From One-Step Accuracy to Global Accuracy
Forward Euler replaces a solution curve by a succession of straight segments. Each step is a small approximation, but the errors from many steps can accumulate. The key question is how the error at one step affects the next.
For a fixed time interval, forward Euler has first-order convergence when its maximum error at the grid points is bounded by a constant times the step size \(h\). The reason is that a sufficiently smooth solution has a one-step defect of size proportional to \(h^2\), while stability ensures that earlier errors are not amplified too quickly. There are about \(1/h\) steps across a fixed interval, so the accumulated error is proportional to \(h\), not \(h^2\).
Set-Up and Assumptions
Consider the initial-value problem \(y'=f(t,y)\), \(y(a)=y_0\), on a fixed interval \([a,b]\). Use the uniform grid \(t_n=a+nh\), where \(h=(b-a)/N>0\), for \(n=0,\ldots,N\). Forward Euler computes \(Y_0=y_0\) and then applies \(Y_{n+1}=Y_n+h f(t_n,Y_n)\).
Two conditions support the argument. First, suppose the exact solution has a bounded second derivative on the interval: \(|y''(t)|\le M\). Taylor’s theorem then gives a one-step defect bounded by \(Mh^2/2\). Second, assume that perturbations in a single Euler step satisfy a stability estimate with some constant \(K\ge 0\): for values in the relevant range,
This estimate says that one Euler step can increase the difference between two starting values by at most the factor \(1+Kh\). A common sufficient condition is that the derivative of \(f\) with respect to its second variable is bounded on the relevant range; the mean value theorem then bounds the change in \(f(t,y)\) as \(y\) changes. Here we use the stability estimate directly.
Why the One-Step Defect Is Quadratic
Start one Euler step at the exact value \(y(t_n)\), rather than at the computed value \(Y_n\). The exact solution satisfies \(y'=f(t,y)\), so Taylor’s theorem gives
The Euler step from that exact starting value would be \(y(t_n)+h f(t_n,y(t_n))\). Thus \(r_{n+1}\) is the difference between the exact next value and that one-step Euler prediction. It is bounded by a constant times \(h^2\), uniformly across the interval. This is the local defect before errors from earlier steps enter.
Deriving the Error Recurrence
Subtract the Euler update from the exact solution’s one-step relation. Since \(e_n=y(t_n)-Y_n\), the result is
The stability assumption bounds the first two terms together: they are the difference between an Euler step starting at \(y(t_n)\) and one starting at \(Y_n\). Therefore
This recurrence displays the two sources of error at each step. Existing error is carried forward with a controlled amplification factor, and a new defect of size at most \(Mh^2/2\) is added.
From the Recurrence to First-Order Convergence
To prove the theorem, write \(C=M/2\), so that \(|e_{n+1}|\le(1+Kh)|e_n|+Ch^2\). Since \(e_0=0\), repeated application gives, for each \(n\ge1\),
For \(x\ge0\), \(1+x\le e^x\), so \((1+Kh)^n\le e^{Knh}\). Also \(nh\le T\), and \(n\le T/h\). Consequently,
The same bound holds for \(e_0=0\), so it holds for the maximum over all grid points. The constants \(M\), \(T\), and \(K\) do not depend on \(h\). That independence is essential: it is what makes the estimate a convergence result as the grid is refined.
The proof also explains the order. There are \(N=T/h\) steps, and each contributes a defect of order \(h^2\). If those contributions are amplified by at most a bounded factor over the fixed interval, their total has order \(Nh^2=(T/h)h^2=Th\).
Worked Example: An Equation with an Exact Global Error
Worked Example: An Equation with an Exact Global Error
Apply forward Euler to \(y'=t\), \(y(0)=0\), on \([0,1]\) with \(N=4\), so \(h=1/4\). The exact solution is \(y(t)=t^2/2\), since its derivative is \(t\) and \(y(0)=0\).
At each grid point \(t_n=n/4\), the update is \(Y_{n+1}=Y_n+t_nh\). Starting from \(Y_0=0\), the successive computations are
The exact endpoint value is \(y(1)=1/2\), so the endpoint error is \(1/2-3/8=1/8\). In fact, the error at every grid point can be computed: \(Y_n=h^2\sum_{k=0}^{n-1}k=h^2n(n-1)/2\), while \(y(t_n)=n^2h^2/2\). Subtracting gives \(e_n=nh^2/2=t_nh/2\), which is nonnegative and at most \(h/2\). At \(n=4\), this gives \(e_4=(1)(1/4)/2=1/8\), matching the endpoint calculation. This example has a global error exactly proportional to \(h\).
Worked Example: Local Defects and Their Propagation
Worked Example: Local Defects and Their Propagation
Consider \(y'=y\), \(y(0)=1\), for \(0\le t\le1\), using \(h=1/2\). The exact solution is \(y(t)=e^t\). Forward Euler gives \(Y_0=1\), then \(Y_1=1+(1/2)(1)=3/2\), and finally \(Y_2=3/2+(1/2)(3/2)=9/4\).
The exact endpoint is \(e\), so the endpoint error is \(e-9/4\). To see how the recurrence accounts for it, the one-step defect starting at the exact value at \(t_n\) is \(r_{n+1}=e^{t_n}(e^h-1-h)\). With \(h=1/2\),
Here the Euler update multiplies an existing error by \(1+h=3/2\). Thus \(e_1=r_1=e^{1/2}-3/2\), and
This agrees with subtracting the computed endpoint \(Y_2=9/4\) from the exact endpoint \(e\). The recurrence shows both the first defect and how the second step carries that error forward while adding another defect.
Worked Example: A Contracting Step
Worked Example: A Contracting Step
Consider \(y'=-y\), \(y(0)=1\), on \([0,1]\), again with \(h=1/2\). The exact solution is \(y(t)=e^{-t}\). The Euler update is \(Y_{n+1}=Y_n-(1/2)Y_n=(1/2)Y_n\), so \(Y_0=1\), \(Y_1=1/2\), and \(Y_2=1/4\). The exact endpoint is \(e^{-1}\), giving \(e_2=e^{-1}-1/4\).
In this case, the error multiplier per step is \(1-h=1/2\). The exact one-step defect is \(r_{n+1}=e^{-t_n}(e^{-h}-1+h)\). Therefore
The recurrence reproduces the endpoint error exactly. Here each step reduces an existing error by a factor of \(1/2\); stability does not require errors to grow, only that their amplification be controlled.
A Common Order-Counting Mistake
A common mistake is to conclude that because the one-step defect is \(O(h^2)\), the global error must also be \(O(h^2)\). That ignores how many steps are taken. For instance, over an interval of length \(1\) with \(h=1/100\), there are \(100\) steps. If each defect is bounded by \(Ch^2\), adding the bounds gives \(100C(1/100)^2=C/100=Ch\), not \(Ch^2\).
This addition gives the right order when error propagation is controlled. Stability matters because defects from earlier steps are carried forward; they cannot simply be counted as though they disappear. The recurrence makes that propagation explicit and shows why a uniform stability bound still yields an overall error of order \(h\).
What First-Order Convergence Does and Does Not Say
The conclusion \(E_{\max}\le C_T h\) is an upper bound as the step size becomes small. It does not say that the error is exactly \(C_T h\), that the error has the same sign at every grid point, or that halving \(h\) must halve the error in every individual computation. The constant depends on the interval, the solution’s smoothness, and the stability bound. If the solution lacks the required smoothness or the numerical steps do not satisfy a suitable stability estimate, this argument does not establish first-order convergence.
First-order convergence is useful because it gives a precise meaning to the accuracy of the method under stated assumptions. It also gives a benchmark for interpreting computations at smaller step sizes. The key reasoning is the combination of a quadratic one-step defect and controlled propagation across a number of steps proportional to \(1/h\).
Check Your Understanding
Use the recurrence and the examples to answer these questions.
- Why does a local defect bounded by a constant times \(h^2\) lead to a global error of order \(h\) over a fixed interval?
- What role does the stability factor \(1+Kh\) play in the error recurrence?
- For \(y'=t\), \(y(0)=0\), what is the Euler error at a grid point \(t_n\) with step size \(h\)?
- In the example \(y'=y\), how does the second step combine the error already present with the new one-step defect?
- Why does a one-step error estimate alone not establish first-order convergence?
