Differential equation – Part III Fourier series

Fourier Series

To solve the ODE ${y^{\prime\prime}} + a{y^{\prime}} + by = f\left( t \right)$, in which $f\left( t \right)$ is often the combination of ${e^t}$, $\sin t$ and $\cos t$, then any reasonable $f\left( t \right)$ which is periodic, can be discontinuous but not terribly discontinuous, can be written as Fourier Series.

$f\left( t \right) = {c_0} + \sum\limits_{n = 1}^\infty {{a_n}\cos nt + {b_n}\sin nt}$

Theorem:

$u\left( t \right),v\left( t \right)$ are functions with period $2\pi$ on the real number, they are orthogonal (perpendicular) on $\left[ { – \pi ,\pi } \right]$ if satisfying

$\int_{ – \pi }^\pi {u\left( t \right)v\left( t \right)} dt = 0$

$\left\{ \begin{gathered} \sin nt \qquad n = 1,2, \cdot \cdot \cdot ,\infty \hfill \\ \cos mt \qquad m = 0,1, \cdot \cdot \cdot ,\infty \hfill \\ \end{gathered} \right.$

any two distinct ones of those are orthogonal on $\left[ { – \pi ,\pi } \right]$.

$\int_{ – \pi }^\pi {\sin nt\cos mt} dt = 0 \qquad \int_{ – \pi }^\pi {{{\sin }^2}nt} dt = \pi \qquad \int_{ – \pi }^\pi {{{\cos }^2}mt} dt = \pi$

Proof:

We can use trig identities, complex exponential, however the two methods only work on trigonometric functions. The generalized method is to use ODE.

$\sin nt$ and $\cos mt$, $m \ne n$, satisfy

${u^{\prime\prime}} + {n^2}u = 0 \to u_n^{\prime\prime} = – {n^2}{u_n}$

${v^{\prime\prime}} + {m^2}v = 0 \to v_m^{\prime\prime} = – {m^2}{v_m}$

$\int_{ – \pi }^\pi {u_n^{\prime\prime}{v_m}} dt = \left. {u_n^{\prime}{v_m}} \right|_{ – \pi }^\pi – \int_{ – \pi }^\pi {u_n^{\prime}v_m^{\prime}} dt = – \int_{ – \pi }^\pi {u_n^{\prime}v_m^{\prime}} dt$

$\int_{ – \pi }^\pi {u_n^{\prime\prime}{v_m}} dt = – {n^2}\int_{ – \pi }^\pi {{u_n}{v_m}} dt$

$\int_{ – \pi }^\pi {v_m^{\prime\prime}{u_n}} dt = – {m^2}\int_{ – \pi }^\pi {{v_m}{u_n}} dt$

The first equation is symmetric while the second and the third are not symmetric, in fact the ${{u_n}}$ and ${{u_n}}$ are equally important. Thus, the only answer is

$\int_{ – \pi }^\pi {{u_n}{v_m}} dt = 0,\ m \ne n$

Calculate the coefficients

$f\left( t \right) = \cdot \cdot \cdot + {a_k}\cos kt + \cdot \cdot \cdot + {a_n}\cos nt + \cdot \cdot \cdot$

$\int_{ – \pi }^\pi {f\left( t \right)} \cos ntdt = \cdot \cdot \cdot + \int_{ – \pi }^\pi {{a_k}\cos kt} \cos ntdt + \cdot \cdot \cdot + \int_{ – \pi }^\pi {{a_n}} {\cos ^2}ntdt + \cdot \cdot \cdot = \int_{ – \pi }^\pi {{a_n}} {\cos ^2}ntdt$

$\int_{ – \pi }^\pi {f\left( t \right)} \cos ntdt = {a_n}\pi \to {a_n} = \frac{1}{\pi }\int_{ – \pi }^\pi {f\left( t \right)} \cos ntdt$

${b_n} = \frac{1}{\pi }\int_{ – \pi }^\pi {f\left( t \right)} \sin ntdt$

$f\left( t \right) = {c_0} + \cdot \cdot \cdot + {a_n}\cos nt + \cdot \cdot \cdot$

$\int_{ – \pi }^\pi {f\left( t \right)} dt = 2\pi {c_0} + \cdot \cdot \cdot + \int_{ – \pi }^\pi {{a_n}\cos nt} dt + \cdot \cdot \cdot = 2\pi {c_0}$

${c_0} = \frac{1}{{2\pi }}\int_{ – \pi }^\pi {f\left( t \right)} dt = \frac{1}{{2\pi }}\int_{ – \pi }^\pi {f\left( t \right)} \cos 0tdt = \frac{{{a_0}}}{2}$

Example:

${a_n} = 0$

${b_n} = – \frac{1}{\pi }\int_{ – \pi }^0 {\sin nt} dt + \frac{1}{\pi }\int_0^\pi {\sin nt} dt = \frac{{1 – \cos n\pi }}{n} + \frac{{1 – \cos n\pi }}{n} = \frac{2}{n}\left( {1 – \cos n\pi } \right)$

Extension of  Fourier Series

$f\left( t \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos nt + {b_n}\sin nt}$

${a_n} = \frac{1}{\pi }\int_{ – \pi }^\pi {f\left( t \right)} \cos ntdt,{b_n} = \frac{1}{\pi }\int_{ – \pi }^\pi {f\left( t \right)} \sin ntdt$

Theorem: since there are formulas for ${a_n}$ and ${b_n}$,

$f\left( t \right) = g\left( t \right) \to F.S.forf\left( t \right) = F.S.forg\left( t \right)$

1. shorten calculation: use evenness & oddness
2. extend the range of Fourier Series

If $f\left( t \right)$ is even, $f\left( { – t} \right) = f\left( t \right)$

$f\left( t \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos nt} ,{b_n} = 0$

If $f\left( t \right)$ is odd, $f\left( { – t} \right) = -f\left( t \right)$

$f\left( t \right) = \sum\limits_{n = 1}^\infty {{b_n}\sin nt} ,{a_n} = 0$

Proof:

$f\left( { – t} \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos nt + {b_n}\sin \left( { – nt} \right)}$

$f\left( t \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos nt + {b_n}\sin nt}$

Thus, ${b_n} = – {b_n}$ for all $n$, the only answer is ${b_n} = 0$.

Example:

\begin{align*}
{b_n} &= \frac{2}{\pi }\int_0^\pi {t\sin nt} dt = \frac{2}{\pi }\left[ {\left. { – \frac{{t\cos nt}}{n}} \right|_0^\pi – \int_0^\pi {\cos nt} dt} \right] \hfill \\
&= \frac{2}{\pi }\left[ {\left. { – \frac{\pi }{m}{{\left( { – 1} \right)}^n}} \right|_0^\pi \left. { + \frac{{\sin nt}}{n}} \right|_0^\pi } \right] = \frac{2}{m}{\left( { – 1} \right)^{n + 1}} \hfill \\
\end{align*}

$F.S.forf\left( t \right) = 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { – 1} \right)}^{n + 1}}}}{m}\sin nt}$

Theorem:

If $f$ is continuous at $t_0$, then $f\left( t \right) = sum\ of\ its\ F.S.\ at{t_0}$.

If $f$ is discontinuous at $t_0$, then $F.S.\ at{t_0}$ converges to the mid point of the jump.

Extension #1: period $2L$

$f\left( t \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos n\frac{\pi }{L}t + {b_n}\sin n\frac{\pi }{L}t}$

${a_n} = \frac{1}{L}\int_{ – L}^L {f\left( t \right)} \cos n\frac{\pi }{L}tdt,{b_n} = \frac{1}{L}\int_{ – L}^L {f\left( t \right)} \sin n\frac{\pi }{L}tdt$

$t = \frac{L}{\pi }u,u = \frac{\pi }{L}t$

Extension #2: periodic extension

If $f\left( t \right)$ is defined on $\left[ {0,L} \right]$

Particular solution with Fourier Series

${x^{\prime\prime}} + \omega _0^2x = f\left( t \right),f\left( t \right) = \sin \omega t,\cos \omega t \to {x_p} = \frac{{\sin \omega t,\cos \omega t}}{{\omega _0^2 – {\omega ^2}}}$

$f\left( t \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos {\omega _n}t + {b_n}\sin {\omega _n}t} ,{\omega _n} = \frac{{n\pi }}{L}$

$f\left( t \right) = s\left( t \right) + \frac{1}{2},s\left( t \right) = \frac{1}{2}g\left( u \right),g\left( u \right) = \frac{4}{\pi }\sum\limits_{odd} {\frac{{\sin nu}}{n}}$

$f\left( t \right) = \frac{1}{2} + \frac{1}{2}\frac{4}{\pi }\sum\limits_{odd} {\frac{{\sin n\pi t}}{n}}$

${x_p} = \frac{{{a_0}}}{{\omega _0^2}} + \sum\limits_{n = 1}^\infty {\frac{{{a_n}\cos {\omega _n}t}}{{\omega _0^2 – \omega _n^2}} + \frac{{{b_n}\sin {\omega _n}t}}{{\omega _0^2 – \omega _n^2}}}$

${x_p} = \frac{1}{{2\omega _0^2}} + \frac{2}{\pi }\sum\limits_{odd}^\infty {\frac{{\sin n\pi t}}{{\omega _0^2 – {{\left( {n\pi } \right)}^2}}}}$

Assume natural frequency ${\omega _0} = 10$, calculate ${x_p}\left( t \right)$

\begin{align*}
{x_p}\left( t \right) & \approx 0.005 + 0.6\left( {\frac{{\sin \pi t}}{{91}} + \frac{{\sin 3\pi t}}{{45}} – \frac{{\sin 5\pi t}}{{600}}} \right) \hfill \\
&  \approx 0.005 + 0.01\sin \pi t + 0.015\sin 3\pi t – 0.001\sin 5\pi t \hfill \\
\end{align*}

If ${\omega _0} = 9$, near-resonance occurs for frequency $3\pi$. Frequencies in $f\left( t \right)$ are hidden, but system picks out resonance with the frequency closest to its natural frequency. Here ${\omega _0} = 10$ and $n = 3$ is the closest to resonance as the coefficient is largest.

Another method to get the coefficients

Use differential of F.S. term by term and assume solution of form

${x_p} = {c_0} + \sum\limits_{n = 1}^\infty {{c_n}\sin n\pi t}$

substitute into the ODE

$\omega _0^2{c_0} + \sum\limits_{n = 1}^\infty {{c_n}\left( {\omega _0^2 – {{\left( {n\pi } \right)}^2}} \right)\sin n\pi t} = \frac{1}{2} + \frac{2}{\pi }\sum\limits_{odd} {\frac{{\sin n\pi t}}{n}}$

by equating coefficients, we can get $c_n$

${c_0} = \frac{1}{{2\omega _0^2}},{c_n} = \frac{2}{\pi }\frac{1}{{\omega _0^2 – {{\left( {n\pi } \right)}^2}}}$

Laplace transform

$\sum\limits_{n = 1}^\infty {{a_n}{x^n}} = A\left( x \right)$

$1 \sim \frac{1}{{1 – x}} = 1 + x + {x^2} + {x^3} + \cdot \cdot \cdot ,\frac{1}{{n!}} \sim {e^x} = \sum {\frac{{{x^n}}}{{n!}}}$

Continuous analog:

$\int_0^\infty {a\left( x \right){x^t}} dx = A\left( x \right) \to \int_0^\infty {f\left( t \right){e^{ – st}}} dt = F\left( s \right)$

Laplace transform:

$L\left( {f\left( t \right)} \right) = F\left( s \right)$

Linear transform:

$L\left( {f + g} \right) = L\left( f \right) + L\left( g \right),L\left( {cf} \right) = cL\left( f \right)$

Commonly used:

\begin{align*}
&1 \mapsto \frac{1}{s},s > 0 \qquad \qquad &\sin at \mapsto \frac{a}{{{s^2} + {a^2}}},s > 0 \\
&{e^{at}} \mapsto \frac{1}{{s – a}} \qquad \qquad  &\cos at \mapsto \frac{s}{{{s^2} + {a^2}}},s > 0 \\
&{e^{at}}f\left( t \right) \mapsto F\left( {s – a} \right),s > a   &{t^n} \mapsto \frac{{n!}}{{{s^{n + 1}}}},s > 0
\end{align*}

Derivation:

$\int_0^\infty {{e^{ – st}}} dt = \mathop {\lim }\limits_{R \to \infty } \int_0^\infty {{e^{ – st}}} dt = \mathop {\lim }\limits_{R \to \infty } \frac{{{e^{ – sR}} – 1}}{{ – s}} = \frac{1}{s},s > 0$

$\int_0^\infty {{e^{at}}f\left( t \right){e^{ – st}}} dt = \int_0^\infty {{e^{at}}f\left( t \right){e^{ – \left( {s – a} \right)t}}} dt = F\left( {s – a} \right),s > a$

${e^{at}} \mapsto \frac{1}{{s – a}},s > a \qquad {e^{\left( {a + bi} \right)t}} \mapsto \frac{1}{{s – \left( {a + bi} \right)}},s > a$

$\cos at = \frac{{{e^{iat}} + {e^{ – iat}}}}{2},L\left( {\cos at} \right) = \frac{1}{2}\left( {\frac{1}{{s – ia}} + \frac{1}{{s + ia}}} \right) = \frac{s}{{{s^2} + {a^2}}},s > 0$

$\sin at = \frac{{{e^{iat}} – {e^{ – iat}}}}{2},L\left( {\cos at} \right) = \frac{a}{{{s^2} + {a^2}}},s > 0$

$\int_0^\infty {{t^n}{e^{ – st}}} dt = \left. {{t^n}\frac{{{e^{ – st}}}}{{ – s}}} \right|_0^\infty – \int_0^\infty {n{t^{n – 1}}\frac{{{e^{ – st}}}}{{ – s}}} dt = 0 – 0 + \frac{n}{s}\int_0^\infty {{t^{n – 1}}{e^{ – st}}} dt = \frac{n}{s}L\left( {{t^{n – 1}}} \right)$

$\mathop {\lim }\limits_{t \to \infty } {t^n}\frac{{{e^{ – st}}}}{{ – s}} = – \frac{1}{s}\mathop {\lim }\limits_{t \to \infty } \frac{{{t^n}}}{{{e^{st}}}} = – \frac{1}{s}\mathop {\lim }\limits_{t \to \infty } \frac{{n{t^{n – 1}}}}{{s{e^{st}}}} = \cdot \cdot \cdot = 0$

$L\left( {{t^n}} \right) = \frac{n}{s}L\left( {{t^{n – 1}}} \right) = \cdot \cdot \cdot = \frac{{n\left( {n – 1} \right) \cdot \cdot \cdot 1}}{{{s^n}}}L\left( {{t^0}} \right) = \frac{{n!}}{{{s^{n + 1}}}}$

inverse Laplace transform

${L^{ – 1}}\left( {\frac{1}{{s\left( {s + 3} \right)}}} \right) = {L^{ – 1}}\left( {\frac{{1/3}}{s} + \frac{{ – 1/3}}{{s + 3}}} \right) = \frac{1}{3} – \frac{1}{3}{e^{ – 3t}}$

Solve ODE with Laplace transform

$f\left( t \right) \mapsto F\left( s \right) = \int_0^\infty {f\left( t \right){e^{ – st}}} dt,s > 0$

\begin{align*}
&1 \mapsto \frac{1}{s},s > 0 \qquad \qquad &\sin at \mapsto \frac{a}{{{s^2} + {a^2}}},s > 0 \\
&{e^{at}} \mapsto \frac{1}{{s – a}} \qquad \qquad  &\cos at \mapsto \frac{s}{{{s^2} + {a^2}}},s > 0 \\
&{e^{at}}f\left( t \right) \mapsto F\left( {s – a} \right),s > a   &{t^n} \mapsto \frac{{n!}}{{{s^{n + 1}}}},s > 0
\end{align*}

$f\left( t \right)$ of “exponential type”

In order to prevent $f\left( t \right)$ goes up too fast that $\int_0^\infty {f\left( t \right){e^{ – st}}} dt \approx \infty$, the limitation on  $f\left( t \right)$  is

$\left| {f\left( t \right)} \right| \le c{e^{kt}},c > 0,k > 0$

${t^n} \le M{e^t}$ for some $M$ and all $t$, as ${t^n}/{e^t}$ starts at zero, ends at zero and is continuous between zero and infinity, it will definitely has the maximum value.

Note #1: $\frac{1}{t}$ is not an exponential type, because $\int_0^\infty {\frac{1}{t}} dt$ does not converge.

Note #2: ${e^{{t^2}}}$ is not an exponential type, because no matter how big $k$ is, ${e^{{t^2}}} > {e^{kt}}$.

Initial value problem IVP

${y^{\prime\prime}} + A{y^{\prime}} + by = h\left( t \right),y\left( 0 \right) = {y_0},{y^{\prime}}\left( 0 \right) = y_0^{\prime}$

Solution:

Laplace transform on both sides: $q\left( s \right)\bar Y = p\left( s \right) \to \bar Y = \frac{{p\left( s \right)}}{{q\left( s \right)}}$

Inverse Laplace transform: $y\left( t \right)$

\begin{align*}
L\left( {{f^{\prime}}\left( t \right)} \right) &= \int_0^\infty {{f^{\prime}}\left( t \right){e^{ – st}}} dt = \left. {{e^{ – st}}f\left( t \right)} \right|_0^\infty + \int_0^\infty {s{e^{ – st}}f\left( t \right)} dt\\
& = 0 – f\left( 0 \right) + s\int_0^\infty {{e^{ – st}}f\left( t \right)} dt = sF\left( s \right) – f\left( 0 \right)
\end{align*}

since ${f\left( t \right)}$ is an exponential type,

$\left| {f\left( t \right)} \right| \le c{e^{kt}} \to \mathop {\lim }\limits_{t \to \infty } \frac{{f\left( t \right)}}{{{e^{st}}}} = 0,$

$L\left( {{f^{\prime\prime}}\left( t \right)} \right) = sL\left( {{f^{\prime}}\left( t \right)} \right) – {f^{\prime}}\left( 0 \right) = s\left[ {sF\left( s \right) – f\left( 0 \right)} \right] – {f^{\prime}}\left( 0 \right)$

${f^{\prime}}\left( t \right) \mapsto sF\left( s \right) – f\left( 0 \right),{f^{\prime\prime}}\left( t \right) \mapsto {s^2}F\left( s \right) – sf\left( 0 \right) – {f^{\prime}}\left( 0 \right)$

Example

${y^{\prime\prime}} – y = {e^{ – t}},y\left( 0 \right) = 1,{y^{\prime}}\left( 0 \right) = 0$

${s^2}\bar Y – s \cdot 1 – 0 – \bar Y = \frac{1}{s}$

$\bar Y = \frac{{{s^2} + s + 1}}{{{{\left( {s + 1} \right)}^2}\left( {s – 1} \right)}} = \frac{{ – 1/2}}{{{{\left( {s + 1} \right)}^2}}} + \frac{{1/4}}{{s + 1}} + \frac{{3/4}}{{s – 1}}$

$y\left( t \right) = \underbrace { – \frac{1}{2}t{e^{ – t}}}_{{y_p}} + \underbrace {\frac{1}{4}{e^{ – t}} + \frac{3}{4}{e^t}}_{{y_c}}$

Convolution

$f\left( t \right) * g\left( t \right) = \int_0^t {f\left( u \right)} g\left( {u – t} \right)du$

Proof:

$F\left( s \right) = \int_0^\infty {{e^{ – st}}f\left( t \right)} dt,G\left( s \right) = \int_0^\infty {{e^{ – st}}g\left( t \right)} dt$

$F\left( s \right)G\left( s \right) = \int_0^\infty {{e^{ – st}}\left( {f\left( t \right) * g\left( t \right)} \right)} dt \to f\left( t \right) * g\left( t \right) \mapsto F\left( s \right)G\left( s \right)$

\begin{align*}
F\left( s \right)G\left( s \right) &= \int_0^\infty {{e^{ – su}}f\left( u \right)} du \cdot \int_0^\infty {{e^{ – sv}}g\left( v \right)} dv = \int_0^\infty {\int_0^\infty {{e^{ – s\left( {u + v} \right)}}f\left( u \right)g\left( v \right)} } dudv\\
&= \int_0^\infty {\int_0^t {{e^{ – st}}f\left( u \right)g\left( {t – u} \right)} } dudt = \int_0^\infty {{e^{ – st}}\left( {\int_0^t {f\left( u \right)g\left( {t – u} \right)du} } \right)} dt\\
&= \int_0^\infty {{e^{ – st}}\left( {f\left( t \right) * g\left( t \right)} \right)} dt
\end{align*}

It should be noted about the limits of the

$dudt = \frac{{\partial \left( {u,v} \right)}}{{\partial \left( {u,t} \right)}}dudt,J = \frac{{\partial \left( {u,v} \right)}}{{\partial \left( {u,t} \right)}} = \left| {\begin{array}{*{20}{c}} {\frac{{\partial u}}{{\partial u}}}&{\frac{{\partial u}}{{\partial t}}}\\ {\frac{{\partial v}}{{\partial u}}}&{\frac{{\partial v}}{{\partial t}}} \end{array}} \right| = 1$

Example:

\begin{align*}
{t^2} * t &= \int_0^t {{u^2}\left( {t – u} \right)du} = \frac{{{t^4}}}{{12}}\\
&= {L^{ – 1}}\left( {\frac{{2!}}{{{s^3}}} \cdot \frac{1}{{{s^2}}}} \right) = {L^{ – 1}}\left( {\frac{2}{{{s^5}}}} \right) = 2 \cdot \frac{{{t^4}}}{{4!}} = \frac{{{t^4}}}{{12}}
\end{align*}

$f\left( t \right) * 1 = \int_0^t {f\left( t \right) \cdot 1du} = \int_0^t {f\left( t \right)du}$

Application:

the dumping rate of radioactive waste is $f\left( t \right)$ in years, and the left amount due to decaying is ${A_0}{e^{ – kt}}$ assuming $k$ is fixed, $A_0$ is the initial amount of waste and $t$ is the amount of time left. By the time $t$, how much waste is left?

In the time period $\left[ {{u_i},{u_{i + 1}}} \right]$, the amount of left is

$f\left( {{u_i}} \right) \cdot \Delta u \cdot {e^{ – k\left( {t – {u_i}} \right)}} = \int_0^t {f\left( u \right){e^{ – k\left( {t – u} \right)}}du} = f\left( t \right) * {e^{ – kt}}$

Dealing with discontinuity with Laplace transform

Jump discontinuity

$u\left( t \right)$ is the unit step and $u\left( 0 \right)$ is not defined shown as the black line. Translate $u\left( t \right)$ to the right shown as the blue line, we get

${u_a}\left( t \right) = u\left( {t – a} \right)$

Thus, the “unit box” shown as the red line is then

${u_{ab}}\left( t \right) = {u_a}\left( t \right) – {u_b}\left( t \right) = u\left( {t – a} \right) – u\left( {t – b} \right)$

To make the inverse Laplace transform unique, we need to multiply a unit step with the $f\left( t \right)$, as we do not care about the formula of  $f\left( t \right)$ when  $t<0$.

$L\left( {f\left( t \right)} \right) = F\left( s \right),{L^{ – 1}}\left( {F\left( s \right)} \right) = u\left( t \right)f\left( y \right)$

We want a formula $L\left( {f\left( {t – a} \right)} \right)$ in terms of $L\left( {f\left( {t} \right)} \right)$, however this formula does not exist because we do not need the line $t<0$ for $L\left( {f\left( {t} \right)} \right)$, but we need the part $-a<t<0$ to get $L\left( {f\left( {t – a} \right)} \right)$, as figured.

\begin{align*}
u\left( {t – a} \right)f\left( {t – a} \right) &\mapsto {e^{ – as}}F\left( s \right) \qquad  &{\rm{the \ right \ formula}}\\
u\left( {t – a} \right)f\left( t \right) &\mapsto {e^{ – as}}L\left( {f\left( {t + a} \right)} \right) \qquad  &{\rm{t – axis \ translation \ formula}}
\end{align*}

Proof:

\begin{align*}
\int_0^\infty {{e^{ – st}}u\left( {t – a} \right)f\left( {t – a} \right)dt} &= \int_{ – a}^\infty {{e^{ – s\left( {{t_1} + a} \right)}}u\left( {{t_1}} \right)f\left( {{t_1}} \right)d{t_1}} = {e^{ – sa}}\int_{ – a}^\infty {{e^{ – s{t_1}}}u\left( {{t_1}} \right)f\left( {{t_1}} \right)d{t_1}} \\
&= {e^{ – sa}}\int_0^\infty {{e^{ – s{t_1}}}u\left( {{t_1}} \right)f\left( {{t_1}} \right)d{t_1}} = {e^{ – sa}}F\left( s \right)
\end{align*}

We get $u\left( {t – a} \right)f\left( {t – a} \right) \mapsto {e^{ – as}}F\left( s \right)$ by replace $t \to t + a$ to get the right side, so

$u\left( {t – a} \right)f\left( t \right) = u\left( {t – a} \right)f\left( {t – a + a} \right) \mapsto {e^{ – as}}L\left( {f\left( {t + a} \right)} \right)$

Example:

\begin{align*}
u\left( {t – 1} \right){t^2} \mapsto {e^{ – s}}L\left( {{{\left( {t + 1} \right)}^2}} \right) &= {e^{ – s}}L\left( {{t^2} + 2t + 1} \right)\\
&= {e^{ – s}}\left( {\frac{2}{{{s^3}}} + \frac{2}{{{s^2}}} + \frac{1}{s}} \right)
\end{align*}

There are three terms in the result, as we can see from the curve, there are two parts and one discontinuous point $t=1$, as indicate by the result ${e^{ – s}}$.

${L^{ – 1}}\left( {\frac{{1 + {e^{ – \pi s}}}}{{{s^2} + 1}}} \right) = {L^{ – 1}}\left( {\frac{1}{{{s^2} + 1}} + \frac{{{e^{ – \pi s}}}}{{{s^2} + 1}}} \right) = u\left( t \right)\sin \left( t \right) + u\left( {t – \pi } \right)\sin \left( {t – \pi } \right) = f\left( t \right)$

f\left( t \right) = \left\{ \begin{align*} \sin t \qquad \qquad \qquad \qquad &0 \le t \le \pi \\ \sin t + \sin \left( {t – \pi } \right) = 0 \qquad &t \ge \pi \end{align*} \right.

It should be noted that ${L^{ – 1}}\left( {\frac{1}{{{s^2} + 1}}} \right) = u\left( t \right)\sin \left( t \right)$, in which the $u\left( t \right)$ can not be omitted because there is an exponential type in this example, which means the function has been translated along the x-axis.

Dirac delta function

impulse of  $f\left( t \right)$ over the time period $\left[ {a,b} \right]$ is defined as

$\int_a^b {f\left( t \right)dt}$

$\frac{1}{h}{u_{0h}}\left( t \right) = \frac{1}{h}\left[ {u\left( t \right) – u\left( {t – h} \right)} \right] \mapsto \frac{1}{h}\left[ {\frac{1}{s} – \frac{{{e^{ – hs}}}}{s}} \right]$

$\mathop {\lim }\limits_{h \to 0} \frac{{1 – {e^{ – hs}}}}{{hs}} = \mathop {\lim }\limits_{u \to 0} \frac{{1 – {e^{ – u}}}}{u} = \frac{{{e^{ – u}}}}{1} = 1$

Dirac delta function:

$\delta \left( t \right) \mapsto 1 \qquad \int_0^\infty {\delta \left( t \right)dt} = 1$

$u\left( t \right)f\left( t \right) * \delta \left( t \right) \mapsto F\left( s \right) \cdot 1,{u^{\prime}}\left( t \right) = \delta \left( t \right)$

Example: kick a ball with impulse $A$ at $t=\pi /2$

${y^{\prime\prime}} + y = A\delta \left( {t – \frac{\pi }{2}} \right),y\left( 0 \right) = 1,{y^{\prime}}\left( 0 \right) = 0$

${s^2}\bar Y – s + \bar Y = A{e^{ – \pi /2}} \cdot 1 \to \bar Y = \frac{s}{{{s^2} + 1}} + \frac{{A{e^{ – \pi /2}}}}{{{s^2} + 1}}$

$y = u\left( t \right)\cos t + u\left( {t – \pi /2} \right) \cdot A\sin \left( {t – \pi /2} \right)$

y = \left\{ \begin{align*} \cos t \qquad \qquad \qquad \qquad & 0 \le t \le \pi /2\\ \cos t + A\sin \left( {t – \pi /2} \right) = \left( {1 – A} \right)\cos t \qquad & t \ge \pi /2 \end{align*} \right.

Note that the solution to this ODE is continuous, the curve is described as follow whit different values of $A$, which means after the time $t=\pi /2$ with the impulse $A$, the motion of the ball would stop or keep going or go backward depending on the magnitude of $A$.

Solution to ODE

${y^{\prime\prime}} + a{y^{\prime}} + by = f\left( t \right),y\left( 0 \right) = 0,{y^{\prime}}\left( 0 \right) = 0$

${s^2}\bar Y + as\bar Y + b\bar Y = F\left( s \right) \to \bar Y = \frac{{F\left( s \right)}}{{{s^2} + as + b}}$

$y\left( t \right) = f\left( t \right) * w\left( t \right) = \int_0^u {f\left( u \right)w\left( {t – u} \right)du}$

${\rm{weight \ function \ of \ system}} \quad w\left( t \right) \mapsto W\left( s \right) = \frac{1}{{{s^2} + as + b}} \quad {\rm{transfer \ function}}$

What is $w\left( t \right)$, really? Assume a ball kicked by impulse at $t=0$ with the ODE

${y^{\prime\prime}} + a{y^{\prime}} + by = \delta \left( t \right),y\left( 0 \right) = 0,{y^{\prime}}\left( 0 \right) = 0$

$\bar Y = \frac{1}{{{s^2} + as + b}} \to y\left( t \right) = w\left( t \right)$