Q56P - Mathematical Methods in Physical Sciences

Question

Find the inverse Laplace transform of the following functions by using (7.16) $\frac{\mathbf{1}}{{\mathbf{p}}^{\mathbf{4}}\mathbf{-}\mathbf{1}}$.

Step-by-Step Solution

Answer

The required inverse Laplace transformation is $\left(z+1\right)\mathit{F}\left(z\right){\mathit{e}}^{\mathbf{z}\mathbf{t}}\mathbf{=}\mathbf{-}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{t}}}{\mathbf{4}}$.

Step 1: Determine the residue of the poles.

So now we find the residue at all poles as,

 

Residue at $\mathit{z}\mathbf{=}\mathit{i}$,  

 $\begin{array}{rcl}\left(z-i\right)F\left(z\right)e^{zt}& =& \left(z-i\right)\frac{e^{zt}}{\left(z-1\right)\left(z+1\right)\left(z+i\right)\left(z-i\right)}\\ \left(z-i\right)F\left(z\right)e^{zt}& =& \frac{e^{zt}}{\left(z-1\right)\left(z+1\right)\left(z+i\right)}\\ \left(z-i\right)F\left(z\right)e^{zt}& =& \frac{e^{it}}{\left(i-1\right)\left(i+1\right)\left(i+i\right)}\\ \left(z-i\right)F\left(z\right)e^{zt}& =& \frac{e^{it}}{4i}\\ & & \end{array}$

 

 

Residue at $\begin{array}{rcl}z & =& - i\end{array}$,   

 $\begin{array}{rcl}\left(z+i\right)F\left(z\right)e^{zt}& =& \left(z+i\right)\frac{e^{zt}}{\left(z-1\right)\left(z+1\right)\left(z+i\right)\left(z-i\right)}\\ \left(z+i\right)F\left(z\right)e^{zt}& =& \frac{e^{zt}}{\left(z-1\right)\left(z+1\right)\left(z-i\right)}\\ \left(z+i\right)F\left(z\right)e^{zt}& =& \frac{e^{it}}{\left(-i-1\right)\left(-i+1\right)\left(-i-i\right)}\\ \left(z+i\right)F\left(z\right)e^{zt}& =& -\frac{e^{-it}}{4i}\end{array}$

 

Step 2: Determine the Laplace transform.

$\frac{1}{p^4-1}$  is given by sum of residue at all poles by Laplace transform,

 $$\begin{gathered} f\left(t\right)=\frac{e^t}{4}-\frac{e^{-t}}{4}+\frac{e^{it}}{4i}-\frac{e^{-it}}{4i} \\ f\left(t\right)=\frac{1}{2}\left(\frac{e^t-e^{-t}}{2}\right)+\frac{1}{2}\left(\frac{e^{it}-e^{-it}}{2}\right) \\ f\left(t\right)=\frac{1}{2}\sin ht+\frac{1}{2}\sin t \end{gathered}$$

  

 

Hence,  $\mathit{f}\left(t\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{t}$ 

Step 3: Determine the poles using inverse transformation.

Using convolution, to find the inverse transform of    $\frac{1}{p^4-1}$

 

Rewrite it as above equation,

 

  $F\left(z\right)e^{zt}=\frac{e^{zt}}{z^4-1}$

 

Determine the poles of $F\left(z\right)e^{zt}$ by factoring the denominator as,

 

  $$\begin{gathered} F\left(z\right)e^{zt}=\frac{e^{zt}}{z^4-1} \\ F\left(z\right)e^{zt}=\frac{e^{zt}}{\left(z^2-1\right)\left(z^2+1\right)} \\ F\left(z\right)e^{zt}=\frac{e^{zt}}{\left(z-1\right)\left(z+1\right)\left(z+i\right)\left(z-i\right)} \end{gathered}$$

 

Simple poles at $z=\pm 1$  and  $z=\pm i$ has the above equation 

Step 4: Determine the residue with simple poles.

So now we find the residues at simple all poles as:

 

Residue at,  $\begin{array}{rcl}\mathit{z}& \mathbf{=}& \mathbf{1}\end{array}$

 $\begin{array}{rcl}\left(z-1\right)\mathit{F}\left(z\right){\mathit{e}}^{\mathbf{z}\mathbf{t}}& \mathbf{=}& \left(z-1\right)\frac{{\mathbf{e}}^{\mathbf{z}\mathbf{t}}}{\left(z-1\right)\left(z+1\right)\left(z+i\right)\left(z-i\right)}\\ \mathbf{(}\mathbf{z}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{F}\mathbf{\left(}\mathbf{z}\mathbf{\right)}{\mathit{e}}^{\mathbf{zt}}& \mathbf{=}& \frac{{\mathbf{e}}^{\mathbf{zt}}}{\mathbf{(}\mathbf{z}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{z}\mathbf{+}\mathbf{i}\mathbf{)}\mathbf{(}\mathbf{z}\mathbf{-}\mathbf{i}\mathbf{)}}\\ \mathbf{(}\mathbf{z}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{F}\mathbf{\left(}\mathbf{z}\mathbf{\right)}{\mathit{e}}^{\mathbf{zt}}& \mathbf{=}& \frac{{\mathbf{e}}^{\mathbf{t}}}{\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{i}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{i}\mathbf{)}}\\ \mathbf{(}\mathbf{z}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{F}\mathbf{\left(}\mathbf{z}\mathbf{\right)}{\mathit{e}}^{\mathbf{zt}}& \mathbf{=}& \frac{{\mathbf{e}}^{\mathbf{t}}}{\mathbf{4}}\end{array}$

  

 

Residue at, $\begin{array}{rcl}\mathit{z}& \mathbf{=}& \mathbf{-}\mathbf{1}\end{array}$  

 

  $\begin{array}{rcl}\left(z+1\right)\mathit{F}\left(z\right){\mathit{e}}^{\mathbf{z}\mathbf{t}}& \mathbf{=}& \left(z+1\right)\frac{{\mathbf{e}}^{\mathbf{z}\mathbf{t}}}{\left(z-1\right)\left(z+1\right)\left(z+i\right)\left(z-i\right)}\\ \left(z+1\right)\mathit{F}\left(z\right){\mathit{e}}^{\mathbf{z}\mathbf{t}}& \mathbf{=}& \frac{{\mathbf{e}}^{\mathbf{z}\mathbf{t}}}{\left(z-1\right)\left(z+i\right)\left(z-i\right)}\\ \left(z+1\right)\mathit{F}\left(z\right){\mathit{e}}^{\mathbf{z}\mathbf{t}}& \mathbf{=}& \frac{{\mathbf{e}}^{\mathbf{-}\mathbf{t}}}{\left(-1-1\right)\left(-1+i\right)\left(-1-i\right)}\\ \left(z+1\right)\mathit{F}\left(z\right){\mathit{e}}^{\mathbf{z}\mathbf{t}}& \mathbf{=}& \mathbf{-}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{t}}}{\mathbf{4}}\end{array}$