Monthly Archives: September 2011

Lie Group and Lie Algebra Representations

Given a matrix Lie group $G$, a representation $\Pi$ of $G$ is a Lie group homomorphism $\Pi: G\longrightarrow\mathrm{GL}(V)$, where $V$ is a finite dimensional vector space and the general linear group $\mathrm{GL}(V)$ is the set of all linear isomorphisms of $V$. For each $g\in G$, $\Pi(g): V\longrightarrow V$ is a linear operator on $V$.

If $\mathfrak{g}$ is a Lie algebra, a representation of $\mathfrak{g}$ is a Lie algebra homomorphism $\pi: \mathfrak{g}\longrightarrow\mathrm{gl}(V)$, where $\mathrm{gl}(V)$ is the Lie algebra of $\mathrm{GL}(V)$.

If $\Pi$ or $\pi$ is a one-to-one homomorphism, then the representation is called faithful.

One may understand a representation as the action of a Lie group or a Lie algebra on the vector space $V$.

Example. [Trivial Representation] Let $G$ be a matrix Lie group. Define the trivial representation of $G$ by $$\Pi: G\longrightarrow\mathrm{GL}(1;\mathbb{C});\ A\longmapsto I.$$ This is an irreducible representation since $\mathbb C$ has no nontrivial subspace. If $\mathfrak{g}$ is a Lie algebra, the trivial representation of $\mathfrak{g}$, $\pi: \mathfrak{g}\longrightarrow\mathrm{gl}(1;\mathbb{C})$ is defined by $\pi(X)=0$ for all $X\in\mathfrak{g}$. This is also an irreducible representation.

Example. [The Adjoint Representation] Let $G$ be a matrix Lie group with Lie algebra $\mathfrak{g}$. The adjoint mapping $\mathrm{Ad}: G\longrightarrow\mathrm{GL}(\mathfrak{g})$ is defined by $$\mathrm{Ad}_A(X)=AXA^{-1}$$ for $A\in G$. We claim that $AXA^{-1}\in\mathfrak{g}$ for $A\in G$ and $X\in\mathfrak{g}$, so that $\mathrm{Ad}_A:\mathfrak{g}\longrightarrow\mathfrak{g}$. First note that for any invertible matrix $A$, $(AXA^{-1})^m=AX^mA^{-1}$. So, \begin{align*}
e^{AXA^{-1}}&=\sum_{m=0}^\infty\frac{(AXA^{-1})^m}{m!}\\
&=A\sum_{m=0}^\infty\frac{X^m}{m!}A^{-1}\\
&=Ae^XA^{-1}.
\end{align*}
Now for $A\in G$ and $X\in\mathfrak{g}$,
\begin{align*}
e^{tAXA^{-1}}&=e^{A\cdot tX\cdot A^{-1}}\\
&=Ae^{tX}A^{-1}\in G
\end{align*} and hence $AXA^{-1}\in\mathfrak{g}$. Note that $\mathrm{Ad}: G\longrightarrow\mathrm{GL}(\mathfrak{g})$ is a Lie group homomorphism. So $\mathrm{Ad}$ can be considered as a representation of $G$ acting on the Lie algebra $\mathfrak{g}$. We call $\mathrm{Ad}$ the adjoint representation of $G$. We can also define the adjoint representation of the Lie algebra $\mathfrak{g}$ as follows:
$$\mathrm{ad}:\mathfrak{g}\longrightarrow\mathrm{gl}(\mathfrak{g});\ \mathrm{ad}_X(Y)=[X,Y].$$ $\mathrm{ad}$ is a Lie algebra homomorphism.

Let $V$ be a finite dimensional real vector space. Then complexification of $V$, $V_{\mathbb{C}}$ is the space of formal linear combination $v_1+iv_2$ with $v_1,v_2\in V$. This is again a real vector space. If we define $$i(v_1+iv_2)=-v_2+iv_1,$$ then $V_{\mathbb{C}}$ becomes a complex vector space. For example, the complexification $\mathfrak{su}(2)_{\mathbb{C}}$ of the Lie algebra $\mathfrak{su}(2)$ is $\mathfrak{sl}(2;\mathbb{C})$.

Some Representations of $\mathrm{SU}(2)$

Let $V_m$ be the space of homogeneous polynomials in two variables with total degree $m\geq 0$
$$f(z_1,z_2)=a_0z_1^m+a_1z_1^{m-1}z_2+a_2z_1^{m-2}z_2^2+\cdots+a_mz_2^m.$$ Then $V_m$ is an $(m+1)$-dimensional complex vector space. Define $\Pi_m:\mathrm{SU}(2)\longrightarrow\mathrm{GL}(V_m)$ by $$[\Pi_m(U)f](z)=f(U^{-1}z).$$
Let us write $U^{-1}=\begin{pmatrix}
U_{11}^{-1} & U_{12}^{-1}\\
U_{21}^{-1} & U_{22}^{-1}
\end{pmatrix}$. Then $U^{-1}z=\begin{pmatrix}
U_{11}^{-1}z_1+U_{12}^{-1}z_2\\
U_{21}^{-1}z_1+U_{22}^{-1}z_2
\end{pmatrix}$, where $z=\begin{pmatrix}z_1\\z_2\end{pmatrix}\in\mathbb{C}^2$. So $[\Pi_m(U)f](z_1,z_2)$ is written as
$$[\Pi_m(U)f](z_1,z_2)=\sum_{k=0}^ma_k(U_{11}^{-1}z_1+U_{12}^{-1}z_2)^{m-k}(U_{21}^{-1}z_1+U_{22}^{-1}z_2)^k.$$
We now show that $\Pi_m$ is indeed a Lie group homomorphism: \begin{align*}
\Pi_m(U_1)[\Pi_m(U_2)f](z)&=\Pi_m(U_2f)(U_1^{-1}z)\\
&=f(U_2^{-1}U_1^{-1}z)\\
&=f((U_1U_2)^{-1}z)\\
&=[\Pi_m(U_1U_2)f](z).
\end{align*} Therefore, $\Pi_m$ is a finite dimensional complex representation of $\mathrm{SU}(2)$. Note that each $\Pi_m$ is irreducible and that every finite dimensional irreducible representation of $\mathrm{SU}(2)$ is equivalent to one and only one of the $\Pi_m$’s.

Now we compute the corresponding representation $\pi_m$ of the Lie algebra $\mathfrak{su}(2)$. $\pi_m$ can be computed as $$\pi_m(X)=\frac{d}{dt}\Pi_m(e^{tX})|_{t=0}.$$ So,
\begin{align*}
[\pi_m(X)f](z)&=\frac{d}{dt}[\Pi_m(e^{tX})f](z)|_{t=0}\\
&=\frac{d}{dt}f(e^{-tX}z)|_{t=0}.
\end{align*} Let $z(t)$ be a curve in $\mathbb{C}^2$ defined as $z(t)=e^{-tX}z$, so that $z(0)=z$. Write $z(t)=(z_1(t),z_2(t))$, where $z_i(t)\in\mathbb{C}$, $i=1,2$. By the chain rule, \begin{align*}
\pi_m(X)f&=\frac{\partial f}{\partial z_1}\frac{dz_1}{dt}|_{t=0}+\frac{\partial f}{\partial z_2}\frac{dz_2}{dt}|_{t=0}\\
&=-\frac{\partial f}{\partial z_1}(X_{11}z_1+X_{12}z_2)-\frac{\partial f}{\partial z_2}(X_{21}z_1+X_{22}z_2),\ \ \ \ \ \mbox{(1)}\end{align*}
since $\frac{dz}{dt}|_{t=0}=-Xz$.

Every finite dimensional complex representation of the Lie algebra $\mathfrak{su}(2)$ extends uniquely to a complex linear representation of the complexification of $\mathfrak{su}(2)$ and the complexification of $\mathfrak{su}(2)$ is isomorphic to $\mathfrak{sl}(2;\mathbb{C})$. Thus, the representation $\pi_m$ of $\mathfrak{su}(2)$ extends to a representation of $\mathfrak{sl}(2;\mathbb{C})$. Note that the Lie algebra $\mathfrak{sl}(2;\mathbb{C})$ is the set of all $2\times 2$ trace-free complex matrices, i.e. matrices of the form $\begin{pmatrix}\alpha & \beta\\\gamma & -\alpha\end{pmatrix}$ where $\alpha$, $\beta$ and $\gamma$ are complex numbers. So any element in $\mathfrak{sl}(2;\mathbb{C})$ can be uniquely written as $\alpha H+\beta X+\gamma Y$, where $H=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}$, $X=\begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix}$, $Y=\begin{pmatrix}
0 & 0\\
1 & 0
\end{pmatrix}$.
Let us calculate $\pi_m$ for the basis members $H$, $X$, and $Y$.
$$(\pi_m(H)f)(z)=-\frac{\partial f}{\partial z_1}z_1+\frac{\partial f}{\partial z_2}z_2$$ so
$$\pi_m(H)=-z_1\frac{\partial}{\partial z_1}+z_2\frac{\partial}{\partial z_2}.$$
Applying $\pi_m(H)$ to a basis element $z_1^kz_2^{m-k}$, we obtain
\begin{align*}
\pi_m(H)z_1^kz_2^{m-k}&=-kz_1^kz_2^{m-k}+(m-k)z_1^kz_2^{m-k}\\
&=(m-2k)z_1^kz_2^{m-k}.
\end{align*}
This means that $z_1^kz_2^{m-k}$ is an eigenvector for $\pi_m(H)$ with eigenvalue $m-2k$. In particular, $\pi_m(H)$ is diagonalizable. Using (1) again we also obtain
$$\pi_m(X)=-z_2\frac{\partial}{\partial z_1},\ \pi_m(Y)=-z_1\frac{\partial}{\partial z_2}$$
and
\begin{align*}
\pi_m(X)z_1^kz_2^{m-k}&=-kz_1^{k-1}z_2^{m-k+1},\ \ \ \ \ \mbox{(2)}\\
\pi_m(Y)z_1^kz_2^{m-k}&=(k-m)z_1^{k+1}z_2^{m-k-1}.\ \ \ \ \ \mbox{(3)}
\end{align*}

Proposition. The representation $\pi_m$ is an irreducible representation of $\mathfrak{sl}(2;\mathbb{C})$.

Proof. Suppose that $W$ is a nonzero invariant subspace of $V_m$. We claim that $W=V_m$. Since $W\ne \{0\}$, there exists $w\in W$ with $w\ne 0$. $w$ can be uniquely written as
$$w=a_0z_1^m+a_1z_1^{m-1}z_2+a_2z_1^{m-2}z_2^2+\cdots+a_mz_2^m$$ with at least one of the $a_k$’s nonzero. Let $k_0$ be the smallest value of $k$ for which $a_k\ne 0$ and consider $\pi_m(X)^{m-k_0}w$. Since $\pi_m(X)$ lowers the power of $z_1$ by 1, $\pi_m(X)^{m-k_0}w$ will kill all the terms in $w$ except $a_{k_0}z_1^{m-k_0}z_2^{k_0}$. On the other hand,
$$\pi_m(X)^{m-k_0}(z_1^{m-k_0}z_2^{k_0})=(-1)^{m-k_0}(m-k_0)!z_2^m.$$ Since $\pi_m(X)^{m-k_0}(z_1^{m-k_0}z_2^{k_0})$ is a multiple of $z_2^m$ and $W$ is invariant, $W$ must contain $z_2^m$. It follows from (2) that $\pi_m(Y)^kz_2^m$ is a nonzero multiple of $z_1^kz_2^{m-k}$ for $0< k\leq m$. Hence, $W$ must contain $z_1^kz_2^{m-k}$, $0< k\leq m$. Since $W$ contains all the basis members of $V_m$, $z_1^kz_2^{m-k}$, $0\leq k\leq m$, then $W=V_m$.

The Lie Algebra of the Orthogonal Group $\mathrm{O}(n)\ (\mathrm{SO}(n))$

It can be easily shown that
$${\rm SO}(2)=\left\{\left(\begin{array}{cc}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{array}
\right): \theta\in[0,2\pi)\right\}\cong{\rm S}^1=\{e^{i\theta}:
\theta\in[0,2\pi)\}.$$Let $\gamma(t)=\left(\begin{array}{cc}
\cos\theta(t) & -\sin\theta(t)\\
\sin\theta(t) & \cos\theta(t)
\end{array}
\right)\in\mathrm{SO}(2)$ with $\theta(0)=0$ and $\dot\theta(0)\ne 0$. Then $\gamma(t)$ be a differentiable (regular) curve in ${\rm SO}(2)$ such that
$\gamma(0)=I$. Thus
$$\dot{\gamma}(0)=\left(\begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right)\left(\frac{d\theta}{dt}\right)_{t=0}$$
is a tangent vector to $\mathrm{SO}(2)$ at the identity $I$. Hence, the tangent space of ${\rm SO}(2)$ at $I$ is a line i.e. ${\rm SO}(2)$ is a one-dimensional Lie group. (We already know that ${\rm SO}(2)$ is a one-dimensional Lie group since it is identified with the unit circle ${\rm S}^1$.)

Remark. $\dot\gamma(0)=\left(\begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right)$ is a skew-symmetric matrix, i.e., $\dot\gamma(0)+{}^t\dot\gamma(0)=0$.

Let $\gamma: (-\epsilon,\epsilon)\buildrel{\rm
diff}\over\longrightarrow{\rm O}(n)$ such that $\gamma(0)=I$. Then $\dot{\gamma}(0)$ is a tangent vector to ${\rm O}(n)$ at $I$. Since $\gamma(t)\in{\rm O}(n)$, $$\gamma(t)\cdot{}^t\gamma(t)=I$$ for each $t\in(-\epsilon,\epsilon)$. Thus,
$$\dot{\gamma}(0)\cdot{}^t\gamma(0)+\gamma(0)\cdot\dot{{}^t\gamma}(0)=0.$$ Since ${}^t\gamma(0)=\gamma(0)=I$, $$\dot{\gamma}(0)+\dot{{}^t\gamma}(0)=\dot{\gamma}(0)+{}^t\dot{\gamma}(0)=0.$$ Hence, we see that any tangent vector to ${\rm O}(n)$ at $I$ is represented as a skew-symmetric $n\times n$ matrix. Conversely, we want to show that every skew-symmetric $n\times n$ matrix is a tangent vector to ${\rm O}(n)$ at $I$.

Suppose that $A$ is a $n\times n$ skew-symmetric matrix. As discussed here,
$$e^{At}=I+At+\frac{(At)^2}{2!}+\cdots+\frac{(At)^n}{n!}+\cdots=I+At+\frac{A^2}
{2!}t^2+\cdots+\frac{A^n}{n!}t^n+\cdots$$
is an $n\times n$ matrix.

If $AB=BA$, then by Cauchy’s Theorem,
$$\left(\sum_{k=0}^\infty\frac{A^k}{k!}\right)\left(\sum_{l=0}^\infty\frac{B^l}
{l!}\right)=\sum_{m=0}^\infty\sum_{p=0}^m\frac{A^{m-p}B^p}{(m-p)!p!}=\sum_{m=0}^\infty\frac{(A+B)^m}{m!}.$$ This implies that $e^Ae^B=e^{A+B}$ if $AB=BA$. In particular, $e^{A}e^{-A}=e^0=I$ so that $e^A$ is non-singular. If $A$ is skew-symmetric, then ${}^t(e^{At})=e^{{}^tAt}=e^{-At}$ and so $e^{At}\cdot{}^t(e^{At})=I$, i.e., $e^{At}\in{\rm O}(n)$. Now, $\displaystyle\frac{de^{At}}{dt}=Ae^{At}$ and $\dot{e^{At}}(0)=A$, i.e., the skew-symmetric matrix $A$ is a tangent vector to ${\rm O}(n)$ at $I$.

Proposition. The tangent space of ${\rm O}(n)$ or ${\rm SO}(n)$ at $I$ is the set of all $n\times n$ skew-symmetric matrices. Denote by ${\mathfrak o}(n)$ (${\mathfrak s\mathfrak o}(n)$) the tangent space of ${\rm O}(n)$ (${\rm SO}(n)$, respectively) at $I$. Note that $\dim{\mathfrak o}(n)=\displaystyle\frac{1}{2}n(n-1)$. This can be easily shown.

Definition. The tangent space ${\mathfrak o}(n)$ (${\mathfrak s\mathfrak o}(n)$) to the Lie group ${\rm O}(n)$ (${\rm SO}(n)$, respectively) at $I$ is called the Lie algebra of ${\rm O}(n)$ (${\rm SO}(n)$, respectively).

Matrix Lie Groups

Definition. A group $(G,\cdot,{}^{-1},e)$ is a Lie group if $G$ is also a differentiable manifold and the binary operation $\cdot: G\times G\longrightarrow G$ and the unary operation (inverse) ${}^{-1}: G\longrightarrow G$ are smooth maps.

A subgroup of a Lie group is not necessarily a Lie subgroup.

Theorem. [C. Chevalley] Every closed subgroup of a Lie group is a Lie subgroup.

Examples of Lie Groups.

  1. Let $M(m,n)=\{m\times n-\mbox{matrices over}\ \mathbb{R}\}\cong\mathbb{R}^{mn}$. Let $A=(a_{ij})\in M(m,n)$. Define an identification map\begin{align*}M(m,n)&\longrightarrow\mathbb{R}^{mn}\\(a_{ij})&\longmapsto(a_{11},\cdots,a_{1n};\cdots;a_{m1},\cdots,a_{mn}).\end{align*} We can naturally define topology on $M(m,n)$ by the identification map. $M(m,n)$ is covered by a single chart and the identification map is the coordinate map.
  2. The General Linear Group ${\rm GL}(n)$: Let $\mathrm{GL}(n)=\{\mbox{non-singular}\ n\times n-\mbox{matrices}\}$. Define a map\begin{align*}\mathrm{GL}(n)&\longrightarrow\mathbb{R}\\A&\longmapsto\det A.\end{align*} This map is onto and continuous since $\det A$ is a polynomial function of entries $a_{ij}$ of $A$. $\mathrm{GL}(n)=\det^{-1}(\mathbb{R}-\{0\})$is an open subset of $\mathbb{R}^{n^2}$, so that it is a submanifold of $\mathbb{R}^{n^2}$. This group is called the general linear group. The set of all $n\times n$ non-singular real (complex) matrices is denoted by $\mathrm{GL}(n;\mathbb{R})$ ($\mathrm{GL}(n;\mathbb{C})$, resp.). More generally, the set $n\times n$ non-singular matrices whose entries are the elements of a field $F$ is denoted by $\mathrm{GL}(n;F)$ or $\mathrm{GL}(V)$ where $V$ is the vector space isomorphic to $F^n$. Note that $\mathrm{GL}(V)$ is also the set of all linear isomorphisms of $V$.
  3. The Orthogonal Group $\mathrm{O}(n)$: The orthogonal group $\mathrm{O}(n)$ is defined to be the set $$\mathrm{O}(n)=\{n\times n-\mbox{orthogonal matrices}\},$$ i.e., $$A\in\mathrm{O}(n)\Longleftrightarrow A\cdot{}^tA=I,$$ where ${}^tA$ is the transpose of $A$ and $I$ is the $n\times n$ identity matrix.
  4. The Special Orthogonal Group $\mathrm{SO}(n)$: The special orthogonal group is defined to be the following subgroup of $\mathrm{O}(n)$: $$\mathrm{SO}(n)=\{A\in\mathrm{O}(n): \det A=1\}.$$
  5. The Special Linear Group $\mathrm{SL}(n)$: The special linear group is defined to be the following subgroup of $\mathrm{GL}(n)$ $$\mathrm{SL}(n)=\{A\in\mathrm{GL}(n): \det A=1\}.$$
  6. The Unitary Group $\mathrm{U}(n)$: The unitary group $\mathrm{U}(n)$ is the set of all $n\times n$-unitary matrices, i.e. $$\mathrm{U}(n)=\{U\in\mathrm{GL}(n;\mathbb{C}): UU^\ast=I\},$$ where $U^\ast={}^t\bar U$. Physicists often write $U^\ast$ as $U^\dagger$. $\mathrm{U}(n)$ is a Lie subgroup of $\mathrm{GL}(n;\mathbb{C})$.
  7. The Special Unitary Group $\mathrm{SU}(n)$: The special unitary group $\mathrm{SU}(n)$ is a Lie subgroup of $\mathrm{U}(n)$ and $\mathrm{SL}(2;\mathbb{C})$ $$\mathrm{SU}(n)=\{U\in\mathrm{SL}(2;\mathbb{C}):UU^\ast=I\}.$$

Proposition. For any $n\times n$ real or complex matrix $X$,
$$e^X:=\sum_{m=0}^\infty\frac{X^m}{m!}$$ converges and is a continuous function.

Proof. For the proof of the proposition click here.

Definition. Let $G$ be a matrix Lie group. The Lie algebra of $G$, denoted by $\mathfrak{g}$, is the set of all matrices $X$ such that $e^{tX}\in G$ for all $t\in\mathbb{R}$.

Definition. A function $A:\mathbb{R}\longrightarrow\mathrm{GL}(n;\mathbb{C})$ is called a one-parameter subgroup of $\mathrm{GL}(n;\mathbb{C})$ if

  1. $A$ is continuous;
  2. $A(0)=I$;
  3. $A(t+s)=A(t)A(s)$ for all $t,s\in\mathbb{R}$.

Theorem. If $A$ is a one-parameter subgroup of $\mathrm{GL}(n;\mathbb{C})$, then there exists uniquely an $n\times n$-complex matrix $X$ such that $A(t)=e^{tX}$ for all $t\in\mathbb{R}$.

In differential geometry, the Lie algebra $\mathfrak{g}$ is defined to be the tangent space $T_eG$ to $G$ at the identity $e$. The two definitions coincide if $G$ is $\mathrm{GL}(n;\mathbb{C})$ or its Lie subgroup. If $X\in\mathfrak{g}$ then by definiton $e^{tX}\in G$ for all $t\in\mathbb{R}$. The one-parameter subgroup  $\{e^{tX}:t\in\mathbb{R}\}$ of $G$ can be regarded as a differentiable curve $\gamma:\mathbb{R}\longrightarrow G$ such that $\gamma(0)=e$ where $e$ is the $n\times n$ identity matrix $I$. Thus $\dot\gamma(0)=X$ is the tangent vector to $G$ at the identity $e$, i.e. $X\in T_eG$. Conversely, $X\in T_eG$. Let $\{\phi_t:G\longrightarrow G\}_{t\in\mathbb{R}}$ be the flow generated by $X$, i.e.
$$\frac{d}{dt}\phi_t(p)=X_{\phi_t(p)}.$$ Then $\phi_t$ is smooth, $\phi_0=e$, and $\phi_t\circ \phi_s=\phi_{t+s}$. That is, $\{\phi_t:G\longrightarrow G\}_{t\in\mathbb{R}}$ is a one-parameter subgroup of $\mathrm{GL}(n;\mathbb{C})$. Hence by the above Theorem, there exists uniquely an $n\times n$-complex matrix $Y$ such that $A(t)=e^{tY}$. Since $\dot A(0)=Y$, $Y=X$ i.e. $A(t)=e^{tX}\in G\leq\mathrm{GL}(n;\mathbb C)$. Therefore $X\in\mathfrak{g}$.

Physicists’ convention: In the physics literature, the exponential map $\exp:\mathfrak{g}\longrightarrow G$ is usually given by $X\longmapsto e^{iX}$ instead of $X\longmapsto e^X$. The reason for that comes from quantum mechanics and it will be discussed later.

References:

[1] Andrew Baker, Matrix Groups, An Introduction to Lie Group Theory, Springer 2001

[2] Brian C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Springer-Verlag 2004

Electrostatic Potential in a Hollow Cylinder

An electrostatic field $E$ (i.e. an electric field produced only by a static charge) is a conservative field, i.e. there exists a scalar potential $\psi$ such that $E=-\nabla\psi$. This is clear from Maxwell’s equations. Since there is no change of the magnetic field $B$ in time, $\nabla\times E=0$. If there is no charge present in a region, $\nabla\cdot E=0$. Together with $E=-\nabla\psi$, we obtain the Laplace equation $\nabla^2\psi=0$. Thus the Laplace equation can be used to find the electrostatic potential $\psi(\rho,\varphi,z)$ in a hollow cylinder with radius $a$ and height $l$ ($0\leq z\leq l$).

Using the separation of variables, we find the mode
\begin{align*}
\psi_{km}(\rho,\varphi,z)&=P_{km}(\rho)\Phi_m(\varphi)Z_k(z)\\
&=J_m(k\rho)[a_m\sin m\varphi+b_m\cos m\varphi][c_1e^{kz}+c_2e^{-kz}].
\end{align*}
The boundary conditions are:
$$\psi(\rho,\varphi,l)=\psi(\rho,\varphi),$$
where $\psi(\rho,\varphi)$ is a potential distribution. Elsewhere on the surface $\psi=0$. Now we find electrostatic potential
$$\psi(\rho,\varphi,z)=\sum_{k,m}\psi_{km}$$
inside the cylinder. From the boundary condition $\psi(\rho,\varphi,0)=0$, we find $c_1+c_2=1$. So we choose $c_1=-c_2=\frac{1}{2}$ and thereby $c_1e^{kz}+c_2e^{-kz}\sinh kz$. Since $\psi=0$ on the lateral surface of the cylinder, $\psi(a,\varphi,z)=0$. This implies that $J_m(ka)=0$. If we write the $n$-th Bessel zero as $a_{mn}$, then $k_{mn}a=a_{mn}$ or $k_{mn}=\frac{a_{mn}}{a}$. Hence,
$$\psi(\rho,\varphi,z)=\sum_{m=0}^\infty\sum_{n=1}^\infty J_m\left(\alpha_{mn}\frac{\rho}{a}\right)[a_m\sin m\varphi+b_m\cos m\varphi]\sinh\left(\alpha_{mn}\frac{z}{a}\right).$$
Finally using the boundary condition
$$\psi(\rho,\varphi)=\sum_{m=0}^\infty\sum_{n=1}^\infty J_m\left(\alpha_{mn}\frac{\rho}{a}\right)[a_m\sin m\varphi+b_m\cos m\varphi]\sinh\left(\alpha_{mn}\frac{1}{a}\right)$$ and the orthogonality of $\sin m\varphi$ and $\cos m\varphi$, we can determine the coefficients $a_m$ and $b_m$ as
\begin{align*}\left\{\begin{aligned}a_{mn}\\b_{mn}\end{aligned}\right\}=\frac{2}{\pi a^2\sinh\left(\alpha_{mn}\frac{1}{a}\right)J_{m+1}^2(\alpha_{mn})}\int_0^{2\pi}\int_0^a\psi(\rho,\varphi)&J_m\left(\alpha_{mn}\frac{\rho}{a}\right)\\
&\left\{\begin{aligned}
\sin m\varphi\\
\cos m\varphi
\end{aligned}\right\}\rho d\rho d\varphi.\end{align*}

Bessel Functions of the First Kind $J_n(x)$ II: Orthogonality

To accommodate boundary conditions for a finite interval $[0,a]$, we need to consider Bessel functions of the form $J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)$. For $x=\frac{\alpha_{\nu m}}{a}\rho$, Bessel’s equation (9) in here can be written as
$$\rho^2\frac{d^2}{d\rho^2}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)+\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)+\left(\frac{\alpha_{\nu m}^2\rho}{a^2}-\frac{\nu^2}{\rho}\right)J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)=0.\ \ \ \ \ \mbox{(10)}$$ Changing $\alpha_{\nu m}$ to $\alpha_{\nu n}$, $J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)$ satisfies
$$\rho^2\frac{d^2}{d\rho^2}J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)+\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)+\left(\frac{\alpha_{\nu n}^2\rho}{a^2}-\frac{\nu^2}{\rho}\right)J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)=0.\ \ \ \ \ \mbox{(11)}$$
Multiply (10) by $J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)$ and (11) by $J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)$ and subtract:
\begin{align*}
J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\frac{d}{d\rho}&\left[\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\right]-J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\frac{d}{d\rho}\left[\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\right]\\&=\frac{\alpha_{\nu n}^2-\alpha_{\nu m}^2}{a^2}\rho J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right).\end{align*}
Integrate this equation with respect to $\rho$ from $\rho=0$ to $\rho=a$:
\begin{align*}\int_0^\rho J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\frac{d}{d\rho}&\left[\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\right]d\rho\\&-\int_0^\rho J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\frac{d}{d\rho}\left[\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\right]d\rho\\&=\frac{\alpha_{\nu n}^2-\alpha_{\nu m}^2}{a^2}\int_0^\rho J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\rho d\rho.\ \ \ \ \ \mbox{(12)}\end{align*}
Using Integration by Parts, we have
\begin{align*}
\int_0^\rho &J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\frac{d}{d\rho}\left[\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\right]d\rho\\&=\left[\rho J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\right]_0^a-\int_0^a \rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)dJ_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right).\end{align*}
Thus (12) can be written as
\begin{align*}\left[\rho J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\right]_0^a-\left[\rho J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\right]_0^a\\=\frac{\alpha_{\nu n}^2-\alpha_{\nu m}^2}{a^2}\int_0^\rho J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\rho d\rho.\ \ \ \ \ \mbox{(13)}\end{align*}
Clearly the LHS of (13) vanishes at $\rho=0$. (Here we consider only $\nu=\mbox{integer}$ case.) It also vanishes at $\rho=a$ if we choose $\alpha_{\nu n}$ and $\alpha_{\nu m}$ to be $n$-th and $m$-th zeros of $J_\nu$. Therefore, for $m\ne n$
$$\int_0^\rho J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\rho d\rho=0.\ \ \ \ \ \mbox{(14)}$$
(14) gives us orthogonality over the interval $[0,a]$.

For $m=n$, we have the normalization integral
$$\int_0^a\left[J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\right]^2\rho d\rho=\frac{a^2}{2}[J_{\nu+1}(\alpha_{\nu m})]^2.\ \ \ \ \ \mbox{(15)}$$

MAT 101 Online Lecture Notes: 2.4 Analyzing Graphs of Quadratic Functions

There are two important topics in this section: graphing the quadratic function $f(x)=ax^2+bx+c$ and finding the (absolute) maximum or the minimum value of $f(x)=ax^2+bx+c$.

First the sign of the leading coefficient $a$ tells us some information about the graph. If $a>0$ then the tail of the graph goes up, i.e. the graph is a smiling face $\smile$. If $a<0$ then the tail of the graph goes down, i.e. the graph is a frowning face $\frown$.

Using the completing the square $f(x)=ax^2+bx+c$ can be written as
$$f(x)=a(x-h)^2+k,$$
where $h=-\frac{b}{2a}$ and $k=f(h)=f\left(-\frac{b}{2a}\right)$. The ordered pair $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$ is called the vertex of the parabola $f(x)$ and the vertical line $x=-\frac{b}{2a}$ is called the axis of symmetry (this is the vertical line that divides the graph of $f(x)$ into two halves). If $a>0$, then $f\left(-\frac{b}{2a}\right)$ is the absolute minimum value of $f(x)$. If $a<0$, then $f\left(-\frac{b}{2a}\right)$ is the absolute maximum value of $f(x)$.

How to sketch the graph of $f(x)=a(x-h)^2+k$?

Your textbook is telling you to sketch the graph of $f(x)=a(x-h)^2+k$ using transformations that you learned in section 1.7 (here and here). In principle, it is right to use transformations but in practice there is an easier way to do. All you need is the sign of $a$, the vertex $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$, and the $y$-intercept $c$. (Although not required, it would be better if you know $x$-intercepts as well.)

Example. Let $f(x)=x^2+7x-8$.

(a) Find the vertex.

Solution. $-\frac{b}{2a}=-\frac{7}{2}$ and
\begin{align*}
f\left(-\frac{b}{2a}\right)&=f\left(-\frac{7}{2}\right)\\
&=\left(-\frac{7}{2}\right)^2+7\left(-\frac{7}{2}\right)-8\\
&=-\frac{81}{4}.
\end{align*}

(b) Find the axis of symmetry.

Solution. The axis of symmetry is the vertical line $x=-\frac{b}{2a}=-\frac{7}{2}$.

(c) Determine whether there is a maximum or minimum value and find that value.

Solution. Since $a=1>0$, there is a minimum and the minimum value is the $y$-coordinate of the vertex $f\left(-\frac{7}{2}\right)=-\frac{81}{4}$.

(d) Graph the function.

Solution. Since $a=1>0$, the graph is a parabola that opens up (smiling face). Also note that the $y$-intercept of $f(x)$ is $-8$. In fact, we can extract more information since $f(x)$ can be easily factored as $(x+8)(x-1)$, so the $x$-intercepts are $x=-8,1$.

MAT 101 Online Lecture Notes: 2.3 Quadratic Equations, Functions and Models

Main topic in this section is solving a quadratic equation $ax^2+bx+c=0$. There are three ways to solve a quadratic equation. The first one is

1. By Factoring: This is a typical method to solve a quadratic equation whenever the polynomial $ax^2+bx+c$ can be easily factored. Here is an example.

Example. Solve the quadratic equation $x^2-3x-4=0$ by factoring.

Solution. The polynomial $x^2-3x-4$ is factored as $(x-4)(x+1)$. So the equation is $(x-4)(x+1)=0$. This means that $x-4=0$ or $x+1=0$, i.e. we obtain two real solutions $x=-1$ or $x=4$.

Example. Solve the quadratic equation $x^2-3=0$.

Solution 1. Recall the factorization formula $(a^2-b^2)=(a+b)(a-b)$. Now
\begin{align*}
x^2-3&=x^2-(\sqrt{3})^2\\
&=(x+\sqrt{3})(x-\sqrt{3}).
\end{align*}
Thus our equation becomes $(x+\sqrt{3})(x-\sqrt{3})=0$ whose solutions are $x=\pm\sqrt{3}$.

Solution 2. The quadratic equation can be written as $x^2=3$. Solving this equation for $x$, we obtain $x=\pm\sqrt{3}$.

Next method is

2. By Completing the Square:

This is a method that can be used to solve any quadratic equation. First note that $$x^2+bx+\left(\frac{b}{2}\right)^2=\left(x+\frac{b}{2}\right)^2.\ \ \ \ \ \mbox{(1)}$$

Example. Solve the equation $x^2-6x-10=0$ by completing the square.

Solution. By adding 10 to each side of the equation, we obtain
$$x^2-6x=10.\ \ \ \ \ \mbox{(2)}$$ Note that half of the coefficient of $x$ is $\frac{-6}{2}=-3$. Add $(-3)^2$ to each side of (2):
$$x^2-6x+(-3)^2=10+(-3)^2.\ \ \ \ \ \mbox{(3)}$$ Now notice that the LHS of (3) is exactly the same form as the LHS of the forumla (1). Hence, the quation (3) becomes
$$(x-3)^2=19.$$ Solving this for $x-3$, we obtain $x-3=\pm\sqrt{19}$. That is, $x=3\pm\sqrt{19}$.

While completing the square can be a useful tool for some other things, I do not strongly recommend this method because there is more convenient method of solving quadratic equations.

3. By Quadratic Formula:

If you apply the method by completing the square to solve the quadratic equation $ax^2+bx+c=0$, we obtain the quadratic formula
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\ \ \ \ \ \mbox{(4)}.$$

Example. Solve the quadratic equation $3x^2+2x-7=0$.

Solution. $a=3$, $b=2$, and $c=-7$. Thus
\begin{align*}
x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
&=\frac{-2\pm\sqrt{2^2-4(3)(-7)}}{2(3)}\\
&=\frac{-1\pm\sqrt{22}}{3}.
\end{align*}

The expression inside radical $b^2-4ac$ is called the discriminant. Using the discriminant, we can tell the following without solving the equation itself.

Theorem. For $ax^2+bx+c=0$ with $a\ne 0$,

  • If $b^2-4ac>0$, then the equation has two distinct real solutions.
  • If $b^2-4ac=0$, then the equation has only one real solution (which is $x=-\frac{b}{2a}$).
  • If $b^2-4ac<0$, then the equation has two complex solutions that are conjugate of each other.

Cylindrical Resonant Cavity

In this lecture, we discuss cylindrical resonant cavity as an example of the applications of Bessel functions.

Recall that electromagnetic waves in vacuum space can be described by the following four equations, called Maxwell’s equations (in vacuum)
\begin{align*}
\nabla\cdot B=0,\\
\nabla\cdot E=0,\\
\nabla\times B=\epsilon_0\mu_o\frac{\partial E}{\partial t},\\
\nabla\times E=-\frac{\partial B}{\partial t},
\end{align*}
where $E$ is the magnetic field and $B$ the magnetic induction, $\epsilon_0$ the electric permittivity, and $\mu_o$ the magnetic permeability.
Now,
\begin{align*}
\nabla\times(\nabla\times E)&=-\frac{\partial}{\partial t}(\nabla\times B)\\
&=-\epsilon_0\mu_0\frac{\partial^2E}{\partial t^2}.
\end{align*}

Resonant cavity is an electromaginetic resonator in which waves oscillate inside a hollow space (device). For more details see Wikipedia entry for Resonator, in particular for Cavity Resonator.

Here we consider a cylindrical resonant cavity. In the interior of a resonant cavity, electromagnetic waves oscillate with a time dependence $e^{-i\omega t}$, i.e. $E(t,x,y,z)$ can be written as $E=e^{-i\omega t}P(x,y,z),$ where $P(x,y,z)$ is a vector-valued function in $\mathbb R^3$. One can easily show that $\frac{\partial^2E}{\partial t^2}=-\omega^2E$ or
$$\nabla\times(\nabla\times E)=\alpha^2E,$$
where $\alpha^2=\epsilon_0\mu_0\omega^2$. On the other hand,
\begin{align*}
\nabla\times(\nabla\times E)&=\nabla\nabla\cdot E-\nabla\cdot\nabla E\\
&=-\nabla^2E.
\end{align*}
Thus, the electric field $E$ satisfies the Helmholtz equation
$$\nabla^2E+\alpha^2E=0.$$
Suppose that the cavity is a cylinder with radius $a$ and height $l$. Without loss of generality we may assume that the end surfaces are at $z=0$ and $z=l$. Let $E=E(\rho,\varphi,z)$. Using separation of variables in cylindrical coordinate system, we find that the $z$-component $E_z(\rho,\varphi,z)$ satisfies the scalar Helmholtz equation
$$\nabla^2E_z+\alpha^2E_z=0,$$
where $\alpha^2=\omega^2\epsilon_0\mu_0=\frac{\omega^2}{c^2}$. The mode of $E_z$ is obtained as

$$(E_z)_{mnk}=\sum_{m,n}J_m(\gamma_{mn}\rho)e^{\pm im\varphi}[a_{mn}\sin kz+b_{mn}\cos kz].\ \ \ \ \ \mbox{(1)}$$ Here $k$ is a separation constant. Consider the boundary conditions: $\frac{\partial E_z}{\partial z}(z=0)=\frac{\partial E_z}{\partial z}(z=l)=0$ and $E_z(\rho=a)=0$. The boundary conditions $\frac{\partial E_z}{\partial z}(z=0)=\frac{\partial E_z}{\partial z}(z=l)=0$ result that $a_{mn}=0$ and $$k=\frac{p\pi}{l},\ p=0,1,2,\cdots.$$ The boundary condition $E_z(\rho=a)=0$ results $$\gamma_{mn}=\frac{\alpha_{mn}}{a},$$ where $\alpha_{mn}$ is the $n$th zero of $J_m$. Thus the mode (1) is written as
$$(E_z)_{mnp}=\sum_{m,n}b_{mn}J_m\left(\frac{\alpha_{mn}}{a}\rho\right)e^{\pm im\varphi}\cos\frac{p\pi}{l}z,\ \ \ \ \ \mbox{(2)}$$ where $p=0,1,2,\cdots$. In physics, the mode (2) is called the transverse magnetic mode or shortly TM mode of oscillation.

We have \begin{align*}\gamma^2&=\alpha^2-k^2\\&=\frac{\omega^2}{c^2}-\frac{p^2\pi^2}{l^2}.\end{align*} Hence the TM mode has resonant frequencies
$$\omega_{mnp}=c\sqrt{\frac{\alpha_{mn}^2}{a^2}+\frac{p^2\pi^2}{l^2}},\ \left\{\begin{aligned}
m&=0,1,2,\cdots\\
n&=1,2,3,\cdots\\
p&=0,1,2,\cdots.\end{aligned}
\right.
$$
For more details about transverse mode, click here and here.

Bessel Functions of the First Kind $J_n(x)$ I: Generating Function, Recurrence Relation, Bessel’s Equation

Let us begin with the generating function

$$g(x,t) = e^\frac{x}{2}\left(t-\frac{1}{t}\right).$$
Expanding this function in a Laurent series, we obtain
$$e^\frac{x}{2}\left(t-\frac{1}{t}\right) = \sum_{n=-\infty}^\infty J_n(x)t^n.$$
The coefficient of $t^n$, $J_n(x)$, is defined to be a Bessel function of the first kind of order $n$.
Now, we determine $J_n(x)$.
\begin{align*}
e^{\frac{x}{2}t}e^{-\frac{x}{2t}}&=\sum_{r=0}^\infty\left(\frac{x}{2}\right)^r\frac{t^r}{r!} \sum_{s=0}^\infty(-1)^s\left( \frac{x}{2}\right)^s \frac{t^{-s}}{s!}\\
&=\sum_{r=0}^\infty\sum_{s=0}^\infty\frac{(-1)^s}{r!s!}\left(\frac{x}{2}\right)^{r+s}t^{r-s}.
\end{align*}
Set $r=n+s$. Then for $n\ge 0$ we obtain
$$J_n(x)=\sum_{s=0}^\infty \frac{(-1)^s}{s!(n+s)!}\left(\frac{x}{2}\right)^{n+2s}.$$

Bessel Functions

Now we find $J_{-n}(x)$. In the above series for $J_n(x)$, we obtain
$$J_{-n}(x)=\sum_{s=0}^\infty\frac{(-1)^s}{s!(s-n)!} \left(\frac{x}{2}\right)^{2s-n}.$$
However, $(s-n)!\rightarrow\infty$ for $s=0,1,\cdots,(n-1)$. So the series may be considered to begin at $s=n$. Replacing $s$ by $s+n$, we obtain
$$J_{-n}(x)=\sum_{s=0}^ \infty\frac{(-1)^{s+n}}{s!(s+n)!}\left( \frac{x}{2} \right)^{n+2s}.$$ Note that $J_n(x)$ and $J_{-n}(x)$ satisfy the relation
$$J_{-n}=(-1)^nJ_n(x).$$

Let us differentiate the generating function $g(x,t)$ with respect to $t$:
\begin{align*}
\frac{\partial g(x,t)}{\partial t} &=\frac{x}{2}\left(1+ \frac{1}{t^2}\right) e^{\frac{x}{2}}\left(t-\frac{1}{t}\right)\\
&=\sum_{n=-\infty}^\infty n J_n(x) t^{n-1}.
\end{align*}
Replace $e^{\frac{x}{2}}\left(t-\frac{1}{t}\right)$ by $\sum_{n=-\infty}^\infty J_n(x) t^n$.
Then
\begin{align*}
\sum_{n=-\infty}^\infty \frac{x}{2} (1+ \frac{1}{t^2}) J_n(x) t^{n}&=\sum_{n=-\infty}^\infty \frac{x}{2} [J_n(x) t^n + J_n(x) t^{n-2}]\\
&=\sum_{n=-\infty}^\infty \frac{x}{2} [J_{n-1}(x) + J_{n+1}(x)] t^{n-1}.
\end{align*}
Thus,
$$\sum_{n=-\infty}^\infty \frac{x}{2} [J_{n-1}(x) + J_{n+1}(x)] t^{n-1}=\sum_{n=-\infty}^\infty n J_n(x) t^n$$
or we obtain the recurrence relation,
$$J_{n-1}(x) + J_{n+1}(x) = \frac{2n}{x} J_n(x).\ \ \ \ \ \mbox{(1)}$$
Now we differentiate $g(x,t)$ with respect to $x$:
\begin{align*}
\frac{\partial g(x,t)}{\partial x}&=\frac{1}{2}\left(1-\frac{1}{t}\right)e^{\frac{x}{2}}\left(t-\frac{1}{t}\right)\\
& =\sum_{n=-\infty}^\infty J_n’(x)t^n.
\end{align*}
This leads to the recurrence relation
$$J_{n-1}(x) – J_{n+1}(x) = 2 J_n’(x).\ \ \ \ \ \mbox{(2)}$$
As a special Case of this recurrence relation, we obtain,
$$J_{0}’(x)=-J_1(x).$$
Adding (1) and (2), we have
$$J_{n-1}(x)=\frac{n}{x}J_n(x) + J_n’(x).\ \ \ \ \ \mbox{(3)}$$
Multiplying (3) by $x^n$:
\begin{align*}
x^n J_{n-1}(x) & = n x^{n-1} J_n(x) + x^n J_n’(x)\\
& = \frac{d}{dx}[ x^n J_n(x)].
\end{align*}
Subtracting (2) from (1), we have
$$J_{n+1}(x) = \frac{n}{x} J_n(x) – J_n’(x).\ \ \ \ \ \mbox{(4)}$$
Multiplying (4) by $-x^{-n}$:
\begin{align*}
-x^{-n} J_{n+1}(x) & = -n x^{-n-1} J_n(x) + x^{-n} J_n’(x)\\
& = \frac{d}{dx}[x^{-n} J_n(x)].
\end{align*}

Using recurrence relations, we can show that the Bessel functions $J_n(x)$ are the solutions of the Bessel’s differential equation. The recurrence relation (3) can be written as
$$x J_n’(x) = x J_{n-1}(x) – n J_n(x).\ \ \ \ \ \mbox{(5)}$$
Differentiating this equation with respect to $x$, we obtain
$$J_n’(x) + x J_n^{\prime\prime}(x) = J_{n-1}(x) + x J_{n-1}’(x) – n J_n’(x)$$
or
$$x J_n^{\prime\prime}(x) + (n+1) J_n’(x) – x J_{n-1}’(x) – J_{n-1}(x) = 0.\ \ \ \ \ \mbox{(6)}$$
Subtracting (5) times $n$ from (6) times $x$ results the equation
$$x^2 J_n^{\prime\prime}(x) + x J_n’(x) – n^2 J_n(x) + x(n-1) J_{n-1}(x) – x^2 J_{n-1}’(x) = 0.\ \ \ \ \ \mbox{(7)}$$
Replace $n$ by $n-1$ in (4) and multiply the resulting equation by $x^2$ to get the equation
$$x^2 J_n(x) = x (n-1) J_{n-1}(x) – x^2 J_{n-1}’(x).\ \ \ \ \ \mbox{(8)}$$
With the equation (8), the equation (7) can be written as
$$x^2 J_n^{\prime\prime}(x) + x J_n’(x) + (x^2 – n^2) J_n(x) = 0.\ \ \ \ \ \mbox{(9)}$$
This is Bessel’s equation. Hence the Bessel functions $J_n(x)$ are the solutions of Bessel’s equation.

Modeling a Vibrating Drumhead III

In the previous discussion, we finally obtained the solution of the vibrating drumhead problem:
$$u(r,\theta,t)=\sum_{n=0}^\infty\sum_{m=1}^\infty J_n(\lambda_{nm}r)\cos(n\theta)[A_{nm}\cos(\lambda_{nm} ct)+B_{nm}\sin(\lambda_{nm}ct)].$$
In this lecture, we determine the Fourier coefficients $A_{nm}$ and $B_{nm}$ using the initial conditions $u(r,\theta,0)$ and $u_t(r,\theta,0)$. Before we go on, we need to mention two types of orthogonalities: the orthogonality of cosine functions and the orthogonality of Bessel functions. First note that
$$\int_0^{2\pi}\cos(n\theta)\cos(k\theta)d\theta=\left\{\begin{array}{ccc}0 & \mbox{if} & n\ne m,\\\pi & \mbox{if} & n=m.\end{array}\right.$$
The reason this property is called an orthogonality is that if $V$ is the set of all (Riemann) integrable real-valued functions on the interval $[a,b]$, then $V$ forms a vector space over $\mathbb R$. This vector space is indeed an inner product space with the inner product $$\langle f,g\rangle=\int_a^bf(x)g(x)dx\ \mbox{for}\ f,g\in V.$$
Bessel functions are orthogonal as well in the following sense:
$$\int_0^1J_n(\lambda_{nm}r)J_n(\lambda_{nl}r)rdr=\left\{\begin{array}{ccc}0 & \mbox{if} & m\ne l,\\\frac{1}{2}[J_{n+1}(\lambda_{nm})]^2 & \mbox{if} & m=l.\end{array}\right.$$

From the solution $u(r,\theta,t)$, we obtain the initial position of the drumhead:
$$u(r,\theta,0)=\sum_n\sum_mJ_n(\lambda_{nm}r)\cos(n\theta)A_{nm}.$$
On the other hand, $u(r,\theta,0)=f(r,\theta)$. Multiply
$$\sum_n\sum_mJ_n(\lambda_{nm}r)\cos(n\theta)A_{nm}=f(r,\theta)$$
by $\cos(k\theta)$ and integrate with respect to $\theta$ from $0$ to $2\pi$:
$$\sum_n\sum_mJ_n(\lambda_{nm}r)A_{nm}\int_0^{2\pi}\cos(n\theta)\cos(k\theta)d\theta=\int_0^{2\pi}f(r,\theta)\cos(k\theta)d\theta.$$ The only nonvanishing term of the above series is when $n=k$, so we obtain
$$\pi\sum_mJ_k(\lambda_{km}r)A_{km}=\int_0^{2\pi}f(r,\theta)\cos(k\theta)d\theta.$$ Multiply this equation by $J_k(\lambda_{kl}r)$ and integrate with respect to $r$ from $0$ to $1$:
$$\pi\sum_mA_{km}\int_0^1J_k(\lambda_{km}r)J_k(\lambda_{kl}r)rdr=\int_0^{2\pi}\int_0^1f(r,\theta)\cos(k\theta)J_k(\lambda_{kl}r)rdrd\theta.$$ The only nonvanishing term of this series is when $m=l$. As a result we obtain:
$$A_{kl}=\frac{1}{\pi L_{kl}}\int_0^{2\pi}\int_0^1f(r,\theta)\cos(k\theta)J_k(\lambda_{kl}r)rdrd\theta$$
or
$$A_{nm}=\frac{1}{\pi L_{nm}}\int_0^{2\pi}\int_0^1f(r,\theta)\cos(n\theta)J_n(\lambda_{nm}r)rdrd\theta,\ n,m=1,2,\cdots$$
where
$$L_{nm}=\int_0^1J_n(\lambda_{nm}r)^2rdr=\frac{1}{2}[J_{n+1}(\lambda_{nm})]^2, n=0,1,2,\cdots.$$
For $n=0$ we obtain
$$A_{0m}\frac{1}{2\pi L_{0m}}\int_0^1f(r,\theta)J_0(\lambda_{0m}r)rdrd\theta,\ m=1,2,\cdots.$$
Using
$$u_t(r,\theta,0)=\sum_n\sum_mJ_n(\lambda_{nm}r)\cos(n\theta)B_{nm}\lambda_{nm}c=g(r,\theta),$$
we obtain
\begin{align*}
B_{nm}&=\frac{1}{c\pi L_{nm}\lambda_{nm}}\int_0^{2\pi}\int_0^1g(r,\theta)\cos(n\theta)J_n(\lambda_{nm}r)rdrd\theta,\ n,m=1,2,\cdots,\\
B_{0m}&=\frac{1}{2c\pi L_{nm}\lambda_{nm}}\int_0^{2\pi}\int_0^1g(r,\theta)J_0(\lambda_{0m}r)rdrd\theta,\ m=1,2,\cdots.
\end{align*}
Unfortunately at this moment I do not know if I can make an animation of the solution using an open source math software package such as Maxima or Sage. I will let you know if I find a way. In the meantime, if any of you have an access to Maple, you can download a Maple worksheet I made here and run it for yourself. In the particular example in the Maple worksheet, I used $f(r,\theta)=J_0(2.4r)+0.10J_0(5.52r)$ and $g(r,\theta)=0$. For an animation of the solution, click here.