2807 Exercises

Example 1. Suppose that $F(s)$ is the Laplace transform of $f(t)$, and let $a > 0$. Find a formula for the Laplace transform of $g(t) = f(at)$ in terms of $F(s)$, by using the integral definition and making a change of variable. Verify your formula by using formulas and rules to compute both $\mathcal{L}(f(t))$ and $\mathcal{L}(f(at))$ with $f(t) = t^n$.
Solution. $$G(s)=\int_0^\infty f(at)e^{-st}dt$$ make the substitution $u = at$. Then $du = adt$, $$\begin{aligned} G(s)&=\int_0^\infty f(u)e^{-su/a}\frac{1}{a}du\\ &=\frac{1}{a}\int_0^\infty f(u)e^{-\frac{s}{a}u}du\\ &=\frac{1}{a}F(\frac{s}{a}) \end{aligned}$$ For example, take $f(t) = t^n$, so $F(s)=\frac{n!}{s^{n+1}}$, and $g(t)=a^nt^n$, so $G(s)=\frac{a^nn!}{s^{n+1}}$. $$\begin{aligned} \frac{1}{a}F(\frac{s}{a})&=\frac{1}{a}\frac{n!}{(\frac{s}{a})^{n+1}}\\ &=\frac{a^nn!}{s^{n+1}}\\ &=G(s) \end{aligned}$$

Example 2. Use your calculus skills: Show that if $h(t) = f(t) ∗ g(t)$ then $H(s) = F(s)G(s)$. Do this by writing $F(s) = \int_0^\infty f(x)e^{-sx}dx$ and $G(s) = \int_0^\infty g(y)e^{-sy}dy$; expressing the product as a double integral; and changing coordinates using $x = t - \tau, y = \tau$.
Solution. $$F(s)G(s)=\int_0^\infty\int_0^\infty f(x)e^{-sx} g(y)e^{-sy} dxdy=\iint_R f(x)g(y)e^{-(x+y)s} dxdy$$ where $R$ is the first quadrant.
We can use the substitution is $x = t - \tau, y = \tau$
To convert to these coordinates, note that the Jacobian is $$\det\begin{bmatrix} \frac{\partial x}{\partial t}&\frac{\partial x}{\partial \tau}\\ \frac{\partial y}{\partial t}&\frac{\partial y}{\partial \tau} \end{bmatrix}=\det\begin{bmatrix} 1&-1\\ 0&1 \end{bmatrix}=1$$ For fixed $t, \tau$ ranges over numbers between $0$ and $t$, and $t$ ranges over positive numbers.
Since $x+y=t$, $$\begin{aligned} F(s)G(s)&=\int_0^\infty\int_0^t f(t-\tau)g(\tau)e^{-st}d\tau dt\\ &=\int_0^\infty(\int_0^t f(t-\tau)g(\tau)d\tau)e^{-st} dt\\ &=\int_0^\infty (f(t)*g(t))e^{-st} dt\\ &=\int_0^\infty h(t)e^{-st} dt\\ &=H(s) \end{aligned}$$