# Laplace transform of the square wave to solve PDE

Solve $$y'' + 3y' +2y = r(t)$$ given $y(0)=0$ and $y'(0) = 0$ where $r(t)$ is the square wave, $$r(t) = u(t-1) - u(t-2)$$

I'm just going to type out the answer as I read it and tell you which parts I understand and which parts I have trouble intuitively grasping, I hope this is OK.

SOLUTION

Using Laplace transforms, $$\begin{array} \ L[y'' +3y'+2y] = L[r(t)]\\ \implies L[y'' +3y' +2y] = L[u(t-1) - u(t-2)]\\ \implies [s^2 Y(s) - s y(0) - y'(0)] + 3[s Y(s) - y(0)] +2 Y(s) = \color{green}{{1 \over s} [e^{-s} - e^{-2s}]}\\ \end{array}$$

The last part in $\color{green}{\text{green}}$ I guess comes from this result for the $\color{green}{\text{Dirac's Delta Function}}$: $$L[\delta (t-a)] = e^{-as}$$ (Because the square wave is a step function?)

Now,

$$\ Y(s) [s^2 +3s +2] = {1 \over s} [e^{-s} - e^{-2s}] = \frac{1}{(s{^2} +3s +2 s)s} [e^{-s} - e^{-2s}]\\$$

$$F(s)= \frac{1}{(s{^2} +3s +2 s)s} = {1 \over 2s} - {1 \over {s+1}} + {1 \over s(s+2)}$$

$$f(t) = L^{-1}[F(s)] = L^{-1}[{1 \over 2s} - {1 \over {s+1}} + {1 \over s(s+2)}] \implies f(t) = {1 \over 2} - e^t + {1 \over 2} e^{-2t}$$

Up to this point I think I can grasp what's going on. But after this, I'm not 100% sure I follow:

$$Y(s) = L^{-1} [ F(s) e^{-s} - F(s) e^{-2s}] = \color{blue}{f(t-1)u(t-1) - f(t-2)u(t-2)}$$

$$\color{red}{ Y(s) = \begin{cases} 0, &0 < t< 1 \\[2ex] {1 \over 2} - e^{-(t-1)} + {1 \over 2} e^{-2(t-1)}, &1< t<2 \\[2ex] {1 \over 2} e^{-(t-1)} - {1 \over 2} e^{-2(t-1)} - {1 \over 2} e^{(t-2)} + {1 \over 2} e^{2(t-2)}, &t>2 \\[2ex] \end{cases}}$$

I have absolutely no idea how $\color{blue}{this}$ became $\color{red}{this}$

I've looked at this result for a few days now but can't seem to get the intuition right. That's why I came here. I need to understand how this answer became this set of cases. I would appreciate your help very much!

• you must have $y(t)$ and not $Y(s)$ in the last part with the inverse Laplace transform. – alexjo Apr 27 '16 at 9:08
• The Laplace transfor of $u(t-a)$ comes from direct integration $\int_a^\infty e^{-st}\mathrm d t$. – alexjo Apr 27 '16 at 9:10
• oooh I see! I think this helps a bit @alexjo I was copying these result's from a classmate's book, he must have mis-written it – Siyanda Apr 27 '16 at 9:10
• This is the 600,000th question on this site. – wythagoras Apr 27 '16 at 9:14

For $t<1$ we have $u(t-1)=0$ and for $t<2$ we have $u(t-2)=0$. So for $t<1$ we have $f(t-1)u(t-1)=0$ and $f(t-2)u(t-2)=0$ and then $y(t)=0$.
For $t> 1$ we have $u(t-1)=1$ and for $t<2$ we have $u(t-2)=0$. So for $1<t<2$ we have $f(t-1)u(t-1)=f(t-1)$ and $f(t-2)u(t-2)=0$ and then $$y(t)=f(t-1)={1 \over 2} - e^{-(t-1)} + {1 \over 2} e^{-2(t-1)}$$.
For $t> 2$ we have $u(t-1)=1$ and for $t>2$ we have $u(t-2)=1$. So for $t>2$ we have $f(t-1)u(t-1)=f(t-1)$ and $f(t-2)u(t-2)=f(t-2)$ and then $$y(t)=f(t-1)-f(t-2)={1 \over 2} - e^{-(t-1)} + {1 \over 2} e^{-2(t-1)}-\left[{1 \over 2} - e^{-(t-1)} + {1 \over 2} e^{-2(t-1)}\right]={1 \over 2} e^{-(t-1)} - {1 \over 2} e^{-2(t-1)} - {1 \over 2} e^{(t-2)} + {1 \over 2} e^{2(t-2)}$$.
$$y(t) = L^{-1} [ F(s) e^{-s} - F(s) e^{-2s}] = (t-1)u(t-1) - f(t-2)u(t-2)$$
$u(t-t_0)$ is the unit step function. It has the value $0$ for all values of $t<t_0$ and has the value $1$ for all values of $t>t_0$. Using this definition you can see that for $0<t<1$ all unit step functions vanish. Resulting in $y(t)=0$. For $1<t<2$ only $f(t-1)$ will appear. For $t>2$ $f(t-1)-f(t-2)$ is the result.