Question

In Problems 19-22 proceed as in Example 3 and find the convolution $\mathbf{f}\mathbf{*}\mathbf{g}$of the given functions. After integrating find the Laplace transform of $\mathbf{f}\mathbf{*}\mathbf{g}$.

$\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{t}\mathbf{,}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}$

Short answer

The convolution of the given function is and the Laplace transform of the convolution is $\frac{24}{s^5}$.

Step-by-step solution

Convolution of two functions

"If f and g are piecewise continuous on the interval $[0,\infty ]$, then the convolution of f and g, denoted by the symbol $\mathrm{f}*\mathrm{g}$is a function defined by the integral ."

$\mathbf{f}\mathbf{*}\mathbf{g}\mathbf{=}{\int }_{\mathbf{0}}^{\mathbf{t}}\mathbf{f}\mathbf{\left(}\mathbf{\tau }\mathbf{\right)}\mathbf{g}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{\tau }\mathbf{)}\mathbf{d\tau }$

Apply the convolution of two functions

The given functions are, $f\left(t\right)=4t$and $g\left(t\right)=3t^2$

By the definition of fand g, the values are,$f\left(\tau \right)=4\tau$and$g(t-\tau )=3(t-\tau {)}^2$.

The convolution of the given functions is computed as follows.

$\begin{array}{l}f*g={\int }_0^{t}f\left(\tau \right)g(t-\tau )d\tau \\ ={\int }_0^t\left(4\tau \right)\left[3{(t-\tau )}^2\right]d\tau \\ ={\int }_0^t\left(4\tau \right)\left[3\left(t^2+{\tau }^2-2t\tau \right)\right]d\tau \\ =12{\int }_0^t\left(t^2\tau +{\tau }^3-2t{\tau }^2\right)d\tau \end{array}$

On further simplification,

$\begin{array}{l}f*g=12{\int }_0^t\left(t^2\tau +{\tau }^3-2t{\tau }^2\right)d\tau \\ =12{\left[\frac{t^2{\tau }^2}{2}+\frac{{\tau }^4}{4}-\frac{2t{\tau }^3}{3}\right]}_0^t\\ =12\left[\frac{t^4}{2}+\frac{t^4}{4}-\frac{2t^4}{3}\right]\\ =t^4\end{array}$

So, the convolution of the given functions is $t^4$.

$\begin{array}{l}f*g={\int }_0^{t}f\left(\tau \right)g(t-\tau )d\tau \\ ={\int }_0^t\left(4\tau \right)\left[3{(t-\tau )}^2\right]d\tau \\ ={\int }_0^t\left(4\tau \right)\left[3\left(t^2+{\tau }^2-2t\tau \right)\right]d\tau \\ =12{\int }_0^t\left(t^2\tau +{\tau }^3-2t{\tau }^2\right)d\tau \end{array}$

Apply Laplace transform

Now, the given function's Laplace transform is obtained as,

$\begin{array}{l}L\{f*g\}=L\left\{t^4\right\}\\ =\frac{4!}{s^{4+1}}\\ =\frac{24}{s^5}\end{array}$

So, the convolution's Laplace transform is $\frac{24}{s^5}$

Hence, the convolution of the given function is $t^4$and the Laplace transform of the convolution is $\frac{24}{s^5}$