Green's function impulse

Question

Solve $y''(t)-a^2y(t)=\delta(t-t'), \ y(0)=0, \ y'(0)=0, \ t'>0.$

I am not sure how to solve an equation that has $\delta(t-t')$ as a solution. The book doesn't really elaborate further than stating that it is an impulse. I am sure I can use Green's Function, since this question is in that section.

Answer 1

If you want to solve $y''-a^{2}y=f$ subject to $y(0)=y'(0)$, the Green function solution is $$ y(t)=\frac{1}{2a}\left[e^{at}\int_{0}^{t}e^{-au}f(u)\,du-e^{-at}\int_{0}^{t}e^{au}f(u)\,du\right]. $$ You can check that $$ y'(t) = \frac{1}{2}\left[e^{at}\int_{0}^{t}e^{-au}f(u)\,du+e^{-at}\int_{0}^{t}e^{au}f(u)\,du\right],\\ y''(t)=\frac{a}{2}\left[e^{at}\int_{0}^{t}e^{-au}f(u)\,dt-ae^{-at}\int_{0}^{t}e^{au}f(u)\,du\right]+f(t). $$ That is, $y''-a^{2}y=f$. And it is easy to check that $y(0)=y'(0)=0$ from the explict expressions given above for $y$, $y'$.

Let $f(u)=\delta(u-t')$. Then $$ \begin{align} y(t) & = \left\{\begin{array}{ll} 0, & 0 \le t < t',\\ \frac{1}{2a}[e^{at}e^{-at'}-e^{-at}e^{at'}] & t > t' \end{array}\right. \\ & = \left\{\begin{array}{ll} 0, & 0 \le t < t',\\ \frac{1}{a}\sinh(a(t-t')) & t > t' \end{array}\right. . \end{align} $$ For problems such as this one, you can approximate $\delta(t-t')$ as a square pulse centered at $t'$ whose integral is $1$. So, for example, the pulse could be concentrated in the time interval $t'-\frac{1}{2n} \le t \le t'+\frac{1}{2n}$ of height $n$. For large enough $n$, the length of time of pulse is immeasurably short, and the response $y$ will be as close as you want to the response given above. So, $\int \delta(t-t')f(t)\,dt = f(t')$ is as close to accurate as you want provided $f$ is continuous near $t'$ and the interval of integration includes $t'$ in its interior. And that results in a response $y(t)$ which is as close as you want to the one given above.

So the response stays at $0$ because of $y(0)=y'(0)=0$, but it immediately starts growing when you first thump it very sharply. It grows without bound because there is an unstable positive feedback. A continuous driving signal can be viewed as a succession of small thumps spaced very closely together in time whose amplitudes are given by the driving function $f$, and the response is then a superposition of these fundamental solutions.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ The differential equation is equivalent to: \begin{align}\left\{\begin{array}{rcl} {\rm y}''\pars{t} - a^{2}{\rm y}\pars{t} & = & 0\,,\qquad t \not= t' \\[2mm] \lim_{t\ \to\ t'^{+}}{\rm y}'\pars{t} -\lim_{t\ \to\ t'^{-}}{\rm y}'\pars{t} & = & 1 \\[2mm] {\rm y}\pars{0} = 0\,,\qquad{\rm y}'\pars{0} & = & 0 \end{array}\right.\qquad\qquad {\Large t'\ >\ 0} \end{align}

It's clear that $\ds{{\rm y}\pars{t}}$ vanishes out identically in $\ds{\pars{0,t'}}$. Then, $$ {\rm y}\pars{t}=\Theta\pars{t - t'} \bracks{{\rm A}\pars{t'}\sinh\pars{at} + {\rm B}\pars{t'}\cosh\pars{at}} $$ which satisfies:

\begin{align} \left\{\begin{array}{rcrcl} \sinh\pars{at'}{\rm A}\pars{t'} & + & \cosh\pars{at'}{\rm B}\pars{t'} & = & 0 \\[2mm] \cosh\pars{at'}a{\rm A}\pars{t'} & + & \sinh\pars{at'}a{\rm B}\pars{t'} & = & 1 \end{array}\right. \end{align}

such that \begin{align} {\rm A}\pars{t'}&=-\,{1 \over a}\verts{% \begin{array}{rr}0 & \cosh\pars{at'} \\[2mm] 1 & a\sinh\pars{at'}\end{array}} ={\cosh\pars{at'} \over a} \\[5mm] {\rm B}\pars{t'}&=-\,{1 \over a}\verts{% \begin{array}{rr}\sinh\pars{at'} & 0 \\[2mm] a\cosh\pars{at'} & 1\end{array}} =-\,{\sinh\pars{at'} \over a} \end{align}

\begin{align} &\color{#66f}{\large{\rm y}\pars{t}} =\Theta\pars{t - t'}\bracks{{\cosh\pars{at'} \over a}\,\sinh\pars{at} - {\sinh\pars{at'} \over a}\,\cosh\pars{at}} \\[5mm]&=\color{#66f}{\large% \Theta\pars{t - t'}\,{\sinh\pars{a\bracks{t - t'}} \over a} =\left\lbrace \begin{array}{lcrcl} 0 & \mbox{if} & t & < & t' \\[2mm] {\sinh\pars{a\bracks{t - t'}} \over a} & \mbox{if} & t & > & t' \end{array}\right.} \end{align}

Let's check the solution: \begin{align} \color{#c00000}{{\rm y}'\pars{t}}& =\delta\pars{t - t'}\,{\sinh\pars{a\bracks{t - t'}} \over a} +\Theta\pars{t - t'}\,{\cosh\pars{a\bracks{t - t'}}a \over a} \\&=\color{#c00000}{\Theta\pars{t - t'}\cosh\pars{a\bracks{t - t'}}} \\[5mm] \color{#c00000}{{\rm y}''\pars{t}}& =\delta\pars{t - t'}\cosh\pars{a\bracks{t - t'}} +\Theta\pars{t - t'}\sinh\pars{a\bracks{t - t'}}a \\&=\delta\pars{t - t'} +a^{2}\braces{\Theta\pars{t - t'}\,{\sinh\pars{a\bracks{t - t'}} \over a}} =\color{#c00000}{\delta\pars{t - t'} + a^{2}{\rm y}\pars{t}} \\[5mm]&\boxed{\quad\ds{\imp\quad{\rm y}''\pars{t} - a^{2}{\rm y}\pars{t} = \delta\pars{t - t'}}\quad} \\&\mbox{with}\quad \braces{\begin{array}{rcl} {\rm y}\pars{0}&=&\Theta\pars{-t'}\,{\sinh\pars{-at'} \over a} = 0 \\[2mm] {\rm y}'\pars{0}&=&\Theta\pars{-t'}\cosh\pars{-at'} = 0 \end{array}}\quad\mbox{since}\quad t' > 0. \end{align}

Answer 3

Yet another approach is to use the Laplace transform method. Let $Y(s) = \mathcal{L}[y]$ then, by taking the Laplace transform of both sides and using the initial conditions, we have $$s^2Y(s) - a^2Y(s) = e^{-t's} \implies Y(s) = \frac{e^{-t's}}{s^2 - a^2}.$$ Recall that $$\mathcal{L}^{-1}\left[\frac{a}{s^2 - a^2}\right] = \sinh(at)$$ and the shifting formula gives $$\mathcal{L}^{-1}[e^{-t's}F(s)] = f(t-t')u(t-t')$$ where $u$ is the Heaviside step function. Thus $$y(t) = \mathcal{L}^{-1}\left[\frac{e^{-t's}}{s^2 - a^2}\right] =\frac{1}{a}\sinh(a(t-t'))u(t-t')$$ which gives the same result as the answers by T.A.E and Felix Marin.