6.2 Laplace Transform of the Derivatives of \(f(t)\)
Suppose that \(f(t)\) is continuous for all \(t\geq 0\), satisfies \(\big |f(t)\big | \leq Me^{\alpha t}\), for \(f'(t)\) that is piecewise continuous only every finite interval in the range \(t\geq 0\). Then the Laplace transform of the derivative \(f'(t)\) exists when \(s>\alpha \), and \[\mathcal {L}\{f'(t)\} = s\hspace {0.1cm} \mathcal {L}\{f\} - f(0)\]
By applying this to \(f''(t)\), we get \begin {align*} \mathcal {L}\{f''(t)\} & = s\hspace {0.1cm} \mathcal {L}\{f'(t)\} - f'(0)\\ & = s\{s\hspace {0.1cm} \mathcal {L}\{f\} -f(0)\} -f'(0)\\\\ \implies \hspace {1cm} \mathcal {L}\{f''(t)\} & = s^2\mathcal {L}\{f(t)\} - s\hspace {0.1cm} f(0) - f'(0) \end {align*}
\[\text {Similarly}\hspace {1cm} \mathcal {L}\{f'''(t)\} = s^3\mathcal {L}\{f(t)\} - s^2f(0) - sf'(0) - f''(0)\]
\[\text {By induction}\hspace {0.4cm} \mathcal {L}\{f^{(n)}\} = s^n \mathcal {L}\{f\} - s^{n-1}f(0)- s^{n-2}f'(0) \cdots \cdots \cdots f^{(n-1)}(0)\]
Solution. Here \(f(t) =t^2,\hspace {0.4cm} f(0) = 0, \hspace {0.4cm} f'(t) = 2t,\hspace {0.4cm} f'(0) = 0,\hspace {0.4cm} f''(t) =2\) and \(\mathcal {L}\{2\} =2\mathcal {L}\{1\} = \dfrac {2}{s}\).
\(\mathcal {L}\{f''\} = \mathcal {L}\{2\}\), but \(\mathcal {L}\{f''\} =s^2\mathcal {L}\{f(t)\} - sf(0) - f'(0)\).
\begin {align*} s^2\mathcal {L}\{f\} - s f(0) - f'(0) & = \mathcal {L}\{2\}\\ s^2 \mathcal {L}\{f\} & = \frac {2}{s}\\\\ \implies \hspace {0.4cm} \mathcal {L}\{f\} & = \frac {2}{s^3} \end {align*}
\[\therefore \hspace {0.5cm} \mathcal {L}\{t^2\} = \frac {2}{s^3}\]
Find \(\mathcal {L}\{t\}\) using the definition of Laplace transform.
Answer: \(\displaystyle {\mathcal {L}\{t\} = \int ^{\infty }_0t e^{-st}dt = \frac {1}{s^2}}\)
| \(f(t)\) | \(\mathcal {L}\{f(t)\}\) | ||
| 1. | 1 | \(1/s\) | |
| 2. | \(t\) | \(1/s^2\) | |
| 3. | \(t^2\) | \(2/s^3\) | |
| 4. | \(t^n\hspace {0.2cm} (n =0,1,2,\cdots \cdots )\) | \(n!/s^{n+1}\) | |
| 5. | \(t^{\alpha }\) (\(\alpha \) is positive) | \(\Gamma (\alpha + 1 )/s^{\alpha + 1}\) | |
| 6. | \(e^{at}\) | \(1/s-a\) | |
| 7. | \(\cos \omega t\) | \(s/(s^2 + \omega ^2)\) | |
| 8. | \(\sin \omega t\) | \(\omega / (s^2 +\omega ^2)\) | |
| 9. | \(\cosh \omega t\) | \(s/(s^2 - \omega ^2)\) | |
| 10. | \(\sinh \omega t\) | \(\omega / (s^2 - \omega ^2)\) |
- 1.
- Solve \(y'' - y = t\), \(\hspace {0.5cm} y(0) = 1, \hspace {0.5cm} y'(0) = 1\)
- 2.
- Solve the initial value problem \[y''+ y = 2t,\hspace {0.5cm} y\bigg (\dfrac {\pi }{4}\bigg )=\dfrac {\pi }{2},\hspace {0.5cm} y'\bigg (\dfrac {\pi }{4}\bigg ) = 2-\sqrt {2}\]
Solution.
Part 1
Taking the Laplace transform over the equation, we get \begin {align*} \mathcal {L}\{y''\} - \mathcal {L}\{y\} = \mathcal {L}\{t\}\\ s^2 \mathcal {L}\{y\} -s y(0) - y'(0) - \mathcal {L}\{y\} & = \frac {1}{s^2} \end {align*}
Let \(\displaystyle {\mathcal {L}\{y\} = Y(s)}\), then \(\displaystyle {s^2y - s - 1 -y = \frac {1}{s^2}}\) \begin {align*} \implies \hspace {0.5cm}(s^2 - 1)y & = \frac {1}{s^2}+ s + 1\\\\ Y(s) & = \frac {1}{s^2(s^2-1)} + \frac {s + 1}{s^2 - 1}\\\\ \implies \hspace {0.5cm} Y(s) & = \frac {1}{s^2(s^2-1)} + \frac {1}{s-1} \end {align*}
Split \(\dfrac {1}{s^2(s^2 - 1)}\) into partial fractions to get \(\dfrac {1}{s^2(s^2 -)} = \dfrac {1}{s^2 -1} - \dfrac {1}{s^2}\)
\[\implies \hspace {0.5cm} Y(s) = \frac {1}{s^2 - 1} - \frac {1}{s^2} + \frac {1}{s - 1}\]
\begin {align*} y(t) & = \mathcal {L}^{-1}\big [Y(s)\big ]\\ & = \mathcal {L}^{-1}\Bigg [\frac {1}{s^2 - 1}-\frac {1}{s^2} + \frac {1}{s -1}\Bigg ] \end {align*}
\[y(t) = \sinh t - t + e^t\]
Part 2
\begin {align*} \mathcal {L}\{y''\} + \mathcal {L}\{y\} & = 2\mathcal {L}\{t\}\\ s^2Y -sY(0) - Y'(0) + Y & = \frac {2}{s^2}\\\\ (s^2 +1)Y & = \frac {2}{s^2} + sY(0) + Y'(0) \end {align*}
\[\implies \hspace {0.5cm} Y = \frac {2}{s^2 (s^2 + 1)} + \frac {s Y(0)}{s^2 + 1} + \frac {Y'}{s^2 + 1}\]
\[Y(s) = s\big [\frac {1}{s^2}-\frac {1}{s^2 + 1}\big ] + Y(0)\frac {s}{s^2 + 1} + Y'(0)\frac {1}{s^2 + 1}\]
\begin {align*} y(t) & = \mathcal {L}^{-1}\{Y(s)\}\\ & = 2t - 2\sin t + Y(0) \cos t + Y'(0)\sin t\\ & = 2t + (Y'(0) - 2)\sin t + Y(0) \cos t\\\\ \implies \hspace {0.5cm} y'(t) & = 2 + (Y'(0) - 2)\cos t - Y(0)\sin t \end {align*}
Applying given initial conditions \[y\bigg (\dfrac {\pi }{4}\bigg ) = \frac {\pi }{2} + \big (Y'(0) -2\big )\cdot \frac {1}{\sqrt {2}} + Y(0)\cdot \frac {1}{\sqrt {2}} = \frac {\pi }{2}\] \[\implies \hspace {0.5cm} \frac {Y'(0) -2}{\sqrt {2}}+ \frac {Y(0)}{\sqrt {2}} = 0\hspace {0.5cm}\cdots \cdots \cdots \cdots \hspace {0.5cm} (1)\]
\[y'\bigg (\dfrac {\pi }{4}\bigg ) = 2 + \frac {\big (Y'(0) -2\big )}{\sqrt {2}} - \frac {Y(0)}{\sqrt {2}} = 2\sqrt {2}\] \[\implies \hspace {0.5cm} \frac {Y'(0) - 2}{\sqrt {2}} - \frac {Y(0)}{\sqrt {2}} = -\sqrt {2}\] \[\implies \hspace {0.5cm} Y'(0) -2 - Y(0) = -2\implies Y'(0) - Y(0) = 0\hspace {0.5cm}\cdots \cdots \cdots \hspace {0.5cm} (2)\]
Solving (1) and (2) simultaneously, we get \(Y'(0) = 1,\hspace {0.5cm} Y(0) = Y'(0) = 1\) \[\therefore \hspace {0.5cm} y(t) = 2t - \sin t + \cos t\]
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.