MATH 308 Lecture 20

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Lecture Notes

Section 6.2

Exercise 11

$y^{″}+3t{y}^{\prime }-6y=1y(0)=0y^{\prime }(0)=0$

No way to solve this... until now.

$\begin{array}{rl}{ℒ}\left\{y^{″}\right\}+3{ℒ}\left\{t{y}^{\prime }\right\}-6{ℒ}\left\{y\right\}& ={ℒ}\left\{1\right\}\\ s^2{ℒ}\left\{y\right\}-y^{\prime }(0)-sy(0)+3(-\frac{\mathrm{d}}{\mathrm{d}s}\left({ℒ}\left\{y^{\prime }\right\}\right))-6{ℒ}\left\{y\right\}& =\frac{1}{s}\\ (s^2-9){ℒ}\left\{y\right\}-3s\frac{\mathrm{d}}{\mathrm{d}s}{ℒ}\left\{y\right\}& =\frac{1}{s}\end{array}$

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

$\begin{array}{rl}-3s{Y}^{\prime }(s)+(s^2-9)Y(s)& =\frac{1}{s}\\ Y^{\prime }(s)-\frac{s^2-9}{3s}Y(s)& =-\frac{1}{3s^2}\\ \mu (s)& ={\mathrm{e}}^{-\int \frac{s^2-9}{3s}\mathrm{d}s}\\ & =s^3{\mathrm{e}}^{-\frac{s^2}{6}}\\ \frac{\mathrm{d}}{\mathrm{d}s}(s^3{\mathrm{e}}^{-\frac{s^2}{6}}Y(s))& =-\frac{1}{3}s{\mathrm{e}}^{-\frac{s^2}{6}}\\ s^3{\mathrm{e}}^{-\frac{s^2}{6}}Y(s)& ={\mathrm{e}}^{-\frac{s^2}{6}}+C\\ Y(s)& =\frac{{\mathrm{e}}^{-\frac{s^2}{6}}}{s^3{\mathrm{e}}^{-\frac{s^2}{6}}}+\frac{C{\mathrm{e}}^{\frac{s^2}{6}}}{s^3}\end{array}$

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

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

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

$\begin{array}{rlr}{u}_c(t)& =\{\begin{array}{ll}0& t<c\\ 1& t\ge c\end{array}& c\ge 0\end{array}$

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:

$f(t)=\{\begin{array}{ll}t& 0\le t<3\\ 1-t& 3\le t<5\\ t^2& 5\le t<6\\ 3& t\ge 6\end{array}$

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

${ℒ}\{u_c(t)f(t-c)\}={\mathrm{e}}^{-cs}{ℒ}\{f(t)\}={\mathrm{e}}^{-cs}F(s)s>a$

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

$u_c(t)f(t-c)={{ℒ}}^{-1}\{{\mathrm{e}}^{-cs}F(s)\}$

Exercise 3

Find the Laplace transform of the function

$\begin{array}{rl}f(t)& =\{\begin{array}{ll}0& t<\pi \\ t-\pi & pi\le t\le 2\pi \\ 0& t\ge 2\pi \end{array}\\ & =(t-\pi )u_{\pi }(t)+(\pi -t)u_{2\pi }(t)\\ {ℒ}\{f(t)\}& ={ℒ}\{(t-\pi )u_{\pi }(t)\}+{ℒ}\{(\pi -t)u_{2\pi }\}\\ {ℒ}\{(t-\pi )u_{\pi }(t)\}& ={\mathrm{e}}^{-\pi s}{ℒ}\{g(t)\}g(t)=tg(t-\pi )=t-\pi \\ {ℒ}\{(\pi -t)u_{2\pi }\}& ={\mathrm{e}}^{-2\pi s}{ℒ}\{h(t)\}\\ h(t-2\pi )& =\pi -t\\ & =\pi -(t-2\pi +2\pi )\\ & =-(t-2\pi )-\pi \\ h(T)& =-T-\pi \\ {ℒ}\{f(t)\}& ={\mathrm{e}}^{-\pi s}{ℒ}\left\{t\right\}+{\mathrm{e}}^{-2\pi s}{ℒ}\{-t-\pi \}\end{array}$