MATH 308 Lecture 20

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Lecture Notes

Section 6.2

Exercise 11

$y''+3t\,y'-6y=1\quad y(0)=0\quad y'(0)=0$

No way to solve this... until now.

${\begin{aligned}{\mathcal {L}}\left\{y''\right\}+3{\mathcal {L}}\left\{t\,y'\right\}-6{\mathcal {L}}\left\{y\right\}&={\mathcal {L}}\left\{1\right\}\\s^{2}\,{\mathcal {L}}\left\{y\right\}-y'(0)-s\,y(0)+3\left(-{\frac {\mathrm {d} }{\mathrm {d} s}}\left({\mathcal {L}}\left\{y'\right\}\right)\right)-6{\mathcal {L}}\left\{y\right\}&={\frac {1}{s}}\\\left(s^{2}-9\right)\,{\mathcal {L}}\left\{y\right\}-3s\,{\frac {\mathrm {d} }{\mathrm {d} s}}\,{\mathcal {L}}\left\{y\right\}&={\frac {1}{s}}\end{aligned}}$

Notice that we have a differential equation: let $Y(s)={\mathcal {L}}\left\{y\right\}(s)$

${\begin{aligned}-3s\,Y'(s)+\left(s^{2}-9\right)\,Y(s)&={\frac {1}{s}}\\Y'(s)-{\frac {s^{2}-9}{3s}}\,Y(s)&=-{\frac {1}{3s^{2}}}\\\mu (s)&=\mathrm {e} ^{-\int {\frac {s^{2}-9}{3s}}\,\mathrm {d} s}\\&=s^{3}\,\mathrm {e} ^{-{\frac {s^{2}}{6}}}\\{\frac {\mathrm {d} }{\mathrm {d} s}}\left(s^{3}\,\mathrm {e} ^{-{\frac {s^{2}}{6}}}\,Y(s)\right)&=-{\frac {1}{3}}\,s\,\mathrm {e} ^{-{\frac {s^{2}}{6}}}\\s^{3}\,\mathrm {e} ^{-{\frac {s^{2}}{6}}}\,Y(s)&=\mathrm {e} ^{-{\frac {s^{2}}{6}}}+C\\Y(s)&={\frac {\mathrm {e} ^{-{\frac {s^{2}}{6}}}}{s^{3}\,\mathrm {e} ^{-{\frac {s^{2}}{6}}}}}+{\frac {C\,\mathrm {e} ^{\frac {s^{2}}{6}}}{s^{3}}}\end{aligned}}$

$C$ must be 0 since ${\mathcal {L}}\left\{y\right\}$ must converge to 0 when $s$ approaches infinity.

${\begin{aligned}{\mathcal {L}}\left\{y\right\}=Y(s)&={\frac {1}{s^{3}}}\\&={\frac {1}{2}}\,{\mathcal {L}}\left\{t^{2}\right\}\\y&={\frac {t^{2}}{2}}\end{aligned}}$

This solution satisfies the initial conditions and the differential equation.

Section 6.3

Lecture Notes

Heaviside Function

The unit step function or Heaviside function is the function $u_{c}$ defined by

${\begin{aligned}u_{c}(t)&={\begin{cases}0&t<c\\1&t\geq c\end{cases}}&c\geq 0\end{aligned}}$

Application: string suddenly breaks in a physics problem.

Exercise 1

Sketch graph of $g(t)=u_{2}(t)+3u_{4}(t)-7u_{5}(t)$

in Maple:

> restart: with(plots): with(plottools):
> u := (s,t) -> piecewise(t<s, 0, t>=s, 1):
> plot(u(2,t) + 3*u(4,t) - 7*u(5,t), t=0..10, discont=true);

Sketch graph of $h(t)=(t-1)\,u_{2}(t)+2(t-3)\,u_{3}(t)$

Exercise 2

Express the following functions in terms of step functions:

$f(t)={\begin{cases}t&0\leq t<3\\1-t&3\leq t<5\\t^{2}&5\leq t<6\\3&t\geq 6\end{cases}}$

Something happens at $t=\{3,5,6\}$, so we can use $u_{c}(t)$ to "turn on" different parts of the function at each "action" point.

$f(t)=A+B\,u_{3}(t)+C\,u_{5}(t)+D\,u_{6}(t)$

• Between 0 and 3, $f(t)=t$, so $A=t$
• Between 3 and 5, $f(t)=1-t$, so $B=1-2t$
• Between 5 and 6, $f(t)=t^{2}$, so $C=t^{2}+t-1$
• After 6, $f(t)=3$, so $D=3-t^{2}$

$f(t)=t+(1-2t)\,u_{3}(t)+(t^{2}+t-1)\,u_{5}(t)+(3-t^{2})\,u_{6}(t)$

Theorem 6.3.1

If $F(s)={\mathcal {L}}\left\{f(t)\right\}$ exists for $s>a\geq 0$, and if $c$ is a positive constant, then

${\mathcal {L}}\left\{u_{c}(t)\,f(t-c)\right\}=\mathrm {e} ^{-c\,s}\,{\mathcal {L}}\left\{f(t)\right\}=\mathrm {e} ^{-c\,s}\,F(s)\quad s>a$

Conversely, if $f(t)={\mathcal {L}}^{-1}\left\{F(s)\right\}$, then

$u_{c}(t)\,f(t-c)={\mathcal {L}}^{-1}\left\{\mathrm {e} ^{-c\,s}\,F(s)\right\}$

Exercise 3

Find the Laplace transform of the function

${\begin{aligned}f(t)&={\begin{cases}0&t<\pi \\t-\pi &pi\leq t\leq 2\pi \\0&t\geq 2\pi \end{cases}}\\&=(t-\pi )\,u_{\pi }(t)+(\pi -t)\,u_{2\pi }(t)\\{\mathcal {L}}\left\{f(t)\right\}&={\mathcal {L}}\left\{(t-\pi )\,u_{\pi }(t)\right\}+{\mathcal {L}}\left\{(\pi -t)\,u_{2\pi }\right\}\\{\mathcal {L}}\left\{(t-\pi )\,u_{\pi }(t)\right\}&=\mathrm {e} ^{-\pi \,s}\,{\mathcal {L}}\left\{g(t)\right\}\quad \quad g(t)=t\quad g(t-\pi )=t-\pi \\{\mathcal {L}}\left\{(\pi -t)\,u_{2\pi }\right\}&=\mathrm {e} ^{-2\pi \,s}\,{\mathcal {L}}\left\{h(t)\right\}\\h(t-2\pi )&=\pi -t\\&=\pi -(t-2\pi +2\pi )\\&=-(t-2\pi )-\pi \\h(T)&=-T-\pi \\{\mathcal {L}}\left\{f(t)\right\}&=\mathrm {e} ^{-\pi \,s}\,{\mathcal {L}}\left\{t\right\}+\mathrm {e} ^{-2\pi \,s}\,{\mathcal {L}}\left\{-t-\pi \right\}\end{aligned}}$