【微分方程導論】筆記9

Theorem 1

(the Laplace transform of the first derivative of $\textit{\textbf{f}}$ )
Suppose:
(1) $f$ is continuous on some interval and of exponential order $\alpha$ ; and
(2) $f'$ is piecewise continuous on some interval.
Then $\mathcal{L}\{f'(t)\}$ exists for $s > \alpha$ and $\displaystyle \mathcal{L}{f'(t)} = s\mathcal{L}{f(t)} - f(0).$

Proof

First, by definition, we write

$\displaystyle \mathcal{L}\{f'(t)\} = \int^\infty_0 e^{-st}f'(t)dt = \lim_{R\rightarrow\infty} \int^R_0e^{-st}f'(t)dt$

Next, we need to evaluate the definite integral $\int^R_0e^{-st}f'(t)dt$ . Because $f'$ is piecewise continuous on $[0,R]$ , we can partition $[0,R]$ by a finite number of points $0 = t_0 < t_1 < \cdots < t_n = R$ , in each of which $f'$ is continuous. Thus, the integral is

$\displaystyle \begin{aligned} &\quad\int^R_0 e^{-st}f'(t)dt \\ &=\int^{t_1}_{t_0}e^{-st}f'(t)dt + \int^{t_2}_{t_1}e^{-st}f'(t)dt + \cdots + \int^{t_n}_{t_{n-1}}e^{-st}f'(t)dt\\ &=\sum^n_{i=1} \int^{t_i}_{t_{i-1}} e^{-st}f'(t)dt\\ &=\sum^n_{i=1} \left\{ \Big[ e^{-st}f(t)\Big]^{t_i}_{t_{i-1}} + \int^{t_i}_{t_{i-1}} se^{-st}f(t)dt \right\} \\ &=\sum^n_{i=1}\Big[ e^{-st}f(t)\Big]^{t_i}_{t_{i-1}} + s\sum^n_{i=1}\int^{t_i}_{t_{i-1}}e^{-st}f(t)dt\\ &= \Big\{ \big[ e^{-st_1}f(t_1) - e^{-st_0}f(t_0)\big] + \big[ e^{-st_2}f(t_2) - e^{-st_1}f(t_1)\big]\\ &\quad\quad + \cdots + \big[ e^{-st_n}f(t_n) - e^{-st_{n-1}}f(t_{n-1}) \big] \Big\}\\ &\quad + s \Bigg[ \int^{t_1}_{t_0} e^{-st}f(t) dt + \int^{t_2}_{t_1} e^{-st}f(t) dt \\ &\quad\quad + \cdots + \int^{t_n}_{t_{n-1}} e^{-st}f(t)dt \Bigg] \\ &=\Big[e^{-st_n}f(t_n)-e^{-st_0}f(t_0)\Big] + s\int^{t_n}_{t_0} e^{-st}f(t) dt\\ &=\Big[e^{-sR}f(R)-e^{-s\cdot 0}f(0)\Big] + s\int^{R}_{0} e^{-st}f(t) dt.\\ \end{aligned}$

Therefore,

$\displaystyle \begin{aligned} \mathcal{L}\{f'(t)\} &= \lim_{R\rightarrow\infty}\int^R_0e^{-st}f'(t)dt \\ &= \lim_{R\rightarrow\infty}\left[e^{-sR}f(R)-f(0) + s\int^{R}_{0} e^{-st}f(t) dt\right]\\ &= \lim_{R\rightarrow\infty}e^{-sR}f(R) - f(0) + s\int^\infty_{0} e^{-st}f(t) dt.\\ \end{aligned}$

The integral in the third term is, by definition, the Laplace transform of $f$ . We still need to examine if the limit in the first term exists as $R\rightarrow\infty$ .

Because $f$ is of exponential $\alpha$ , there exists $t_0, M > 0$ , such that $e^{-\alpha t}|f(t)| \leq M$ for $t \geq t_0$ , which implies

$\displaystyle |f(t)| \leq Me^{\alpha t} ,\quad t \geq t_0$

Take $t = R \leq t_0$

$\displaystyle e^{-sR}|f(R)| \leq Me^{\alpha R}e^{-sR} = Me^{-(s-\alpha) R}$

When $s - \alpha > 0$ , we see that $\displaystyle \lim_{R\rightarrow\infty} e^{-(s-\alpha)R} = 0$ , and $\displaystyle \lim_{R\rightarrow\infty}e^{-sR}|f(R)| =0$ , and thus $\displaystyle \lim_{R\rightarrow\infty}e^{-sR}f(R) =0$ .

Finally, we confirm $\mathcal{L}{f'(t)}$ exists for $s > \alpha$ and

$\displaystyle \mathcal{L}{f'(t)} = - f(0) +s\mathcal{L}{f(t)}.$ ◪

Theorem 2

(the Laplace transform of the $\textit{\textbf{n}}$ th derivative of $\textit{\textbf{f}}$ )
Suppose: (1) $f, f', \ldots, f^{(n-1)}$ are continuous on some interval and of exponential order $\alpha$ ; and (2) $f^{(n)}$ is piecewise continuous on some interval.
Then $\mathcal{L}\{f^{(n)}(t)\}$ exists for $s > \alpha$ and
$\displaystyle \begin{aligned} \mathcal{L}{f^{(n)}(t)} &= s^n\mathcal{L}{f(t)} - s^{n-1}f(0) - \cdots- f^{(n-1)}(0) \\ &= s^n\mathcal{L}{f(t)} - \sum^{n-1}_{k=0}s^{n-k}f^{(k)}(0). \end{aligned}$

Proof

$\displaystyle \begin{aligned} \mathcal{L}\{f^{(n)}(t)\} &= \mathcal{L}\left\{\frac{d}{dt}\left(f^{(n-1)}(t)\right)\right\} \\ &= s\mathcal{L}\{f^{(n-1)}(t)\} - f^{(n-1)}(0)\\ &= s\left[s\mathcal{L}\{f^{(n-2)}(t)\} - f^{(n-2)}(0)\right]- f^{(n-1)}(0) \\ &= s^2\mathcal{L}\{f^{(n-2)}(t)\} - sf^{(n-2)}(0) - f^{(n-1)}(0) \\ &= \cdots \\ &= s^n\mathcal{L}\{f(t)\} - s^{n-1}f(0) - \cdots- f^{(n-1)}(0). \end{aligned}$

◪

Example 1

Solve the I.V.P. $y'' -y- 2y=0, y(0) = 1, y'(0) = 0.$

Solution
Take $\mathcal{L}$ on both sides of the equation

$\begin{aligned} &\quad\mathcal{L}\{y''\} - \mathcal{L}\{y'\} + 2 \mathcal{L}\{y\} = \mathcal{L}\{0\} \\ \Rightarrow & [s^2 \mathcal{L}\{y\}-sy(0)-y'(0)] - [s \mathcal{L}\{y\}-y(0)] - 2 \mathcal{L}\{y\} = 0 \end{aligned}$

Letting $\ \mathcal{L}\{y(t)\} = Y(s)$ and substituting the initial conditions $y(0) = 1, y'(0) = 0$ into the above transformed equation gives

$\displaystyle s^2Y - s -s Y +1 -2Y = 0,$

which is an algebraic equation with the solution

$\displaystyle Y(s)=\frac{s-1}{s^2-s-2}.$

This is a fraction and it is not easy to find its inverse Laplace transform directly. But the denominator $(s^2-s-2)$ can be factorized into $(s-2)(s+1)$ . So we use the method of partial fraction decomposition by letting

$\displaystyle \frac{s-1}{s^2-s-2} = \frac{A}{s-2}+\frac{B}{s+1},$

so that $A(s+1) + B(s-2) = (A+B)s+(A-2B) = s - 1$ , which leads to $A+B = 1, A-2B = -1 \Rightarrow A=\frac{1}{3}, B= \frac{2}{3}$ . Thus,

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

Note that this differential equation is linear and homogeneous and thus can be immediately solved by writing its characteristic equation, which is more convenient. But we’ll see the utility of the Laplace transform later. ❏

Example 2

Solve the I.V.P. $y'' + y = \sin t, y(0) = 2, y'(0) = 1.$

Solution
Take $\mathcal{L}$ on both sides of the equation and let $\ \mathcal{L}\{y(t)\} = Y(s)$ :

$\displaystyle \begin{aligned} &\quad (s^2Y-2s-1)+Y=\frac{2}{s^2+4}\\&\Rightarrow (s^2+1)Y = \frac{2}{s^2+4}+2s+1\\&\Rightarrow Y = \frac{1}{s^2+1}\frac{2s^3+s^2+8s+6}{s^4+4}.\end{aligned}$

We use the method of partial fraction decomposition by letting

$\displaystyle \frac{2s^3+s^2+8s+6}{(s^4+4)(s^2+1)} = \frac{As+B}{s^2+1}+\frac{Cs+D}{s^2+4},$

so that $(As+B)(s^2+4) + (Cs+D)(s^2+1) = 2s^3+s^2+8s+6$ , which leads to

$\displaystyle \left\{ \begin{matrix} A+C&=2\\ B+D&=1\\ C+4A&=8\\ D+4B&=6 \end{matrix} \right. \Rightarrow \left\{\begin{matrix}A=2\\ B=\frac{5}{3}\\ C=0\\ D=-\frac{2}{3}\end{matrix}\right.$

and then

$\displaystyle Y(s) = \frac{2s+(3/5)}{s^2+1} + \frac{-2/3}{s^2+4}.$

Such fractions are recognizable — they are the Laplace transform of sine and cosine functions. Taking the inverse Laplace transform, we have

$\displaystyle \begin{aligned} y(t) &= \mathcal{L}^{-1}\{Y(s)\}\\ &=2\mathcal{L}^{-1}\left\{\frac{s}{s^2+1}\right\}+\frac{3}{5}\mathcal{L}^{-1}\left\{\frac{1}{s^2+1}\right\}-\frac{1}{3}\mathcal{L}^{-1}\left\{\frac{2}{s^2+4}\right\}\\ &= 2\cos t + \frac{3}{5}\sin t - \frac{1}{3}\sin 2t. \end{aligned}$

Note that this differential equation is as well applicable to the theory of linear homogeneous equations; and the particular solution can be solved by the method of undetermined coefficients. ❏

Theorem 3

(translation of $\textit{\textbf{s}}$ )
Supposed the function $f(t)$ is such that $\mathcal{L}\{f(t)\} = F(s)$ exists for $s > \alpha$ , then
$\displaystyle\mathcal{L}\{e^{at}f(t)\} = F(s-a)$ .

Proof

$\displaystyle \begin{aligned} F(s) &= \int_0^\infty e^{-st} f(t) dt \\ F(s-a) &= \int_0^\infty e^{-(s-a)t} f(t) dt \\ &=\int_0^\infty e^{-st}e^{at} f(t) dt\\ &=\mathcal{L}\{e^{at}f(t)\}. \end{aligned}$

◪

Example 3

Find $\mathcal{L}\{te^{3t}\}$ .

Solution
We compare this form with the formula in Theorem 3 and see that $f(t) = t$ . Since we know

$\displaystyle F(s)=\mathcal{L}\{f(t)\}=\mathcal{L}\{t\} = \frac{1}{s^2},\quad s>0,$

so we have

$\displaystyle F(s-3)=\mathcal{L}\{e^{3t} f(t)\}=\mathcal{L}\{t\} = \frac{1}{(s-3)^2},\quad s>0.$

❏

Example 4

Find $\displaystyle \mathcal{L}^{-1}\left\{\frac{1}{s^2-4s+5}\right\}$ .

Solution
We complete the square for denominator,

$\displaystyle \frac{1}{(s-2)^2+1},$

and compare this form with the formula in Theorem 3 and see that $F(s) = 1/(s^2+1)$ . Since we know

$\displaystyle \mathcal{L}^{-1}\{F(s)\} = \mathcal{L}^{-1}\{1/(s^2+1)\} = \sin t$

so we have

$\displaystyle \begin{aligned} \mathcal{L}^{-1}\{F(s-2)\}&=\mathcal{L}^{-1}\left\{\frac{1}{(s-2)^2+1}\right\}\\&=e^{2t}f(t) = e^{2t}\sin t.\end{aligned}$

❏

Theorem 4

(differentiation of $\textbf{\textit{F}(\textit{s})}$ )
Supposed the function $f(t)$ is such that $\mathcal{L}\{f(t)\} = F(s)$ exists for $s > \alpha$ , then
$\displaystyle \mathcal{L}\{t^nf(t)\} = (-1)^n \frac{d^n}{ds^n} F(s)$ .

Proof
We repeatedly diffentiate $F(s)$ :

$\displaystyle \begin{aligned} F(s)&=\int_0^\infty e^{-st}f(t)dt,\\ \frac{d}{ds}F(s) &= \int_0^\infty (-t)e^{-st}f(t)dt = -\mathcal{L}\{tf(t)\},\\ \frac{d^2}{ds^2}F(s) &= \int_0^\infty (-t)^2e^{-st}f(t)dt = (-1)^2\mathcal{L}\{t^2f(t)\}, \ldots \end{aligned}$

By the mathematical induction, we can prove that

$\displaystyle \frac{d^n}{ds^n}F(s)=\int_0^\infty (-t)^n e^{-st}f(t)dt = (-1)^n\mathcal{L}\{t^nf(t)\}.$

❏

Example 5

Find $\mathcal{L}\{t^2 \sin 3t\}$ .

Solution
We compare this form with the formula in Theorem 3 and see that $f(t) = \sin 3t$ . Since we know

$\displaystyle F(s)=\mathcal{L}\{f(t)\}=\mathcal{L}\{\sin 3t\} = \frac{3}{s^2+3^2},\quad s>0,$

so we have

$\displaystyle \begin{aligned} \mathcal{L}\{t^2 f(t)\}&=\mathcal{L}\{t^2 \sin 3t\} \\ &= (-1)^2\frac{d^2}{ds^2}\frac{3}{s^2+3^2} \\ &=\frac{d}{ds}\left[-\frac{3\cdot 2s}{(s^2+9)^2}\right] \\ &= \frac{18(s^2-3)}{(s^2+9)^3},\quad s>0. \end{aligned}$

❏

A blue-sky wanderer

【微分方程導論】筆記9

Theorem 1