4.5 Numerical Time Stepping and Euler-Type Methods

An Equation You Cannot Solve

A falling object encounters air resistance that grows with the square of its speed. The faster it goes, the harder the air pushes back. Newton's second law gives:

$$m\frac{dv}{dt} = mg - cv^2$$

This is a perfectly well-defined differential equation. It tells you the acceleration at every instant, given the current velocity. It is an initial-value problem of exactly the kind you studied in Section 4.2. And yet --- try to solve it by separation of variables. Try substitution. Try every technique from calculus class.

You will not find a closed-form solution in terms of elementary functions.

This is not a failure of technique. It is a fundamental fact: most differential equations that arise in physics cannot be solved with pencil and paper. The models from Section 4.3 --- constant acceleration, linear drag --- were the exceptions, not the rule. Real-world forces produce equations that resist every algebraic method you know.

But the equation still means something. At every instant, it tells you the acceleration. If you know the current velocity, you can compute the current acceleration. And if you know the acceleration, you can estimate the velocity a tiny moment later. Then you can repeat: compute the new acceleration from the new velocity, take another tiny step, and repeat again.

A computer can do this thousands of times per second. The result is not a formula --- it is a table of numbers, a list of positions and velocities at a sequence of times. But that table is the solution, built one small step at a time.

How, exactly, does this work?

Prediction

Before you read on: Suppose you use this step-by-step method to solve the simple free-fall problem $dv/dt = -g$, starting from rest, using a time step of $\Delta t = 0.1$ s. After one second (10 steps), will the numerical velocity be:

(a) Exactly equal to the true velocity $v = -gt = -10$ m/s

(b) Slightly too large in magnitude (more negative than $-10$ m/s)

(c) Slightly too small in magnitude (less negative than $-10$ m/s)

Commit to your answer. The answer will reveal something important about how the method works.

The Guiding Question

How can we build an approximate solution to a differential equation one small time step at a time?

You already know the answer in spirit. The derivative tells you the slope. The slope tells you where to go next. You just need to turn that idea into a recipe --- a systematic procedure that a person or a computer can follow without thinking.

Exploration: Building a Solution by Hand

Let's start with the simplest possible case: an object dropped from rest under constant gravitational acceleration. The differential equation is:

$$\frac{dv}{dt} = -g = -10 \text{ m/s}^2$$

You know the exact solution: $v(t) = -gt = -10t$ m/s. That is the whole point --- we will use a problem where we know the answer, so we can see exactly what the numerical method does and where it goes wrong.

Here is the idea. You stand at time $t = 0$ with velocity $v = 0$. The equation tells you the acceleration right now: $a = -10$ m/s$^2$. If the acceleration stayed at this value for a short time $\Delta t$, the velocity would change by $a \cdot \Delta t$. So you estimate:

$$v_{\text{new}} \approx v_{\text{old}} + a \cdot \Delta t$$

Then you move to the next moment, compute the acceleration there, and repeat.

For constant acceleration, the acceleration never changes, so this recipe should work perfectly. Let's see.

[Interactive: Euler Stepper. Students choose a differential equation from a menu ($dv/dt = -g$, $dv/dt = -bv$, $dv/dt = g - cv^2$) and a time step $\Delta t$ via a slider (0.5 s, 0.2 s, 0.1 s, 0.05 s, 0.01 s). The system builds the solution one step at a time: at each step, a small right triangle appears on the $v$-vs-$t$ graph, with horizontal leg $\Delta t$ ("run") and vertical leg $a_n \cdot \Delta t$ ("rise"). The numerical solution is drawn as a staircase of straight-line segments. When an exact solution exists, it is overlaid as a smooth curve. A readout shows the current values of $t$, $v$, $a$, and the error $v_{\text{numerical}} - v_{\text{exact}}$. Guided prompts: "Halve the time step. What happens to the error at $t = 1$ s? Halve it again. Do you see a pattern?"]

Spend a few minutes with the interactive. Start with $dv/dt = -g$ and a large time step like $\Delta t = 0.5$ s. Watch the triangles march across the graph. Then switch to $dv/dt = -bv$ and notice something new: the triangles get smaller as the velocity decays. The slope changes at every step because the acceleration depends on the current velocity.

Before you read on: When you used the interactive with $dv/dt = -bv$ and a time step of 0.2 s, did the numerical solution lie above or below the exact exponential decay curve? Why might that be?

Concept Reveal: Euler's Method

What you just did has a name. It is called Euler's method, after Leonhard Euler, who described it in 1768. The idea is strikingly simple:

Given a differential equation $\frac{dv}{dt} = a(v, t)$ and an initial condition $v(0) = v_0$, compute the solution at discrete times $t_0, t_1, t_2, \ldots$ separated by a fixed time step $\Delta t$:

$$v_{n+1} = v_n + a_n \cdot \Delta t$$

That is it. One line. The acceleration $a_n$ is evaluated at the current values of $v_n$ and $t_n$, and the result gives the velocity at the next time step.

If you also want position, you add a second update:

$$x_{n+1} = x_n + v_n \cdot \Delta t$$

Together, these two equations march the state of the system forward in time: compute the acceleration, update the velocity, update the position, advance the clock, repeat.

Why Does It Work?

The definition of the derivative says:

$$\frac{dv}{dt} = \lim_{\Delta t \to 0} \frac{v(t + \Delta t) - v(t)}{\Delta t}$$

If you rearrange this (dropping the limit, which is what introduces the approximation), you get:

$$v(t + \Delta t) \approx v(t) + \frac{dv}{dt} \cdot \Delta t$$

This is Euler's method. It is the definition of the derivative, run in reverse --- not as a way to find the slope, but as a way to use the slope to step forward. The differential equation provides $dv/dt$ at every point. Euler's method uses that information to build the solution one piece at a time.

The approximation improves as $\Delta t$ shrinks, because the dropped limit becomes more nearly valid. In the limiting case $\Delta t \to 0$, you recover the exact solution. But you would also need infinitely many steps, which is not practical. The art of numerical methods is choosing $\Delta t$ small enough for acceptable accuracy, but large enough to finish in a reasonable time.

The Algorithm as a Table

Formulas are compact, but they can obscure the rhythm of the computation. A table makes the step-by-step logic visible. Each row is one tick of the clock. Each column is one quantity. You fill in the table left to right, top to bottom, and the solution assembles itself.

Here is the template:

The key discipline: at each step, the acceleration is computed from the current velocity (and possibly time), using the differential equation. You never look ahead. You never use future values. Everything flows forward from the present state.

Worked Example: Constant Acceleration ($dv/dt = -g$)

Let's fill in this table completely for the simplest case. An object is dropped from rest at height $x_0 = 0$ (measuring downward as positive is tempting, but let's keep the standard convention: $x$ points upward). The differential equation is:

$$\frac{dv}{dt} = -g = -10 \text{ m/s}^2$$

Initial conditions: $v_0 = 0$ m/s, $x_0 = 100$ m (dropped from 100 m above the ground).

Time step: $\Delta t = 0.2$ s.

Step 0. We start with the given initial conditions. The acceleration is $a_0 = -10$ m/s$^2$ (constant, from the DE).

Step 1. Update the velocity: $v_1 = v_0 + a_0 \Delta t = 0 + (-10)(0.2) = -2$ m/s. Update the position: $x_1 = x_0 + v_0 \Delta t = 100 + (0)(0.2) = 100$ m. Compute the new acceleration: $a_1 = -10$ m/s$^2$ (still constant).

Step 2. Update: $v_2 = v_1 + a_1 \Delta t = -2 + (-10)(0.2) = -4$ m/s. Position: $x_2 = x_1 + v_1 \Delta t = 100 + (-2)(0.2) = 99.6$ m. Acceleration: $a_2 = -10$ m/s$^2$.

Continuing this process for 5 steps:

Checking Against the Exact Solution

For constant acceleration, the exact solution is $v(t) = v_0 - gt$ and $x(t) = x_0 + v_0 t - \frac{1}{2}gt^2$.

At $t = 1.0$ s: - Exact velocity: $v = 0 - 10(1) = -10$ m/s. - Euler velocity: $-10$ m/s.

The velocities match exactly. This is because the acceleration is constant --- the Euler step uses a constant slope, and the true solution is a straight line in the $v$-vs-$t$ graph. A straight-line approximation to a straight line is exact.

But look at position: - Exact position: $x = 100 + 0 - \frac{1}{2}(10)(1)^2 = 95$ m. - Euler position: $96$ m.

The Euler position is too high by 1 m. The object has not fallen far enough in the numerical solution.

Why? Because the position update $x_{n+1} = x_n + v_n \Delta t$ uses the velocity at the beginning of each interval. But the object is accelerating --- its velocity at the end of the interval is more negative than at the beginning. By using the beginning-of-interval velocity, we underestimate how far the object falls during each step. The error accumulates over many steps.

Return to the Prediction

Now go back to your prediction. The velocity came out exactly right --- answer (a) --- because constant acceleration produces a linear velocity function, and Euler's method handles linear functions perfectly. But the position was too large (the object didn't fall far enough), which corresponds to an error that would appear in a position-based version of the prediction. If the prediction had asked about position, the answer would be (c): the displacement is too small in magnitude.

This is a general pattern. Euler's method uses the slope at the start of each step. When the slope is changing (as it does for position, whose "slope" is the velocity, which is changing), the method systematically falls behind.

Pause and check: Can you explain, in one sentence, why the Euler velocity is exact but the Euler position has error in the constant-acceleration example? If not, re-read the two paragraphs above before moving on.

Faded Example: Velocity-Dependent Acceleration ($dv/dt = -bv$)

Now let's try an equation where the acceleration is not constant. Consider an object experiencing linear drag:

$$\frac{dv}{dt} = -bv$$

with $b = 1$ s$^{-1}$, initial velocity $v_0 = 10$ m/s, and time step $\Delta t = 0.5$ s. We will track velocity only (no position for now) to keep the focus on how changing acceleration affects the method.

The exact solution is $v(t) = v_0 e^{-bt} = 10e^{-t}$ (from Section 4.3). We can check our numerical answer against it.

Step 0. Given: $t_0 = 0$, $v_0 = 10$ m/s. Acceleration: $a_0 = -bv_0 = -(1)(10) = -10$ m/s$^2$.

Step 1. $v_1 = v_0 + a_0 \Delta t = 10 + (-10)(0.5) = 5.0$ m/s. Time: $t_1 = 0.5$ s. Acceleration: $a_1 = -bv_1 = -(1)(5.0) = -5.0$ m/s$^2$.

Step 2. $v_2 = v_1 + a_1 \Delta t =$ ?

Fill in the remaining steps yourself.

What Happens with a Smaller Step?

Try the same problem with $\Delta t = 0.25$ s for four steps (reaching $t = 1.0$ s). Does the error at $t = 1.0$ s get better? By roughly how much?

Why Euler's Method Has Error: A Geometric View

The source of error has a clean geometric interpretation. Consider the $v$-vs-$t$ graph for $dv/dt = -bv$. The exact solution is a decaying exponential --- a curve that bends upward (concave up). Euler's method replaces this curve with a sequence of straight-line segments, each tangent to the curve at the start of the interval.

When a curve is concave up (bending upward), the tangent line lies below the curve. So each straight-line segment underestimates where the curve goes next. The numerical solution falls below the true solution.

When a curve is concave down (bending downward), the tangent line lies above the curve, and the numerical solution overshoots.

This is why the direction of the error depends on the shape of the solution curve. For exponential decay ($v = v_0 e^{-bt}$), the curve bends upward, and Euler undershoots. For exponential growth ($v = v_0 e^{bt}$), the curve bends downward in the sense relevant to the approximation, and Euler's method would lag behind the rapidly growing true solution.

Pause and think: Sketch the $v$-vs-$t$ graph for exponential decay. Draw a tangent line at $t = 0$. Does the tangent line lie above or below the curve at $t = 0.5$? This single picture explains all of the error in the faded example above.

Connection: From Section 4.1 to Here

In Section 4.1, you learned that a differential equation is a motion-generating rule: given the current state, it tells you what happens next. Euler's method is the most literal way to follow that rule. It says: compute the current acceleration, pretend it holds for a short time, step forward, and repeat.

There is nothing clever about it. No algebraic tricks, no inspired substitutions, no insight required beyond "read off the slope, take a step." That simplicity is both its strength and its limitation. It always works --- you can apply it to any differential equation, whether or not a closed-form solution exists. But it requires small steps for accuracy, and it accumulates error with every step.

The analytical methods of Sections 4.2 and 4.3 give exact answers when they work. Euler's method gives approximate answers that can be made as accurate as you like by taking smaller steps. These are complementary tools, not competing ones. For the quadratic drag equation $m\frac{dv}{dt} = mg - cv^2$ that opened this section, Euler's method is not a backup plan --- it is the primary method of solution.

Independent Problem: Quadratic Drag ($dv/dt = g - cv^2/m$)

Now apply Euler's method to an equation that truly cannot be solved by elementary methods. An object falls from rest through a medium with quadratic drag:

$$\frac{dv}{dt} = g - \frac{c}{m}v^2$$

Let $g = 10$ m/s$^2$ and $c/m = 0.1$ m$^{-1}$. Use $\Delta t = 1$ s and initial velocity $v_0 = 0$ (measuring downward as positive here, so $v$ and $g$ are both positive).

Build the table for 5 steps. Describe what happens to the acceleration as the velocity increases.

Multiple Representations

We have been working primarily with tables, but Euler's method can be understood through several lenses. Consider the linear drag problem $dv/dt = -v$ with $v_0 = 10$ m/s and $\Delta t = 0.5$ s.

As a table: each row gives the state of the system at one moment.

As a graph: the numerical solution is a staircase of straight-line segments on the $v$-vs-$t$ plot. Each segment has a slope equal to the acceleration at the start of that interval. The segments connect to form a piecewise-linear approximation to the smooth exponential curve.

As a formula: for this specific equation, $v_{n+1} = v_n - v_n \Delta t = v_n(1 - \Delta t)$. So $v_n = v_0(1 - \Delta t)^n$. With $\Delta t = 0.5$: $v_n = 10(0.5)^n$. This is a geometric sequence approximating the exponential $10e^{-t}$.

As physical interpretation: at each moment, the drag force is proportional to the current velocity. Euler's method says: hold that drag force constant for $\Delta t$ seconds, let it decelerate the object, then reassess. It is like a coach who checks in every $\Delta t$ seconds and adjusts the training plan based on the current state --- rather than continuously monitoring.

All four representations describe the same computational process. Fluency means being able to move between them.

Spaced Retrieval

Before moving to practice, test your recall of earlier material.

Recall prompt 1: In Section 4.1, you saw that a differential equation is a motion-generating rule. In your own words, what does that mean?

Recall prompt 2: What is the difference between a differential equation and an initial-value problem? (Section 4.2)

Recall prompt 3: In Section 4.4, you learned about linearization. How is linearization similar to what Euler's method does? (Hint: both replace a curve with a straight line.)

Practice Layers

Layer 1: Concrete --- Complete an Euler Method Table

Problem 1. An object is launched upward with initial velocity $v_0 = 20$ m/s from ground level ($x_0 = 0$). The only acceleration is gravity: $dv/dt = -10$ m/s$^2$. Use Euler's method with $\Delta t = 0.5$ s to fill in the table for 5 steps. Track both velocity and position.

Layer 2: Pattern --- Compare Euler Results with Different Step Sizes

Problem 2. For the linear drag equation $dv/dt = -2v$ with $v_0 = 8$ m/s, compute the Euler velocity at $t = 1$ s using three different step sizes:

(a) $\Delta t = 0.5$ s (2 steps to reach $t = 1$)

(b) $\Delta t = 0.25$ s (4 steps to reach $t = 1$)

(c) $\Delta t = 0.1$ s (10 steps to reach $t = 1$)

The exact answer is $v(1) = 8e^{-2} \approx 1.08$ m/s. For each step size, compute the error and compare.

Layer 3: Structure --- Why Does Euler Lag Behind Exponential Growth?

Problem 3. Consider the equation $dv/dt = kv$ with $k > 0$ and $v_0 > 0$. The exact solution is exponential growth: $v(t) = v_0 e^{kt}$.

(a) Explain, using the geometric tangent-line argument, why Euler's method always underestimates the true velocity for this equation.

(b) Does the error get better or worse as time goes on? Why?

Layer 4: Creation --- Build a Numerical Solution

Problem 4. Consider a velocity-dependent acceleration that has no elementary closed-form solution. For example:

$$\frac{dv}{dt} = 10 - 0.5v - 0.02v^2$$

with $v_0 = 0$ and $\Delta t = 0.5$ s.

(a) Build an Euler method table for 10 steps (reaching $t = 5$ s).

(b) Describe the qualitative behavior: Does the velocity approach a limiting value? If so, estimate it from your table.

(c) Find the terminal velocity analytically by setting $dv/dt = 0$ and solving the quadratic. Compare with your numerical estimate.

Reflection

Why is numerical solving not just a "backup" for when you can't do algebra --- but a fundamental tool in its own right?

Consider what you have seen in this section. The quadratic drag equation, the combined linear-and-quadratic drag equation --- these are not exotic, artificial problems. They are the equations that arise naturally when you model real physical situations. The equations that can be solved analytically are the special cases. Numerical methods are not a last resort; they are the default tool for the vast majority of problems in physics and engineering.

Moreover, a numerical solution is not a lesser kind of answer. A table of velocities at discrete times tells you everything a formula does: how fast the object moves at each moment, when it reaches terminal velocity, how long the transient lasts. You can plot it, interpolate it, and extract physical insight from it. The answer is a function --- it just happens to be represented as a list of numbers rather than a symbolic expression.

In many ways, numerical methods are closer to the physics than analytical methods are. A differential equation says: "Given the current state, here is the next instant." Euler's method follows that instruction literally. It is physics as recipe, physics as algorithm, physics as the universe actually computes itself --- one moment at a time.

Chapter Summary: Differential Equations of Motion

This chapter began with a question: How does a physical rule about acceleration become a complete prediction of motion? Over five sections, we built the answer.

Section 4.1: From acceleration laws to differential equations. When acceleration depends on velocity or position --- not just time --- the equation that governs motion becomes a differential equation. The equation $dv/dt = f(v, t)$ is a motion-generating rule: it tells you the next instant from the current instant. This is a fundamentally different object from the kinematic equations of Chapter 2, where $a(t)$ was given and could be integrated directly.

Section 4.2: Solving initial-value problems. A differential equation alone has infinitely many solutions. Initial conditions select one. Together, the DE and the initial conditions form an initial-value problem --- the complete mathematical model of a specific motion. Every constant of integration corresponds to a measurable piece of physical information.

Section 4.3: Acceleration models and their consequences. Different forms of acceleration laws produce qualitatively different motions. Constant acceleration gives parabolas. Linear velocity-dependence gives exponential decay or growth. Position-dependence gives oscillation. The form of the law predicts the character of the motion, even before you solve the equation.

Section 4.4: Approximation and linearization. When exact equations are too complex, linearization replaces the equation with a simpler one that is accurate near a chosen point. This is a principled scientific move, not a shortcut. The key skill is knowing when the approximation is trustworthy and when it breaks down.

Section 4.5: Numerical time stepping. When no analytical method works, Euler's method builds the solution one step at a time: compute the acceleration, update the velocity, advance the clock, repeat. The accuracy improves with smaller steps. This is not a fallback --- it is the primary method for the majority of equations in physics.

The hero concept of this chapter is the initial-value problem: the idea that an acceleration law plus starting conditions generates a unique, complete motion. This is the mathematical expression of determinism in classical mechanics. Given the rule and the starting point, the future is determined. The methods differ --- analytical when possible, numerical when necessary --- but the structure is always the same: a rule, a starting state, and a unique trajectory.

Chapter-End Retrieval

Close your notes. Answer these from memory.

1. What is an initial-value problem? What are its two components, and why are both necessary?

2. Why does the form of the acceleration law determine the character of the motion? Give an example of two different acceleration laws that produce qualitatively different motions.

3. What is Euler's method? Write the update equations and explain, in one sentence, why the method works.

4. When would you use an analytical approach to solve a differential equation, and when would you use a numerical approach? Is one always better than the other?

5. What is the relationship between the time step $\Delta t$ and the accuracy of Euler's method?

After you have attempted all five, check your answers against the chapter summary above.

Looking Ahead

Chapters 1 through 3 asked: given the motion, describe it. Chapter 4 reversed the question: given the rule for acceleration, predict the motion. You learned to translate physical scenarios into differential equations, solve them analytically when possible, approximate them when useful, and compute them numerically when necessary.

But one enormous question has been left unanswered: where does the acceleration law come from?

We have treated $dv/dt = -g$, $dv/dt = -bv$, and $dv/dt = g - cv^2$ as given rules. We have never asked why the acceleration takes one form and not another. What determines whether a falling object experiences constant acceleration, linear drag, or quadratic drag? What physical quantity controls the acceleration?

The answer is force. Chapter 5 introduces Newton's laws of motion --- the laws that connect force to acceleration. Newton's second law, $\vec{F} = m\vec{a}$, is the bridge between the physical world (forces: gravity, friction, tension, drag) and the mathematical world (differential equations of motion). Every differential equation you solved in Chapter 4 was secretly a statement about forces. In Chapter 5, you will learn where those statements come from.

The mathematical machinery is ready. You know how to solve initial-value problems, how to classify acceleration models, how to linearize, and how to compute numerically. What you need now are the rules themselves --- the force laws that nature provides. That is the subject of dynamics, and it begins in the next chapter.