The Laplace transform · Differential Equations

Intuition

The Laplace transform turns a function of time into a function of a new variable, s, by weighting it with a decaying exponential and adding up the result over all positive time. What makes it worth doing is what it does to derivatives: differentiation in time becomes multiplication by s, so a linear differential equation with constant coefficients becomes an algebraic equation, with its initial conditions built in. Solve the algebra, transform back, and the problem is done.

Logarithms turn multiplication into addition, so a hard multiplication becomes an easy addition and a look-up in a table. The Laplace transform turns differentiation into multiplication in the same way, and its table of transforms is the look-up.

plane diagram: the-weight-of-the-transform

The transform of the constant $1$ is the area under the weight $e^{-st}$ , which is $\tfrac1s$ : $2$ , $1$ and $\tfrac12$ for these three values of $s$ . The larger $s$ , the faster the weight dies and the less of a function's future the transform sees.

An integral against a decaying exponential

For a function $f$ defined for $t\ge0$ , its Laplace transform is the function $F$ below, defined at every $s$ where the improper integral converges. When $f$ is piecewise continuous and of exponential order, for some constants and , the integral converges for every . The transform is linear: .

F(s)=\mathcal{L}\{f\}(s)=\int_{0}^{\infty}e^{-st}f(t)\,dt

First transforms

$\mathcal{L}\{1\}=\int_{0}^{\infty}e^{-st}\,dt=\tfrac1s$ for .

The transform of an exponential

\mathcal{L}\{e^{at}\}(s)=\frac{1}{s-a}\qquad(s>a)

Combine the two exponentials into one, with rate s minus a, and integrate it up to a finite time T. When s exceeds a, the exponential at T dies as T grows, so only the value at zero survives, and it gives one over s minus a. For s at or below a the integral diverges, which is why the transform is defined only to the right of a.

Proof steps

\mathcal{L}\{e^{at}\}=\int_{0}^{\infty}e^{-(s-a)t}\,dt

Combine the two exponentials.

=\lim_{T\to\infty}\Bigl[\frac{-e^{-(s-a)t}}{s-a}\Bigr]_{0}^{T}

An improper integral is a limit of proper ones.

s>a\ \Longrightarrow\ e^{-(s-a)T}\to0

The exponential dies as T grows when s exceeds a.

\mathcal{L}\{e^{at}\}=0-\frac{-1}{s-a}=\frac{1}{s-a}

Only the lower limit contributes.

Applications

Practice

Weight and Add Up

Multiply f by the decaying exponential and integrate over all positive time. The answer is a function of s.

F(s)=\int_{0}^{\infty}e^{-st}f(t)\,dt

Try it

What is $\mathcal{L}\{1\}$ ?