Read corners and jumps as generalized derivatives

Try

A recorded position starts at zero, rises with slope 2 for one time unit, then falls with slope −1 for two time units and stays at zero. Its corners change velocity instantly. Does that mean the position itself jumps? Draw the one-sided position values and slopes separately. The position is continuous; its slope jumps by 2, −3 and 1 at times 0, 1 and 3.

Now suppose a recording offset is introduced at time 2. After that instant every position is shifted by J. This is a mathematical jump model, not a claim that a physical object teleports. It introduces a different generalized derivative from the original corners.

Predict

Predict which derivative contains an impulse when position has a corner, and which derivative contains one when position itself jumps. Then predict the sign of the impulse if the offset is J = −1. An impulse coefficient is a signed weight; drawing a very tall finite line does not define it.

Experiment

Separate corners from position jumps

Temporary analytic teaching values. Controls do not save evidence or award understanding. Record a prediction and explanation in the lesson practice.

Position jump J at time 2 -101 Reset inspector

Position x(t) = 2r(t) − 3r(t−1) + r(t−3) + J H(t−2), with r(t) = max(t,0). At time 2: left limit 1, right limit 1. A single assigned point value does not alter a distribution.

First derivative: ordinary slope 2H(t) − 3H(t−1) + H(t−3), plus 0δ(t−2). Second derivative: 2δ(t) − 3δ(t−1) + δ(t−3), plus 0δ′(t−2). These are signed distribution coefficients, not finite graph heights.

• At 0: slopes 0 → 2; slope jump 2; position jump 0.
• At 1: slopes 2 → -1; slope jump -3; position jump 0.
• At 2: slopes -1 → -1; slope jump 0; position jump 0.
• At 3: slopes -1 → 0; slope jump 1; position jump 0.

The δ′ term acts on a smooth test function φ as 0φ′(2); the first-derivative δ term acts as 0φ(2). At J = 0 the original graph is continuous, but corners still produce second-derivative impulses.

Start at J = 0 and recover the three slope jumps from the graph. Change J to −1. The ordinary slope remains −1 on both sides of time 2, but the one-sided positions differ. Explain why a graph of only the ordinary slope would now lose information. Change to J = 1 and compare the action of the second derivative on a test function with positive derivative at time 2.

Understand

Define differentiation through what it does to a test

A smooth compactly supported test function φ is zero outside a bounded interval and has derivatives of every order. An ordinary locally integrable function f acts on it by integrating fφ. The distributional derivative Df acts by −∫fφ′. A point impulse at a acts by δₐ(φ) = φ(a), and its derivative acts by δₐ′(φ) = −φ′(a). These definitions are developed in distributions and fundamental solutions; here we apply them to a recorded graph.

For a piecewise C¹ function with finitely many jumps, split −∫fφ′ into its smooth intervals. Integration by parts makes each interval contribute its ordinary derivative integral and two endpoint terms. At a shared endpoint a, the terms combine as [f(a+) − f(a−)]φ(a). Outer terms vanish because φ has compact support. Therefore:

Changing the assigned value f(a) at a single point changes no integral. One-sided limits, not the drawn open/closed dot convention, determine the jump contribution. Apply the same argument to the piecewise ordinary derivative, assuming f is piecewise C² with one-sided derivative limits:

A continuous corner contributes a delta to the second derivative. A position jump contributes a delta-prime as well as its first-derivative delta. There is no operation of replacing delta-prime by a large ordinary value at a point.

Assemble the graph without losing its delays

Write H for the unit step and r(t) = max(t,0) for the ramp. The inspector’s position is:

The ramp coefficients are changes of slope: 0 → 2 at zero, 2 → −1 at one, and −1 → 0 at three. Their sum is zero, so the final slope vanishes. Differentiating gives:

For a test with φ(0) = 1, φ(1) = 2, φ(3) = −1 and φ′(2) = 4, the second derivative acts as −5 − 4J. In particular J = −1 gives −1, not −9. This numerical sign check uses the definition of delta-prime rather than a picture of acceleration.

The converse construction needs initial information. Integrating the three delta sources twice gives the three ramps, but any affine function has zero second derivative. Zero prehistory removes that ambiguity. Adding a position jump requires a delta-prime source in the second-order equation. A finite physical force impulse in the usual mass model instead jumps velocity while position stays continuous.

Read causal convolution limits from support

Let w be a causal impulse response, zero before time zero. A unit-height pulse applied from time 1 to 3 is q = H(t−1) − H(t−3). For rest initial conditions, the response is the convolution of w and q. Intersect the pulse’s support with past times before changing variables. Using elapsed time τ = t−s:

This requires a locally integrable kernel and uses ordinary pulse integrals. For example w(τ) = exp(−τ) on nonnegative τ gives 1 − exp(−(t−1)) during the pulse and exp(−(t−3)) − exp(−(t−1)) after it. The response stays continuous at both switches. A pulse integrates a range of impulse responses; it is not a single impulse at its midpoint.

Apply

Translate a new graph z = r(t−1) − r(t−2) − 2H(t−3) into its first and second generalized derivatives. Verify every coefficient by listing one-sided values and slopes. Then reconstruct the forcing in Dz + 2z = g. This transfer changes both the times and the type of equation, so the original diagram alone is insufficient.

Continue

Return to steps and impulses for the physical velocity-jump interpretation, or convolution for a general causal input. Continue to fundamental matrices and coordinate changes to preserve all initial-state information in a vector system.

Source scope: generalized graph differentiation and finite-pulse convolution are assessed in MIT 18.03SC’s final materials, Fall 2011, consulted 28 October 2026. This graph, proof sequence and practice are original.