Moving Earlier in Time with Euler’s Method
Euler’s method is usually introduced as a way to approximate a solution at later times from its value at an initial time. The same update formula can also move toward earlier times: choose a negative step length, evaluate the slope at the current point, and apply the update.
The arithmetic is simple, but the consequences are not always symmetric. A negative step changes how the numerical updates amplify errors, and a method that damps errors while moving forward may amplify them while moving backward. The direction of the step must therefore be considered both in the calculation and in assessing reliability.
Here, “forward Euler” names the explicit update rule, not the direction in which time must increase. If \(h<0\), then \(t_{n+1}<t_n\). Starting with data at a later time \(T\), repeated updates approximate the solution at decreasing times. The slope is evaluated at the current time and current numerical value, not at the new time.
Setting Up a Backward Calculation
Suppose the known value is \(y(T)=Y_0\), and the desired time is \(T-L\), where \(L>0\). If the interval is divided into \(N\) equal steps, choose \(h=-L/N\). Then
The time decreases by \(|h|\) at every update. The minus sign is already part of \(h\), so it must be used in the product \(h f(t_n,Y_n)\). Do not separately subtract a slope and also use a negative step: that would apply the sign twice.
Worked Example: Two Negative Steps for a Time-Dependent Slope
Approximate the solution of \(y'=t\), with \(y(2)=5\), at times \(1.5\) and \(1\), using \(h=-0.5\).
At the initial point, \(t_0=2\), \(Y_0=5\), and \(f(t_0,Y_0)=2\). Thus
At the next point, \(t_1=1.5\), so the slope is \(f(t_1,Y_1)=1.5\). The second update is
To check the approximations, the exact solution follows from integrating \(y'=t\): \(y(t)=t^2/2+C\). The condition \(y(2)=5\) gives \(5=2+C\), so \(C=3\). Therefore \(y(1.5)=1.125+3=4.125\), and \(y(1)=0.5+3=3.5\). Euler gives \(4\) and \(3.25\), respectively; substituting those values into the two Euler updates reproduces them exactly. The discrepancy from the exact values is numerical error, not an additional sign adjustment.
Why the Error Still Has First-Order Global Size
Negative steps do not change the basic local accuracy of Euler’s formula. To see why, let \(y(t)\) be an exact solution with bounded second derivative on the interval, and consider one signed step \(h\), which may be negative. Taylor’s formula about the starting time gives
if \(|y''(s)|\le M\) throughout the step. Since \(y'(t)=f(t,y(t))\), the exact solution advances by \(h f(t,y(t))\) plus a remainder bounded by \(M h^2/2\). The bound involves \(h^2\), so its size is the same for either direction. This is the local truncation error: the one-step discrepancy when the step starts from the exact value.
To understand accumulated error, let \(e_n=Y_n-y(t_n)\) be the numerical error at step \(n\), and assume \(f\) is Lipschitz in its second argument on the region of interest: for some \(K\ge 0\), \(|f(t,u)-f(t,v)|\le K|u-v|\). Denote the one-step Taylor remainder at \(t_n\) by \(R_n\). Subtracting the exact one-step relation from the Euler update gives
Taking absolute values, applying the Lipschitz condition, and using \(|R_n|\le M h^2/2\) yields
In particular, the negative sign of \(h\) does not justify dropping its magnitude in the error estimate. If the starting value is exact, \(e_0=0\), repeated application of the inequality gives, for \(K>0\),
On an interval of fixed length \(L\), \(n|h|\le L\), and \((1+K|h|)^n\le e^{Kn|h|}\le e^{KL}\). Hence the global error is bounded by a constant times \(|h|\). If \(K=0\), the recurrence instead gives \(|e_n|\le nMh^2/2\le LM|h|/2\), with the same order. Thus, under these smoothness and Lipschitz assumptions, backward Euler updates using the explicit formula have first-order global accuracy as the step size tends to zero. This accuracy statement does not guarantee stability for a particular step size.
Stability Can Change with the Direction
For the scalar equation \(y'=\lambda y\), one Euler step multiplies the current value by \(1+h\lambda\). The exact solution over that step multiplies it by \(e^{\lambda h}\). For a real negative \(\lambda\), forward steps can damp the numerical value when \(|1+h\lambda|<1\). If the same equation is stepped backward with \(h<0\), then \(h\lambda>0\), so \(1+h\lambda>1\): each step amplifies a perturbation in the numerical value. The exact solution also grows when followed backward in time. Euler’s multiplier can nevertheless produce growth at a different rate, affecting errors and the quality of the approximation.
Worked Example: A Decaying Equation Followed Backward
Use two steps of size \(h=-0.25\) to approximate the solution of \(y'=-2y\) at \(t=0.5\), given \(y(1)=1\).
At \(t_0=1\), \(Y_0=1\), the slope is \(f(t_0,Y_0)=-2\). Therefore
At the new point, \(f(t_1,Y_1)=-2(1.5)=-3\). The next update gives
The exact solution is \(y(t)=e^{-2(t-1)}\), so \(y(0.5)=e\), approximately \(2.718\) to three decimal places. The numerical multiplier at each backward step is \(1+h\lambda=1+(-0.25)(-2)=1.5\), confirming the computed amplification \(1\cdot1.5\cdot1.5=2.25\). In contrast, with a forward step of \(0.25\), the multiplier is \(1+0.25(-2)=0.5\), which damps values. The equation decays forward and grows backward; the step direction changes the numerical behaviour accordingly.
The reverse direction can have the opposite stability character for a growing equation. For example, with \(y'=2y\), the backward-step multiplier for \(h=-0.25\) is \(1+(-0.25)(2)=0.5\), which damps numerical perturbations. Whether that damping also gives an accurate answer depends on the step size and the solution’s local curvature; stability and accuracy are related but distinct questions.
Worked Example: A Growing Equation Followed Backward
Worked Example: Stepping Backward with a Damping Multiplier
Approximate the solution of \(y'=2y\) at \(t=0.5\), given \(y(1)=4\), using \(h=-0.25\).
The first slope is \(2(4)=8\). Hence
At \(t_1=0.75\), the slope is \(2(2)=4\). Thus
The exact solution satisfying the data is \(y(t)=4e^{2(t-1)}\). At \(t=0.5\), it equals \(4e^{-1}\), approximately \(1.472\) to three decimal places. The two Euler multipliers give \(4(0.5)(0.5)=1\), agreeing with the updates. This example has a damping multiplier in the backward direction, yet the approximation still differs noticeably from the exact value. Damping by itself is not an accuracy guarantee.
A Negative Step Does Not Undo an Euler Step
A common mistake is to assume that stepping forward and then using the same formula with a negative step must return to the starting value. That is true for reversing an exact solution map, but it is generally false for Euler’s approximate map.
For a concrete check, take \(y'=y\), start with \(y(0)=1\), and step forward by \(h=0.2\). The result is \(Y_1=1+0.2(1)=1.2\) at \(t=0.2\). Now use \(h=-0.2\) from that numerical value. The slope is \(1.2\), so the new value is \(1.2+(-0.2)(1.2)=0.96\), not \(1\). The product of the two Euler multipliers is \(1.2(0.8)=0.96\), not \(1\). The correction is to treat the negative step as a new approximation to the reverse-time solution, not as an exact undo operation.
Practical Checks Before Interpreting a Backward Approximation
A reliable calculation begins by checking the setup. Confirm that the known value is at the later starting time, that the time grid decreases, and that the signed step is used exactly once in each update. Then consider the differential equation’s behaviour in the reverse direction: a solution that decays forward may grow backward, and perturbations can grow with it.
- Write the actual times \(t_n\) before computing the slopes.
- Evaluate \(f\) at the current pair \((t_n,Y_n)\), then multiply by the negative \(h\).
- Check whether the scalar stability multiplier \(1+h\lambda\), when applicable, damps or amplifies values.
- Compare with an exact solution or a refined computation when one is available; stability alone does not measure accuracy.
Negative-step Euler is useful when a solution value is known at a later time and an earlier value is needed. Its update is the ordinary explicit Euler formula with a signed step. The local Taylor error remains proportional to \(h^2\), and under standard smoothness and Lipschitz conditions the global error remains first order in \(|h|\). What can change sharply is stability: following the equation backward may amplify perturbations that forward integration damps.
Check Your Understanding
For each question, pay attention to the signed time step as well as the slope.
- For \(y'=f(t,y)\), write one explicit Euler update that moves from \(t_n\) to an earlier time.
- Starting from \(t=3\), what is the next time after one step with \(h=-0.2\)?
- For \(y'=-3y\) and \(h=-0.1\), what is the Euler multiplier, and does it amplify or damp a perturbation?
- Why does the one-step Taylor error remain bounded by a constant times \(h^2\) when \(h<0\)?
- Does applying a negative Euler step generally undo a positive Euler step exactly? Give the reason.
