Método de las series para obtener la transformada inversa de Laplace. Laplace.

Introducción

Si $F(s)$ tiene un desarrollo en serie de potencias de los recíprocos de $s$ dado por

$\displaystyle F(s) = \frac{a_0}{s} + \frac{a_1}{s^2} + \frac{a_2}{s^3} + \frac{a_3}{s^4} + \cdots$

entonces, dentro de algunas condiciones, se puede invertir (es decir, aplicando la transformada inversa de Laplace en ambos miembros) término por término para obtener lo siguiente

$\displaystyle f(t) = a_0 + a_1 t + \frac{a_2 t}{2!} + \frac{a_3 t}{3!} + \cdots$

Problemas resueltos

Problema 1. Calcular $\displaystyle \mathcal{L}^{-1} \left[\frac{e^{-1/s}}{s} \right]$ .

Solución. Al tomar la función dentro del símbolo $\mathcal{L}^{-1}$ , se aplica el desarrollo de series.

$\displaystyle \mathcal{L}^{-1} \left[\frac{e^{-1/s}}{s} \right] \rightarrow \frac{e^{-1/s}}{s}$

$\displaystyle \frac{e^{-1/s}}{s} = \frac{1}{s} e^{-1/s}$

$\displaystyle \frac{e^{-1/s}}{s} = \frac{1}{s} \left(1 - \frac{1}{1!} \frac{1}{s} + \frac{1}{2!} \frac{1}{s^2} - \frac{1}{3!} \frac{1}{s^3} + \cdots \right)$

$\displaystyle \frac{e^{-1/s}}{s} = \frac{1}{s} - \frac{1}{1!} \frac{1}{s^2} + \frac{1}{2!} \frac{1}{s^3} - \frac{1}{3!} \frac{1}{s^4} + \cdots$

Aplicando la transformada inversa de Laplace en ambos miembros,resulta

$\displaystyle \mathcal{L}^{-1} \left[\frac{e^{-1/s}}{s} \right] = \mathcal{L}^{-1} \left[\frac{1}{s} - \frac{1}{1!} \frac{1}{s^2} + \frac{1}{2!} \frac{1}{s^3} - \frac{1}{3!} \frac{1}{s^4} + \cdots \right]$

$\displaystyle \mathcal{L}^{-1} \left[\frac{e^{-1/s}}{s} \right] = \mathcal{L}^{-1} \left[\frac{1}{s} \right] - \frac{1}{1!} \mathcal{L}^{-1} \left[\frac{1}{s^2} \right] + \frac{1}{2!} \mathcal{L}^{-1} \left[\frac{1}{s^3}\right] - \frac{1}{3!} \mathcal{L}^{-1}\left[\frac{1}{s^4}\right] + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[\frac{e^{-1/s}}{s} \right] = 1 - \frac{1}{1!} \frac{t}{1!} + \frac{1}{2!} \frac{t^2}{2!} - \frac{1}{3!} \frac{t^3}{3!} + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[\frac{e^{-1/s}}{s} \right] = 1 - \frac{1}{(1!)^2} t + \frac{1}{(2!)^2} t^2 - \frac{1}{(3!)^2} t^3 + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[\frac{e^{-1/s}}{s} \right] = 1 - t + \frac{1}{1^2 \ 2^2} t^2 - \frac{1}{1^2 \ 2^2 \ 3^2} t^3 + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[\frac{e^{-1/s}}{s} \right] = 1 - (t^{1/2})^2 + \frac{1}{1^2 \ 2^2} (t^{1/2})^4 - \frac{1}{1^2 \ 2^2 \ 3^2} (t^{1/2})^6 + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[\frac{e^{-1/s}}{s} \right] = 1 - \frac{2^2}{2^2} (t^{1/2})^2+ \frac{1}{1^2 \ 2^2} \frac{2^4}{2^4} (t^{1/2})^4 - \frac{1}{1^2 \ 2^2 \ 3^2} \frac{2^6}{2^6} (t^{1/2})^6 + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[\frac{e^{-1/s}}{s} \right] = 1 - \frac{2^2 (t^{1/2})^2}{2^2} + \frac{1}{1^2 \ 2^2} \frac{2^4 (t^{1/2})^4}{2^4} - \frac{1}{1^2 \ 2^2 \ 3^2} \frac{2^6 (t^{1/2})^6}{2^6} + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[\frac{e^{-1/s}}{s} \right] = 1 - \frac{(2t^{1/2})^2}{2^2} + \frac{1}{1^2 \ 2^2} \frac{(2t^{1/2})^4}{2^4} - \frac{1}{1^2 \ 2^2 \ 3^2} \frac{(2t^{1/2})^6}{2^6} + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[\frac{e^{-1/s}}{s} \right] = 1 - \frac{(2t^{1/2})^2}{2^2} + \frac{(2t^{1/2})^4}{2^2 \ 4^2} - \frac{(2t^{1/2})^6}{2^2 \ 4^2 \ 6^2} + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[\frac{e^{-1/s}}{s} \right] = J_0 \left(2 \sqrt{t} \right)$

Finalmente

$\displaystyle \therefore \mathcal{L}^{-1} \left[\frac{e^{-1/s}}{s} \right] = J_0 \left(2 \sqrt{t} \right)$

Problema 2. Resolver $\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right]$ .

Solución. Al tomar la función dentro del símbolo $\mathcal{L}^{-1}$ , se aplica el desarrollo de series.

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] \rightarrow e^{-\sqrt{s}}$

$\displaystyle e^{-\sqrt{s}} = 1 - s^{1/2} + \frac{1}{2!} s^{2/2} - \frac{1}{3!} s^{3/2} + \frac{1}{4!} s^{4/2} - \frac{1}{5!} s^{5/2} + \cdots$

$\displaystyle e^{-\sqrt{s}} = 1 - s^{1/2} + \frac{1}{2!} s - \frac{1}{3!} s^{3/2} + \frac{1}{4!} s^{2} - \frac{1}{5!} s^{5/2} + \cdots$

Aplicando la transformada inversa de Laplace en ambos miembros

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \mathcal{L}^{-1} \left[1 - s^{1/2} + \frac{1}{2!} s - \frac{1}{3!} s^{3/2} + \frac{1}{4!} s^{2} - \frac{1}{5!} s^{5/2} + \cdots \right]$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \mathcal{L}^{-1} [1] - \mathcal{L}^{-1} [s^{1/2}] + \frac{1}{2!} \mathcal{L}^{-1} [s] - \frac{1}{3!} \mathcal{L}^{-1} [s^{3/2}] + \frac{1}{4!} \mathcal{L}^{-1} [s^{2}] - \frac{1}{5!} \mathcal{L}^{-1} [s^{5/2}] + \cdots$

Para todo valor $n$ entero positivo o cero, se utiliza la siguiente fórmula

$\displaystyle \mathcal{L}^{-1} \left[s^{n + 1/2} \right] = \frac{1}{\Gamma \left(-n - \frac{1}{2} \right)} t^{-n-3/2}$

$\displaystyle \mathcal{L}^{-1} \left[s^{n + 1/2} \right] = \frac{(-1)^{n+1}}{\sqrt{\pi}} \left(\frac{1}{2} \right) \left(\frac{3}{2} \right) \left(\frac{5}{2} \right) \cdots \left(\frac{2n+1}{2} \right)t^{-n-3/2}$

Entonces

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \mathcal{L}^{-1} [s^0] - \mathcal{L}^{-1} [s^{1/2}] + \frac{1}{2!} \mathcal{L}^{-1} [s] - \frac{1}{3!} \mathcal{L}^{-1} [s^{3/2}] + \frac{1}{4!} \mathcal{L}^{-1} [s^{2}] - \frac{1}{5!} \mathcal{L}^{-1} [s^{5/2}] + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \mathcal{L}^{-1} [s^0] - \mathcal{L}^{-1} [s^{0+1/2}] + \frac{1}{2!} \mathcal{L}^{-1} [s] - \frac{1}{3!} \mathcal{L}^{-1} [s^{1 + 1/2}] + \frac{1}{4!} \mathcal{L}^{-1} [s^{2}] - \frac{1}{5!} \mathcal{L}^{-1} [s^{2+1/2}] + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = 0 - \frac{(-1)^{0+1}}{\Gamma \left(-0-\frac{1}{2} \right)} t^{0-3/2} + \frac{1}{2!} (0) - \frac{1}{3!} \frac{(-1)^{1+1}}{\Gamma \left(-1-\frac{1}{2} \right)} t^{-1-3/2} + \frac{1}{4!} (0) - \frac{1}{5!} \frac{(-1)^{2+1}}{\Gamma \left(-2-\frac{1}{2} \right)} t^{-2-3/2} + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = - \frac{(-1)^{1}}{\Gamma \left(-\frac{1}{2} \right)} t^{-3/2} - \frac{1}{3!} \frac{(-1)^{2}}{\Gamma \left(-\frac{3}{2} \right)} t^{-5/2} - \frac{1}{5!} \frac{(-1)^{3}}{\Gamma \left(-\frac{5}{2} \right)} t^{-7/2} + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \frac{1}{\Gamma \left(-\frac{1}{2} \right)} t^{-3/2} - \frac{1}{3!} \frac{1}{\Gamma \left(-\frac{3}{2} \right)} t^{-5/2} + \frac{1}{5!} \frac{1}{\Gamma \left(-\frac{5}{2} \right)} t^{-7/2} + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \left(\frac{1}{2} \right) \frac{1}{\sqrt{\pi}} t^{-3/2} - \frac{1}{3!} \left(\frac{1}{2}\right) \left(\frac{3}{2} \right) \frac{1}{\sqrt{\pi}} t^{-5/2} + \frac{1}{5!} \left(\frac{1}{2} \right) \left(\frac{3}{2} \right) \left(\frac{5}{2} \right) \frac{1}{\sqrt{\pi}} t^{-7/2} + \cdots$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \left(\frac{1}{2} \right) \frac{1}{\sqrt{\pi}} \left[ t^{-3/2} - \frac{1}{3!} \left(\frac{3}{2} \right) t^{-5/2} + \frac{1}{5!} \left(\frac{3}{2} \right) \left(\frac{5}{2} \right) t^{-7/2} + \cdots \right]$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \left(\frac{1}{2} \right) \frac{1}{\sqrt{\pi}} t^{-3/2} \left[1 - \frac{1}{3!} \left(\frac{3}{2} \right) t^{-1} + \frac{1}{5!} \left(\frac{3}{2} \right) \left(\frac{5}{2} \right) t^{-2} + \cdots \right]$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \left(\frac{1}{2} \right) \frac{1}{\sqrt{\pi}} t^{-3/2} \left[1 - \frac{1}{3 \cdot 2 \cdot 1} \left(\frac{3}{2} \right) t^{-1} + \frac{1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \left(\frac{3}{2} \right) \left(\frac{5}{2} \right) t^{-2} + \cdots \right]$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \left(\frac{1}{2} \right) \frac{1}{\sqrt{\pi}} t^{-3/2} \left[1 - \frac{1}{2 \cdot 1} \left(\frac{1}{2} \right) t^{-1} + \frac{1}{4 \cdot 2 \cdot 1} \left(\frac{1}{2} \right) \left(\frac{1}{2} \right) t^{-2} + \cdots \right]$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \left(\frac{1}{2} \right) \frac{1}{\sqrt{\pi}} t^{-3/2} \left[1 - \frac{1}{2^2} t^{-1} + \frac{1}{4 \cdot 2^3} t^{-2} + \cdots \right]$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \left(\frac{1}{2} \right) \frac{1}{\sqrt{\pi}} t^{-3/2} \left[1 - \frac{1}{2^2} t^{-1} + \frac{1}{2^5} t^{-2} + \cdots \right]$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \frac{1}{2 \sqrt{\pi} t^{3/2}} \left[1 - \frac{1}{2^2 \ t} + \frac{1}{2^5 \ t^2} + \cdots \right]$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \frac{1}{2 \sqrt{\pi} t^{3/2}} \left[1 - \frac{1}{1 \cdot 2^2 t} + \frac{1}{2 \cdot 2^4 t^2} + \cdots \right]$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \frac{1}{2 \sqrt{\pi} t^{3/2}} \left[1 - \frac{1}{1 \cdot 2^2 t} + \frac{1}{2 \cdot (2^2 t)^2} + \cdots \right]$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \frac{1}{2 \sqrt{\pi} t^{3/2}} \left[1 - \frac{1}{1!} \frac{1}{2^2 t} + \frac{1}{2!} \frac{1}{(2^2 t)^2} + \cdots \right]$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \frac{1}{2 \sqrt{\pi} t^{3/2}} \left[1 - \frac{1}{1!} \left(\frac{1}{2^2 t} \right) + \frac{1}{2!} \left( \frac{1}{2^2 t} \right)^2 + \cdots \right]$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \frac{1}{2 \sqrt{\pi} t^{3/2}} \left[1 - \frac{1}{1!} \left(\frac{1}{4t} \right) + \frac{1}{2!} \left( \frac{1}{4t} \right)^2 + \cdots \right]$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \frac{1}{2 \sqrt{\pi} t^{3/2}} \left(e^{-\frac{1}{4t}} \right)$

$\displaystyle \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \frac{1}{2 \sqrt{\pi} \ t^{3/2} \ e^{\frac{1}{4t}}}$

Finalmente

$\displaystyle \therefore \mathcal{L}^{-1} \left[e^{-\sqrt{s}} \right] = \frac{1}{2 \sqrt{\pi} \ t^{3/2} \ e^{\frac{1}{4t}}}$

Método de las series para obtener la transformada inversa de Laplace. Laplace.

Introducción

Problemas resueltos

Publicado por Cesar Reyes

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Método de las series para obtener la transformada inversa de Laplace. Laplace.

Introducción

Problemas resueltos

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Publicado por Cesar Reyes

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