Exercises — Existence and uniqueness theorem — Picard-Lindelöf (statement)

Here $f(x,y)=3y+\sin x$. The starting point is $(x_0,y_0)=(0,2)$. $\frac{\partial f}{\partial y}=3.$ The $y$-derivative is the constant $3$, so $|f(x,y_1)-f(x,y_2)|=|3y_1-3y_2|=3|y_1-y_2|$. That is exactly the Lipschitz inequality with $L=3$, and it holds for all $(x,y)$. So $f$ is globally Lipschitz in $y$. (The $\sin x$ term depends only on $x$, so it cancels in the difference — Lipschitz is about $y$, not $x$.)

$f(x,y)=\sqrt{|y|}=|y{|}^{1/2}$ is continuous everywhere, so existence is fine (Peano, Peano existence theorem). But $\partial f/\partial y=\frac{1}{2}|y{|}^{-1/2}\to \infty$ as $y\to 0$, so $f$ is not Lipschitz in $y$ near $y=0$. Hence Picard–Lindelöf does not guarantee uniqueness. Existence still holds; uniqueness is the casualty.

The rectangle is $R=\{(x,y):|x|\le 1,|y|\le 2\}$ (since $x_0=y_0=0$).

• $M={\max }_R|x^2+y^2|=1^2+2^2=5$ (both terms maxed at the corner).
• ${\partial }_{y}f=2y$, so $L={\max }_R|2y|=2\cdot 2=4$.
• $h=\min \left(a,\frac{b}{M}\right)=\min \left(1,\frac{2}{5}\right)=\frac{2}{5}=\mathrm{0.4.}$ So a unique solution is guaranteed on $|x|\le 0.4$. (Compare parent example 1 with $b=1$: enlarging $b$ can shrink $h$ because $M$ grows faster — see the L2 trap.)

The rectangle here is $R=\{(x,y):|x|\le 10,|y-1|\le b\}$. On $|y-1|\le b$ the largest $|y|$ is $1+b$, so $M=(1+b{)}^2$. The $b/M$ constraint is $g(b)=\frac{b}{(1+b{)}^2}.$ Maximise: $g^{\prime }(b)=\frac{(1+b{)}^2-b\cdot 2(1+b)}{(1+b{)}^4}=\frac{(1+b)-2b}{(1+b{)}^3}=\frac{1-b}{(1+b{)}^3}.$ Setting $g^{\prime }(b)=0$ gives $b=1$. Why this is a maximum (not a min): the denominator $(1+b{)}^3>0$ for $b\ge 0$, so the sign of $g^{\prime }$ is the sign of the numerator $1-b$. For $0<b<1$ we have $1-b>0$ so $g^{\prime }>0$ (rising); for $b>1$ we have $1-b<0$ so $g^{\prime }<0$ (falling). A sign change $+\to -$ at $b=1$ is exactly a maximum. Then $M=(1+1{)}^2=4$ and $b/M=\frac{1}{4}$. Since $a=10\gg \frac{1}{4}$, the min is $b/M$, so $h_{\max }=\frac{1}{4}=0.25\text{at }b=1.$ Beauty check: the true solution $y=\frac{1}{1-x}$ blows up at $x=1$, so no finite rectangle can push $h$ past that. Our $h=0.25$ is a safe local guarantee, not the maximal interval.

$f(t,y)=2ty$, $y_0(x)=1$. $y_1=1+{\int }_0^{x}2t\cdot 1dt=1+x^2.$ $y_2=1+{\int }_0^{x}2t(1+t^2)dt=1+{\int }_0^x(2t+2t^3)dt=1+x^2+\frac{x^4}{2}.$ $y_3=1+{\int }_0^{x}2t\left(1+t^2+\frac{t^4}{2}\right)dt=1+{\int }_0^x\left(2t+2t^3+t^5\right)dt=1+x^2+\frac{x^4}{2}+\frac{x^6}{6}.$ The pattern is the partial sums of $\sum _{k=0}^{\infty }\frac{(x^2{)}^k}{k!}=e^{x^2}$. So the iterates converge to the unique solution $y=e^{x^2}$ — exactly what separation of variables gives ($\ln y=x^2$). This is Picard iteration (method of successive approximations) literally building the Taylor series.

$y\equiv 0$: then $y^{\prime }=0$ and $y^{2/3}=0$; both sides equal $0$, and $y(0)=0$. ✔ (valid for all $x$.) $y=(x/3{)}^3$: $y^{\prime }=3\cdot \frac{1}{3}{\left(\frac{x}{3}\right)}^2={\left(\frac{x}{3}\right)}^2$. Now check the right side for each sign of $x$:

• $x\ge 0$: then $(x/3{)}^3\ge 0$ and $y^{2/3}={\left[(x/3{)}^3\right]}^{2/3}=(x/3{)}^2$. Matches $y^{\prime }$. ✔
• $x<0$: then $(x/3{)}^3<0$. Reading $y^{2/3}$ as $((x/3{)}^3{)}^{2/3}=|x/3{|}^2=(x/3{)}^2$ (squaring kills the sign), it still equals $y^{\prime }=(x/3{)}^2$. So the same analytic formula solves the IVP on both sides — the cubic curve passes through the origin from below-left to above-right, and $y\equiv 0$ overlaps it there.

Two distinct solutions through the same point ⟹ uniqueness fails. Cause: ${\partial }_y(y^{2/3})=\frac{2}{3}{y}^{-1/3}\to \infty$ as $y\to 0$, so no finite $L$ near $y=0$.

The figure below plots both solution curves. The blue flat line is $y\equiv 0$; the orange curve is $y=(x/3{)}^3$ drawn for both $x<0$ and $x>0$. They cross at the red dot $(0,0)$ — that single shared point carrying two solutions is precisely the non-uniqueness the theorem forbids when Lipschitz holds. The gray arrow marks where ${\partial }_{y}f$ blows up (at $y=0$), the missing ingredient.

General uniqueness. Both solve the integral form, so subtracting, $y(x)-\tilde{y}(x)={\int }_{x_0}^x[f(t,y)-f(t,\tilde{y})]dt.$ Let $\varphi (x)=|y(x)-\tilde{y}(x)|\ge 0$. Take absolute values and use Lipschitz $|f(t,y)-f(t,\tilde{y})|\le L\varphi (t)$: $\varphi (x)\le L{\int }_{x_0}^x\varphi (t)dt.$ Grönwall (with zero constant term) gives $\varphi (x)\le 0\cdot e^{L(x-x_0)}=0$. Since $\varphi \ge 0$, $\varphi \equiv 0$, so $y\equiv \tilde{y}$. Uniqueness. ✔ Perturbation bound. For $f(x,y)=2y+g(x)$, $L=2$. Two solutions $y,\tilde{y}$ with $|y(x_0)-\tilde{y}(x_0)|=\epsilon$ obey the general Grönwall estimate $\varphi (x)\le \epsilon e^{L(x-x_0)}$. At $x=x_0+1$: $\varphi (x_0+1)\le \epsilon e^{2\cdot 1}=\epsilon e^2\approx 7.389\epsilon .$ (This is sharp: $y-\tilde{y}$ satisfies $(y-\tilde{y}{)}^{\prime }=2(y-\tilde{y})$, exactly $\epsilon e^{2(x-x_0)}$.)

For linear ODEs the Lipschitz constant is $L=|p|\le P$, a finite constant independent of $y$ — the box height $b$ never forces $L$ up. Why finite-time blow-up is the only obstruction (the continuation argument). Picard–Lindelöf first gives a local solution on a small interval. You can then restart the theorem at the right endpoint, using the value there as a new initial condition, extending the solution a bit further; repeat. This process either continues forever (covering the whole interval) or halts — and the only way it can halt at some finite $x^{\ast }$ is if the solution runs off to $\pm \infty$ as $x\to x^{\ast }$ (it escapes every box, so no new box can be built). In symbols: a maximal solution that stops at finite $x^{\ast }$ must have $|y(x)|\to \infty$ there (Maximal interval of existence). So if we can bound $|y|$ on every finite interval, blow-up is impossible and continuation never halts. Grönwall supplies exactly that bound: $|y(x)|\le (|y_0|+\int |q|)e^{P|x-x_0|}$, finite for every finite $x$. With no blow-up available, the continuation runs to the full domain where $p,q$ are defined. Contrast: $y^{\prime }=y^2$ is not globally Lipschitz (${\partial }_y=2y$ unbounded); Grönwall gives no such bound, and the solution $y=\frac{1}{1-x}$ from $y(0)=1$ genuinely satisfies $|y|\to \infty$ at $x=1$ — the continuation halts there. Unbounded ${\partial }_{y}f$ is what permits finite-time blow-up.

(a) On $R$, $|x|\le 1$ and $|y|\le 2$, so $M=\max |x+y|=1+2=3$. ${\partial }_{y}f=1\Rightarrow L=1$. Then $h=\min \left(1,\frac{1}{3}\right)=\frac{1}{3}.$ (b) Recall (from the reminders above) the Picard operator $T[u](x)=y_0+{\int }_{x_0}^{x}f(t,u(t))dt$, acting on continuous functions with the sup-norm $\Vert u\Vert ={\max }_{|x|\le h}|u(x)|$. Its contraction constant is $Lh$: for two candidates $u,v$, $\Vert T[u]-T[v]\Vert \le Lh\Vert u-v\Vert .$ Here $Lh=1\cdot \frac{1}{3}=\frac{1}{3}<1$, so $T$ shrinks distances — it is a genuine contraction in the sup-norm on $|x|\le \frac{1}{3}$, and the Banach fixed-point theorem directly gives a unique fixed point (the solution). (c) $y_0=1$. First iterate: $y_1=1+{\int }_0^x(t+y_0)dt=1+{\int }_0^x(t+1)dt=1+x+\frac{x^2}{2}.$ Second iterate: here $f(t,y_1(t))=t+y_1(t)=t+\left(1+t+\frac{t^2}{2}\right)$. Collecting like terms, the constant is $1$, the $t$-terms $t+t=2t$, and the quadratic is $\frac{t^2}{2}$, so the integrand is $1+2t+\frac{t^2}{2}$. Hence $y_2=1+{\int }_0^x\left(1+2t+\frac{t^2}{2}\right)dt=1+x+x^2+\frac{x^3}{6}.$ (d) Solve the linear ODE $y^{\prime }-y=x$. Integrating factor $e^{-x}$: $(y{e}^{-x}{)}^{\prime }=x{e}^{-x}$, so integrating the right side (by parts) $\int x{e}^{-x}dx=-(x+1)e^{-x}$, giving $y{e}^{-x}=-(x+1)e^{-x}+C$, i.e. $y=C{e}^x-x-1$. Apply $y(0)=1$: at $x=0$ the term $-x-1$ equals $-(0)-1=-1$, so $y(0)=C\cdot 1-1=C-1$. Setting $C-1=1$ gives $C=2$. Thus $y(x)=2e^x-x-1.$ At $x=h=\frac{1}{3}$: $y(\frac{1}{3})=2e^{1/3}-\frac{1}{3}-1=2e^{1/3}-\frac{4}{3}\approx \mathrm{1.4600.}$

The rectangle is $R=\{(x,y):|x|\le a,|y|\le b\}$ with $a$ large. On $|y|\le b$, $M=\max |1+y^2|=1+b^2$. Optimise $g(b)=\frac{b}{1+b^2}$: $g^{\prime }(b)=\frac{(1+b^2)-b\cdot 2b}{(1+b^2{)}^2}=\frac{1-b^2}{(1+b^2{)}^2}.$ Setting $g^{\prime }(b)=0$ gives $b^2=1$, so $b=1$ (taking $b>0$). Why a maximum: the denominator $(1+b^2{)}^2>0$, so $\mathrm{sign}{g}^{\prime }=\mathrm{sign}(1-b^2)$. For $0<b<1$, $1-b^2>0\Rightarrow g^{\prime }>0$ (rising); for $b>1$, $1-b^2<0\Rightarrow g^{\prime }<0$ (falling). Sign change $+\to -$ at $b=1$ ⟹ maximum. Then $M=1+1=2$ and $h=\frac{b}{M}=\frac{1}{2}=0.5$. Since $\pi /2\approx 1.5708$, our best local guarantee $h=0.5$ is well inside the true blow-up time — safe but conservative, as expected of a local theorem.

The figure below shows this geometrically. The dashed gray box is $R$ (with $b=1$). The green wedge is the pair of lines of slope $\pm M$ from the start — the solution is trapped inside it, and it exits the box's height exactly at $x=h=0.5$ (blue vertical line). The orange curve is the true solution $y=\tan x$, which keeps climbing and blows up only at the red dotted line $x=\pi /2\approx 1.57$. The gap between the blue and red lines is the difference between what the theorem guarantees and what actually happens.

::: answer) Which quantity bounds slope? ::: $M={\max }_R|f|$ Which quantity controls convergence and uniqueness? ::: the Lipschitz constant $L$ Formula for the guaranteed half-interval? ::: $h=\min (a,b/M)$ Continuity alone gives which conclusion? ::: existence only (Peano) Tool that turns "two solutions" into "one solution"? ::: Grönwall's inequality