--- title: Existence, Uniqueness, and Euler's Method module: First-Order Equations moduleNumber: 2 lessonNumber: 5 order: 205 summary: > Existence and uniqueness can be settled before any attempt to solve. The existence-uniqueness theorem gives sufficient conditions on f, and a standard example shows what fails when they do not hold. Picard's successive approximations build the solution as the limit of an iteration, and Euler's method turns the same tangent-line idea into a numerical procedure for the equations no formula reaches. topics: [First-Order Equations] draft: false sources: - book: Boyce ref: "§2.8 The Existence and Uniqueness Theorem; §2.7 Numerical Approximations: Euler's Method; §2.4 (Theorem 2.4.2)" - book: Simmons ref: "Ch. 13 §69 The Method of Successive Approximations; §70 Picard's Theorem" --- Most first-order initial value problems have no elementary solution, so two questions come before any attempt to solve one: whether a solution exists, and whether it is unique. Both matter: a model with no solution predicts nothing, and a model with several solutions predicts nothing definite. The existence-uniqueness theorem answers both under mild hypotheses, and its proof supplies a method — Picard iteration — that also underlies the numerical approach when no formula is available. ## The existence and uniqueness theorem For the general nonlinear problem the guarantee is local and rests on the continuity of $f$ together with its partial derivative in $y$. > **Theorem (Existence and uniqueness).** Let $f$ and $\partial f / \partial y$ be > continuous on a rectangle $\alpha < t < \beta$, $\gamma < y < \delta$ containing > $(t_0, y_0)$. Then there is an interval $|t - t_0| < h$ on which the initial > value problem > > $$ > y' = f(t, y), \qquad y(t_0) = y_0 > $$ > > has exactly one solution $y = \phi(t)$. Two features distinguish this from the [linear theorem](/differential-equations/first-order/linear-first-order-integrating-factors). First, it is _local_: the solution is promised only on some interval $|t - t_0| < h$, possibly small, and the theorem gives no estimate of $h$. Second, the hypotheses are _sufficient but not necessary_. Continuity of $f$ alone secures existence; uniqueness needs the extra control on $\partial f/\partial y$. When the equation is linear, $f = -p(t)y + g(t)$ has $\partial f/\partial y = -p(t)$, and the two continuity requirements reduce to continuity of $p$ and $g$, recovering the stronger global linear statement. A geometric corollary of uniqueness: two solution curves cannot intersect. A crossing point would be an initial condition satisfied by two different solutions, contradicting the theorem. This is what forbids solutions from touching an equilibrium in finite time in the [phase-line analysis](/differential-equations/first-order/autonomous-and-population-dynamics). ## When uniqueness fails The condition on $\partial f/\partial y$ cannot be dropped. Consider $$ y' = y^{1/3}, \qquad y(0) = 0. $$ Here $f = y^{1/3}$ is continuous everywhere, so a solution exists, but $\partial f/\partial y = \tfrac13 y^{-2/3}$ blows up at $y = 0$ — exactly the initial point. Uniqueness is not guaranteed, and indeed it fails. Separating variables gives $y = \left(\tfrac{2}{3}t\right)^{3/2}$ for $t \geq 0$, but the constant function $y \equiv 0$ is also a solution, as is its negative. Worse, for any $t_0 > 0$ the function that stays at zero until $t_0$ and then lifts off along $\pm\left(\tfrac23(t - t_0)\right)^{3/2}$ is a solution too. Infinitely many solution curves pass through the origin, fanning out into a **solution funnel**. $$ % caption: When the hypothesis on the y-derivative fails, uniqueness can fail: % infinitely many solutions of y' = y^{1/3} leave the origin, some immediately, % others after resting at zero for a while. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (-0.4,0) -- (5.2,0) node[right] {$t$}; \draw[->, black] (0,-1.8) -- (0,1.9) node[above] {$y$}; % the zero solution \draw[black, thick] (0,0) -- (5.0,0); % immediate lift-off, upper and lower \draw[thick] plot[domain=0:5, samples=50] (\x, {0.32*pow(\x,1.5)}); \draw[thick] plot[domain=0:5, samples=50] (\x, {-0.32*pow(\x,1.5)}); % delayed lift-off from t0 = 2.2 \draw[thick] plot[domain=2.2:5, samples=40] (\x, {0.32*pow(\x-2.2,1.5)}); \draw[thick] plot[domain=2.2:5, samples=40] (\x, {-0.32*pow(\x-2.2,1.5)}); \fill[acc] (0,0) circle (2pt); \node[acc, anchor=north east] at (0,-0.1) {$(0$, $0)$}; \node[black, anchor=south] at (3.6,0.02) {rest, then lift o\/f\/f}; \end{tikzpicture} $$ Away from the $t$-axis the derivative $\partial f/\partial y$ is finite and continuous, so through any point _not_ on the axis there is a unique solution. The failure is confined to where the hypothesis breaks. ## The interval of existence The number $h$ in the theorem is genuinely local, and its size can be estimated from a bound on $f$. Suppose $f$ and $\partial f/\partial y$ are continuous on the rectangle $R: |t| \le a$, $|y| \le b$ (translating the initial point to the origin), and let $M$ bound $|f|$ on $R$. Each iterate in the construction below has slope at most $M$ in magnitude, so its graph stays inside a bow-tie region of slopes $\pm M$ through the origin. That region remains inside $R$ as long as $|t| \le b/M$, giving $$ h = \min\!\left(a, \frac{b}{M}\right). $$ The solution is certain to exist at least on $|t| < h$. It may extend much further, but the theorem alone promises only this. $$ % caption: The existence rectangle R and the bow-tie of slope M through the % origin; the solution is guaranteed as far as the bow-tie stays inside R, which % fixes h = min(a, b/M). \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % rectangle R \draw[black, thick] (-3,-1.6) rectangle (3,1.6); \node[black, anchor=north east] at (3,1.6) {$R$}; \draw[black] (0,-1.6) node[below]{}; \node[black, anchor=north] at (3,-1.6) {$t = a$}; \node[black, anchor=east] at (-3,1.35) {$y = b$}; % axes \draw[black] (-3,0) -- (3,0); \draw[black] (0,-1.6) -- (0,1.6); % bow-tie: lines of slope M=0.8 through origin, shaded region \fill[acc!12] (0,0) -- (2,1.6) -- (-2,1.6) -- (0,0) -- (2,-1.6) -- (-2,-1.6) -- cycle; \draw[black, thick] (-2,-1.6) -- (2,1.6); \draw[black, thick] (-2,1.6) -- (2,-1.6); \draw[black, dashed] (2,-1.6) -- (2,1.6); \node[black, anchor=north] at (2,-1.65) {$t = \tfrac{b}{M}$}; \fill[black] (0,0) circle (1.8pt); \end{tikzpicture} $$ ## Picard's successive approximations The proof of existence is constructive. Integrating $y' = f(t,y)$ from the initial point converts the initial value problem $y' = f(t,y)$, $y(0) = 0$ into the equivalent **integral equation** $$ \phi(t) = \int_0^t f\big(s, \phi(s)\big)\,\d s. $$ Any solution of one solves the other. Picard's idea is to solve the integral equation by iteration: start from a guess $\phi_0(t) = 0$ and feed each approximation back through the integral to produce the next.[^simmons-picard] $$ \phi_{n+1}(t) = \int_0^t f\big(s, \phi_n(s)\big)\,\d s, \qquad n = 0, 1, 2, \dots $$ Each iterate satisfies the initial condition; none generally satisfies the equation, but the sequence converges to one that does. > **Worked example.** For $y' = 2t(1 + y)$, $y(0) = 0$, the integral equation is > $\phi(t) = \int_0^t 2s(1 + \phi(s))\,\d s$. Starting from $\phi_0 = 0$ and > feeding each iterate back through the integral, > > $$ > \phi_1 = \int_0^t 2s\,\d s = t^2, > \qquad > \phi_2 = \int_0^t 2s(1 + s^2)\,\d s = t^2 + \frac{t^4}{2}, > $$ > > $$ > \phi_3 = \int_0^t 2s\!\left(1 + s^2 + \frac{s^4}{2}\right)\d s = t^2 + \frac{t^4}{2} + \frac{t^6}{2\cdot 3}. > $$ > > By induction $\phi_n(t) = \sum_{k=1}^{n} \dfrac{t^{2k}}{k!}$, the partial sum of a > series converging for all $t$ with limit > > $$ > \phi(t) = \sum_{k=1}^{\infty}\frac{t^{2k}}{k!} = e^{t^2} - 1, > $$ > > the exact solution. The iterates agree with it on a widening interval as $n$ > grows. $$ % caption: Picard iterates for y' = 2t(1+y): each successive approximation tracks % the true solution over a wider interval, converging to it everywhere. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (-0.2,0) -- (3.0,0) node[right] {$t$}; \draw[->, black] (0,-0.2) -- (0,2.9) node[above] {$y$}; % true solution 0.5(e^{t^2}-1) \draw[acc, very thick] plot[domain=0:1.35, samples=60] (\x, {0.5*(exp(\x*\x)-1)}); \node[acc, anchor=east] at (1.32,2.4) {limit}; % phi_2 = t^2 + t^4/2 (closer to the limit) \draw[black, thick] plot[domain=0:1.35, samples=40] (\x, {0.5*(\x*\x + \x*\x*\x*\x/2)}); \node[black, anchor=west] at (1.4,1.55) {$n=2$}; % phi_1 = t^2 (farthest) \draw[black, thick] plot[domain=0:1.35, samples=40] (\x, {0.5*\x*\x}); \node[black, anchor=west] at (1.4,0.85) {$n=1$}; \end{tikzpicture} $$ The convergence proof rests on a **Lipschitz condition**: if $\partial f/\partial y$ is bounded by $K$ on the rectangle, then $|f(t, y_1) - f(t, y_2)| \le K|y_1 - y_2|$, which controls how fast successive iterates separate and forces the series of differences to converge. This same condition can replace continuity of $\partial f/\partial y$ in the theorem's hypotheses. ## Euler's method Picard iteration proves solutions exist but rarely computes them past a few steps. For actual numbers, Euler's method (1768) turns the tangent line into a stepping rule. The solution through $(t_0, y_0)$ has known slope $f(t_0, y_0)$ there, so its tangent line $$ y = y_0 + f(t_0, y_0)(t - t_0) $$ approximates it for a short step. Advancing to $t_1 = t_0 + h$ gives an approximate value $y_1$; from $(t_1, y_1)$ the slope $f(t_1, y_1)$ gives the next segment, and so on. With uniform step $h$ the rule is $$ y_{n+1} = y_n + f(t_n, y_n)\,h, \qquad n = 0, 1, 2, \dots $$ The result is a polygonal path of linked tangent segments approximating the true curve. $$ % caption: Euler's method links tangent segments: at each step the slope f(t_n, % y_n) extends the path by one interval, and the polygon drifts from the true % solution as errors accumulate. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (-0.2,0) -- (5.6,0) node[right] {$t$}; \draw[->, black] (0,-0.2) -- (0,3.2) node[above] {$y$}; % true solution (a concave curve) \draw[black, very thick] plot[domain=0:5.2, samples=60] (\x, {2.9*(1 - exp(-0.55*\x))}); \node[black, anchor=south] at (4.7,2.55) {true solution}; % euler polygon under it (steps of h=1.3) \coordinate (e0) at (0,0); \coordinate (e1) at (1.3,{0 + 1.3*1.6}); % recompute simple stepping visually: slopes decreasing \coordinate (p0) at (0,0); \coordinate (p1) at (1.3,1.5); \coordinate (p2) at (2.6,2.35); \coordinate (p3) at (3.9,2.85); \coordinate (p4) at (5.2,3.15); \draw[acc, very thick] (p0) -- (p1) -- (p2) -- (p3) -- (p4); \foreach \p in {p0,p1,p2,p3,p4} \fill[acc] (\p) circle (2pt); \node[black, anchor=north west] at (1.3,1.45) {$y_1$}; \node[black, anchor=north west] at (2.6,2.3) {$y_2$}; % step markers \draw[black, dashed] (1.3,0) -- (1.3,1.5); \draw[black, dashed] (2.6,0) -- (2.6,2.35); \node[black, anchor=north] at (0.65,-0.05) {$h$}; \end{tikzpicture} $$ > **Worked example.** Apply Euler's method to $y' = 3 - 2t - 0.5y$, $y(0) = 1$, > with step $h = 0.2$. Each step reads the slope $f(t_n, y_n) = 3 - 2t_n - 0.5y_n$ > and advances $y_{n+1} = y_n + h\,f(t_n, y_n)$. The first slope is > $f(0, 1) = 2.5$, so $y_1 = 1 + 0.2(2.5) = 1.5$; then $f(0.2, 1.5) = 1.85$ gives > $y_2 = 1.5 + 0.2(1.85) = 1.87$, and so on. The exact solution > $y = 14 - 4t - 13e^{-t/2}$ supplies the error at each node. > > | $t_n$ | slope $f(t_n, y_n)$ | Euler $y_n$ | exact $y(t_n)$ | error | > | --- | --- | --- | --- | --- | > | $0.0$ | $2.500$ | $1.0000$ | $1.0000$ | $0.0000$ | > | $0.2$ | $1.850$ | $1.5000$ | $1.4371$ | $0.0629$ | > | $0.4$ | $1.265$ | $1.8700$ | $1.7565$ | $0.1135$ | > | $0.6$ | $0.739$ | $2.1230$ | $1.9694$ | $0.1536$ | > | $0.8$ | $0.265$ | $2.2707$ | $2.0858$ | $0.1849$ | > | $1.0$ | $-$ | $2.3236$ | $2.1151$ | $0.2085$ | > > The error grows at every node: the global truncation error accumulates in > proportion to the number of steps taken. Two kinds of error accumulate: - **Local truncation error** is the mistake made in a single step, from replacing the curve by its tangent. For Euler's method it is proportional to $h^2$. - **Global truncation error** is the total deviation at a fixed $t$ after many steps. Because the number of steps grows as $1/h$, the global error is proportional to $h$: Euler's method is **first order**. Halving the step size roughly halves the error. First-order accuracy is coarse. The [Runge-Kutta methods](/differential-equations/numerical/euler-and-runge-kutta) sample the slope at several points per step to reach far higher order at comparable cost, but every one of them is a refinement of the tangent-stepping idea Euler introduced. ## Summary - The **existence-uniqueness theorem** guarantees a single local solution when $f$ and $\partial f/\partial y$ are continuous near the initial point; continuity of $f$ alone gives existence but not uniqueness. - Uniqueness can genuinely **fail**: $y' = y^{1/3}$, $y(0) = 0$ has infinitely many solutions, precisely because $\partial f/\partial y$ is unbounded there. - The **interval of existence** is only local, bounded below by $h = \min(a, b/M)$ from a bound $M$ on $f$ over a rectangle. - **Picard iteration** proves existence by converging the sequence $\phi_{n+1} = \int_0^t f(s, \phi_n)\,\d s$ to the solution, under a Lipschitz condition. - **Euler's method** computes solutions numerically by linked tangent steps $y_{n+1} = y_n + f(t_n, y_n)h$, with global error proportional to $h$. [^simmons-picard]: **Simmons**, _Differential Equations_, Ch. 13 §69–§70 — the method of successive approximations and Picard's theorem: reformulating the initial value problem as an integral equation and iterating to the solution under a Lipschitz condition. The existence rectangle, non-uniqueness example, and Euler's method follow Boyce §2.8, §2.4, and §2.7.