Kansas State University

Discussion

Theorem If $\lim_{t\to\infty}e^{-st}f(t)=0$, then ${\mathcal L}\{f'(t)\}=s{\mathcal L}\{f(t)\}-f(0)$

Proof $$ \begin{align} {\mathcal L}\{f'(t)\}&=\int_0^{\infty}f'(t)e^{-st}\,dt \\ &=f(t)e^{-st}\biggl|_0^{\infty}+s\int_0^{\infty}f(t)e^{-st}\,dt \\ &=-f(0)+s{\mathcal L}\{f(t)\}, \end{align} $$ where we have integrated by parts.

Note that for $s$ sufficiently large, $e^{-st}\to0$ very quickly as $t\to\infty$, so the hypothesis that $e^{-st}f(t)\to0$ as $t\to\infty$ will usually hold. In this class, we won't worry about assumptions on the growth of $f(t)$, but do remember they may be an issue in later classes.

Corollary $$ \begin{align} {\mathcal L}\{f''(t)\}&=s^2{\mathcal L}\{f(t)\}-sf(0)-f'(0) \\ {\mathcal L}\{f^{(n)}(t)\}&=s^n{\mathcal L}\{f(t)\}-s^{n-1}f(0) -s^{n-2}f'(0)-\cdots-f^{(n-1)}(0) \end{align} $$ Proof To prove the result for $f''(t)$ we apply the theorem twice. $$ \begin{align} {\mathcal L}\{f''(t)\}&={\mathcal L}\{(f')'(t)\} \\ &=s{\mathcal L}\{f'(t)\}-f'(0) \\ &=s\bigl(s{\mathcal L}\{f(t)\}-f(0)\bigl)-f'(0) \\ &=s^2{\mathcal L}\{f(t)\}-sf(0)-f'(0) \end{align} $$ The result for $f^{(n)}(t)$ is proven by applying the theorem $n$ times, or more properly by "induction" (if you have questions about proving things by induction, stop by my office).

So the Laplace transform converts differentiation into multiplication. Thus it converts differential equations into algebraic equations. But solving algebraic equations is easy. Consider the initial value problem $$ \begin{align} x' &= x \\ x(0) &= 1 \end{align} $$ We take the Laplace transform of both sides of the equation to obtain $$ s{\mathcal L}\{x(t)\} - x(0) = {\mathcal L}\{x(t)\} $$ We now use $x(0)=1$ and solve to find $$ {\mathcal L}\{x(t)\} = 1/(s-1) $$ Finally, we look in the table to find that the function whose Laplace transform is $1/(s-1)$ is $$x(t)=e^t$$ which is the solution.

I mentioned earlier that the Laplace transform is only one of many transform techniques. The basic idea of all transform techniques is exactly what is happening here. You have some operator, differentiation in our case, that is difficult to work with. You find a transform that converts that operator into something easier to work with, multiplication in this case. You then solve the simplified problem and undo the transform. The last step is usually the tricky one.

Paradigm:

STEP 1: Take the Laplace transform of both sides. $$ \begin{align} {\mathcal L}\{y''+2y'+5y\}&={\mathcal L}\{y''\}+2{\mathcal L}\{y'\}+5{\mathcal L}\{y\} \\ &=s^2{\mathcal L}\{y\}-sy(0)-y'(0)+2\bigl(s{\mathcal L}\{y\}-y(0)\bigl)+5{\mathcal L}\{y\} \\ &=s^2{\mathcal L}\{y\}-2s+2(s{\mathcal L}\{y\}-2)+5{\mathcal L}\{y\} \\ &=(s^2+2s+5){\mathcal L}\{y\}-2s-4 \\ {\mathcal L}\{0\}&=0 \end{align} $$ So our equation becomes $$ (s^2+2s+5){\mathcal L}\{y\}-2s-4=0 $$ STEP 2: Solve for ${\mathcal L}\{y\}$. $$ {\mathcal L}\{y\}=\frac{2s+4}{s^2+2s+5} $$ STEP 3: Take the inverse Laplace transform to find the solution. $$ y={\mathcal L}^{-1}\left\{\frac{2s+4}{s^2+2s+5}\right\} $$ Now the denominator is a quadratic which doesn't factor into linear terms (i.e. it has no real roots). So we complete the square to get $$ \frac{2s+4}{s^2+2s+5}=\frac{2s+4}{(s+1)^2+2^2} $$ Now in our table we find the following lines with a denominator similar to this $$ \begin{align} e^{at}\sin(bt)\qquad&\qquad\frac{b}{(s-a)^2+b^2} \\ e^{at}\cos(bt)\qquad&\qquad\frac{s-a}{(s-a)^2+b^2} \end{align} $$ If we let $a=-1$ and $b=2$ then we will get the denominator we are looking for. We now pick a multiple of $\displaystyle\frac{s+1}{(s+1)^2+2^2}$ with the right $s$ term and then use a multiple of $\displaystyle\frac{2}{(s+1)^2+2^2}$ to get the right constant term. $$ \begin{align} \frac{2s+4}{(s+1)^2+2^2}&=\frac{2(s+1)+2}{(s+1)^2+2^2} \\ &=2\frac{s+1}{(s+1)^2+2^2}+\frac{2}{(s+1)^2+2^2} \end{align} $$ and so taking the inverse Laplace transform gives $$ \begin{align} y&={\mathcal L}^{-1}\left\{2\frac{s+1}{(s+1)^2+2^2}+\frac{2}{(s+1)^2+2^2}\right\} \\ &=2{\mathcal L}^{-1}\left\{\frac{s+1}{(s+1)^2+2^2}\right\} +{\mathcal L}^{-1}\left\{\frac{2}{(s+1)^2+2^2}\right\} \\ &=2e^{-t}\cos(2t)+e^{-t}\sin(2t) \end{align} $$ Usually when we solve initial value problems we find the general solution and then plug in the initial values to solve for the constants. When using Laplace transforms the initial values get plugged in right at step 1 and we get the unique solution to the initial value problem without having to find the general solution. Also note that while we have written $y$ as a function of $t$ in our solution, the name of the independent variable doesn't matter. If the problem had originally been posed with $y$ as a function of $x$ then the solution would be $y=2e^{-x}\cos(2x)+e^{-x}\sin(2x)$.

EXAMPLE: Find the general solution of $y''+4y'+4y=e^{-t}$.

Step 1: $$ \begin{align} {\mathcal L}\{y''\}+4{\mathcal L}\{y'\}+4{\mathcal L}\{y\}&={\mathcal L}\{e^{-t}\} \\ s^2{\mathcal L}\{y\}-sy(0)-y'(0)+4\bigl(s{\mathcal L}\{y\}-y(0)\bigl)+4{\mathcal L}\{y\}&=\frac{1}{s+1} \\ (s^2+4s+4){\mathcal L}\{y\}-sy(0)-y'(0)-4y(0)&=\frac{1}{s+1} \end{align} $$ Step 2: $$ {\mathcal L}\{y\}=\frac{1}{(s+1)(s^2+4s+4)}+ \frac{sy(0)+y'(0)+4y(0)}{s^2+4s+4} $$ Step 3: $$ \begin{align} y(t)&={\mathcal L}^{-1}\left\{\frac{1}{(s+1)(s^2+4s+4)}+ \frac{sy(0)+y'(0)+4y(0)}{s^2+4s+4}\right\} \\ &={\mathcal L}^{-1}\left\{\frac{1}{(s+1)(s+2)^2}\right\} + {\mathcal L}^{-1}\left\{\frac{sy(0)+y'(0)+4y(0)}{(s+2)^2}\right\} \end{align} $$ We use partial fractions to evaluate the first inverse Laplace transform. $$ \begin{align} \frac{1}{(s+1)(s+2)^2}&\buildrel{\text{set}}\over{=} \frac{A}{s+1}+\frac{B}{s+2}+\frac{C}{(s+2)^2} \\ 1&=A(s+2)^2+B(s+1)(s+2)+C(s+1) \\ 1&=(A+B)s^2+(4A+3B+C)s+(4A+2B+C) \end{align} $$ $$ \begin{align} 0&=A+B \tag {$s^2$ term} \\ 0&=4A+3B+C \tag {$s$ term} \\ 1&=4A+2B+C \tag {constant term} \end{align} $$ This has solution $A=1$, $B=-1$, $C=-1$. So $$ \begin{align} {\mathcal L}^{-1}\left\{\frac{1}{(s+1)(s+2)^2}\right\} &={\mathcal L}^{-1}\left\{\frac{1}{s+1}\right\} - {\mathcal L}^{-1}\left\{\frac{1}{s+2}\right\} -{\mathcal L}^{-1}\left\{\frac{1}{(s+2)^2}\right\} \\ &=e^{-t}-e^{-2t}-te^{-2t} \end{align} $$ We also use partial fractions to evaluate the second inverse Laplace transform. $$ \begin{align} \frac{sy(0)+y'(0)+4y(0)}{(s+2)^2}&\buildrel{\text{set}}\over{=} \frac{A}{s+2}+\frac{B}{(s+2)^2} \\ sy(0)+y'(0)+4y(0)&=A(s+2)+B \\ y(0)s + \bigl(y'(0)+4y(0)\bigl)&=As+(2A+B) \end{align} $$ $$ \begin{align} y(0)&=A \tag {$s$ term} \\ y'(0)+4y(0)&=2A+B \tag {constant term} \end{align} $$ Solving these two equations in two unknowns we get $A=y(0)$ and $B=y'(0)+2y(0)$. So $$ \begin{align} {\mathcal L}^{-1}\left\{\frac{y(0)s+y'(0)+4y(0)}{(s+2)^2}\right\} &= {\mathcal L}^{-1}\left\{\frac{y(0)}{s+2}\right\} + {\mathcal L}^{-1}\left\{\frac{y'(0)+2y(0)}{(s+2)^2}\right\} \\ &=y(0)e^{-2t}+(y'(0)+2y(0))te^{-2t} \end{align} $$ Adding the two pieces together we get $$ y(t)=e^{-t}-e^{-2t}-te^{-2t}+y(0)e^{-2t}+(y'(0)+2y(0))te^{-2t} $$ This is our general solution. It has two arbitrary constants, $y(0)$ and $y'(0)$. The form of this solution is somewhat more messy than what we would get by using the techniques in the previous chapter, but the constants here have a definite geometric meaning. The answer is the same of course, however you solve the problem. The only thing that changes is the form the answer is written in.