Q5E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the convolution theorem to find the inverse Laplace transform of the given function.

$\frac{\mathbf{1}}{\mathbf{s}\mathbf{(}{\mathbf{s}}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{1}\mathbf{)}}$

Step-by-Step Solution

Answer

The inverse Laplace transform for the given function using the convolution theorem is.

 $L^{-1}\left\{\frac{1}{s(s^2 + 2)}\right\}\left(t\right)=1-\cos t$

 

Step 1: Define convolution theorem

Let $\mathit{f}\left(t\right)$and $\mathit{g}\left(t\right)$be piecewise continuous on $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$and of exponential order $\mathit{\alpha }$and set $\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{ }\mathbf{=}\mathbf{ }\mathit{L}\mathbf{\{}\mathbf{ }\mathit{f}\mathbf{\}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathit{G}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}\mathit{L}\mathbf{\{}\mathbf{ }\mathit{g}\mathbf{\}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$, then,

$\mathit{L}\left\{f*g\right\}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{ }\mathbf{=}\mathbf{ }\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathit{G}\mathbf{\left(}\mathit{s}\mathbf{\right)}$

or

${\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{f\left(s\right)*g\left(s\right)\right\}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{ }\mathbf{=}\mathbf{ }\left(f*g\right)\mathbf{\left(}\mathit{t}\mathbf{\right)}$

 

Step 2: Find the inverse Laplace transform by using the convolution theorem

Consider the given function, $\frac{1}{s\left(s^2+1\right)}$

Let,$F\left(s\right)\frac{1}{s}$

$G\left(s\right)\frac{1}{(s^2 + 1)}$

Since, we know

$f\left(t\right) =L^{-1}\left\{\frac{1}{s}\right\}\left(t\right) =1$

$g\left(t\right) =L^{-1}\left\{\frac{1}{s^2+1}\right\}\left(t\right) =\sin t$

Apply inverse Laplace transform and use convolution theorem to obtain,

$$\begin{gathered} L^{-1}\left\{\frac{1}{s^2+1}\right\}\left(t\right) =L^{-1}\left\{F\left(s\right)G\left(s\right)\right\}\left(t\right) \\ =f\left(t\right)\times g\left(t\right) \\ =1\times \sin t \end{gathered}$$

Thus, use the convolution formula, $(f * g)\left(t\right)={\int }_0^{t}f\left(t-v\right)g\left(v\right)dv$

$$\begin{gathered} L^{-1}\left(\frac{1}{s\left(s^2+1\right)}\right)\left(t\right)={\int }_0^t\sin vdv \\ ={\left[-\cos v\right]}_0^t \\ =1-\cos t \end{gathered}$$

Therefore, $L^{-1}\left(\frac{1}{s\left(s^2+1\right)}\right)\left(t\right)=1-\cos t$

thus the inverse Laplace transform for the given function is $L^{-1}\left(\frac{1}{s\left(s^2+1\right)}\right)\left(t\right)=1-\cos t$