Using laplace transforms to solve a piecewise defined function initial value problem

Question

I want to use laplace transforms to solve the following:

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$$\frac{d^2 y}{dt^2}+16 y = f(t) = \left\{\begin{array} 1 1&t\lt\pi\\0&t\geq \pi\end{array}\right.\text{ with } y(0)=0 \text{ and } \frac{dy}{dt}(0)=0$$

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Can I have some direction on solving this? I have heard I can use a shifting theorem. Is there a way to do this without using said shifting theorem?

A link to the right thing would be sufficient for my thanks!

Answer 1

You can define a piecewise function using the Heaviside Unit Step Function and then take the Laplace Transform of that.

In general, if:

$$f(t)= \left\{\begin{array} fg(t) &0 \le t < a\\h(t) &t \ge a \end{array}\right.$$

then:

$$f(t) = g(t) - g(t)u(t-a) + h(t) u(t-a)$$

In your example, we have:

$$\tag 1 f(t) = 1 - (1)u(t- \pi) + (0) u(t- \pi) = 1 - 1~u(t- \pi)$$

Of course, you can do this other ways and here is an example (use the definition straight off), Laplace transform of unit step function.

The Laplace Transform of $(1)$ is given by:

$$\mathscr{L} (1 - 1~u(t-\pi)) = \dfrac{1}{s} - \dfrac{e^{-\pi s}}{s} = \dfrac{1 - e^{-\pi s}}{s}$$

The Laplace Transform of the other part with initial conditions yields:

• $\mathscr{L} (y''(t)) = s^2y(s) -s y(0) -y'(0) = s^2y(s)$
• $\mathscr{L} (16y(t)) = 16y(s)$

Putting it all together yields:

$$y(s) = \dfrac{1 - e^{-\pi s}}{s(s^2 + 16)}$$

I will assume you can take it from here.

Answer 2

This is the asker trying to build an answer, but it isn't going well. I think maybe I am actually supposed to use the Heaviside function?

This shall be solved using the second shifting property:

$g(t)= \left\{\begin{array}ff(t-a) &t\gt a\\0 &t\lt a \end{array}\right.$

$$\mathcal{L}\{g(t)\}=\int_{-\infty}^\infty e^{-st}g(t) dt$$

First we need to recognise what we are working with. So we want $0\to\pi$ for the bounds of one part of the above integral, and $\pi\to\infty$ for the second bound.

$$\int_{-\infty}^\pi e^{-st}g(t) dt$$