variance $t$, $\mathbb{E}|X_t|^2=1+t$; at $t=1$, this is $2$. The claimed estimate holds because $4e^6-1>3>2$. The exponential estimate is deliberately coarse: its purpose is to give a valid bound from the growth assumption, not an exact moment.

A Common Mistake: Losing the Additive Constant

A frequent error is to apply Gronwall directly to $m(t)=\mathbb{E}|X_t|^2$ after obtaining an inequality involving $1+m(s)$. For example, in the Brownian example, the estimate derived above has the form \[ m(t)\le 3\mathbb{E}|\xi|^2+C\int_0^t(1+m(s))\,ds. \] It is not valid to replace the right-hand side by $3\mathbb{E}|\xi|^2+C\int_0^t m(s)\,ds$: the integral of the $1$ has been deleted.

For a concrete test of that mistaken deletion, take $X_t=B_t$, so $m(t)=t$, and use $K=1,T=1$, hence $C=6$. If one incorrectly asserted $m(t)\le6\int_0^t m(s)\,ds$, then at $t=0.1$ the right-hand side would be $6\int_0^{0.1}s\,ds=0.03$, while $m(0.1)=0.1$. The asserted inequality fails. The correction is to define $u(t)=1+m(t)$, so the constant inside the integral becomes part of the function being bounded. The resulting valid estimate is $u(t)\le(1+3\mathbb{E}|\xi|^2)e^{Ct}$; the prefactor includes the $1$ added to $m$.

How to Use the Estimate

The deterministic inequality and its SDE application are useful whenever a nonnegative continuous quantity is controlled by its own time integral. The method has three practical steps:

Write the complete inequality, keeping additive constants and any forcing terms.
Choose a nonnegative function $u$ that absorbs terms such as $1+m(t)$, and identify the constants $a$ and $b$.
Apply the proved inequality only after checking continuity and finiteness of the quantities involved.

In an SDE argument, the source of the integral inequality must also be justified. Here the drift contribution was bounded by Cauchy–Schwarz, the diffusion contribution by the Itô isometry, and the coefficient terms by linear growth. Gronwall supplies the final exponential bound; it does not replace those preceding estimates or prove that a solution exists.

Check Your Understanding

Use the stated inequalities and verify the constants in each application.

If $u(t)\le5+2\int_0^t u(s)\,ds$, what bound follows at time $t$, including when the initial constant is positive?
For $u(t)\le1+\int_0^t s\,ds+4\int_0^t u(s)\,ds$ on $[0,T]$, identify $a$, $b$, and $f$, and write the forcing-term bound.
In the SDE proposition, explain why the function $u(t)=1+\mathbb{E}|X_t|^2$, rather than just $\mathbb{E}|X_t|^2$, absorbs the term $1+|X_s|^2$.
Which integrability condition on $\sigma(s,X_s)$ is required for the Itô isometry used in the moment estimate?
For $dX_t=dB_t$, $X_0=1$, compare the exact moment $1+t$ with the bound $4e^{6t}-1$ on $[0,1]$.