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Mathematical Methods of Physics: For Undergraduate

Sunil Karna

Section 6.3 Laplace Transforms

The Laplace transform is used to solve differential equations and analyze systems in the field of mathematics and engineering. It transforms a function of time, typically denoted as \(f(t)\text{,}\) into a function of a complex variable, usually denoted as m\(F(s)\text{,}\) where ’s’ is the complex variable. The Laplace transform provides a way to analyze functions in the frequency domain rather than the time domain. Laplace transform helps us in solving the differential equations with boundary values without finding the general solution and the values of the arbitrary constants. It converts differential equation (partial differential equation, PDE or ordinary differential equation, ODE) into algebric equation and hence easy to solve. It transforms a function with continuous time domain to Laplace domain (called s-parameter). Let \(f(t)\) be the function defined for all \(+ ve\) values of \(t\text{,}\) then Laplace Transform of \(f(t)\) is denoted as
\begin{equation*} \mathscr{L}\{f(t)\} = F(s) = \int\limits_{0}^{\infty}e^{-st}f(t)\,dt. \end{equation*}

Some useful formulae:.

  1. \begin{equation*} \mathscr{L}\{1\} = \frac{1}{s} \end{equation*}

    Proof.

    Since
    \begin{equation*} \mathscr{L}\{f(t)\} = \int\limits_{0}^{\infty}e^{-st}f(t)\,dt. \end{equation*}
    \begin{equation*} \therefore \mathscr{L}(1) = \int\limits_{0}^{\infty}e^{-st}.1\,dt = \left.\frac{e^{-st}}{-s}\right\vert_{0}^{\infty} \end{equation*}
    \begin{equation*} =0+\frac{1}{s}=\frac{1}{s}. \end{equation*}
  2. \begin{equation*} \mathscr{L}\{t^{n}\}=\frac{n!}{s^{n+1}} \quad \text{where}\quad n=0,1,2,\cdots \end{equation*}

    Proof.

    \begin{equation*} \mathscr{L}(t^{n})= \int\limits_{0}^{\infty}e^{-st} t^{n}\,dt \end{equation*}
    put \(st=x\) so that \(\,dt = \frac{\,dx}{s}\text{.}\)
    \begin{equation*} \therefore \mathscr{L}(t^{n})= \int\limits_{0}^{\infty}e^{-x} \left(\frac{x}{s}\right)^{n}\frac{\,dx}{s} \end{equation*}
    \begin{equation*} = \frac{1}{s^{n+1}}\int\limits_{0}^{\infty}e^{-x} x^{n}\,dx=\frac{n!}{s^{n+1}} \end{equation*}
  3. \begin{equation*} \mathscr{L}\{e^{at}\}=\frac{1}{s-a}; \quad (s \gt a) \end{equation*}

    Proof.

    \begin{equation*} \mathscr{L}(e^{at})=\int\limits_{0}^{\infty}e^{-st}e^{at}\,dt= \int\limits_{0}^{\infty}e^{-(s-a)t}\,dt \end{equation*}
    \begin{equation*} = \left.\frac{e^{-(s-a)t}}{-(s-a)}\right\vert_{0}^{\infty}=\frac{1}{s-a}. \end{equation*}
  4. \begin{equation*} \mathscr{L}\{\cosh at\} = \frac{s}{s^{2}-a^{2}}; \quad (s^{2} \gt a^{2}) \end{equation*}

    Proof.

    \begin{equation*} \mathscr{L}(\cosh at) =L\left[\frac{e^{at}+e^{-at}}{2}\right]= \frac{1}{2}\left[L(e^{at}) +L(e^{-at})\right] \end{equation*}
    \begin{equation*} =\frac{1}{2}\left[\frac{1}{s-a}+\frac{1}{s+a}\right] = \frac{s}{s^{2}-a^{2}}. \end{equation*}
  5. \begin{equation*} \mathscr{L}\{\sinh at\} = \frac{a}{s^{2}-a^{2}}; \quad (s^{2} \gt a^{2}) \end{equation*}

    Proof.

    \begin{equation*} \mathscr{L}(\sinh at)=L\left[\frac{e^{at}-e^{-at}}{2}\right] =\frac{1}{2}\left[L(e^{at}) - L(e^{-at})\right] \end{equation*}
    \begin{equation*} = \frac{1}{2}\left[\frac{1}{s-a}-\frac{1}{s+a}\right]= \frac{a}{s^{2}-a^{2}}. \end{equation*}
  6. \begin{equation*} \mathscr{L}\{\sin at\} = \frac{a}{s^{2}+a^{2}}; \quad (s \gt 0) \end{equation*}

    Proof.

    \begin{equation*} \mathscr{L}(\sin at) =L\left[\frac{e^{iat}-e^{-iat}}{2i}\right] =\frac{1}{2i}\left[L(e^{iat}) - L(e^{-iat})\right] \end{equation*}
    \begin{equation*} = \frac{1}{2i}\left[\frac{1}{s-ia}-\frac{1}{s+ia}\right] = \frac{a}{s^{2}+a^{2}}. \end{equation*}
  7. \begin{equation*} \mathscr{L}\{\cos at\} = \frac{s}{s^{2}+a^{2}}; \quad (s \gt 0) \end{equation*}

    Proof.

    \begin{equation*} \mathscr{L}(\cos at) =L\left[\frac{e^{iat}+e^{-iat}}{2}\right]=\frac{1}{2}\left[\frac{1}{s-ia}+\frac{1}{s+ia}\right] \end{equation*}
    \begin{equation*} = \frac{s}{s^{2}+a^{2}}. \end{equation*}
  8. (Refer to Subsubsection 6.3.1.2 First-shifting property)
    \begin{equation*} \mathscr{L}\{e^{at}t^{n}\}= \frac{n!}{(s-a)^{n+1}} \quad \left[\because L(t^{n}) = \frac{n!}{s^{n+1}}\right] \end{equation*}
  9. \begin{equation*} \mathscr{L}\left\{e^{at}\cosh bt\right\} = \frac{s-a}{(s-a)^{2}-b^{2}} \end{equation*}
  10. \begin{equation*} \mathscr{L}\left\{e^{at}\sinh bt\right\} = \frac{b}{(s-a)^{2}-b^{2}} \end{equation*}
  11. \begin{equation*} \mathscr{L}\left\{e^{at}\sin bt\right\} = \frac{b}{(s-a)^{2}+b^{2}} \end{equation*}
  12. \begin{equation*} \mathscr{L}\left\{e^{at}\cos bt\right\} = \frac{s-a}{(s-a)^{2}+b^{2}} \end{equation*}
  13. (Refer to Subsubsection 6.3.1.3 Inverse Laplace Transforms)
    \begin{equation*} \mathscr{L}^{-1}\left\{\frac{1}{s}\right\}=1 \end{equation*}
  14. \begin{equation*} \mathscr{L}^{-1}\left\{\frac{1}{s^{n}}\right\}=\frac{t^{n-1}}{!(n-1)};\quad\text{here}\quad n=0,1,2,3,\cdots \end{equation*}
  15. \begin{equation*} \mathscr{L}^{-1}\left\{\frac{1}{s-a}\right\} =e^{at} \end{equation*}
  16. \begin{equation*} \mathscr{L}^{-1}\left\{\frac{s}{s^{2}-a^{2}}\right\} =\cosh at \end{equation*}
  17. \begin{equation*} \mathscr{L}^{-1}\left\{\frac{1}{s^{2}-a^{2}}\right\} =\frac{1}{a}\sinh at \end{equation*}
  18. \begin{equation*} \mathscr{L}^{-1}\left\{\frac{1}{s^{2}+a^{2}}\right\} =\frac{1}{a}\sin at \end{equation*}
  19. \begin{equation*} \mathscr{L}^{-1}\left\{\frac{s}{s^{2}+a^{2}}\right\} =\cos at \end{equation*}
  20. \begin{equation*} \mathscr{L}^{-1}\left\{\frac{1}{(s-a)^{2}+b^{2}}\right\} =\frac{1}{b}e^{at}\sin bt \end{equation*}
  21. \begin{equation*} \mathscr{L}^{-1}\left\{\frac{s-a}{(s-a)^{2}+b^{2}}\right\} =e^{at}\cos bt \end{equation*}
  22. \begin{equation*} \mathscr{L}^{-1}\left\{\frac{1}{(s-a)^{2}-b^{2}}\right\} =\frac{1}{b}e^{at}\sinh bt \end{equation*}
  23. \begin{equation*} \mathscr{L}^{-1}\left\{\frac{s-a}{(s-a)^{2}-b^{2}}\right\} =e^{at}\cosh bt \end{equation*}
  24. \begin{equation*} \mathscr{L}^{-1}\left\{\frac{1}{(s^{2}+a^{2})^{2}}\right\} =\frac{1}{2a^{3}}(\sin at-at\cos at) \end{equation*}
  25. \begin{equation*} \mathscr{L}^{-1}\left\{\frac{s}{(s^{2}+a^{2})^{2}}\right\} =\frac{1}{2a} t\sin at \end{equation*}
  26. \begin{equation*} \mathscr{L}^{-1}\left\{\frac{s^{2}-a^{2}}{(s^{2}+a^{2})^{2}}\right\} = t\cos at \end{equation*}
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