Inverse Laplace Transform

The inverse Laplace transform recovers the original time-domain function from its transform, primarily using partial fraction decomposition and table lookups.

Definition and Uniqueness

Definition4.5Inverse Laplace transform

The inverse Laplace transform of $F(s)$ is the function $f(t)$ such that $\mathcal{L}\{f\} = F$ . When it exists for piecewise continuous functions of exponential order, the inverse is unique (up to values at points of discontinuity). It is given by the Bromwich integral:

$f(t) = \frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}e^{st}F(s)\,ds$

where $\gamma$ is greater than the real part of all singularities of $F$ .

Theorem4.5Lerch's theorem

If $F(s) = G(s)$ for all $s > s_0$ , and both $f, g$ are piecewise continuous of exponential order with $\mathcal{L}\{f\} = F$ and $\mathcal{L}\{g\} = G$ , then $f(t) = g(t)$ at all points of continuity.

Partial Fraction Method

RemarkStrategy for rational transforms

Most inverse Laplace transforms in practice involve rational functions $F(s) = P(s)/Q(s)$ with $\deg P < \deg Q$ . The method:

Factor $Q(s)$ into linear and irreducible quadratic factors.
Decompose $F$ into partial fractions.
Invert each term using the table of standard transforms.

ExamplePartial fractions example

Find $\mathcal{L}^{-1}\left\{\frac{3s+2}{(s-1)(s^2+4)}\right\}$ .

Partial fractions: $\frac{3s+2}{(s-1)(s^2+4)} = \frac{A}{s-1} + \frac{Bs+C}{s^2+4}$ .

$3s+2 = A(s^2+4) + (Bs+C)(s-1)$ . Setting $s=1$ : $5 = 5A$ , so $A = 1$ .

Comparing $s^2$ : $0 = A + B$ , so $B = -1$ . Comparing constants: $2 = 4A - C = 4 - C$ , so $C = 2$ .

$F(s) = \frac{1}{s-1} + \frac{-s+2}{s^2+4} = \frac{1}{s-1} - \frac{s}{s^2+4} + \frac{2}{s^2+4}.$

Inverting: $f(t) = e^t - \cos 2t + \sin 2t$ .

The Bromwich Integral and Residues

Theorem4.6Inverse via residues

If $F(s)$ is rational with poles $s_1, \ldots, s_n$ , then

$f(t) = \sum_{k=1}^n \text{Res}\left(e^{st}F(s), s_k\right).$

ExampleInverse via residues

For $F(s) = \frac{1}{(s-1)^2(s+2)}$ :

At $s = 1$ (pole of order 2): $\text{Res} = \frac{d}{ds}\frac{e^{st}}{s+2}\big|_{s=1} = \frac{te^t \cdot 3 - e^t}{9} = \frac{(3t-1)e^t}{9}$ .

At $s = -2$ (simple pole): $\text{Res} = \frac{e^{-2t}}{(-2-1)^2} = \frac{e^{-2t}}{9}$ .

$f(t) = \frac{(3t-1)e^t + e^{-2t}}{9}.$

Applications to Systems

RemarkSystems of ODEs via Laplace transforms

For a system $\mathbf{x}' = A\mathbf{x} + \mathbf{g}(t)$ with $\mathbf{x}(0) = \mathbf{x}_0$ , the Laplace transform gives

$s\mathbf{X}(s) - \mathbf{x}_0 = A\mathbf{X}(s) + \mathbf{G}(s) \implies \mathbf{X}(s) = (sI - A)^{-1}(\mathbf{x}_0 + \mathbf{G}(s)).$

The solution is $\mathbf{x}(t) = \mathcal{L}^{-1}\{(sI-A)^{-1}\}\mathbf{x}_0 + \mathcal{L}^{-1}\{(sI-A)^{-1}\mathbf{G}(s)\}$ . The matrix $\mathcal{L}^{-1}\{(sI-A)^{-1}\} = e^{At}$ is the matrix exponential.

Example2x2 system via Laplace

Solve $x' = 3x - y$ , $y' = x + y$ , $x(0) = 1$ , $y(0) = 0$ .

$(sI - A)\mathbf{X} = \mathbf{x}_0$ : $\begin{pmatrix}s-3 & 1 \\ -1 & s-1\end{pmatrix}\begin{pmatrix}X\\Y\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix}$ .

$\det(sI-A) = (s-3)(s-1)+1 = s^2-4s+4 = (s-2)^2$ . So $X = \frac{s-1}{(s-2)^2} = \frac{1}{s-2} + \frac{1}{(s-2)^2}$ .

$x(t) = e^{2t} + te^{2t} = (1+t)e^{2t}$ . Similarly $Y = \frac{1}{(s-2)^2}$ , so $y(t) = te^{2t}$ .