Solution. By induction, if \(\phi _k = \sum _{j \leq k}t^j/j!\) then \(\phi _{k+1} = 1 + \int _0^t\phi _k = 1 + \sum _{j \leq k}t^{j+1}/(j + 1)!\), the next partial sum. On \(R\), \(M = 1 + b\) (the maximum of \(|x|\) for \(|x - 1| \leq b\)), so \(\alpha = \min (a, b/(1 + b))\), which is less than \(1\) for every \(b\). The theorem guarantees existence only on \(|t| < 1\), although the solution \(e^t\) exists for all \(t\); global existence for linear equations is proved in the next section.

Solution. \(M = \max (t^2 + x^2) = 2\), and \(L = \max |\partial f/\partial x| = \max |2x| = 2\). Then \(\alpha = \min (1, 1/2) = \frac 12\). On \(|t| \leq \frac 12\), \(L|t| \leq 1\), and the error bound is \(\frac ML\frac {1}{(k + 1)!}e = \frac {e}{(k + 1)!}\). Since \(9! = 362\,880\) gives \(7.5 \times 10^{-6}\) and \(10! = 3\,628\,800\) gives \(7.5 \times 10^{-7}\), \(k + 1 = 10\), that is \(k = 9\) iterates, suffice.

Solution. \(x \equiv 0\) is a solution, and so is \(x(t) = 0\) for \(t \leq c\), \((t - c)^2/4\) for \(t > c\), for every \(c \geq 0\): on \(t > c\), \(x' = (t - c)/2 = \sqrt {x}\). Near \(x = 1\), \(\sqrt {|x|}\) is continuously differentiable, so it is Lipschitz on a rectangle around \((0, 1)\) (Lemma 2.2.2) and the theorem applies. The failure at \(x = 0\) comes from \(\frac {d}{dx}\sqrt x = \frac {1}{2\sqrt x}\) being unbounded there.

Solution. \(x(t) = x(t_0) + \int _{t_0}^tA(s)x(s)\,ds\), so \(u = \|x\|\) satisfies \(u(t) \leq \|x(t_0)\| + \int _{t_0}^t\|A(s)\|u(s)\,ds\). The general form of Theorem 2.4.1, with \(c \equiv \|x(t_0)\|\) and \(k = \|A\|\), gives the bound. In particular a solution of a linear system cannot blow up on a bounded interval, which is the key to global existence in the next section.

Solution. \(L = 1\). By Theorem 2.4.2 with \(\varepsilon = 0\), \(|x(t) - y(t)| \leq 0.01e^t \leq 0.01e^2 = 0.074\) on \([0, 2]\). With equal initial values and \(\varepsilon = 10^{-8}\), the error at \(t = 2\) is at most \(10^{-8}(e^2 - 1) = 6.4 \times 10^{-8}\).

Solution. From \(\phi _0 = 0\), every iterate is \(0\). From \(\phi _0 = t^3\), \(\phi _1 = \int _0^t3(s^3)^{2/3}\,ds = \int _0^t3s^2\,ds = t^3\), so \(t^3\) is also a fixed point. A contraction has only one fixed point, so the integral operator is not a contraction in any metric of the kind used; the reason is the failure of the Lipschitz condition at \(y = 0\) (Example 2.2.3(c)).

Solution. For \(t \geq t_0\), \begin {align*} \|T\phi (t) - T\psi (t)\| &\leq L\int _{t_0}^te^{2L(s - t_0)}e^{-2L(s - t_0)}\|\phi (s) - \psi (s)\|\,ds\\ &\leq L\|\phi - \psi \|_\ast \int _{t_0}^te^{2L(s - t_0)}\,ds \leq \frac 12\|\phi - \psi \|_\ast e^{2L(t - t_0)} . \end {align*}

Multiplying by \(e^{-2L(t - t_0)}\) and taking the supremum gives the result; \(t < t_0\) is the same. Since \(\|\cdot \|_\ast \) lies between \(e^{-2L\alpha }\|\cdot \|_\infty \) and \(\|\cdot \|_\infty \), the space is still complete, and Banach’s theorem gives a unique fixed point on the whole interval \(J\), with no restriction \(L\alpha < 1\).