JOVANA
Library Glossary Getting Started Three Levels Fields How it works Mission
Join the mission
All guides

The Laplace Transform

Multiply a function by a decaying exponential, integrate it away to infinity, and a function of time t becomes a function of a new variable s. One integral, and the calculus of differential equations turns into the algebra of polynomials in s.

An integral that changes what a function is made of

You have spent a whole rung learning to *solve* differential equations — guessing forms, building characteristic equations, varying parameters. The Laplace transform offers a completely different deal. Instead of solving a differential equation in its own language of calculus, it carries the whole problem across a bridge into a new world where the equation is no longer differential at all — it is just ordinary algebra. You solve the easy algebra there, then carry the answer back. The bridge is a single integral, and learning to walk across it is the entire subject of this rung.

Here is the bridge itself. Given a function f(t) defined for t >= 0 — think of t as time, with t = 0 the moment you switch the system on — its Laplace transform is F(s) = integral from 0 to infinity of e^{-s t} f(t) dt. That is the whole definition of the Laplace transform. You take your function of time, multiply it by the decaying exponential e^{-s t}, and integrate the product all the way out to infinity. What comes out no longer depends on t at all — t has been integrated away. It depends only on the new variable s, so we write the result as F(s) and call s the transform variable.

Notice the shape of this operation. The integral runs over t and produces an answer that is a function of s — exactly the pattern you met when the Gamma function turned an integral over t into a function of its parameter. The transform is a machine that eats a whole function and spits out a whole function: feed it f(t), receive F(s). We will write that as L{f(t)} = F(s), and you should picture two parallel worlds — the t-domain where functions live in time, and the s-domain where their transforms live. Everything we do is move a problem from the first to the second and back.

The first transforms, computed by hand

The definition is an improper integral, so let us actually evaluate one and feel it work. Take the simplest function, f(t) = 1 for all t >= 0. Then F(s) = integral from 0 to infinity of e^{-s t} dt. The antiderivative of e^{-s t} is -(1/s) e^{-s t}, and as t runs to infinity, e^{-s t} dies to zero — *provided s is positive*, so the exponential really decays. At the top the term vanishes; at t = 0 it equals -(1/s). Subtracting, F(s) = 0 - (-(1/s)) = 1/s. So L{1} = 1/s. One constant function in time has become one simple reciprocal in s.

Now the single most important pair. Take f(t) = e^{a t}, a growing or decaying exponential. Then e^{-s t} f(t) = e^{-s t} e^{a t} = e^{-(s - a) t}, and the integral is exactly the one we just did but with s replaced by s - a. So L{e^{a t}} = 1/(s - a), valid this time when s > a so that s - a > 0 keeps the exponential decaying. Stare at this: the exponential, the function that *is* its own derivative up to a constant and was the hero of every constant-coefficient ODE, transforms into the dead-simple algebraic object 1/(s - a). That single fact — exponentials become simple poles in s — is the secret engine behind the transform's power over differential equations.

L{e^{a t}} = integral_0^infinity e^{-s t} e^{a t} dt
           = integral_0^infinity e^{-(s - a) t} dt

              [   -1               ]^{t -> infinity}
           = [ ------- e^{-(s-a)t} ]
              [  s - a             ]_{t = 0}

           = 0  -  ( -1/(s - a) )      ( needs s - a > 0 )
           = 1/(s - a)

  Set a = 0  ->  L{1} = 1/s.
  Replace a by i*omega and take parts  ->  L{cos(omega t)} = s/(s^2 + omega^2),
                                          L{sin(omega t)} = omega/(s^2 + omega^2).
The exponential pair, computed straight from the definition. Every other elementary pair — constants, sines, cosines — falls out of this one by choosing a special value of a.

Where the integral is allowed to live: convergence

You felt the catch already: each computation came with a fine-print condition — s > 0 for the constant, s > a for the exponential. This is not a nuisance; it is a real and necessary part of the definition. An improper integral out to infinity only gives a finite number if the integrand shrinks fast enough. The factor e^{-s t} is our shrinking agent, and whether it wins depends on how fast f(t) grows. The set of s for which the defining integral converges is called the region of convergence, and outside it the transform F(s) simply does not exist as a number.

The pattern is wonderfully simple. For a function of exponential order — one that grows no faster than some C e^{a t} for large t — the region of convergence is a half-plane, every s with real part Re(s) > a. The threshold a is called the abscissa of convergence; it sits just to the right of wherever f(t) is fastest-growing. For f(t) = e^{3 t} the region is Re(s) > 3; for a bounded function like sin(t) it is Re(s) > 0; for a polynomial, which grows slower than any exponential, it is again Re(s) > 0. Picture a vertical line in the s-plane sweeping rightward: the transform lives on everything to the right of the line where f stops being tameable.

The table of pairs, and why it is enough

In practice you almost never compute the defining integral. Once a handful of basic transforms are known, you assemble everything else from them, because the transform inherits linearity straight from the integral: L{c_1 f + c_2 g} = c_1 L{f} + c_2 L{g}. So you keep a short table of transform pairs and read it both directions. The core entries are exactly the ones we earned above and their close relatives: powers of t, exponentials, and the trigonometric pair that comes from feeding an imaginary exponent into L{e^{a t}}.

Here are the workhorses, each on its half-plane. L{1} = 1/s and more generally L{t^n} = n!/s^{n+1} for a whole number n — and for non-integer powers the factorial becomes the Gamma function, L{t^p} = Gamma(p+1)/s^{p+1}, a lovely place where two of this Volume's threads meet. Then L{e^{a t}} = 1/(s - a); L{cos(omega t)} = s/(s^2 + omega^2); and L{sin(omega t)} = omega/(s^2 + omega^2). A growing or decaying oscillation, e^{a t} times a sine or cosine, just shifts the cosine pair: L{e^{a t} cos(omega t)} = (s - a)/((s - a)^2 + omega^2). That last move — multiplying by e^{a t} in time becomes replacing s by s - a — is the first shifting theorem, the single most-used shortcut in the whole table.

Why this turns calculus into algebra

Now the payoff that justifies the whole bridge. Apply the transform to a derivative and integrate by parts once: L{f'(t)} = s F(s) - f(0). Look hard at what happened — *differentiation in the t-world became multiplication by s in the s-world*, and the initial value f(0) drops out of the boundary term, baked right in for free. This is the transform of derivatives, and applied twice it gives L{f''(t)} = s^2 F(s) - s f(0) - f'(0). Every derivative you had to chase with calculus becomes a power of s you can chase with algebra.

So picture a linear ODE like y'' + 3 y' + 2 y = f(t) with known starting values. Transform every term. Each derivative turns into a power of s times Y(s) minus initial data; the equation becomes (s^2 + 3 s + 2) Y(s) = (something built from f and the initial values). There are no more derivatives anywhere — just s and Y(s), tied together by an ordinary algebraic equation. You solve it the way you solved for x in middle school: divide. Y(s) = (right-hand side)/(s^2 + 3 s + 2). The differential equation has dissolved into a fraction.

  1. Transform. Apply L to both sides of the differential equation; every derivative becomes a power of s and the initial conditions enter automatically through the derivative rule.
  2. Solve the algebra. The result is one ordinary equation in Y(s) with no derivatives left — isolate Y(s) by dividing, exactly as you would solve for any unknown.
  3. Invert. Massage Y(s) with partial fractions into pieces that match the table, then read each piece back into a function of t — that f(t) is your answer, initial conditions already satisfied.

That three-step rhythm — transform, solve the algebra, invert — is the spine of everything ahead. The next guides flesh out each step and widen its reach: how to transform switches and sudden impulses (Heaviside steps and the Dirac delta) so that even discontinuous forcing goes through cleanly, how the convolution theorem handles an arbitrary input, and how inversion really works as a contour integral back in the complex s-plane. But the heart of the method is already in your hands: one integral builds a bridge to a world where differentiation is just multiplication by s, and a small table carries you both ways across it.