6.2 Laplace Transform of the Derivatives of \(f(t)\)
Theorem 6.2.1. Suppose that \(f(t)\) is continuous for \(t\geq 0\) and of exponential order \(\alpha \), \(|f(t)|\leq Me^{\alpha t}\), and that \(f'(t)\) is piecewise continuous on every finite interval. Then the Laplace transform of \(f'\) exists for \(s>\alpha \), and \[\mathcal {L}\{f'(t)\} = s\, \mathcal {L}\{f\} - f(0) .\]
Proof. Integrating by parts, \[\int _0^T e^{-st}f'(t)\,dt = \Big [e^{-st}f(t)\Big ]_0^T + s\int _0^Te^{-st}f(t)\,dt .\] As \(T\to \infty \), \(|e^{-sT}f(T)|\leq Me^{-(s-\alpha )T}\to 0\), so the right side tends to \(-f(0) + s\mathcal {L}\{f\}\). (If \(f'\) has jumps, the integration by parts is done on each interval of continuity, and the boundary terms cancel because \(f\) is continuous.) □
Applying the theorem to \(f'\) in place of \(f\), \[\mathcal {L}\{f''(t)\} = s\, \mathcal {L}\{f'(t)\} - f'(0) = s\big (s\, \mathcal {L}\{f\} -f(0)\big ) -f'(0) = s^2\mathcal {L}\{f(t)\} - s\, f(0) - f'(0) ,\] and by induction, under the corresponding hypotheses on \(f,\dots ,f^{(n-1)}\), \[\mathcal {L}\{f^{(n)}\} = s^n \mathcal {L}\{f\} - s^{n-1}f(0)- s^{n-2}f'(0) - \cdots - f^{(n-1)}(0) .\]
Example 6.2.2. Find \(\mathcal {L}\{t^2\}\) using Theorem 6.2.1.
Solution. Here \(f(t) = t^2\), \(f(0) = 0\), \(f'(t) = 2t\), \(f'(0) = 0\) and \(f''(t) = 2\), with \(\mathcal {L}\{2\} = \frac {2}{s}\). So \[s^2\mathcal {L}\{t^2\} - s\cdot 0 - 0 = \mathcal {L}\{2\} = \frac 2s,\qquad \mathcal {L}\{t^2\} = \frac {2}{s^3} .\] In the same way, by induction, \(\mathcal {L}\{t^n\} = \frac {n!}{s^{n+1}}\).
Exercise 6.2.3. Find \(\mathcal {L}\{t\}\) using the definition of the Laplace transform.
Answer: \(\mathcal {L}\{t\} = \int ^{\infty }_0t e^{-st}\,dt = \frac {1}{s^2}\), as in Example 6.0.3.
The transforms found so far are collected below; each holds for \(s\) large enough that the defining integral converges.
| \(f(t)\) | \(\mathcal {L}\{f(t)\}\) | |
| 1. | \(1\) | \(\frac {1}{s}\) |
| 2. | \(t\) | \(\frac {1}{s^2}\) |
| 3. | \(t^2\) | \(\frac {2}{s^3}\) |
| 4. | \(t^n\ (n =0,1,2,\dots )\) | \(\frac {n!}{s^{n+1}}\) |
| 5. | \(t^{\alpha }\) (\(\alpha >-1\)) | \(\frac {\Gamma (\alpha + 1)}{s^{\alpha + 1}}\) |
| 6. | \(e^{at}\) | \(\frac {1}{s-a}\) |
| 7. | \(\cos \omega t\) | \(\frac {s}{s^2 + \omega ^2}\) |
| 8. | \(\sin \omega t\) | \(\frac {\omega }{s^2 +\omega ^2}\) |
| 9. | \(\cosh \omega t\) | \(\frac {s}{s^2 - \omega ^2}\) |
| 10. | \(\sinh \omega t\) | \(\frac {\omega }{s^2 - \omega ^2}\) |
- 1.
- Solve \(y'' - y = t\), \(y(0) = 1\), \(y'(0) = 1\).
- 2.
- Solve the initial value problem \[y''+ y = 2t,\qquad y\left (\frac {\pi }{4}\right )=\frac {\pi }{2},\qquad y'\left (\frac {\pi }{4}\right ) = 2-\sqrt {2} .\]
Solution.
- 1.
- Transforming the equation, with \(Y = \mathcal {L}\{y\}\), \[s^2 Y -s y(0) - y'(0) - Y = \frac {1}{s^2},\qquad \text {so}\qquad (s^2 - 1)Y = \frac {1}{s^2}+ s + 1 .\] Hence \[Y(s) = \frac {1}{s^2(s^2-1)} + \frac {s + 1}{s^2 - 1} = \frac {1}{s^2(s^2-1)} + \frac {1}{s-1} .\] By partial fractions in \(s^2\), \(\frac {1}{s^2(s^2 - 1)} = \frac {1}{s^2 -1} - \frac {1}{s^2}\), so \[Y(s) = \frac {1}{s^2 - 1} - \frac {1}{s^2} + \frac {1}{s - 1},\qquad y(t) = \sinh t - t + e^t .\] As a check, \(y(0) = 1\) and \(y'(0) = 1 - 1 + 1 = 1\).
- 2.
- The conditions are given at \(\frac {\pi }{4}\) rather than \(0\), so write \(y(0) = y_0\) and \(y'(0) = y_1\), unknown for now. Transforming, \[s^2Y - sy_0 - y_1 + Y = \frac {2}{s^2},\qquad Y = \frac {2}{s^2 (s^2 + 1)} + y_0\frac {s}{s^2 + 1} + y_1\frac {1}{s^2 + 1} .\] Since \(\frac {2}{s^2(s^2 + 1)} = 2\left (\frac {1}{s^2} - \frac {1}{s^2 + 1}\right )\), \[y(t) = 2t - 2\sin t + y_0 \cos t + y_1\sin t = 2t + (y_1 - 2)\sin t + y_0\cos t ,\] with \(y'(t) = 2 + (y_1 - 2)\cos t - y_0\sin t\). At \(t = \frac {\pi }{4}\), where \(\sin t = \cos t = \frac {1}{\sqrt 2}\), the two conditions become \[\frac {\pi }{2} + \frac {y_1 - 2 + y_0}{\sqrt {2}} = \frac {\pi }{2},\qquad 2 + \frac {y_1 - 2 - y_0}{\sqrt {2}} = 2 - \sqrt 2 ,\] that is, \(y_1 + y_0 = 2\) and \(y_1 - y_0 = 0\). So \(y_0 = y_1 = 1\) and \[y(t) = 2t - \sin t + \cos t .\]
Solution. Transforming, \((s^2Y - 3s - 1) + 4(sY - 3) + 3Y = 0\), so \[(s^2 + 4s + 3)Y = 3s + 13,\qquad Y = \frac {3s + 13}{(s + 1)(s + 3)} = \frac {5}{s + 1} - \frac {2}{s + 3},\] and \(y(t) = 5e^{-t} - 2e^{-3t}\). As a check, \(y(0) = 3\) and \(y'(0) = -5 + 6 = 1\).
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