Laplace of derivatives — key property for solving ODEs: Visual walkthrough

Step 0 — The four marks we are allowed to use

Before any algebra, let us agree on what the marks on the page mean, anchored to a picture.

Definition The cast of characters

$t$ — time, running left to right, starting at $0$ . Picture the horizontal axis.
$f (t)$ — some signal, e.g. a wobble that dies away. Picture a curve above the $t$ -axis.
$s$ — the Laplace-domain variable: a dial we turn, controlling how fast the curtain closes. Think of $s$ as a decay rate. Turn $s$ up and the curtain slams shut quickly; turn it down and it stays open longer. It is not time — it lives in a different world (the " $s$ -domain") and is the input to the output function $F (s)$ . Throughout we take $s > 0$ (later: $s$ large enough).
$e^{- s t}$ — a shrinking curtain built from $s$ . For $s > 0$ this starts at $1$ (when $t = 0$ ) and slides down toward $0$ as $t$ grows. It is our "weighting": it pays a lot of attention near $t = 0$ and almost none far to the right.
$\int_{0}^{\infty} (\dots) d t$ — "add up the area under the curve from $t = 0$ forever." The tall skinny $\int$ is a stretched letter S for Sum.

(That is four distinct marks — $t$ , $f$ , $s$ /its curtain $e^{- s t}$ , and the sum $\int$ — the curtain being just $s$ dressed up.)

Look at the figure: the cyan curve is $f (t)$ , the amber curve is the shrinking curtain $e^{- s t}$ , and the white shaded region is what happens when you multiply them — that shaded area is $F (s)$ .

WHY a curtain at all? Because $f (t)$ might not settle down as $t \to \infty$ , so its raw area could be infinite. The curtain $e^{- s t}$ crushes the far-away part to zero, making the total area finite — provided the dial $s$ is turned high enough to beat $f$ 's growth rate $c$ . This crushing is the secret ingredient we cash in at Step 4.

Step 1 — Ask the real question: what is the area of $f^{'} (t)$ under the curtain?

WHAT. We want $L {f^{'} (t)}$ — the same machine, but fed the slope $f^{'} (t)$ instead of $f (t)$ .

$L {f^{'} (t)} = \int_{0}^{\infty} curtain e^{- s t} slope of f f^{'} (t) d t$

WHY. Solving an ODE means we keep meeting $f^{'}$ , $f^{''}$ , ... We desperately want to express their transforms using $F (s)$ (the transform of plain $f$ ) so the calculus collapses into algebra. So the goal is: get $f^{'}$ out of the integral and $f$ back in.

PICTURE. $f^{'} (t)$ is the steepness of the cyan curve at each moment. Where $f$ climbs, $f^{'}$ is positive (above the axis); where $f$ falls, $f^{'}$ is negative (below). We are weighting that steepness by the curtain and adding it up.

Step 2 — Pick the tool: integration by parts (and why this tool)

We have a product being integrated: (curtain) $\times$ (a derivative). The one rule built exactly for "an integral containing a derivative" is Integration by Parts.

WHY not substitution or anything else? Substitution rewrites a variable; it cannot relocate a derivative between two factors. By-parts is the only elementary move that says "let me differentiate this piece instead of that piece." Since our whole aim is to un-prime $f$ , by-parts is forced.

We choose: $u = e^{- s t} (curtain), d v = f^{'} (t) d t (the derivative) .$

WHY this pairing? We set $d v = f^{'} (t) d t$ so that its antiderivative is $v = f (t)$ — the prime cancels itself when we integrate. That is the whole trick. Then: $d u = \frac{d}{d t} e^{- s t} d t = - s e^{- s t} d t, v = f (t) .$

The figure shows the split: amber is $u = e^{- s t}$ (which we will differentiate, gaining a $- s$ ), cyan is $d v = f^{'} (t) d t$ (which we will integrate, giving back the original $f$ ).

Step 3 — Apply the rule: the equation splits into two pieces

WHAT. Substitute our $u, v, d u, d v$ into $\int u d v = uv - \int v d u$ :

$L {f^{'} (t)} = boundary term uv [e^{- s t} f (t)]_{0}^{\infty} - - \int v d u \int_{0}^{\infty} f (t) (- s e^{- s t}) d t$

Read it term-by-term:

$[e^{- s t} f (t)]_{0}^{\infty}$ — the boundary term. It means "value of $e^{- s t} f (t)$ at the top end $\infty$ , minus its value at the bottom end $0$ ." A pure edge quantity, no integral left.
$- \int_{0}^{\infty} f (t) (- s e^{- s t}) d t$ — the leftover integral, but notice $f$ appears now with no prime — exactly what we wanted.

WHY. The prime is gone from the integral; it got paid for by (a) a boundary term and (b) a factor $- s$ dragged out front.

PICTURE. The single shaded region of Step 2 has been cut into two accounts: a thin edge contribution at $t = 0$ and $t = \infty$ (the boundary), and a bulk area that now contains plain $f$ .

Step 4 — Cash in the curtain: what the boundary term actually equals

Now we evaluate $[e^{- s t} f (t)]_{0}^{\infty}$ at its two ends. This is where every case must be checked.

Top end, $t \to \infty$ : $lim_{t \to \infty} e^{- s t} f (t) = 0.$ WHY it is zero: by our exponential-order fine print, $∣ f (t) ∣ \leq M e^{c t}$ , so $∣ e^{- s t} f (t) ∣ \leq M e^{- (s - c) t} \to 0$ whenever the dial $s > c$ . The curtain wins. (Degenerate warning below shows what breaks if $s \leq c$ .)

Bottom end, $t = 0$ : $e^{- s \cdot 0} f (0) = e^{0} f (0) = 1 \cdot f (0) = f (0) .$ WHY: at $t = 0$ the curtain is fully open ( $e^{0} = 1$ ), so it hides nothing — the raw starting value $f (0)$ shows through.

Therefore $[e^{- s t} f (t)]_{0}^{\infty} = top 0 - bottom f (0) = - f (0) .$

The degenerate case, drawn. What if the dial $s$ is set too low — below the growth rate $c$ ? Then $e^{- s t} f (t) = e^{- (s - c) t} \cdot (\dots)$ has a positive exponent and blows up instead of dying. The boundary never settles, and the whole transform integral diverges.

Step 5 — Recognise the leftover integral and assemble the result

WHAT. Clean up the leftover integral. Pull the constant $- s$ (and its minus, giving $+ s$ ) outside: $- \int_{0}^{\infty} f (t) (- s e^{- s t}) d t = + s \int_{0}^{\infty} e^{- s t} f (t) d t .$

The key recognition: that surviving integral is, symbol for symbol, the definition of $F (s)$ from Step 0: $\int_{0}^{\infty} e^{- s t} f (t) d t = F (s) .$

WHY this is the punchline. We started wanting $f^{'}$ expressed via $F (s)$ . Here it is — the integral folded back into $F (s)$ on its own. Substituting both pieces:

$L {f^{'} (t)} = Step 4 boundary - f (0) + Step 5 bulk s F (s) = s F (s) - f (0)$

Term-by-term of the final formula:

$s F (s)$ — "multiply the transform by $s$ ." Differentiation became multiplication. The $s$ came from differentiating the curtain in Step 2.
$- f (0)$ — the fee for starting the sum at $t = 0$ ; the initial value, courtesy of the fully-open curtain.

Step 6 — Do the exact same move twice: the second derivative

WHAT. We never need a new idea for $f^{''}$ . Just relabel $g = f^{'}$ and reuse Step 5's result on $g$ : $L {g^{'} (t)} = s L {g} - g (0) .$

Now unpack $g$ :

$g = f^{'} \Rightarrow g^{'} = f^{''}$ , so the left side is $L {f^{''}}$ .
$L {g} = L {f^{'}} = s F (s) - f (0)$ (Step 5 again).
$g (0) = f^{'} (0)$ — the starting slope.

Substitute: $L {f^{''}} = s (s F (s) - f (0)) - f^{'} (0) = multiply twice s^{2} F (s) - fee 1 s f (0) - fee 2 f^{'} (0) .$

WHY it looks like that. Each application of the rule adds one power of $s$ and subtracts one initial value, and the powers of $s$ ride down $s^{1}, s^{0}$ as the derivative order of the initial value rides up $f (0), f^{'} (0)$ .

The one-picture summary

One figure, the whole story: feed $f^{'}$ into the machine → by-parts splits it → the top edge dies (curtain wins, provided the dial $s > c$ ) → the bottom edge donates $- f (0)$ → the bulk folds back into $F (s)$ scaled by $s$ → out comes $s F (s) - f (0)$ .

Recall Feynman retelling — the walkthrough in plain words

We wanted the "area" of a signal's slope under a shrinking curtain, where a dial $s$ sets how fast the curtain shuts. There is exactly one calculus trick that moves a derivative from one thing onto another — integration by parts — so we used it, choosing the slope as the piece to integrate so its derivative-ness cancels and the plain signal comes back. That trick handed us two bills: a leftover area and a boundary charge measured only at the two ends. At the far end $t = \infty$ the curtain has crushed everything to zero — as long as we turned the dial past the signal's growth rate. At the near end $t = 0$ the curtain is fully open, so the signal's starting height $f (0)$ shows through — with a minus, because boundaries subtract bottom from top. The leftover area, once we pull the constant out front, is literally the definition of $F (s)$ again, now multiplied by $s$ . Glue the pieces: $s F (s) - f (0)$ . Differentiation turned into "times $s$ ", and the price of starting at zero is your initial condition $f (0)$ — which is exactly the thing that makes solving initial value problems so clean. Do the trick twice and you get $s^{2} F (s) - s f (0) - f^{'} (0)$ ; the pattern is just the same move, echoing.

Recall Self-check clozes

The tool that moves a derivative between factors is integration by parts. The variable $s$ is the ==Laplace-domain dial / decay rate that sets how fast the curtain $e^{- s t}$ closes. The top boundary vanishes because the curtain $e^{- s t}$ decays faster than $f$ grows, i.e. $∣ e^{- s t} f ∣ \leq M e^{- (s - c) t} \to 0$ when the dial $s > c$ ==. The term $- f (0)$ appears because ==at $t = 0$ the curtain equals $1$ , so the raw starting value shows through, and boundaries subtract the bottom end==.

Which symbol turned differentiation into multiplication, and where did it come from?

The factor

s

, produced by differentiating the curtain

e^{- s t}

during integration by parts.

State the first-derivative rule and name each term.

L {f^{'}} = s F (s) - f (0)

;

s F (s)

= "differentiation is multiply-by-

s

- f (0)

= boundary fee at

t = 0

Why does the second-derivative rule need no new idea?

You apply the first-derivative rule to

g = f^{'}

, then substitute

L {f^{'}} = s F (s) - f (0)

and

g (0) = f^{'} (0)

For what

s

is the rule valid, and why?

For

s

greater than

f

's growth rate

c

(region of convergence), so the boundary term at

\infty

vanishes and the integral converges.

Connections

Parent topic (Hinglish)
Laplace Transform — Definition and Existence
Integration by Parts
Exponential Order and Convergence of Integrals
Solving Initial Value Problems with Laplace
Inverse Laplace Transform and Partial Fractions
Linearity of the Laplace Transform
Fourier Transform — comparison (no boundary term)