Q4E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the convolution theorem to obtain a formula for the solution to the given initial value problem, where g(t) is piecewise continuous on $\mathbf{1}\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$and of exponential order.

$\mathbf{[}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{ }\mathbf{+}\mathbf{ }\mathit{y}\mathbf{ }\mathbf{=}\mathbf{ }\mathit{g}\mathbf{(}\mathit{t}\mathbf{)}\mathbf{ }\mathbf{;}\mathbf{ }\mathit{y}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{,}\mathbf{ }\mathit{y}\mathbf{\text{'}}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}$

Step-by-Step Solution

Answer

The solution to the given initial value problem by using a convolution theorem to obtain a formula is.

$y\left(t\right) ={\int }_0^t\sin (t - v) g\left(v\right)dv+\sin t$

 

Step 1: Define convolution formula

Let f(t) and g(t) be continuous on the interval,$\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$ the convolution can be denoted as *. Thus, the convolution for the function f(t) and g(t) be defined as,

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

 

Step 2: Apply Laplace transform and use convolution formula to obtain a solution

Consider the given equation,

$y\text{'}\text{'} + y = g\left(t\right);y\left(0\right) = 0,quad y\text{'}\left(0\right) = 1$

 Apply Laplace transform,

$L\{y\text{'}\text{'} + y\}=L\left[g\right(t\left)\right]$

$$\begin{gathered} s^{2}y\left(s\right) - sy\left(0\right) - y\text{'}\left(0\right) + y\left(s\right) = g\left(s\right) \\ {s}^{2}y\left(s\right) - 1 + y\left(s\right) = g\left(s\right) \\ \left(s^2+1\right) y\left(s\right) - 1 = g\left(s\right) \\ y\left(s\right)=\frac{g\left(s\right)}{s^2+1}+\frac{1}{s^2+1} \end{gathered}$$

Take, inverse Laplace transform, we get

$L^{-1}\left[y\right(s\left)\right]=L^{-1}\left[\frac{g\left(s\right)}{\left(s^2+1\right)}\right]+L^{-1}\left[\frac{1}{\left(s^2+1\right)}\right]....\left(1\right)$

Use the convolution formula,

$L^{-1}\left[\int \left(s\right)g\left(s\right)\right]={\int }_0^{t}f(t - v) g\left(v\right)dv$

Since,

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

f(t)=sint

Substitute in the formula,

$L^{-1}\left[\frac{g\left(s\right)}{s^2+1}\right]={\int }_0^t[\sin (t - v\left)g\right(v)dv$

Hence, equation becomes,

$y\left(t\right)={\int }_0^t[\sin (t - v\left)g\right(v)dv+\sin t$

Therefore, the obtained solution for the given initial value problem is.

$y\left(t\right)={\int }_0^t[\sin (t - v\left)g\right(v)dv+\sin t$