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Laplace Transforms & Euler Steps

Useful preparation: Second-Order Equations & Linear Systems

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Meaning & notation

A Laplace transform converts a time-dependent equation into an algebraic equation in a new variable s. Integration by parts turns a derivative into multiplication by s and an initial-value term. Euler stepping approaches the same initial-value problem locally: follow the current tangent for a short time, then recompute the slope at the new approximate point.

Unilateral Laplace transform: For a function f defined for nonnegative time, its transform F(s) is the integral of $\displaystyle \operatorname{exp}\, \left(-st\right)\, f\, \left(t\right)$ from zero to infinity, whenever that improper integral converges.
Derivative rule: If Y is the transform of y and the relevant integrals exist, the transform of y prime is sY-y(0). Initial conditions therefore become part of the algebra.
Explicit Euler method: For y prime=f(t,y), step size h gives y at the next grid point approximately as the current y plus h times the current slope.

Conditions before calculation

Piecewise continuity on finite intervals and an exponential growth bound are sufficient for transform existence for sufficiently large real s. Transform existence is not automatic.
The derivative rule requires suitable regularity of y and convergence of the transforms, with the boundary term at infinity vanishing; use the right-hand initial value at zero.
Euler is a numerical approximation. Under standard smoothness and Lipschitz assumptions it has global error of order h over a fixed finite interval; stability can still force a small step.

Why the method works

1. Keep the initial term

Integration by parts explains the minus sign at zero. Dropping this term solves a different initial-value problem.

\mathcal L\{y'\}=\int_0^\infty e^{-st}y'(t)\,dt=sY(s)-y(0)

2. Solve and invert

After algebraically isolating Y, split it into known transforms. Check the recovered time function in both the original equation and the initial condition.

3. Assess numerical behavior

Euler uses the slope at the old state. For decay y prime=-k y with positive k, the multiplier must have magnitude below one to decay numerically.

y_{n+1}=y_n+h f(t_n,y_n),\qquad |1-kh|<1

Worked examples

EXAMPLE 1

Use a Laplace transform to solve y prime+ $\displaystyle 3\, y = 6$ with $\displaystyle y\, \left(0\right) = 1$ .

Transform both sides, retaining the initial value and the transform of a constant.
$(sY-1)+3Y=\frac6s$
Isolate Y and decompose it into two known transforms.
$Y=\frac{1+6/s}{s+3}=\frac2s-\frac1{s+3}$
Invert and verify the differential equation and initial value.
$y(t)=2-e^{-3t},\qquad y'+3y=3e^{-3t}+6-3e^{-3t}=6,\quad y(0)=1$

Conclusion: The solution starts at 1 and approaches equilibrium 2 for increasing nonnegative time.

EXAMPLE 2

Take two Euler steps for y prime=-3y, $\displaystyle y\, \left(0\right) = 2$ , with $\displaystyle h = 0.1$ .

The current-slope rule gives a multiplier of 0.7.
$y_{n+1}=y_n-0.3y_n=0.7y_n$
Apply the update twice, always using the previous approximate value.
$y_1=1.4,\qquad y_2=0.98,\qquad t_2=0.2$
Compare with the exact exponential at the same time.
$y(0.2)=2e^{-0.6}\approx1.097623,\qquad |y_2-y(0.2)|\approx0.117623$

Conclusion: Euler underestimates this decaying exponential here; the computed value is an approximation, even though the update arithmetic is exact.

Pitfalls & counterexamples

Tempting claim: A stable differential equation stays stable under any Euler step size.

Correction: The discretization has its own stability restriction.

Test the claim: For y prime=-3y, a step $\displaystyle h = 1$ gives multiplier -2, so the numerical magnitudes double while the exact solution decays.

Check your understanding

What is the Laplace transform of y double prime when $\displaystyle y\, \left(0\right) = a$ and y prime(0)=b?

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Compare your reasoning

s squared Y minus sa minus b.

Apply the first-derivative rule twice, retaining both initial values.

Connect this idea

Transforms are especially useful for linear systems and switched forcing. Euler links derivatives to algorithms; better accuracy and stiffness handling motivate higher-order and implicit methods.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.