Inverse Trigonometric Functions

$y=\sin x$ is not a one-to-one function so there cannot be an inverse function of $y=\sin x$. However we we restrict its domain to the closed interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ it becomes a one-to-one function as shown in Figure 1.

Figure 1. The graph of y=sin(x) on [-pi/2,pi/2]

This means that we can consider $y=\sin^{-1}x$, the inverse function of $y=\sin x$. That is to say, $y=\sin^{-1}x$ is the value in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ for which $x=\sin y$. $y=\sin^{-1}x$ is also denoted by $y=\arcsin x$. Similarly,

  1. $y=\cos^{-1}x$ (or $y=\arccos x$) is the value in $[0,\pi]$ for which $x=\cos y$.
  2. $y=\tan^{-1}x$ (or $y=\arctan x$) is the value in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ for which $x=\tan y$.
  3. $y=\cot^{-1}x$ (or $y=arccot x$) is the value in $(0,\pi)$ for which $x=\cot y$.

Remark. In general, $y=\sin^{-1}x$ is an inverse relation of $y=\sin x$ which is multiple-valued. In order to consider differentiation, we require it to be single-valued and when $-\frac{\pi}{2}\leq\sin^{-1}x\leq\frac{\pi}{2}$ we call it the principal value of $\sin^{-1}x$ and denote it by $\mathrm{Sin}^{-1}x$. Throughout this note we will only consider principal values so we won’t be using the traditional notation like $y=\mathrm{Sin}^{-1}x$.

Recall that the graph of $y=f(x)$ and the graph of its inverse function $y=f^{-1}(x)$ are symmetric about the line $y=x$. Using this symmetry one can obtain the graph of an inverse trigonometric function. For example, Figure 2 shows the graph of $y=\sin x$ and the graph of $y=\sin^{-1} x$.

Figure 2. The graphs of y=sin(x) (in red), y=arcsin(x) (in blue) and y=x (in black).
Figure 3. The graph of y=arcsin(x)
Figure 4. The graph of y=arccos(x)
Figure 5. The graph of y=arctan(x) on [-20,20].
Figure 6. The graph of y=arccot(x) on [-20,20].

In addition, $y=\sec^{-1}x$ has domain $|x|\geq 1$ and range $\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right]$.

Figure 7. The graph of y=arcsec(x). The horizontal asymptote is y=pi/2.

$y=\csc^{-1}x$ has domain $|x|\geq 1$ and range $\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]$.

The graph of y=arccsc(x).

Inverse Functions and Their Derivatives

Let $y=f^{-1}(x)$ denote the inverse function of $f(x)$. Then \begin{equation}\label{eq:invfn}x=f(y)\end{equation} Differentiate \eqref{eq:invfn} with respect to $x$. \begin{equation}\label{eq:dinvfn}1=f'(y)\frac{dy}{dx}\end{equation} Solving \eqref{eq:dinvfn} for $\frac{dy}{dx}$ we have \begin{equation}\label{eq:dinvfn2}\frac{dy}{dx}=\frac{1}{f'(f^{-1}(x))}\end{equation}

Example. Let $f(x)=\ln x$. Knowing $f'(x)=\frac{1}{x}$ find the derivative of $f^{-1}(x)=e^x$.

Solution. Using \eqref{eq:dinvfn2} $$\frac{d}{dx}e^x=\frac{1}{\frac{1}{e^x}}=e^x$$

Although knowing the formula \eqref{eq:dinvfn2} is convenient, you can always find the derivative of an inverse function by following the same process of deriving \eqref{eq:dinvfn2}. It’s not actually anymore difficult or complicated than using \eqref{eq:dinvfn2}.

The Derivative of Inverse Trigonometric Functions

Let $f(x)=\sin x$. Then $f^{-1}(x)=\sin^{-1}x$. Using \eqref{eq:dinvfn2} \begin{align*}\frac{d}{dx}\sin^{-1}x&=\frac{1}{\cos(\sin^{-1}x)}\\&=\frac{1}{\sqrt{1-\sin^2(\sin^{-1}x)}}\\&=\frac{1}{\sqrt{1-x^2}}\end{align*} where $|x|<1$. The reason the sign in front of $\sqrt{}$ is positive is that $-\frac{\pi}{2}<\sin^{-1}x<\frac{\pi}{2}$ so $\cos(\sin^{-1}x)> 0$.

Alternative derivation: Let $y=\sin^{-1}x$. Then $x=\sin y$. By implicit differentiation $$1=\cos y\frac{dy}{dx}$$ Hence $$\frac{dy}{dx}=\frac{1}{\cos y}=\frac{1}{\cos(\sin^{-1}x)}=\frac{1}{\sqrt{1-x^2}}$$

If $u$ is a functions of $x$, then $$\frac{d}{dx}\sin^{-1}u=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx},\ |u|<1$$ Similarly we obtain the rest of derivative formulas. \begin{align*}\frac{d}{dx}\cos^{-1}u&=-\frac{1}{\sqrt{1-u^2}}\frac{du}{dx},\ |u|<1\\\frac{d}{dx}\tan^{-1}x&=\frac{1}{1+u^2}\frac{du}{dx}\\\frac{d}{dx}\cot^{-1}x&=-\frac{1}{1+u^2}\frac{du}{dx}\\\frac{d}{dx}\sec^{-1}u&=\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx},\ |u|>1\\\frac{d}{dx}\csc^{-1}x&=-\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx},\ |u|>1\end{align*} In turns out we don’t really have to calculate all these formulas. We just need to calculate for example $\frac{d}{dx}\sin^{-1}$, $\frac{d}{dx}\tan^{-1}x$, $\frac{d}{dx}\sec^{-1}$ the rest can be obtained by inverse function-inverse cofunction identities \begin{align*}\cos^{-1}x&=\frac{\pi}{2}-\sin^{-1}x\\\cot^{-1}x&=\frac{\pi}{2}-\tan^{-1}x\\\csc^{-1}x&=\frac{\pi}{2}-\sec^{-1}x\end{align*}In case you haven’t seen this identities before they can be easily obtained from cofunction identities. For example sine and cosine are cofunctions of each other as you learned in trigonometry, namely $$\sin\left(\frac{\pi}{2}-x\right)=\cos x,\ \cos\left(\frac{\pi}{2}-x\right)=\sin x$$ Let $$\cos\left(\frac{\pi}{2}-y\right)=\sin y=x$$ Then $$\frac{\pi}{2}-y=\cos^{-1}x$$ i.e. the first identity above $$\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x$$

Integration Formulas

For any constant $a\ne 0$,

  1. $\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right)+C$, valid for $u^2<a^2$
  2. $\int\frac{du}{a^2+u^2}=\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C$
  3. $\int\frac{du}{u\sqrt{u^2-a^2}}=\frac{1}{a}\sec^{-1}\left|\frac{u}{a}\right|+C$, valid for $|u|>a>0$

Example. \begin{align*}\int_{\sqrt{2}/2}^{\sqrt{3}/2}\frac{dx}{\sqrt{1-x^2}}&=\left.\sin^{-1}x\right|_{\sqrt{2}/2}^{\sqrt{3}/2}\\&=\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)-\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\\&=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}\end{align*}

Example. \begin{align*}\int\frac{dx}{\sqrt{3-4x^2}}&=\frac{1}{2}\int\frac{du}{\sqrt{3-u^2}}\ (u=2x)\\&=\frac{1}{2}\sin^{-1}\left(\frac{u}{\sqrt{3}}\right)+C\\&=\frac{1}{2}\sin^{-1}\left(\frac{2x}{\sqrt{3}}\right)+C\end{align*}

Example. \begin{align*}\int\frac{dx}{4x-x^2}&=\int\frac{dx}{4-(x-2)^2}\\&=\int\frac{du}{4-u^2}\ (u=x-2)\\&=\sin^{-1}\left(\frac{u}{2}\right)+C\\&=\sin^{-1}\left(\frac{x-2}{2}\right)+C\end{align*}

Example. \begin{align*}\frac{dx}{4x^2+4x+2}&=\int\frac{dx}{4\left(x+\frac{1}{2}\right)^2+1}\\&=\frac{1}{2}\int\frac{du}{u^2+1}\ \left[u=2\left(x+\frac{1}{2}\right)\right]\\&=\frac{1}{2}\tan^{-1}u+C\\&=\frac{1}{2}\tan^{-1}(2x+1)+C\end{align*}

Example. \begin{align*}\int\frac{dx}{e^{2x}-6}&=\int\frac{du}{u\sqrt{u^2-6}}\ (u=e^x)\\&=\frac{1}{\sqrt{6}}\sec^{-1}\left(\frac{e^x}{\sqrt{6}}\right)+C\end{align*} where $e^x>\sqrt{6}$ or equivalently $x>\ln\sqrt{6}\approx 0.8959$.

Integrating Inverses of Functions

Let $y=f^{-1}(x)$. Then $x=f(y)$ and $dx=f'(y)dy$. So with integration by parts, we have \begin{equation}\begin{aligned}\int f^{-1}(x)dx&=\int yf'(y)dy\\&=yf(y)-\int f(y)dy\\&=xf^{-1}(x)-\int f(y)dy\end{aligned}\label{eq:intinvfn}\end{equation} Using \eqref{eq:dinvfn2} we can also rewrite \eqref{eq:intinvfn} as \begin{equation}\label{eq:intinvfn2}\int f^{-1}(x)dx=xf^{-1}(x)-\int\frac{x}{f'(f^{-1}(x))}dx\end{equation}

Using \eqref{eq:intinvfn}\begin{align*}\int\cos^{-1}xdx&=x\cos^{-1}x-\sin y+C\\&=x\cos^{-1}x-\sin(\cos^{-1}x)+C\end{align*}Since $x=\cos y$, $$\sin(\cos^{-1}x)=\sin y=\sqrt{1-x^2}$$ (Recall that $0\leq\cos^{-1}x\leq\pi$ so $\sin^{-1}x\geq 0$.) Hence, \begin{equation}\label{eq:intinvcos}\int\cos^{-1}xdx=x\cos^{-1}x-\sqrt{1-x^2}+C\end{equation} Of course one can obtain \eqref{eq:intinvcos} using \eqref{eq:intinvfn2} though the required calculation is a bit longer. The integral \eqref{eq:intinvcos} can be also found without using \eqref{eq:intinvfn} or \eqref{eq:intinvfn2}. Using integration by parts\begin{align*}\int\cos^{-1}xdx&=x\cos^{-1}x+\int\frac{x}{\sqrt{1-x^2}}dx\\&=x\cos^{-1}x-\sqrt{1-x^2}+C\end{align*} The rest of the integrals of inverse trigonometric functions are given by\begin{align*}\int\sin^{-1}xdx&=x\sin^{-1}x+\sqrt{1-x^2}+C\\\int\tan^{-1}xdx&=x\tan^{-1}x-\frac{1}{2}\ln(1+x^2)+C\\\int\cot^{-1}xdx&=x\cot^{-1}x+\frac{1}{2}\ln(1+x^2)+C\\\int\sec^{-1}xdx&=x\sec^{-1}x-\frac{x}{|x|}\ln|x+\sqrt{x^2-1}|+C\\\int\csc^{-1}xdx&=x\csc^{-1}x+\frac{x}{|x|}\ln|x+\sqrt{x^2-1}|+C\end{align*}

The Laplace Transform: Differential Equations with Variable Coefficients

In this note we study how to solve differential equations with variable coefficients using the Laplace transform. For this we need Derivatives of Transforms. Differentiating
$$f(s)=\int_0^\infty e^{-st}F(t)dt$$
with respect to $s$ we obtain
\begin{align*}
f'(s)&=\int_0^\infty e^{-st}(-tF(t))dt\\
&=\mathcal{L}\{-tF(t)\}
\end{align*}
Continue differentiating to find
\begin{equation}
\label{eq:laplace14}
f^{(n)}(s)=\mathcal{L}\{(-t)^nF(t)\}
\end{equation}
for $n=1,2,\cdots$.

Example. Given that $\mathcal{L}\{\sin kt\}=\frac{k}{s^2+k^2}$, find $\mathcal{L}\{t\sin kt\}$.

Solution. Using \eqref{eq:laplace14}
\begin{align*}
\mathcal{L}\{t\sin kt\}&=-\frac{d}{ds}\frac{k}{s^2+k^2}\\
&=\frac{2ks}{(s^2+k^2)^2}.
\end{align*}

The equation \eqref{eq:laplace14} together with transform of derivative formula allows us to transform differential equations with variable coefficients. For example,
\begin{align*}
\mathcal{L}\{t^nX(t)\}&=(-1)^nx^{(n)}(s)\\
\mathcal{L}\{t^2\dot{X}(t)\}&=\frac{d^2}{ds^2}[sx(s)-X(0)]\\
&=\frac{d}{ds}[x(s)+sx'(s)]\\
&=sx^{\prime\prime}(s)+2x'(s)\\
\mathcal{L}\{t\ddot{X}(t)\}&=-\frac{d}{ds}[s^2x(s)-sX(0)-\dot{X}(0)]\\
&=-s^2x'(s)-2sx(s)+X(0)
\end{align*}

We are now ready to solve differential equations with variable coefficients.

Example. Find the solution of the problem
$$\ddot{X}(t)+t\dot{X}(t)-X(t)=0,\ X(0)=0,\ \dot{X}(0)=1$$

Solution. The transformed equation is
$$s^2x(s)-1-\frac{d}{ds}[sx(s)]-x(s)=0$$
which can be written as the first-order linear differential equation
$$\frac{d}{ds}x(s)+\left(\frac{2}{s}-s\right)x(s)=-\frac{1}{s}$$
The integrating factor is
$$\mu(s)=e^{\int\left(\frac{2}{s}-s\right)ds}=s^2e^{-\frac{1}{2}s^2}$$
and hence the solution $x(s)$ is
\begin{align*}
x(s)&=\frac{\int\mu(s)\left(-\frac{1}{2}\right)ds}{\mu(s)}\\
&=\frac{-\int se^{-\frac{1}{2}s^2}ds}{s^2e^{-\frac{1}{2}s^2}}\\
&=\frac{1}{s^2}+\frac{C}{s^2}e^{\frac{1}{2}s^2}
\end{align*}
where $C$ is a constant. Since $x(s)\to 0$ as $s\to\infty$, $C$ must be $0$. Therefore $x(s)=\frac{1}{s^2}$ and consequently $X(t)=t$.

Example. Solve Bessel’s equation with index zero
$$t\ddot{X}(t)+\dot{X}(t)+tX(t)=0$$
with the initial condition $X(0)=1$.

Solution. The transformed equation is
$$-\frac{d}{ds}[s^2x(s)-s-\dot{X}(0)]+sx(s)-1-\frac{d}{ds}x(s)=0$$
which simplifies to the first-order separable differential equation
$$(s^2+1)x'(s)+sx(s)=0$$
Performing the integrals
$$\int\frac{dx}{x}=-\int\frac{sds}{s^2+1}$$
we find
\begin{align*}
x(s)&=\frac{C}{\sqrt{s^2+1}}\\
&=\frac{C}{s}\left(1+\frac{1}{s^2}\right)^{-\frac{1}{2}}\\
&=\frac{C}{2}\sum_{n=0}^\infty\begin{pmatrix}
-\frac{1}{2}\\
n
\end{pmatrix}\left(\frac{1}{s^2}\right)^n,
\end{align*}
where $C$ is a constant and $s>1$, by the Binomial Theorem.
\begin{align*}
\begin{pmatrix}
-\frac{1}{2}\\
n
\end{pmatrix}&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots\left(-\frac{1}{2}-n+1\right)}{n!}\\
&=\frac{(-1)^n1\cdot 2\cdot 3\cdot 5\cdots(2n-1)}{2^nn!}\\
&=\frac{(-1)^n(2n)!}{(2^nn!)^2}
\end{align*}
So we have
$$x(s)=C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}\frac{(2n)!}{s^{2n+1}}$$
Since $x(s)$ is an infinite sum we cannot directly use the linearity of $\mathcal{L}^{-1}$ to obtain $X(t)$. Nonetheless we can show that
$$X(t)=C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}$$
\begin{align*}
\mathcal{L}\left\{C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}\right\}&=C\mathcal{L}\left\{\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}\right\}\\
&=C\int_0^\infty e^{-st}\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}dt\\
&=C\int_0^\infty e^{-st}\lim_{k\to\infty}\sum_{n=0}^k\frac{(-1)^n}{(2^nn!)^2}t^{2n}dt\\
&=C\lim_{k\to\infty}\sum_{n=0}^k\frac{(-1)^n}{(2^nn!)^2}\int_0^\infty e^{-st}t^{2n}dt\\
&=C\lim_{k\to\infty}\sum_{n=0}^k\frac{(-1)^n}{(2^nn!)^2}\mathcal{L}\{t^{2n}\}\\
&=C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}\frac{(2n)!}{s^{2n+1}}\\
&=x(s)
\end{align*}
By the uniqueness of $\mathcal{L}^{-1}$, we have
$$X(t)=C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}$$
Since $X(0)=1$, we obtain $C=1$ and hence
$$X(t)=\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}$$
This series is denoted by $J_0(t)$ i.e.
\begin{equation}
\begin{aligned}
J_0(t&)=\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}\\
&=1-\frac{t^2}{2^2}+\frac{t^4}{2^2\times 4^2}-\frac{t^6}{2^2\times 4^2\times 6^2}+\cdots
\end{aligned}\label{eq:bessel0}
\end{equation}
One can easily show using, for example, the ratio test that the series in \eqref{eq:bessel0} converges for all $t$. We now have the Laplace transform
\begin{equation}
\mathcal{L}\{J_0(t)\}=\frac{1}{\sqrt{s^2+1}}\ (s>1)
\end{equation}

The differential equation
\begin{equation}
\label{eq:besseleqn}
t^2\ddot{X}(t)+t\dot{X}(t)+(t^2-n^2)X(t)=0
\end{equation}
is called the Bessel’s equation of index $n$. The solution $X(t)$ is
$$X(t)=CJ_n(t),\ n=0,1,2,\cdots$$
where
\begin{equation}
\label{eq:besseln}
J_n(t)=\sum_{k=0}^\infty\frac{(-1)^k}{k!(n+k)!}\left(\frac{t}{2}\right)^{n+2k}
\end{equation}
$J_n(t)$, $n=0,1,2,\cdots$ is called the Bessel function of the first kind. There is another solution of the Bessel’s equation in \eqref{eq:besseleqn} which is linearly independent from $J_n(t)$. It is in a pretty horrible form
\begin{equation}
\begin{aligned}
N_n(x)=&\frac{2}{\pi}\left[\ln\left(\frac{x}{2}\right)+\gamma-\frac{1}{2}\sum_{p=1}^n\frac{1}{p}\right]J_n(x)\\
&-\frac{1}{\pi}\sum_{r=0}^\infty\frac{(-1)^r}{r!(n+r)!}\left(\frac{x}{2}\right)^{n+2r}\sum_{p=1}^r\left[\frac{1}{p}+\frac{1}{p+n}\right]\\
&-\frac{1}{\pi}\sum_{r=0}^{n-1}\frac{(n-r-1)!}{r!}\left(\frac{x}{2}\right)^{-n+2r}
\end{aligned}\label{eq:neumann}
\end{equation}
where $\gamma$ is the Euler-Mascheroni constant defined by
\begin{align*}
\gamma&=\lim_{n\to\infty}\left(\sum_{m=1}^n\frac{1}{m}-\ln n\right)\\
&\approx 0.57721566\cdots
\end{align*}
$N_n(x)$, $n=0,1,2,\cdots$ is called the Bessel function of the second kind or the Neumann function. Hence the general solution of the Bessel’s equation is given by
$$X(t)=AJ_n(x)+BN_n(x)$$
Those who wish to know more about Bessel functions and Neumann functions may refer to the reference [1] below.

Let us now consider $n=1$ case.
$$t^2\ddot{X}(t)+t\dot{X}(t)+(t^2-1)X(t)=0$$
Using \eqref{eq:laplace14} we obtain
\begin{align*}
\mathcal{L}\{t^2\ddot{X}(t)\}&=\frac{d^2}{ds^2}\mathcal{L}\{\ddot{X}(t)\}\\
&=\frac{d^2}{ds^2}[s^2x(s)-sX(0)-\dot{X}(0)]\\
&=s^2x^{\prime\prime}(s)+4sx'(s)+2x(s)\\
\mathcal{L}\{t\dot{X}(t)\}&=-\frac{d}{ds}\mathcal{L}\{\dot{X}(t)\}\\
&=-\frac{d}{ds}[sx(s)-X(0)]\\
&=-x(s)-sx'(s)\\
\mathcal{L}\{t^2X(t)\}&=\frac{d^2}{ds^2}\mathcal{L}\{X(t)\}\\
&=x^{\prime\prime}(s)
\end{align*}
Hence the transformed equation is
\begin{equation}
\label{eq:laplace15}
(s^2+1)x^{\prime\prime}(s)+3sx'(s)=0
\end{equation}
Let $y(s)=x'(s)$. Then \eqref{eq:laplace15} becomes the separable first-order differential equation
$$(s^2+1)y'(s)+3sy(s)=0$$
which can then be written as
$$\frac{dy}{y}=-\frac{3s}{s^2+1}ds$$
Integrating this we find
$$y(s)=\frac{dx}{ds}=\frac{C_1}{(s^2+1)^{\frac{3}{2}}}$$
Using the trigonometric substitution $s=\tan\theta$, we find
$$x(s)=\int\frac{C_1}{(s^2+1)^{\frac{3}{2}}}ds=\frac{C_1s}{\sqrt{s^2+1}}+C_2$$
Since $\lim_{s\to\infty}x(s)=0$, $C_2=-C_1$. Setting $C_1=C$, we have
\begin{equation}
\begin{aligned}
x(s)&=C\left[\frac{s}{\sqrt{s^2+1}}-1\right]\\
&=C[s\mathcal{L}\{J_0(t)\}-J_0(0)]\\
&=C\mathcal{L}\{J_0′(t)\}
\end{aligned}\label{eq:laplace16}
\end{equation}
Hence,
$$X(t)=CJ_0′(t)$$
From \eqref{eq:besseln} we find $J_1(t)=-J_0′(t)$ so
\begin{equation}
\label{eq:laplace17}
X(t)=-CJ_1(t)
\end{equation}
We can obtain the Laplace transform of $J_1(t)$ using \eqref{eq:laplace16} and \eqref{eq:laplace17}.
\begin{align*}
\mathcal{L}\{J_1(t)\}&=1-\frac{s}{\sqrt{s^2+1}}\\
&=\frac{1}{\sqrt{s^2+1}(\sqrt{s^2+1}+s)}
\end{align*}

References:

[1] Mathematical Methods for Physicists, George Arfken, Third Edition, Academic Press, 1985

The Laplace Transform: Convolution

Definition. The convolution $F\ast G$ of $F(t)$ and $G(t)$ is defined by
$$F(t)\ast G(t)=\int_0^t F(\tau)G(t-\tau)d\tau$$

We introduce Convolution Theorem without a proof. For those who are interested a proof can be found in [1] of the References below.

Theorem. Let $\mathcal{L}\{F(t)\}=f(s)$ and $\mathcal{L}\{G(t)\}=g(s)$. Suppose that $F(t)$ and $G(t)$ are piecewise continuous and are of order $e^{\alpha t}$ as $t\to\infty$. Then $\mathcal{L}\{F(t)\ast G(t)\}$ exists when $s>\alpha$ and it is $f(s)g(s)$. Equivalently,
$$\mathcal{L}^{-1}\{f(s)g(s)\}=F(t)\ast G(t)$$

Example. Let $F(t)=t$ and $G(t)=e^{at}$. Then $\mathcal{L}\{F(t)\}=\frac{1}{s^2}$ and $\mathcal{L}\{G(t)\}=\frac{1}{s-a}$. By Convolution Theorem
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{1}{s^2}\frac{1}{s-a}\right\}&=t\ast e^{at}\\
&=\int_0^t\tau e^{a(t-\tau)}d\tau\\
&=\frac{1}{a^2}(e^{at}-at-1)
\end{align*}
Note that partial fractions can be also used to obtain the result.

When $G(t)=F(t)$, we have the formula
$$[f(s)]^2=\mathcal{L}\{F(t)\ast F(t)\}$$

Example.
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{1}{(s^2+k^2)^2}\right\}&=\frac{1}{k^2}\mathcal{L}^{-1}\left\{\frac{k^2}{(s^2+k^2)^2}\right\}\\
&=\frac{1}{k^2}\int_0^t\sin kt\ast \sin kt\\
&=\frac{1}{k^2}\int_0^t\sin k\tau\sin k(t-\tau)d\tau\\
&=\frac{1}{k^3}\{\sin kt-kt\cos kt\}
\end{align*}

Theorem (Properties of Convolution).

  1. $F(t)\ast G(t)=G(t)\ast F(t)$
  2. $F(t)\ast[G(t)+H(t)]=F(t)\ast G(t)+F(t)\ast H(t)$
  3. $F(t)\ast [kG(t)]=k[F(t)\ast G(t)]$ where $k$ is a constant.
  4. $F(t)\ast[G(t)\ast H(t)]=[F(t)\ast G(t)]\ast H(t)$

Proof. We prove only the part 1.
\begin{align*}
F(t)\ast G(t)&=\int_0^t F(\tau)G(t-\tau)d\tau\\
&=\int_0^tF(t-\lambda)G(\lambda)d\lambda\ (\lambda=t-\tau)\\
&=G(t)\ast F(t)
\end{align*}

\begin{align*}
\frac{1}{s}f(s)&=\mathcal{L}\{F(t)\ast 1\}\\
&=\mathcal{L}\left\{\int_0^t F(\tau)d\tau\right\}\\
\frac{1}{s^2}f(s)&=\mathcal{L}\{F(t)\ast 1\ast 1\}\\
&=\mathcal{L}\{(F(t)\ast 1)\ast 1\}\\
&=\mathcal{L}\left\{\int_0^t(F(\tau)\ast 1)d\tau\right\}\\
&=\mathcal{L}\left\{\int_0^t\int_0^{\tau}F(\lambda)d\lambda d\tau\right\}
\end{align*}
Therefore we have the following theorem.

Theorem.
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{1}{s}f(s)\right\}&=\int_0^tF(\tau)d\tau\\
\mathcal{L}^{-1}\left\{\frac{1}{s^2}f(s)\right\}&=\int_0^t\int_0^{\tau}F(\lambda)d\lambda d\tau
\end{align*}

Convolution Theorem can be used for solving non-homogeneous linear differential equations.

Example. Find the general solution of the differential equations
$$\ddot{X}(t)+k^2X(t)=F(t)$$

Solution. Let $\mathcal{L}\{X(t)\}=x(s)$. Then the transformed equation is
$$s^2x(s)-sX(0)-\dot{X}(0)+k^2x(s)=f(s)$$
So we have
$$x(s)=\frac{1}{k}\frac{k}{s^2+k^2}f(s)+\frac{s}{s^2+k^2}X(0)+\frac{1}{k}\frac{k}{s^2+k^2}\dot{X}(0)$$
Therefore by Convolution Theorem the general solution is given by
\begin{equation}
\begin{aligned}
X(t)&=\frac{1}{k}(\sin kt)\ast F(t)+X(0)\cos kt+\frac{\dot{X}(0)}{k}\sin kt\\
&=\frac{1}{k}\int_0^t\sin k(t-\tau)F(\tau)d\tau+C_1\cos kt+C_2\sin kt
\end{aligned}\label{eq:conv}
\end{equation}

Let us redo the problem in the last example in here using \eqref{eq:conv}.

Example. Find the general solution of the differential equation
$$\ddot{x}+4x=8\tan t,\ -\frac{\pi}{2}<x<\frac{\pi}{2}$$

Solution. Using the formula \eqref{eq:conv} with $k=2$
\begin{align*}
x(t)&=\frac{1}{2}\int_0^t\sin 2(t-\tau)(8\tan\tau)d\tau+C_1\cos 2t+C_2\sin 2t\\
&=4\int_0^t\sin 2(t-\tau)\tan\tau d\tau+C_1\cos 2t+C_2\sin 2t\\
&=C_1\cos 2t+C_2\sin 2t-4t\cos 2t+4\sin 2t\ln(\cos t)
\end{align*}

Convolution Theorem can be also used for solving integral equations as shown in the following example.

Example. Solve the integral equation
$$X(t)=at+\int_0^tX(\tau)\sin(t-\tau)d\tau$$

Solution. $X(t)=at+X(t)\ast \sin t$ so its transformed equation is
$$x(s)=\frac{a}{s^2}+x(s)\frac{1}{s^2+1}$$
Hence we have
\begin{align*}
x(s)&=\frac{a}{s^2}\frac{s^2+1}{s^2}\\
&=a\left(\frac{1}{s^2}+\frac{1}{s^4}\right)
\end{align*}
Therefore the solution is given by
$$X(t)=a\left(t+\frac{1}{6}t^3\right)$$

References:

[1] Ruel V. Churchill, Operational Mathematics, McGraw-Hill, 1958

The Laplace Transform: Further Properties

Let $\mathcal{L}\{F(t)\}=f(s)$ i.e. $f(s)=\int_0^\infty e^{-st}F(t)dt$. Then
\begin{equation}
\label{eq:laplace11}
e^{-bs}f(s)=\int_0^\infty e^{-s(t+b)}F(t)dt
\end{equation}
where $b$ is a positive constant. Using the substitution $\tau=t+b$ \eqref{eq:laplace11} can be written as
$$e^{-bs}f(s)=\int_b^\infty e^{-s\tau}F(\tau-b)d\tau$$
Define a function $F_b(\tau)$ by
\begin{equation}
F_b(\tau)=\left\{\begin{aligned}
0,\ &0<\tau<b\\
F(\tau-b),\ &\tau>b
\end{aligned}\right.\label{eq:laplace12}
\end{equation}
Then
$$e^{-bs}f(s)=\int_0^\infty e^{-s\tau}F_b(tau)d\tau$$

Theorem. If $\mathcal{L}\{F(t)\}=f(s)$ then for any positive constant $b$,
$$e^{-bs}f(s)=\mathcal{L}\{F_b(t)\}$$
where $F_b(t)$ is defined by \eqref{eq:laplace12}.

Let
$$S_b(t)=\left\{\begin{aligned}
0,\ &0<t<b\\
1,\ &t>b
\end{aligned}\right.$$
for any positive constant $b$. Define $S_0(t)=1$ for $t>0$. Then $S_b(t)$ is called a \emph{unit step function}. Using this unit step function, $F_b(t)$ can be written as
$$F_b(t)=S_b(t)F(t-b)$$

Example.
\begin{align*}
F_b(t)&=S_b(t)\sin k(t-b)\\
&=\mathcal{L}^{-1}\left\{\frac{ke^{-bs}}{s^2+k^2}\right\}
\end{align*}

\begin{align*}
\mathcal{L}\{F(at)\}&=\int_0^\infty e^{-st}F(at)dt\ (a>0)\\
&=\frac{1}{a}\int_0^\infty e^{-\frac{s}{a}\tau}F(\tau)d\tau\ (\tau=at)\\
&=\frac{1}{a}f\left(\frac{s}{a}\right)
\end{align*}

Theorem. If $\mathcal{L}\{F(t)\}=f(s)$ when $s>\alpha$, then
$$\mathcal{L}\{F(at)\}=\frac{1}{a}f\left(\frac{s}{a}\right)\ (s>a\alpha,\ a>0)$$

Example. Given that
$$\frac{s}{s^2+1}=\mathcal{L}\{\cos t\}$$
\begin{align*}
\frac{s}{s^2+k^2}&=\frac{1}{k^2}\frac{s}{\left(\frac{s}{k}\right)^2+1}\\
&=\frac{1}{k}\frac{\frac{s}{k}}{\left(\frac{s}{k}\right)^2+1}\\
&=\mathcal{L}\{\cos kt\}
\end{align*}

With the translation of transform (equation (1) in here) and the above theorem the effect of a general linear substitution for $s$ can be shown in the following formula.
\begin{align*}
f(as-b)&=f\left[a\left(s-\frac{b}{a}\right)\right]\\
&=\mathcal{L}\left\{\frac{1}{a}e^{\frac{b}{a}t}F\left(\frac{t}{a}\right)\right\}\ (a>0)
\end{align*}

The Laplace Transform: Solving Differential Equations

In this note we study how the Laplace transform can be used to solve differential equations. For now we only consider linear differential equations with constant coefficients. But later on we will see that the Laplace transform is also a powerful tool to solve linear differential equations with variable coefficients.

Example. Find the general solution of the differential equation
$$\ddot{X}(t)+k^2X(t)=0$$

Solution. Using the linearity transform the differential equation.
\begin{equation}
\label{eq:laplace10}
\mathcal{L}\{\ddot{X}(t)\}+k^2\mathcal{L}\{X(t)\}=0
\end{equation}
By the transform of derivative formula
$$\mathcal{L}\{\ddot{X}(t)\}=s^2\mathcal{L}\{X(t)\}-sX(0)-\dot{X}(0)$$
Let $X(0)=A$ and $\dot{X}(0)=B$. Then \eqref{eq:laplace10} becomes an algebraic equation of $\mathcal{L}\{X(t)\}$
$$(s^2+k^2)\mathcal{L}\{X(t)\}-As-B=0$$
Thus
\begin{align*}
\mathcal{L}\{X(t)\}&=\frac{As+B}{s^2+k^2}\\
&=A\frac{s}{s^2+k^2}+\frac{B}{k}\frac{k}{s^2+k^2}
\end{align*}
Hence,
$$X(t)=A\cos kt+B’\sin kt$$
where $B’=\frac{B}{k}$.

Example. Find the solution of the initial value problem
$$\ddot{X}(t)-\dot{X}(t)-6X(t)=2$$
with $X(0)=1$ and $\dot{X}(0)=0$.

Solution. The transform of the differential equation, with the aid of the transform of derivative formula, is
$$[s^2\mathcal{L}\{X(t)\}-sX(0)-\dot{X}(0)]-[s\mathcal{L}\{X(t)\}-X(0)]-6\mathcal{L}\{X(t)\}=\frac{2}{s}$$
This simplifies to
\begin{align*}
(s^2-s-6)\mathcal{L}\{X(t)\}&=\frac{2}{s}+s-1\\
&=\frac{s^2-s+2}{s}
\end{align*}
So
\begin{align*}
\mathcal{L}\{X(t)\}&=\frac{s^2-s+2}{s(s^2-s-6)}\\
&=\frac{s^2-s+2}{s(s-3)(s+2)}\\
&=\frac{A}{s}+\frac{B}{s-3}+\frac{C}{s+2}
\end{align*}
This results in the equation
$$A(s-3)(s+2)+Bs(s+2)+Cs(s-3)=s^2-s+2$$
For $s=0$, we obtain $-6A=2$ i.e $A=-\frac{1}{3}$. For $s=3$, we obtain $15B=8$ i.e. $B=\frac{15}{8}$. For $s=-2$, we obtain $10C=8$ i.e $C=\frac{4}{5}$.
Hence,
$$\mathcal{L}\{X(t)\}=-\frac{1}{3}\frac{1}{s}+\frac{15}{8}\frac{1}{s-3}+\frac{4}{5}\frac{1}{s+2}$$
and therefore
$$X(t)=-\frac{1}{3}+\frac{15}{8}e^{3t}+\frac{4}{5}e^{-2t}$$

Example. Solve the problem
$$\dddot{X}(t)-2\ddot{X}(t)+5\dot{X}(t)=0$$
with $X(0)=0$, $\dot{X}(0)=1$, and $X\left(\frac{\pi}{8}\right)=1$.

Solution. The transformed equation, with the aid ofthe transform of derivative formula, is
\begin{align*}
[s^3\mathcal{L}\{X(t)\}-s^2X(0)-s\dot{X}(0)-\ddot{X}(0)]&-2[s^2\mathcal{L}\{X(t)\}-sX(0)-\dot{X}(0)]\\&
+5[s\mathcal{L}\{X(t)\}-X(0)]=0
\end{align*}
This simplifies, with the initial conditions $X(0)=0$ and $\dot{X}(0)=1$, to the equation
$$(s^3-2s^2+5s)\mathcal{L}\{X(t)\}-s+2-\ddot{X}(0)=0$$
Let $\ddot{X}(0)=C$. Then
\begin{align*}
\mathcal{L}\{X(t)\}&=\frac{C-2+s}{s^3-2s^2+5s}\\
&=\frac{C-2+s}{s(s^2-2s^+5)}\\
&=\frac{C-2+s}{s[(s-1)^2+4]}\\
&=\frac{C-2}{5}\left[\frac{1}{s}-\frac{s-1}{(s-1)^2+4}\right]+\frac{C+3}{10}\frac{2}{(s-1)^2+4}
\end{align*}
Hence
$$X(t)=\frac{C-2}{5}+e^t\left(\frac{C+3}{10}\sin 2t-\frac{C-2}{5}\cos 2t\right)$$
The condition $X\left(\frac{\pi}{8}\right)=1$ results in the equation
$$1=\frac{C-2}{5}+\frac{e^{\frac{\pi}{8}}}{10\sqrt{2}}(-C+7)$$ whose solution is $C=7$. Therefore,
$$X(t)=1+e^t(\sin 2t-\cos 2t)$$