First-Order ODEs & Existence Theorems

A first-order ordinary differential equation (ODE) is an equation of the form

$y'(t) = f(t, y(t)),$

where $f: D \to \mathbb{R}$ is a given function on an open set $D \subseteq \mathbb{R}^2$, and $y: I \to \mathbb{R}$ is the unknown function defined on an interval $I \subseteq \mathbb{R}$. The variable $t$ is the independent variable (often representing time), $y$ is the dependent variable (the state), and $f$ is the right-hand side (the vector field, in 1D).

An initial value problem (IVP) is an ODE together with an initial condition:

$y'(t) = f(t, y(t)), \qquad y(t_0) = y_0,$

where $(t_0, y_0) \in D$ is the specified starting point. A solution is a differentiable function $y: I \to \mathbb{R}$ (with $t_0 \in I$) satisfying both the equation and the initial condition.

The direction field of $y' = f(t, y)$ is the assignment $(t, y) \mapsto (1, f(t, y))$ — a line element of slope $f(t, y)$ at each point. An integral curve is a curve $\gamma(t) = (t, y(t))$ that is tangent to the direction field at every point, i.e., $y'(t) = f(t, y(t))$.

Consider $y' = ky$ with $k > 0$. The direction field has slope $ky$ at $(t, y)$: horizontal along $y = 0$, steeper as $|y|$ increases. Solutions are $y(t) = Ce^{kt}$ — exponentials that grow away from the equilibrium $y = 0$. Each initial condition $y(0) = C$ determines a unique integral curve.

The direction field of $y' = f(t, y)$ is a 1D analog of the gradient field $\nabla L(\theta)$ from The Gradient & Directional Derivatives. Gradient descent $\dot{\theta} = -\nabla L(\theta)$ is a first-order ODE whose direction field is $-\nabla L$. The integral curves are the gradient descent trajectories. This is not an analogy — it is literally the same mathematical structure.

An ODE $y' = f(t, y)$ is separable if it can be written as $y' = g(t)h(y)$ for continuous functions $g$ and $h$. The general solution procedure is:

1. Separate: $\displaystyle\frac{1}{h(y)}\,dy = g(t)\,dt$.
2. Integrate both sides: $H(y) = G(t) + C$ where $H$ and $G$ are antiderivatives.
3. Solve for $y$ if possible, obtaining the solution in explicit form $y = H^{-1}(G(t) + C)$.

The logistic equation $y' = ry(1 - y/K)$ is separable with $g(t) = r$ and $h(y) = y(1 - y/K)$. Partial fractions and integration give the logistic curve

$y(t) = \frac{K}{1 + \left(\frac{K}{y_0} - 1\right)e^{-rt}},$

which appears as the sigmoid activation function in ML when $K = 1$, $r = 1$. The solution interpolates between the unstable equilibrium $y = 0$ and the stable equilibrium $y = K$, approaching the carrying capacity from any positive initial condition.

The equation $y' = -k(y - T_{\text{env}})$ models a body at temperature $y$ cooling toward ambient temperature $T_{\text{env}}$. This is separable: $\frac{dy}{y - T_{\text{env}}} = -k\,dt$. Integrating gives $y(t) = T_{\text{env}} + (y_0 - T_{\text{env}})e^{-kt}$ — an exponential decay toward equilibrium.

When $h(y_0) = 0$, the constant function $y(t) \equiv y_0$ is an equilibrium solution (the direction field is horizontal along $y = y_0$). Division by $h(y)$ during separation is only valid away from these equilibria. The equilibria of the logistic equation are $y = 0$ (unstable) and $y = K$ (stable) — solutions approach $K$ from any positive initial condition.

An ODE is linear of first order if it has the form

$y' + p(t)y = q(t),$

where $p$ and $q$ are continuous functions on an interval $I$. The equation is homogeneous if $q(t) = 0$ and inhomogeneous otherwise.

If $p$ and $q$ are continuous on an interval $I$ containing $t_0$, then the IVP $y' + p(t)y = q(t)$, $y(t_0) = y_0$ has the unique solution

$y(t) = e^{-P(t)}\left(y_0 + \int_{t_0}^{t} q(s)\,e^{P(s)}\,ds\right)$

where $P(t) = \int_{t_0}^{t} p(s)\,ds$. In particular, the solution exists on all of $I$ — there is no finite-time blow-up for linear equations.

Multiply $y' + p(t)y = q(t)$ by $\mu(t) = e^{P(t)}$. The left side becomes $\frac{d}{dt}[\mu(t) y(t)]$ by the product rule (Topic 5, Theorem 3): $\mu y' + \mu p y = \mu y' + \mu' y = (\mu y)'$. So

$(\mu y)' = \mu q.$

Integrate from $t_0$ to $t$:

$\mu(t) y(t) - \mu(t_0) y(t_0) = \int_{t_0}^{t} \mu(s) q(s)\,ds.$

Since $\mu(t_0) = e^0 = 1$, solve for $y(t)$:

$y(t) = e^{-P(t)}\left(y_0 + \int_{t_0}^{t} q(s)\,e^{P(s)}\,ds\right).$

Uniqueness follows from the Picard-Lindelöf theorem (Theorem 3 below) since $f(t, y) = q(t) - p(t)y$ is Lipschitz in $y$ with $L = \sup|p|$. $\square$

Consider $y' + 2y = e^{-t}$, $y(0) = 3$. The integrating factor is $\mu(t) = e^{2t}$. Multiplying: $(e^{2t} y)' = e^{2t} \cdot e^{-t} = e^t$. Integrating: $e^{2t} y = e^t + C$, so $y(t) = e^{-t} + Ce^{-2t}$. From $y(0) = 3$: $C = 2$. The solution is $y(t) = e^{-t} + 2e^{-2t}$ — the forced response $e^{-t}$ plus a transient $2e^{-2t}$ that decays faster.

Theorem 1 guarantees that the solution exists on the entire interval $I$ where $p$ and $q$ are continuous. This is a stronger result than the general Picard-Lindelöf theorem, which only guarantees local existence. The linearity of the equation prevents finite-time blow-up — the growth rate of $y$ is at most exponential, never faster. This global existence result breaks down for nonlinear equations (see Section 7).

The equation $M(t, y)\,dt + N(t, y)\,dy = 0$ is exact on a simply connected region $D$ if there exists a function $\Psi: D \to \mathbb{R}$ with $\frac{\partial \Psi}{\partial t} = M$ and $\frac{\partial \Psi}{\partial y} = N$. The function $\Psi$ is the potential function, and solutions are the level curves $\Psi(t, y) = C$.

Let $M$ and $N$ have continuous first partial derivatives on a simply connected open set $D$. Then $M\,dt + N\,dy = 0$ is exact if and only if

$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}$

on $D$. This is the ODE analog of the conservative field criterion $\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}$ from Line Integrals & Conservative Fields (Theorem 4).

Consider $(2ty + 3)\,dt + (t^2 + 4y)\,dy = 0$. Check: $M_y = 2t = N_t$. So $\Psi_t = 2ty + 3 \Rightarrow \Psi = t^2 y + 3t + g(y)$. Then $\Psi_y = t^2 + g'(y) = t^2 + 4y \Rightarrow g'(y) = 4y \Rightarrow g(y) = 2y^2$. Solution: $t^2 y + 3t + 2y^2 = C$.

Consider $(2y)\,dt + t\,dy = 0$. Check: $M_y = 2 \neq 1 = N_t$. Not exact. Try an integrating factor $\mu(t)$: we need $\frac{M_y - N_t}{N} = \frac{2 - 1}{t} = \frac{1}{t}$ to depend only on $t$. It does: $\mu(t) = e^{\int 1/t\,dt} = t$. Multiplying: $(2ty)\,dt + t^2\,dy = 0$. Now $M_y = 2t = N_t$ — exact. Potential: $\Psi_t = 2ty \Rightarrow \Psi = t^2 y$. Solution: $t^2 y = C$.

Exact equations and conservative fields are the same mathematics in different notation. The dictionary:

The simply connected domain requirement (Topic 15, Definition 5) ensures there are no topological obstructions.

A function $f: D \to \mathbb{R}$ (with $D \subseteq \mathbb{R}^2$) satisfies a Lipschitz condition in $y$ on $D$ if there exists a constant $L \geq 0$ such that

$|f(t, y_1) - f(t, y_2)| \leq L|y_1 - y_2|$

for all $(t, y_1), (t, y_2) \in D$. The constant $L$ is the Lipschitz constant. If $f$ is $C^1$ in $y$ and $|\partial f / \partial y| \leq L$ on $D$, then $f$ is Lipschitz with constant $L$ (by the Mean Value Theorem).

Let $f: D \to \mathbb{R}$ be continuous on an open set $D \subseteq \mathbb{R}^2$ and satisfy a Lipschitz condition in $y$ on $D$ with Lipschitz constant $L$. Let $(t_0, y_0) \in D$. Then there exists $\delta > 0$ such that the IVP

$y' = f(t, y), \qquad y(t_0) = y_0$

has a unique solution $y: [t_0 - \delta, t_0 + \delta] \to \mathbb{R}$.

The IVP $y' = f(t, y)$, $y(t_0) = y_0$ is equivalent to the integral equation

$y(t) = y_0 + \int_{t_0}^{t} f(s, y(s))\,ds.$

This reformulation is crucial: it converts a differential equation (involving derivatives) into an integral equation (involving only continuous functions and Riemann integration). The Picard iteration is defined in integral form, avoiding the need to differentiate the iterates.

The structure of the proof is identical to the IFT proof (Topic 12, §8). Define an operator $T$, show it maps a closed set into itself, show it is a contraction, and invoke completeness. The only difference is the space: the IFT works in $(\mathbb{R}^n, \|\cdot\|)$, while Picard-Lindelöf works in $(C([t_0 - \delta, t_0 + \delta]), \|\cdot\|_\infty)$. The reader who followed the IFT proof already knows the playbook.

The Picard iterates for the IVP $y' = f(t, y)$, $y(t_0) = y_0$ are defined recursively:

$y_0(t) = y_0 \quad\text{(the constant initial guess)}$

$y_{n+1}(t) = y_0 + \int_{t_0}^{t} f(s, y_n(s))\,ds \quad\text{for } n \geq 0.$

We give the full proof via the contraction mapping principle.

Setup. Choose a closed rectangle $R = \{(t, y) : |t - t_0| \leq a,\, |y - y_0| \leq b\} \subseteq D$. Let $M = \max_R |f|$ (which exists since $f$ is continuous on the compact set $R$). Set $\delta = \min(a, b/M)$. Let $X = C([t_0 - \delta, t_0 + \delta])$ with the sup-norm $\|y\|_\infty = \max_{|t - t_0| \leq \delta} |y(t)|$, and let $S = \{y \in X : \|y - y_0\|_\infty \leq b\}$. The set $S$ is closed in the complete metric space $(X, \|\cdot\|_\infty)$.

Step 1 — $T$ maps $S$ into $S$. Define the Picard operator $T: S \to X$ by

$(Ty)(t) = y_0 + \int_{t_0}^{t} f(s, y(s))\,ds.$

For $y \in S$:

$|(Ty)(t) - y_0| = \left|\int_{t_0}^{t} f(s, y(s))\,ds\right| \leq M|t - t_0| \leq M\delta \leq b.$

So $Ty \in S$.

Step 2 — $T$ is a contraction (for small $\delta$). For $y, z \in S$:

$|(Ty)(t) - (Tz)(t)| = \left|\int_{t_0}^{t} [f(s, y(s)) - f(s, z(s))]\,ds\right| \leq \int_{t_0}^{t} L|y(s) - z(s)|\,ds \leq L\delta \|y - z\|_\infty.$

If we further require $\delta \leq 1/(2L)$, then $\|Ty - Tz\|_\infty \leq \frac{1}{2}\|y - z\|_\infty$. So $T$ is a contraction with factor $\lambda = L\delta \leq 1/2$.

Step 3 — Fixed point. The space $(S, \|\cdot\|_\infty)$ is a closed subset of a complete metric space, hence complete. By the contraction mapping theorem (Topic 12, Theorem 6), $T$ has a unique fixed point $y^* \in S$:

$y^*(t) = y_0 + \int_{t_0}^{t} f(s, y^*(s))\,ds.$

This $y^*$ is the unique solution to the IVP on $[t_0 - \delta, t_0 + \delta]$.

Step 4 — Convergence of Picard iterates. The contraction mapping theorem also guarantees that $y_n \to y^*$ uniformly:

$\|y_n - y^*\|_\infty \leq \frac{\lambda^n}{1 - \lambda}\|y_1 - y_0\|_\infty \to 0.$

The convergence is geometric with rate $\lambda = L\delta$. $\square$

Starting from $y_0(t) = 1$:

$y_1(t) = 1 + \int_0^t 1\,ds = 1 + t$

$y_2(t) = 1 + \int_0^t (1 + s)\,ds = 1 + t + \frac{t^2}{2}$

$y_3(t) = 1 + t + \frac{t^2}{2} + \frac{t^3}{6}$

The pattern is $y_n(t) = \sum_{k=0}^{n} t^k/k!$ — the partial sums of $e^t$. The Picard iterates converge to the Taylor series of the exact solution $y = e^t$.

Starting from $y_0(t) = 0$:

$y_1(t) = \int_0^t s\,ds = \frac{t^2}{2}$

$y_2(t) = \int_0^t \left(s + \frac{s^4}{4}\right)ds = \frac{t^2}{2} + \frac{t^5}{20}.$

The iterates grow in complexity — no closed-form solution exists for this Riccati equation, but the Picard iterates provide a convergent sequence of polynomial approximations.

If $y$ is a solution on $(a, b)$ and $\lim_{t \to b^-} (t, y(t))$ exists and lies in $D$, then $y$ can be extended past $b$. Equivalently, if $y$ cannot be extended past $b$, then either $|y(t)| \to \infty$ as $t \to b^-$ or $(t, y(t))$ approaches the boundary of $D$.

Consider $y' = y^2$, $y(0) = 1$. This is separable: $-1/y = t + C$, so $y(t) = 1/(1 - t)$. The solution blows up at $t = 1$: $y(t) \to +\infty$ as $t \to 1^-$. The maximal interval of existence is $(-\infty, 1)$, not $(-\infty, +\infty)$.

The nonlinearity $y^2$ amplifies growth faster than exponential, causing the solution to reach infinity in finite time. Compare with the linear equation $y' = y$, whose solution $y = e^t$ grows but never blows up.

Theorem 1 guarantees that solutions to linear equations $y' + p(t)y = q(t)$ exist on the entire interval where $p$ and $q$ are continuous. This is because linear growth $|f(t, y)| \leq C(1 + |y|)$ prevents finite-time blow-up by Gronwall’s inequality. Nonlinear equations ($y^2$, $y^3$, $\tan y$) can violate this bound and reach infinity in finite time.

If $f$ is continuous on an open set $D \subseteq \mathbb{R}^2$ containing $(t_0, y_0)$, then the IVP $y' = f(t, y)$, $y(t_0) = y_0$ has at least one local solution. (No Lipschitz condition required.) The proof uses the Arzelà-Ascoli theorem (compactness in function spaces) rather than contraction mapping.

Consider $y' = y^{2/3}$, $y(0) = 0$. The function $f(y) = y^{2/3}$ is continuous but not Lipschitz at $y = 0$ (the derivative $f'(y) = \frac{2}{3}y^{-1/3} \to \infty$ as $y \to 0$).

Two solutions through $(0, 0)$:

1. $y(t) \equiv 0$ (the trivial solution).
2. $y(t) = (t/3)^3 = t^3/27$ (verified: $y' = t^2/9 = (t^3/27)^{2/3}$).

In fact, for any $c \geq 0$, the function

$y_c(t) = \begin{cases} 0, & t \leq c \\ ((t-c)/3)^3, & t > c \end{cases}$

is also a solution — there are infinitely many solutions through the origin.

The Lipschitz condition $|f(t, y_1) - f(t, y_2)| \leq L|y_1 - y_2|$ prevents the right-hand side from changing “too fast” as $y$ varies. When $f$ is not Lipschitz (as in $y^{2/3}$ near $y = 0$), the direction field near the initial point is too “flat” — nearby solutions cannot distinguish themselves, and multiple integral curves can merge or branch at the non-Lipschitz point. The Lipschitz condition is the minimal regularity that prevents this pathology.