Why the Region of a Step Matters
A forward Euler calculation uses one slope to advance an approximate solution. That is straightforward when the slope function is defined and regular throughout the region being approximated. Near a point where the slope is undefined or discontinuous, however, a numerical update can be misleading: its arithmetic may be possible even though the step crosses a point at which the differential equation cannot be followed in the usual way.
For an initial-value problem \(y'=f(t,y)\), with current approximation \(y_n\) at time \(t_n\), the forward Euler update with step size \(h>0\) is
The formula evaluates \(f\) at the left endpoint. It does not automatically check what happens between \(t_n\) and \(t_{n+1}\). That is the key issue in this tutorial.
What Counts as a Regular Region?
Think of the differential equation as having a region of the \((t,y)\)-plane in which its slope function is defined and continuous. A point where the formula for \(f\) is undefined is a singular point for that formula. A point or curve where the slope changes discontinuously is a discontinuity. Either can divide the plane into regions where different reasoning applies.
Continuity in a neighborhood of the current point is a useful local condition: nearby points have slopes close to the current slope. For a standard local existence-and-uniqueness guarantee, a common sufficient condition is that \(f\) is continuous in a neighborhood and locally Lipschitz in \(y\) there. The Euler step itself requires only that \(f(t_n,y_n)\) be defined, but that bare requirement does not provide the same assurance about the solution or its approximation.
For screening a proposed step, ask whether the time interval crosses a known singular time or discontinuity. If the issue depends on \(y\), also check whether the predicted movement could reach a state where \(f\) is undefined. A step that ends on or crosses such a boundary should not be treated as an ordinary step within one regular region.
Worked Example: A Step Crossing a Singular Time
Worked Example: A Step Crossing a Singular Time
Consider \(y'=1/(1-t)\), \(y(0)=0\). The slope is undefined at \(t=1\). What happens if one Euler step uses \(h=1.2\)?
At the starting point, the slope is
The update gives
The arithmetic is defined because \(f(0,0)\) exists. But the time interval from \(0\) to \(1.2\) crosses \(t=1\), where the slope is undefined. The update is therefore not a valid approximation to one solution segment through that entire interval.
This is consistent with the exact solution. Since \(y'=1/(1-t)\), integration with \(y(0)=0\) gives \(y(t)=-\ln(1-t)\) for \(t<1\). This solution cannot be continued through \(t=1\) as a finite real-valued solution. The Euler result \(1.2\) does not fix that obstruction.
For a step that stays on the safe side, take \(h=0.4\). Then \(f(0,0)=1\), so \(y_1=0+0.4(1)=0.4\) at \(t_1=0.4\). The step remains in \(t<1\). This check establishes that the step does not cross the singular time; it does not by itself certify a small approximation error.
Check: The computed increment is \(h f(0,0)=0.4(1)=0.4\), so the updated value and time are \(y_1=0.4\) and \(t_1=0.4\), both on the regular side of the boundary.
Why a Discontinuity Calls for a Split
A discontinuity in the slope is different from a pole such as \(1/(1-t)\), but a step can still hide important information by crossing it. If the slope changes at a known time, the interval should be split there. The first step uses the slope on the first side; the next step uses the slope on the other side. This lets the calculation represent the change rather than extending one side’s slope across the whole interval.
A value assigned to the slope at the single switching time does not remove the jump on either side. When the equation specifies a right-hand branch at the switch, the step starting at that time uses that branch. The example below makes this convention explicit.
Worked Example: Splitting a Step at a Slope Discontinuity
Suppose \(y(0)=0\), and
A single step of length \(2\) evaluates the slope at \(t=0\), where it is \(1\). It therefore gives \(y_1=0+2(1)=2\) at \(t=2\). But it has carried the pre-switch slope over the interval after \(t=1\).
Instead, split the calculation at \(t=1\). For the first step, \(h=1\) and \(f(0,0)=1\), so
For the second step, start at \(t=1\). The specified branch there is \(f=4\), so
Here, the split result agrees with the exact change: the slope is \(1\) over a duration of \(1\), then \(4\) over a duration of \(1\), giving total change \(1(1)+4(1)=5\).
Check: The two Euler increments are \(1\) and \(4\), and \(0+1+4=5\). The unsplit update gives \(2\) because it uses only the first branch’s slope across a region where that branch no longer applies.
State-Dependent Singularities
Not every boundary is a fixed time. The slope might be undefined at certain values of \(y\). For example, \(f(t,y)=-2/y\) is undefined at \(y=0\). In such cases, checking only the time endpoints is not enough: the proposed change in \(y\) may reach or pass a state-space boundary.
The straight segment represented by an Euler step joins \((t_n,y_n)\) to \((t_{n+1},y_{n+1})\). If that segment crosses a boundary where \(f\) is undefined, the step should be treated with caution even when the endpoint value itself is not singular. The segment is a screening aid, not a proof of the unknown exact path; it can nevertheless reveal an obvious reason not to take the step as an ordinary update.
Worked Example: An Euler Endpoint Beyond a State Singularity
Consider \(y'=-2/y\) with \(y(0)=1\), and try \(h=1\). At the initial point, the slope exists:
The Euler update is
The endpoint \(y_1=-1\) is not itself a singular value. But the straight segment from \(y=1\) to \(y=-1\) passes through \(y=0\), where \(f\) is undefined. The step crosses the state boundary and should not be accepted as an ordinary step within the domain of this equation.
The exact equation gives an independent check on the obstruction. Multiplying by \(y\) and differentiating \(y^2\) along a solution gives
Using \(y(0)=1\), we obtain \(y^2=1-4t\). The positive solution reaches \(y=0\) when \(1-4t=0\), or \(t=1/4\), well before the proposed endpoint \(t=1\).
A shorter step of \(h=0.2\) gives \(y_1=1+0.2(-2)=0.6\), which stays positive at its endpoint. The predicted straight segment from \(1\) to \(0.6\) also stays away from zero. This avoids the immediate crossing seen with \(h=1\), but, as with any single-step check, it is not a general proof of accuracy.
Check: For \(h=1\), the endpoint calculation is \(1-2=-1\), and the straight segment crosses \(0\). For \(h=0.2\), the calculation is \(1-0.4=0.6\), and the segment stays between \(0.6\) and \(1\).
Why Regularity Supports the Euler Update
The reason to avoid discontinuities and singularities is visible in the exact change over a step. Suppose a solution \(y(t)\) exists on \([t_n,t_n+h]\), where \(h>0\), and write \(y_n=y(t_n)\) for this justification. The differential equation implies
Euler replaces the slope throughout the interval by its starting value, giving the increment \(h f(t_n,y(t_n))\). Subtracting this proposed increment from the exact one shows precisely what must be small for the step to be reasonable:
If the slope along the solution is continuous, its values near the start approach the starting slope as the interval shrinks. Thus the average slope over a sufficiently short interval approaches the value Euler uses. A discontinuity or singularity inside the interval defeats that simple comparison: the integrand may jump or fail to be defined.
A quantitative version follows if the slope along the solution changes at most linearly with time from the initial value. Suppose, for some constant \(K\geq 0\), that
Using the integral expression, the absolute difference between the exact and Euler increments is then bounded by
This bound explains the role of regularity: when the slope changes in a controlled way throughout a step, replacing it by its starting value incurs a controlled local error. If a step crosses an undefined or discontinuous point, the stated bound cannot be applied across that step.
A Practical Boundary Check
Before computing or accepting an Euler step near a suspected problem point, use the following checks. They assess whether the proposed interval stays within a region where the slope formula makes sense; they are not a substitute for a full error analysis.
Identify times or states where \(f(t,y)\) is undefined, and locate any known jumps in its value.
Verify that \(f(t_n,y_n)\) exists. If it does not, the Euler formula cannot be used at that point.
Determine whether the time interval crosses a known time boundary, or whether the predicted state movement reaches or crosses a state boundary.
For a known jump, end the step at the switch and begin a new step using the appropriate branch. Do not step through a singular point as if the slope remained regular.
Sometimes the location of a crossing is not known exactly. A computed endpoint alone cannot guarantee that the true solution stayed in the regular region between steps. In that situation, treat the numerical result cautiously rather than interpreting a successful calculation as evidence that no boundary was crossed.
Common Mistake: “The Formula Returned a Number”
A common error is to check only whether the slope at the starting point is defined. In the first example, \(f(0,0)=1\) exists, so \(h=1.2\) produces the number \(1.2\). But the step crosses \(t=1\), where the slope is undefined. The correction is not to perform the same update more confidently: it is to recognize that the proposed interval leaves the regular domain and keep the step on the valid side of the boundary.
A related mistake is to use a pre-discontinuity slope for a whole step that extends past a known jump. In the piecewise example, that gives \(2\) rather than the split calculation \(5\). The correction is to end a step at the switching time and apply the new branch after the switch. A computable result is not, by itself, proof that the chosen interval was appropriate.
Check Your Understanding
For each question, focus on whether the step remains in a region where the slope function is defined and regular.
- For \(y'=1/(2-t)\), why is a step from \(t=1\) to \(t=3\) problematic even if the starting slope is defined?
- If a slope changes from \(2\) to \(6\) at a known time, why should a step that spans that time be split?
- For \(y'=-1/y\), what state value must an Euler step avoid crossing?
- What is the difference between an Euler update being computable and being appropriate for approximating a solution over the step?
- In the derivation of the Euler increment error, what condition allows the slope difference to be bounded by \(Ks\)?
