Composability of ODEs

A \solve-ode is usually a closed box: hand it a rate dy/dt, a starting value, start time and end time, and get back a trajectory.

A simple ODE

• Define placeholder y
• Define initial condition:
• The main equation:
• Initial time:
• End time:
• Run the ODE

But the rate can be any reactive expression, including one that reads another trajectory.

Two ODEs can be chained: the second one's slope, at every instant, is a value sampled live from the first one's curve.

We use two trivial ODEs, say ODE A and ODE B:

ODE	meaning
	a plain decaying curve,
	its slope is 's current height,

Try to interpret the behavior of :

• Since , the slope of is always positive but decreasing with time.
• Where is tall (near ), climbs steeply
• Where has decayed (near ), flattens out.

's curve is shaped entirely by , which itself comes from an ODE.

---

How to implement this chain

Setup

• Grid size
• Start time
• End time
• Number of steps
• Start value of A
• Start value of B
• Placeholder for A
• Placeholder for B
• Color for A

---

1. Solve ODE A

• Define rate
• Solve A and paint it red

A red decaying curve. Nothing surprising yet.

2. Let ODE B read A

The rate of B is not a formula — it is a point sampled off A's curve:

• Sampling time for ODE B: this is the time variable for B's solver, which sweeps from t-start to t-end.
• Sample at time s: a-pt.y is , and it updates as s moves.
• Solve for , using a-pt.y as its rate.

As ODE B's solver sweeps s from to , a-pt.y re-reads at each step, so 's slope tracks 's height the whole way. That is the only line that matters: a-pt.y belongs to ODE A, but it is ODE B's right-hand side.

---

Where can this be useful?

• Neural ODEs: the backward pass of a neural ODE is itself an ODE, whose rate depends on the values from the forward pass's trajectory.