Let \(u\) be differentiable and suppose \[ u'(t)\le a\,u(t),\qquad t\ge0, \qquad u(0)\le b, \] where \(a,b>0\). Starting from first principles, derive an upper bound for \(u(t)\) for \(t>0\).
Let \(x:\mathbb R\to\mathbb R^n\) satisfy \[ x'(t)=f(x(t)), \] where \(f:\mathbb R^n\to\mathbb R^n\) is continuous and \[ \langle f(y),y\rangle\le\|y\|^2 \] for every \(y\in\mathbb R^n\). Derive an inequality showing that \(\|x(t)\|\) grows at most exponentially.
Solution
1. In Part (1),\[ \boxed{u(t)\le b e^{at}} \qquad (t\ge0). \]
Define \[ v(t)\coloneqq e^{-at}u(t). \] By the product rule, \[ \begin{aligned} v'(t) &= -ae^{-at}u(t)+e^{-at}u'(t)\\ &= e^{-at}\bigl(u'(t)-au(t)\bigr) \le0. \end{aligned} \] Hence \(v\) is nonincreasing on \([0,\infty)\), so \[ e^{-at}u(t) = v(t) \le v(0) = u(0) \le b. \] Multiplying by \(e^{at}>0\) gives \[ u(t)\le be^{at}. \]
2. For the solution in Part (2), the squared norm\[ U(t)\coloneqq\norm{x(t)}^2 \]satisfies\[ U'(t)\le2U(t). \]
Using the differential equation, \[ \begin{aligned} U'(t) &= 2\inner{x'(t)}{x(t)}\\ &= 2\inner{f(x(t))}{x(t)}\\ &\le 2\norm{x(t)}^2\\ &= 2U(t), \end{aligned} \] by the assumed inequality for \(f\).
3. For every \(t\ge0\),\[ \boxed{ \norm{x(t)} \le e^t\norm{x(0)} }. \]
Apply step 1 to \(U\) with \[ a=2, \qquad b=U(0)=\norm{x(0)}^2. \] This gives \[ \norm{x(t)}^2 = U(t) \le U(0)e^{2t} = \norm{x(0)}^2e^{2t}. \] Both sides are nonnegative, so taking square roots yields \[ \norm{x(t)} \le e^t\norm{x(0)}. \]
4. Q.E.D.
Step 1 gives the scalar exponential upper bound, and step 3 gives the requested exponential-growth inequality for the vector solution.