From One-Step Error to Error Across an Interval
A small step size makes an individual Euler step more accurate. But a numerical solution over a whole interval uses many steps, so its total error depends both on the error made in each step and on how errors already present affect later steps.
For Euler’s method, a single step started from the exact solution has an error of size \(h^2\), where \(h\) is the step size. Over an interval of fixed length, however, the number of steps is proportional to \(1/h\). Under suitable smoothness and stability conditions, these local errors accumulate to a global error of size \(h\). This is what it means for Euler’s method to have first-order global accuracy.
Set Up the Error Carefully
Consider the initial-value problem \(y'=f(t,y)\), \(y(a)=y_0\), on a fixed interval \([a,b]\). Let \(y(t)\) be its exact solution. Divide the interval into \(N\) equal steps, so \(h=(b-a)/N\) and \(t_n=a+nh\). Euler’s method computes
Define the signed global error at the \(n\)th grid point by \(e_n=y(t_n)-Y_n\). To control its growth, suppose the following conditions hold throughout the interval and the region containing the exact and numerical values:
- The exact solution has a continuous second derivative, with \(|y''(t)|\le M\) for some constant \(M\).
- The slope function is Lipschitz continuous in \(y\): for the relevant \(t,u,v\), \(|f(t,u)-f(t,v)|\le L|u-v|\), where \(L\ge 0\).
The second condition limits how much a difference in solution values can change the slope. It is the stability control needed to bound the effect of errors already accumulated. The fixed interval and these bounds are essential: the resulting error constant may depend on \(M\), \(L\), and \(b-a\), but not on \(h\).
Why a Single Exact-Start Step Has Error of Size \(h^2\)
Start one Euler step at the exact value \(y(t_n)\). Its prediction is \(y(t_n)+h f(t_n,y(t_n))\). Since \(y'=f(t,y)\), the difference between the exact next value and that prediction is
The bound on \(y''\) gives a precise estimate. Applying the integral form of the Taylor remainder, or integrating the change in \(y'\) over the step, gives
Thus
This is the local one-step error bound when the starting value is exact. In repeated Euler steps, though, a step usually starts from \(Y_n\), not \(y(t_n)\). The difference between those starting values must be included.
Deriving the Global Error Recurrence
Add and subtract the Euler slope evaluated at the exact starting value. Using the definition of \(e_n\), write
These three terms have distinct roles: the local defect of the exact-start step, the error already present at time \(t_n\), and the effect of that starting-value error on the slope. Apply the triangle inequality, the local bound, and the Lipschitz condition:
This recurrence explains the accumulation. Each step adds a fresh error bounded by \(Mh^2/2\), while the factor \(1+Lh\) accounts for possible growth of the error carried into that step.
From the Recurrence to a First-Order Bound
Because the initial value is exact, \(e_0=0\). Repeatedly applying the recurrence gives, for \(L>0\),
Since \(1+Lh\le e^{Lh}\) and \(nh=t_n-a\), this yields the convenient bound
For \(L=0\), the recurrence instead gives \(|e_n|\le nMh^2/2\). Since \(nh=t_n-a\), this becomes \(|e_n|\le M(t_n-a)h/2\). In either case, on a fixed interval there is a constant \(C\), independent of \(h\), such that \(|e_n|\le Ch\) at every grid point. With the initial value exact and the stated bounds, this proves first-order global accuracy.
Worked Example: Local Errors Accumulate Exactly to \(h/2\)
Worked Example: Local Errors Accumulate Exactly to \(h/2\)
Apply Euler’s method to \(y'=t\), \(y(0)=0\), on \([0,1]\), using \(N\) equal steps. Here \(h=1/N\). The exact solution is \(y(t)=t^2/2\), since its derivative is \(t\) and it satisfies the initial condition.
At step \(n\), Euler’s update is \(Y_{n+1}=Y_n+h t_n\), with \(t_n=nh\). Therefore
The exact value at \(t=1\) is \(y(1)=1/2\). The final error is consequently
We can also check the error step by step. On a step starting at \(t_n\), the exact change is \(\bigl((t_n+h)^2-t_n^2\bigr)/2=t_nh+h^2/2\). Euler’s change is \(t_nh\), so the exact-start local defect is \(h^2/2\) on every step. There are \(N=1/h\) steps, giving total error \(N(h^2/2)=h/2\), as found directly.
Finally, \(y''(t)=1\), so \(M=1\), and \(f(t,y)=t\) is independent of \(y\), so \(L=0\). The bound for \(L=0\) gives \(|e_N|\le M(1-0)h/2=h/2\). The computed error equals that bound. This example makes clear how \(N\) local errors of size \(h^2\) can produce a global error of size \(h\).
Worked Example: A Bound for Euler’s Approximation to \(y'=y\)
Worked Example: A Bound for Euler’s Approximation to \(y'=y\)
Consider \(y'=y\), \(y(0)=1\), on \([0,1]\), with \(h=1/4\). The exact solution is \(y(t)=e^t\). Euler’s values are computed by \(Y_{n+1}=Y_n+(1/4)Y_n=(5/4)Y_n\):
The exact endpoint value is \(e\approx2.718281828\), so the endpoint error is
For the bound, \(f(t,y)=y\) has Lipschitz constant \(L=1\). Also \(y''(t)=e^t\le e\) on \([0,1]\), so we may take \(M=e\). The finite geometric-sum bound gives
The actual error, approximately \(0.2769\), is less than the bound, approximately \(0.4898\). The bound guarantees control, not an exact prediction of the error. In particular, a valid upper bound can be substantially larger than the error it controls.
A Common Mistake: Dropping the Existing Error
A tempting but incorrect argument compares the next exact value only with the Euler step starting from the exact value, then adds a slope-difference term. That omits the difference between the exact and numerical starting values at the current step. The missing term is \(e_n\).
Worked Example: Correcting an Incomplete Recurrence
For the problem \(y'=t\), \(y(0)=0\), take \(h=1/4\). The exact solution is \(y(t)=t^2/2\), and Euler’s method gives \(Y_0=0\), \(Y_1=0\), \(Y_2=1/16\). At \(t_2=1/2\), the exact value is \(y(t_2)=1/8\), so \(e_2=1/8-1/16=1/16\).
An incomplete estimate would say that the next error is at most the new local defect \(h^2/2=1/32\), because \(L=0\). But the actual next error is \(e_3=1/2-y(3/4)\text{’s Euler approximation}\). Euler gives \(Y_3=Y_2+(1/4)(1/2)=1/16+1/8=3/16\); the exact value is \(y(3/4)=9/32\). Hence
which is greater than \(1/32\). The corrected recurrence includes the error already present:
Here \(M=1\), \(L=0\), and \(|e_2|=1/16\), so it gives \(|e_3|\le1/16+1/32=3/32\), exactly matching the computed error. This check shows why the starting-value difference cannot be left out.
What “First Order” Does and Does Not Claim
A bound of the form \(|e_n|\le Ch\) says that reducing \(h\) reduces the guaranteed global error proportionally, with the constant \(C\) fixed for the problem and interval. For example, halving \(h\) halves this upper bound. The result does not say the error is exactly \(Ch\), nor that it must be nonzero. Some problems have cancellations or special structure that make Euler’s error smaller; if the slope is constant, Euler’s method follows the exact solution with no error.
The essential reasoning is the balance between two quantities: each exact-start step contributes an error proportional to \(h^2\), while a fixed interval contains approximately \(1/h\) steps. The Lipschitz condition ensures that previously accumulated error does not grow without control as those steps are repeated. Together, these facts give the first-order global bound.
Check Your Understanding
Answer these questions using the definitions and error recurrence from this tutorial.
- How does the local one-step error differ from the global error at a grid point?
- Why does the local bound \(Mh^2/2\) alone not bound the global error after several steps?
- For \(L>0\), what recurrence bounds \(|e_{n+1}|\) in terms of \(|e_n|\), \(M\), \(L\), and \(h\)?
- For \(y'=t\), \(y(0)=0\) on \([0,1]\), what is the global error after \(N\) equal Euler steps?
- Does first-order global accuracy mean that every problem has a nonzero error proportional to \(h\)? Explain briefly.
