13.4: Properties of Laplace transform
- Page ID
- 6554
We have already used the linearity of Laplace transform when we computed \(\mathcal{L} (\cos (\omega t))\). Let’s officially record it as a property.
The Laplace transform is linear. That is, if \(a\) and \(b\) are constants and \(f\) and \(g\) are functions then
\[\mathcal{L} (af + bg) = a \mathcal{L} (f) + b \mathcal{L} (g). \nonumber \]
(The proof is trivial –integration is linear.)
A key property of the Laplace transform is that, with some technical details,
Laplace transform transforms derivatives in \(t\) to multiplication by \(s\) (plus some details).
This is proved in the following theorem.
If \(f(t)\) has exponential type \(a\) and Laplace transform \(F(s)\) then
\[\mathcal{L} (f'(t); s) = sF(s) - f(0), \text{ valid for Re}(s) > a. \nonumber \]
- Proof
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We prove this using integration by parts.
\(\mathcal{L} (f'; s) = \int_{0}^{\infty} f'(t) e^{-st}\ dt = f(t) e^{-st} \vert_{0}^{\infty} + \int_{0}^{\infty} s f(t) e^{-st} \ dt = -f(0) + sF(s).\)
In the last step we used the fact that at \(t = \infty, f(t) e^{-st} = 0\), which follows from the assumption about exponential type.
Equation 13.5.2 gives us formulas for all derivatives of \(f\).
\[\mathcal{L} (f''; s) = s^2 F(s) - sf(0) - f'(0) \nonumber \]
\[\mathcal{L} (f'''; s) = s^3 F(s) - s^2 f(0) - sf'(0) - f''(0) \nonumber \]
\(Proof\). For Equation 13.5.3:
\(\mathcal{L} (f''; s) = \mathcal{L} ((f')'; s) = s \mathcal{L} (f'; s) - f'(0) = s(sF(s) - f(0)) - f'(0) = s^2 F(s) - sf(0) - f'(0). \text{ QED}\)
The proof Equation 13.5.4 is similar. Also, similar statements hold for higher order derivatives.
There is a further complication if we want to consider functions that are discontinuous at the origin or if we want to allow \(f(t)\) to be a generalized function like \(\delta (t)\). In these cases \(f(0)\) is not defined, so our formulas are undefined. The technical fix is to replace 0 by \(0^{-}\) in the definition and all of the formulas for Laplace transform. You can learn more about this by taking 18.031.
If \(f(t)\) has exponential type \(a\), then \(F(s)\) is an analytic function for \(\text{Re} (s) > a\) and
\[F'(s) = -\mathcal{L} (tf(t); s). \nonumber \]
- Proof
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We take the derivative of \(F(s)\). The absolute convergence for \(\text{Re} (s)\) large guarantees that we can interchange the order of integration and taking the derivative.
\(F'(s) = \dfrac{d}{ds} \int_{0}^{\infty} f(t) e^{-st}\ dt = \int_{0}^{\infty} -t f(t) e^{-st}\ dt = \mathcal{L} (-tf(t); s).\)
This proves Equation 13.5.5.
Equation 13.5.5 is called the \(s\)-derivative rule. We can extend it to more derivatives in \(s\): Suppose \(\mathcal{L} (f;s) = F(s)\). Then,
\[\mathcal{L} (tf(t); s) = -F'(s) \nonumber \]
\[\mathcal{L} (t^n f(t); s) = (-1)^n F^{(n)} (s) \nonumber \]
Equation 13.5.6 is the same as Equation 13.5.5 above. Equation 13.5.7 follows from this.
Use the \(s\)-derivative rule and the formula \(\mathcal{L} (1;s) = 1/s\) to compute the Laplace transform of \(t^n\) for \(n\) a positive integer.
Solution
Let \(f(t) = 1\) and \(F(s) = \mathcal{L} (f; s)\). Using the \(s\)-derivative rule we get
\(\mathcal{L} (t;s) = \mathcal{L} (t f; s) = -F' (s) = \dfrac{1}{s^2}\)
\(\mathcal{L} (t^2;s) = \mathcal{L} (t^2 f; s) = (-1)^2 F'' (s) = \dfrac{2}{s^3}\)
\(\mathcal{L} (t^n;s) = \mathcal{L} (t^n f; s) = (-1)^n F^n (s) = \dfrac{n!}{s^{n + 1}}\)
As usual, assume \(f(t) = 0\) for \(t < 0\). Suppose \(a > 0\). Then,
\[\mathcal{L} (f(t - a); s) = e^{-as} F(s) \nonumber \]
- Proof
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We go back to the definition of the Laplace transform and make the change of variables \(\tau = t - a\).
\(\begin{array} {rcl} {\mathcal{L} (f(t - a); s)} & = & {\int_{0}^{\infty} f(t - a) e^{-st}\ dt = \int_{a}^{\infty} f(t - a) e^{-st}\ dt} \\ {} & = & {\int_{0}^{\infty} f(\tau) e^{-s(\tau + a)} \ d \tau = e^{-sa} \int_{0}^{\infty} f(\tau) e^{-s \tau} \ d \tau = e^{-sa} F(s).} \end{array}\)
The properties in Equations 13.5.1-13.5.8 will be used in examples below. They are also in the table at the end of these notes.