Section 6.2 The Laplace Transform and Initial Value Problems
¶Theorem 6.2.1. Laplace transform of dfdt.
Suppose f has a Laplace transform L{f}=F(s) and is controlled by some exponential Keat. Then the Laplace transform of f′ is given by
L{f′(t)}=sL{f(t)}−f(0).
Proof.
Let
\begin{equation*}
\mathcal{L}\left\{ f^{\prime}(t)\right\} =\lim_{B\to\infty}\int_{0}^{B}f^{\prime}(t)e^{-st}dt
\end{equation*}
and we use integration by parts
\begin{align*}
\end{align*}
we have
\begin{align*}
\mathcal{L}\left\{ f^{\prime}(t)\right\} \amp =\lim_{B\to\infty}\left[f(t)e^{-st}\right]_{t=0}^{t=B}+\int_{0}^{B}f(t)se^{-st}dt\\
\amp =\left[0-f(0)\right]+s\int_{0}^{\infty}f(t)e^{-st}dt\\
\amp =-f(0)+s\mathcal{L}\left\{ f(t)\right\} ,
\end{align*}
here we use the condition from Theorem 6.2.1 that says \(\left|f(t)\right|\leq Ke^{at}\) for \(t\geq M\) which implies that \(\lim_{B\to\infty}f(B)e^{-sB}=0\) when \(s>a\text{.}\) Rearranging gives us the desired result.
Corollary 6.2.2.
Suppose f,f′,…,f(n) are nice functions that have Laplace transforms, then
L{f(n)(t)}=snL{f(t)}−sn−1f(0)⋯−sf(n−2)(0)−f(n−1)(0).
Example 6.2.3.
Subsection 6.2.1 Inverse Laplace Transforms
The Inverse Laplace transform L−1 is the function that satisfies L−1{L[f]}=f. In other words,
L−1{F}=f⟺L{f}=F.
Some example inverse transforms.
L−1{1s}=1.
L−1{1s−1}=et
L−1{10s+1}=10L−1{1s−(−1)}=10e−t.
L−1{6s2+7}=6L−1{1s2+(√7)2}=6√7L−1{√7s2+(√7)2}=6√7sin(√7t).
L−1{1s−a}=eat.
We'll typically have rational functions for which we need to find inverse transforms. As an example, let's compute L−1{4(s−1)(s+1)}. Whenever we have linear or irreducible quadratic factors in the denominator, we need to use partial fractions:
4(s−1)(s+1)=A(s−1)+B(s+1),
hence
4=A(s+1)+B(s−1),0⋅s+4=(A+B)s+(A−B)
so that
A+B=0A−B=4
and get A=2,B−2. Thus
4(s−1)(s+1)=2s−1−2s+1.
Therefore:
L−1{4(s−1)(s+1)}=L−1{4(s−1)(s+1)}=L−1{2s−1}−L−1{2s+1}=2et−2e−t.
Example 6.2.4.
Find \(\mathcal{L}^{-1}\left\{ \frac{6}{s\left(s+4\right)}\right\} \text{.}\)
First
\begin{equation*}
\frac{6}{s\left(s+4\right)}=\frac{A}{s}+\frac{B}{\left(s+4\right)}
\end{equation*}
so that
\begin{equation*}
6=A\left(s+4\right)+Bs
\end{equation*}
or
\begin{equation*}
0s+6=\left(A+B\right)s+4A
\end{equation*}
and get
\begin{align*}
\end{align*}
so that \(A=-\frac{3}{2}\) and \(B=\frac{3}{2}\) hence
\begin{align*}
\mathcal{L}^{-1}\left\{ \frac{6}{s\left(s+4\right)}\right\} \amp =\mathcal{L}^{-1}\left\{ \frac{3/2}{s}+\frac{-3/2}{\left(s+4\right)}\right\} \\
\amp =\frac{3}{2}\mathcal{L}^{-1}\left\{ \frac{1}{s}\right\} -\frac{3}{2}\mathcal{L}\left\{ \frac{1}{\left(s+4\right)}\right\} \\
\amp =\frac{3}{2}\cdot1-\frac{3}{2}\mathcal{L}\left\{ \frac{1}{s-(-4)}\right\} \\
\amp =\frac{3}{2}\cdot1-\frac{3}{2}e^{-4t}
\end{align*}
Subsection 6.2.2 Higher order IVPs
We'll now combine the partial fraction technique with the Laplace transform to solve higher order examples.Example 6.2.5.
Solve
\begin{equation*}
y^{\prime}=y-4e^{-t},\,\,\,y(0)=1
\end{equation*}
using Laplace transforms.
Step 1: Find the Laplace Transform of the ODE (The going forwards part):
\begin{align*}
\mathcal{L}\left\{ y^{\prime}\right\} =\mathcal{L}\left\{ y\right\} -4\mathcal{L}\left\{ e^{-t}\right\} \amp \iff s\mathcal{L}\left\{ y\right\} -y(0)=\mathcal{L}\left\{ y\right\} -4\frac{1}{s+1}\\
\amp \iff s\mathcal{L}\left\{ y\right\} -1=\mathcal{L}\left\{ y\right\} -4\frac{1}{s+1}.
\end{align*}
Step 2: Solve for \(\mathcal{L}\left\{ y\right\} \) using algebra: and get
\begin{equation*}
\mathcal{L}\left\{ y\right\} =\frac{1}{s-1}-\frac{4}{\left(s-1\right)\left(s+1\right)}.
\end{equation*}
Step 3: We want to go backwards and invert this. But first let's do partial fractions:
\begin{equation*}
\frac{4}{\left(s-1\right)\left(s+1\right)}=\frac{A}{\left(s-1\right)}+\frac{B}{\left(s+1\right)},
\end{equation*}
hence
\begin{align*}
4 \amp =A\left(s+1\right)+B\left(s-1\right),\\
0\cdot s+4 \amp =\left(A+B\right)s+\left(A-B\right)
\end{align*}
so that
\begin{align*}
A+B \amp =0\\
A-B \amp =4
\end{align*}
and get \(A=2,B=-2\text{.}\) Thus
\begin{equation*}
\frac{4}{\left(s-1\right)\left(s+1\right)}=\frac{2}{s-1}-\frac{2}{s+1}.
\end{equation*}
Step 4: Use the inverse Laplace transform to get
\begin{align*}
y=\mathcal{L}^{-1}\left\{ \mathcal{L}\left\{ y\right\} \right\} \amp =\mathcal{L}^{-1}\left\{ \frac{1}{s-1}\right\} -\mathcal{L}^{-1}\left\{ \frac{4}{\left(s-1\right)\left(s+1\right)}\right\} \\
\amp =\mathcal{L}^{-1}\left\{ \frac{1}{s-1}\right\} -\left(\mathcal{L}^{-1}\left\{ \frac{2}{s-1}\right\} -\mathcal{L}^{-1}\left\{ \frac{2}{s+1}\right\} \right)\\
\amp =e^{t}-\mathcal{L}^{-1}\left\{ \frac{2}{s-1}\right\} +\mathcal{L}^{-1}\left\{ \frac{2}{s+1}\right\} \\
\amp =e^{t}-2e^{t}+2e^{-t}\\
\amp =-e^{t}+2e^{-t}.
\end{align*}
Example 6.2.6.
Solve
\begin{equation*}
y^{\prime}+4y=6,\,\,\,y^{\prime}(0)=0
\end{equation*}
using Laplace transforms.
Step 1: Find the Laplace Transform of the ODE (The going forwards part):
\begin{align*}
\mathcal{L}\left\{ y^{\prime}\right\} +4\mathcal{L}\left\{ y\right\} =\mathcal{L}\left\{ 6\right\} \amp \iff s\mathcal{L}\left\{ y\right\} -y(0)+4\mathcal{L}\left\{ y\right\} =\frac{6}{s}
\end{align*}
Step 2: Solve for \(\mathcal{L}\left\{ y\right\} \) using algebra: and get
\begin{equation*}
\mathcal{L}\left\{ y\right\} =\frac{6}{s\left(s+4\right)}.
\end{equation*}
Step 3: Partial Fractions (We did this already)
\begin{equation*}
\frac{6}{s\left(s+4\right)}=\frac{3/2}{s}+\frac{-3/2}{\left(s+4\right)}
\end{equation*}
Step 4: Use the inverse Laplace transform to get
\begin{align*}
y=\mathcal{L}^{-1}\left\{ \mathcal{L}\left\{ y\right\} \right\} \amp =\mathcal{L}^{-1}\left\{ \frac{3/2}{s}+\frac{-3/2}{\left(s+4\right)}\right\} \\
\amp =\frac{3}{2}\mathcal{L}^{-1}\left\{ \frac{1}{s}\right\} -\frac{3}{2}\mathcal{L}\left\{ \frac{1}{\left(s+4\right)}\right\} \\
\amp =\frac{3}{2}\cdot1-\frac{3}{2}\mathcal{L}\left\{ \frac{1}{s-(-4)}\right\} \\
\amp =\frac{3}{2}\cdot1-\frac{3}{2}e^{-4t}
\end{align*}