# Self-Adjoint Differential Equations II: Hermitian Operators

Let $\mathcal{L}$ be a second-order self-adjoint differential operator. Then $\mathcal{L}u(x)$ may be written as
$$\label{eq:selfadjoint}\mathcal{L}u(x)=\frac{d}{dx}\left[p(x)\frac{du(x)}{dx}\right]+q(x)u(x)$$ as we discussed here. Multiply \eqref{eq:selfadjoint} by $v^\ast$ ($v^\ast$ is the complex conjugate of $v$) and integrate
\begin{align*}
\int_a^bv^\ast\mathcal{L}udx&=\int_a^bv^\ast\frac{d}{dx}\left[p(x)\frac{du(x)}{dx}\right]dx+\int_a^bv^\ast qudx\\
&=\int_a^bv^\ast d\left[p(x)\frac{du(x)}{dx}\right]+\int_a^bv^\ast qudx\\
&=v^\ast p\frac{du}{dx}|_a^b-\int_a^b {v^\ast}^\prime pu’dx+\int_a^bv^\ast qudx
\end{align*}
We may impose
$$\label{eq:bc}v^\ast p\frac{du}{dx}|_a^b=0$$
as a boundary condition.
\begin{align*}
-\int_a^b {v^\ast}^\prime pu’dx&=-\int_a^b {v^\ast}^\prime pdu\\
&=-{v^\ast}^\prime pu|_a^b+\int_a^b u(p{v^\ast}^\prime)’dx
\end{align*}
We may also impose
$$\label{eq:bc2}-{v^\ast}^\prime pu|_a^b=0$$
as a boundary condition. Then
\begin{align*}
\int_a^bv^\ast\mathcal{L}udx&=\int_a^b u(p{v^\ast}^\prime)’dx+\int_a^bv^\ast qudx\\
&=\int_a^b u\mathcal{L}v^\ast dx
\end{align*}

Definition. A self-adjoint operator $\mathcal{L}$ is called a Hermitian operator with respect to the functions $u(x)$ and $v(x)$ if

$$\label{eq:hermitian}\int_a^bv^\ast\mathcal{L}udx=\int_a^b u\mathcal{L}v^\ast dx$$

That is, a self-adjoint operator $\mathcal{L}$ which satisfies the boundary conditions \eqref{eq:bc} and \eqref{eq:bc2} is a Hermitian operator.

Hermitian Operators in Quantum Mechanics

In quantum mechanics, the differential operators need to be neither second-order nor real. For example, the momentum operator is given by $\hat p=-i\hbar\frac{d}{dx}$. Therefore we need an extended notion of Hermitian operators in quantum mechanics.

Definition. The operator $\mathcal{L}$ is Hermitian if
$$\label{eq:hermitian2}\int \psi_1^\ast\mathcal{L}\psi_2 d\tau=\int(\mathcal{L}\psi_1)^\ast\psi_2 d\tau$$
Note that \eqref{eq:hermitian2} coincides with \eqref{eq:hermitian} if $\mathcal{L}$ is real. In terms of Dirac’s braket notation \eqref{eq:hermitian2} can be written as
$$\langle\psi_1|\mathcal{L}\psi_2\rangle=\langle\mathcal{L}\psi_1|\psi_2\rangle$$

The adjoint operator $A^\dagger$ of an operator $A$ is defined by
$$\label{eq:adjoint}\int \psi_1^\ast A^\dagger \psi_2 d\tau=\int(A\psi_1)^\ast\psi_2 d\tau$$ Again in terms of Dirac’s braket notation \eqref{eq:adjoint} can be written as
$$\langle\psi_1|A^\dagger\psi_2\rangle=\langle A\psi_1|\psi_2\rangle$$
If $A=A^\dagger$ then $A$ is said to be self-adjoint. Clearly, self-adjoint operators are Hermitian operators. However the converse need not be true. Although we will not delve into this any deeper here, the difference is that Hermitian operators are always assumed to be bounded while self-adjoint operators are not necessarily restricted to be bounded. That is, bounded self-adjoint operators are Hermitian operators. Physicists don’t usually distinguish self-adjoint operators and Hermitian operators, and often they mean self-adjoint operators by Hermitian operators. In quantum mechanics, observables such as position, momentum, energy, angular momentum are represented by (Hermitian) linear operators and the measurements of observables are given by the eigenvalues of linear operators. Physical observables are regarded to be bounded and continuous, because the measurements are made in a laboratory (so bounded) and points of discontinuity are mathematical points and nothing smaller than the Planck length can be observed. As well-known any bounded linear operator defined on a Hilbert space is continuous.

For those who are interested: This may cause a notational confusion, but in mathematics the complex conjugate $a^\ast$ is replaced by $\bar a$ and the adjoint $a^\dagger$ is replaced by $a^\ast$. Let $\mathcal{H}$ be a Hilbert space. By the Riesz Representation Theorem, it can be shown that for any bounded linear operator $a:\mathcal{H}\longrightarrow\mathcal{H}’$, there exists uniquely a bounded linear operator $a^\ast: \mathcal{H}’\longrightarrow\mathcal{H}$ such that
$$\langle a^\ast\eta|\xi\rangle=\langle\eta|a\xi\rangle$$ for all $\xi\in\mathcal{H}$, $\eta\in\mathcal{H}’$. This $a^\ast$ is defined to be the adjoint of the bounded operator $a$. ${}^\ast$ defines an involution on $\mathcal{B}(\mathcal{H})$, the set of all bounded lineart operators of $\mathcal{H}$ and $\mathcal{B}(\mathcal{H})$ with ${}^\ast$ becomes a C${}^\ast$-algebra. In mathematical formulation of quantum mechanics, observables are represented by self-adjoint operators of the form $a^\ast a$, where $a\in\mathcal{B}(\mathcal{H})$. Note that $a^\ast a$ is positive i.e. its eigenvalues are non-negative.

Definition. The expectation value of an operator $\mathcal{L}$ is
$$\langle\mathcal{L}\rangle=\int \psi^\ast\mathcal{L}\psi d\tau$$
$\langle\mathcal{L}\rangle$ corresponds to the result of a measurement of the physical quantity represented by $\mathcal{L}$ when the physical system is in a state described by $\psi$. The expectation value of an operator should be real and this is guaranteed if the operator is Hermitian. To see this suppose that $\mathcal{L}$ is Hermitian. Then
\begin{align*}
\langle\mathcal{L}\rangle^\ast&=\left[\int \psi^\ast\mathcal{L}\psi d\tau\right]^\ast\\
&=\int\psi\mathcal{L}^\ast\psi^\ast d\tau\\
&=\int(\mathcal{L}\psi)^\ast\psi d\tau\\
&=\int\psi^\ast\mathcal{L}\psi d\tau\ (\mbox{since $\mathcal{L}$ is Hermitian})\\
&=\langle\mathcal{L}\rangle
\end{align*}
That is, $\langle\mathcal{L}\rangle$ is real.

There are three important properties of Hermitian (self-adjoint) operators:

• The eigenvalues of a Hermitian operator are real.
• The eigenfunctions of a Hermitian operator are orthogonal.
• The eigenfunctions of a Hermitian operator form a complete set.

References:

1. G. Arfken, Mathematical Methods for Physicists, 3rd Edition, Academic Press 1985
2. W. Greiner, Quantum Mechanics, An Introduction, 4th Edition, Springer-Verlag 2001
3. P. Szekeres, A Course in Modern Mathematical Physics: Groups, Hilbert Space and Differential Geometry, Cambridge University Press 2004

Let $\mathcal{L}$ be the second-order linear differential operator
$$\mathcal{L}=p_0(x)\frac{d^2}{dx^2}+p_1(x)\frac{d}{dx}+p_2(x)$$
which acts on a function $u(x)$ as
$$\label{eq:ldo}\mathcal{L}u(x)=p_0(x)\frac{d^2u(x)}{dx^2}+p_1(x)\frac{du(x)}{dx}+p_2(x)u(x).$$

Define an adjoint operator $\bar{\mathcal{L}}$ by
\begin{align*}
\bar{\mathcal{L}}&:=\frac{d^2}{dx^2}[p_0u]-\frac{d}{dx}[p_1u]+p_2u\\
&=p_0\frac{d^2u}{dx^2}+(2p_0^\prime-p_1)\frac{du}{dx}+(p_0^{\prime\prime}-p_1^\prime+p_2)u.
\end{align*}
If $\mathcal{L}=\bar{\mathcal{L}}$, $\mathcal{L}$ is said to be self-adjoint. One can immediately see that $\mathcal{L}=\bar{\mathcal{L}}$ if and only if $$\label{eq:self-adjoint}p_0^\prime=p_1.$$ Let $p(x)=p_0(x)$ and $q(x)=p_2(x)$. Then
\begin{align*}
\mathcal{L}=\bar{\mathcal{L}}&=p\frac{d^2u}{dx^2}+\frac{dp}{dx}\frac{du}{dx}+qu\\
&=\frac{d}{dx}\left[p(x)\frac{du(x)}{dx}\right]+qu(x).
\end{align*}
Note that one can transform a non-self-adjoint 2nd-order linear differential operator to a self-adjoint one. The idea is similar to that of finding a integrating factor to transform a non-separable first-order linear differential equation to a separable one.

Suppose that \eqref{eq:ldo} is not self-adjoint, i.e. $p_1\ne p_0′$. Multiply $\mathcal{L}$ by $\frac{f(x)}{p_0(x)}$. Then
$$\mathcal{L}’:=\frac{f}{p_0}\mathcal{L}=f\frac{d^2u}{dx^2}+f\frac{p_1}{p_0}\frac{du}{dx}+f\frac{p_2}{p_0}u.$$
Suppose $\mathcal{L}’$ is self-adjoint. Then by \eqref{eq:self-adjoint}
$$f’=f\frac{p_1}{p_0}.$$
That is,
$$f(x)=\exp\left[\int^x\frac{p_(t)}{p_0(t)}dt\right].$$
If $p_1=p_0′$, then
\begin{align*}
\frac{f(x)}{p_0}&=\frac{1}{p_0}\exp\left[\int^x\frac{p_1}{p_0}dt\right]\\
&=\frac{1}{p_0}\exp\left[\int^x\frac{p_0^\prime}{p_0}dt\right]\\
&=\frac{1}{p_0}\exp(\ln p_0(x))\\
&=\frac{1}{p_0(x)}\cdot p_0\\
&=1
\end{align*}
i.e. $f(x)=p_0(x)$ as expected.

Eigenfunctions, Eigenvalues

From separation of variables or directly from a physical problem, we have second-order linear differential equation of the form
$$\label{eq:sl}\mathcal{L}u(x)+\lambda w(x)u(x)=0,$$
where $\lambda$ is a constant and $w(x)>0$ is a function called a density or weighting function. The constant $\lambda$ is called an eigenvalue and $u(x)$ is called an eigenfunction.

Example. [Schrödinger Equation]

The Schrödinger equation
$$H\psi=E\psi$$
is of the form \eqref{eq:sl}. Recall that $H$ is the Hamiltonian operator
$$H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)$$
where $V(x)$ is a potential. So $H$ is a second-order linear differential operator. The weight function $w(x)=-1$ and $E$ is energy as an eigenvalue. Clearly Schrödinger equation is self-adjoint.

Example. [Legendre's Equations]

Legendre’s equation
$$(1-x^2)y^{\prime\prime}-2xy’+n(n+1)y=0$$ is of the form \eqref{eq:sl}, where $\mathcal{L}y=(1-x^2)y^{\prime\prime}-2xy’$, $w(x)=1$, and $\lambda=n(n+1)$. Since $p_0^\prime=-2x=p_1$, Legendre’s equations are self-adjoint.

# MAT 101 Online Lecture Notes: 3.2 Graphing Polynomial Functions

Polynomial functions have the following important property:

Every polynomial function of degree $n$ has at most $n$ real zeros.”

This property is called the Fundamental Theorem of Algebra.

As an application of this property, we see that a polynomial function of degree $n$ can have at most $n$ $x$-intercepts and at most $(n-1)$ turning points (local maximum and local minimum values).

Example. The function $f(x)=x^4-7x^3+12x^2+4x-16$ has three turning points that are two local minimum values and one local maximum value.

Graphing a Polynomial Function $p(x)$

1. First find all zeros of $p(x)$
2. Considering the even or odd multiplicity of each factor of $p(x)$, we can see the graph is crossing or touching the $x$-axis at each zero.
3. Use the leading-term test to determine the end behavior.
4. Use the $y$-intercept.

Example. Consider $f(x)=x^4-7x^3+12x^2+4x-16$. It can be factored as $f(x)=(x+1)(x-2)^2(x-4)$. (At this moment you don’t have to worry about how we get the facotring. We will discuss this in Sections 3.3 and 3.4.) So we find three zeros $-1$, $2$ (with multiplicity 2), and $4$. We can tell that the graph crosses at $-1$ and $4$ and touches the graph without crossing at $2$. Since the degree is $4$, an even number the graph goes up when $x\to -\infty$ and $x\to\infty$. These findings are all featured in the above graph.

The Intermediate Value Theorem

Let $p(x)$ be a polynomial (with real coefficients). Suppose that $p(a)$ and $p(b)$ have different signs for two distinct numbers $a$ and $b$. Then the graph of $p(x)$ must cross the $x$-axis between $a$ and $b$, i.e. $p(x)$ must have a zero between $a$ and $b$. This property is called the Intermediate Value Theorem.

Example. (a) Use the Intermediate Value Theorem to determine if
$$f(x)=x^3+3x^2-9x-13$$
has a zero between $a=1$ and $b=2$.

Solution. All you have to do is to evaluate $f(x)$ at $x=1$ and $x=2$, i.e. calculate $f(1)$ and $f(2)$ and see if they are different.
\begin{align*}
f(1)&=(1)^3+3(1)^2-9(1)-13=-18,\\
f(2)&=(2)^3+3(2)^2-9(2)-13=-11.
\end{align*}
$f(1)$ and $f(2)$ have the same sign, so the Intermediate Value Theorem won’t tell if $f(x)$ has a zero between $1$ and $2$. The following graph shows that it actually does not.

(b) Does $f(x)$ has a zero between $a=-5$ and $b=-4$?

Solution. $f(-5)=-18$ and $f(-4)=7$. Since their signs are different, by the Intermediate Value Theorem, there must be a zero between $-5$ and $-4$. The following graph confirms it.

# Legendre Functions III: Special Values, Parity, Orthogonality

Special Values

From the generating function
$$g(x,t)=\frac{1}{(1-2xt+t^2)^{1/2}},$$
when $x=1$ we obtain
\begin{align*}
g(1,t)&=\frac{1}{(1-2t+t^2)^{1/2}}\\
&=\frac{1}{1-t}\\
&=\sum_{n=0}^\infty t^n,
\end{align*}
since $|t|<1$. On the other hand,
$$g(1,t)=\sum_{n=0}^\infty P_n(1)t^n.$$
So by comparison we get
$$P_n(1)=1.$$ Similarly, if we let $x=-1$,
$$P_n(-1)=(-1)^n.$$
For $x=0$, the generating function results
$$(1+t^2)^{-1/2}=1-\frac{1}{2}t^2+\frac{3}{8}t^4+\cdots+(-1)^n\frac{1\cdot 3\cdots (2n-1)}{2^nn!}t^{2n}+\cdots.$$
Thus we obtain
\begin{align*}
P_{2n}(0)&=(-1)^n\frac{1\cdot 3\cdots (2n-1)}{2^nn!}=(-1)^n\frac{(2n-1)!!}{(2n)!!},\\
P_{2n+1}(0)&=0,\ n=0,1,2,\cdots.
\end{align*}
Recall that the double factorial !! is defined by
\begin{align*}
(2n)!!&=2\cdot 4\cdot 6\cdots (2n),\\
(2n-1)!!&=1\cdot 3\cdot 5\cdots (2n-1).
\end{align*}

Parity

$g(t,x)=g(-t,-x)$, that is
$$\sum_{n=0}^\infty P_n(x)t^n=\sum_{n=0}^\infty P_n(-x)(-t)^n$$
which results the parity
$$P_n(-x)=(-1)^nP_n(x).\ \ \ \ \ (1)$$
(1) tells that if $n$ is even, $P_n(x)$ is an even function and if $n$ is odd, $P_n(x)$ is an odd function.

Orthogonality

Multiply the Legendre’s diferential equation
$$\frac{d}{dx}[(1-x^2)P_n'(x)]+n(n+1)P_n(x)=0\ \ \ \ \ (2)$$ by $P_m(x)$.
$$P_m(x)\frac{d}{dx}[(1-x^2)P_n'(x)]+n(n+1)P_m(x)P_n(x)=0.\ \ \ \ \ (3)$$
Replace $n$ by $m$ in (2) and then multiply the resulting equation by $P_n(x)$.
$$P_n(x)\frac{d}{dx}[(1-x^2)P_m'(x)]+m(m+1)P_m(x)P_n(x)=0.\ \ \ \ \ (4)$$
Subtract (4) from (3) and integrate the resulting equation with respect to $x$ from $-1$ to 1.
\begin{align*}
\int_{-1}^1&\left\{P_m(x)\frac{d}{dx}[(1-x^2)P_n'(x)]-P_n(x)\frac{d}{dx}[(1-x^2)P_m'(x)]\right\}dx\\
&=[m(m+1)P_m(x)P_n(x)-n(n+1)P_m(x)P_n(x)].\end{align*}
Using integration by parts,
\begin{align*}
\int_{-1}^1P_m(x)\frac{d}{dx}[(1-x^2)P_n'(x)]dx&=\\&(1-x^2)P_m(x)P_n’(x)|_{-1}^1-\int_{-1}^1P_m(x)P_n(x)dx\\
&=-\int_{-1}^1P_m(x)P_n(x)dx.
\end{align*}
Since the integration of the second term inside $\{\ \ \}$ would have the same value, the LHS vanishes.
Hence for $m\ne n$,
$$\int_{-1}^1P_m(x)P_n(x)dx=0.\ \ \ \ \ (5)$$
That is, $P_m(x)$ and $P_n(x)$ are orthogonal for the interval $[-1,1]$.
For $x=\cos\theta$, the orthogonality (5) is given by
$$\int_0^\pi P_n(\cos\theta)P_m(\cos\theta)\sin\theta d\theta=0.$$

Integrate
$$(1-2xt+t^2)^{-1}=\left[\sum_{n=0}^\infty P_n(x)t^n\right]^2$$
with respect to $x$ from $-1$ to $1$. Due to the orthogonality (5), the integration of all the crossing terms in the RHS will vanish, and so we obtain
$$\int_{-1}^1\frac{dx}{1-2xt+t^2}=\sum_{n=0}^\infty \left\{\int_{-1}^1[P_n(x)]^2dx\right\}t^{2n}.$$
\begin{align*}
\int_{-1}^1\frac{dx}{1-2xt+t^2}&=\frac{1}{2t}\int_{(1-t)^2}^{(1+t)^2}\frac{dy}{y}\\
&=\frac{1}{t}\ln\left(\frac{1+t}{1-t}\right)\\
&=\sum_{n=0}^\infty\frac{2}{2n+1}t^{2n}\ (\mbox{since $|t|<1$}).
\end{align*}
Therefore we have the normalizer of Legendre polynomial $P_n(x)$
$$\int_{-1}^1[P_n(x)]^2dx=\frac{2}{2n+1}.$$

Expansion of Functions

Suppose that
$$\sum_{n=0}^\infty a_nP_n(x)=f(x).\ \ \ \ \ (6)$$
Multiply (6) by $P_m(x)$ and integrate with respect to $x$ from $-1$ to 1:
$$\sum_{n=0}^\infty a_n\int_{-1}^1 P_n(x)P_m(x)dx=\int_{-1}^1f(x)P_m(x)dx.$$
By the orthogonality (5), we obtain
$$\frac{2}{2m+1}a_m=\int_{-1}^1f(x)P_m(x)dx\ \ \ \ \ (7)$$
and hence $f(x)$ can be written as
$$f(x)=\sum_{n=0}^\infty\frac{2n+1}{2}\left(\int_{-1}^1 f(t)P_m(t)dt\right)P_n(x).\ \ \ \ \ (8)$$
This expansion in a series of Legendre polynomials is called a Legendre series. Clearly if $f(x)$ is continuous (or integrable) on the interval $[-1,1]$, it can be expanded as a Legendre series.

(7) can be considered as an integral transform, a finite Legendre transform and (8) can be considered as the inverse transform.

Let us consider the integral operator
$$\mathcal{P}_m:=P_m(x)\frac{2m+1}{2}\int_{-1}^1P_m(t)[\ \cdot\ ]dt.\ \ \ \ \ (9)$$
Then
$$\mathcal{P}_mf(t)=a_mP_m(x).$$
The operator (9) projects out the $m$th component of the function $f(x)$.

# Structural Equations

Definition. The dual 1-forms $\theta_1,\theta_2,\theta_3$ of a frame $E_1,E_2,E_3$ on $\mathbb{E}^3$ are defined by
$$\theta_i(v)=v\cdot E_i(p),\ v\in T_p\mathbb{E}^3.$$
Clearly $\theta_i$ is linear.

Example. The dual 1-forms of the natural frame $U_1,U_2,U_3$ are $dx_1$, $dx_2$, $dx_3$ since
$$dx_i(v)=v_i=v\cdot U_i(p)$$
for each $v\in T_p\mathbb{E}^3$.

For any vector field $V$ on $\mathbb{E}^3$,
$$V=\sum_i\theta_i(V)E_i.$$
To see this, let us calculate for each $V(p)\in T_p\mathbb{E}^3$
\begin{align*}
\sum_i\theta_i(V(p))E_i(p)&=\sum_i(V(p)\cdot E_i(p))E_i(p)\\
&=\sum_iV_i(p)E_i(p)\\
&=V(p).
\end{align*}

Lemma. Let $\theta_1,\theta_2,\theta_3$ be the dual 1-forms of a frame $E_1, E_2, E_3$. Then any 1-form $\phi$ on $\mathbb{E}^3$ has a unique expression
$$\phi=\sum_i\phi(E_i)\theta_i.$$

Proof. Let $V$ be any vector field on $\mathbb{E}^3$. Then
\begin{align*}
\sum_i\phi(E_i)\theta_i(V)&=\sum_i\phi(E_i)\theta_i(V)\\
&=\phi(\sum_i\theta_i(V)E_i)\ \mbox{by linearity of $phi$}\\
&=\phi(V).
\end{align*}
Let $A=(a_{ij})$ be the attitude matrix of a frame field $E_1$, $E_2$, $E_3$, i.e.
$$E_i=\sum_ja_{ij}U_j,\ i=1,2,3.\ \ \ \ \ \mbox{(1)}$$
Clearly $\theta_i=\sum_j\theta_i(U_j)dx_j$. On the other hand,
$$\theta_i(U_j)=E_i\cdot U_j=\left(\sum_ka_{ik}U_k\right)\cdot U_j=a_{ij}.$$ Hence the dual formulation of (1) is
$$\theta_i=\sum_ja_{ij}dx_j.\ \ \ \ \ \mbox{(2)}$$

Theorem. [Cartan Structural Equations] Let $E_1$, $E_2$, $E_3$ be a frame field on $\mathbb{E}^3$ with dual 1-forms $\theta_1$, $\theta_2$, $\theta_3$ and connection forms $\omega_{ij}$, $i,j=1,2,3$. Then

1. The First Structural Equations: $$d\theta_i=\sum_j\omega_{ij}\wedge\theta_j.$$
2. The Second Structural Equations: $$d\omega_{ij}=\sum_k\omega_{ik}\wedge\omega_{kj}.$$

Proof. The exterior derivative of (2) is
$$d\theta_i=\sum_jda_{ij}\wedge dx_j.$$ Since $\omega=dA\cdot{}^tA$ and ${}^tA=A^{-1}$ (recall that $A$ is an orthogonal matrix), $dA=\omega\cdot A$, i.e.
$$da_{ij}=\sum_k\omega_{ik}a_{kj}.$$
So,
\begin{align*}
d\theta_i&=\sum_j\left\{\left(\sum_k\omega_{ik}a_{kj}\right)\wedge dx_j\right\}\\
&=\sum_k\left\{\omega_{ik}\wedge\sum_j a_{kj}dx_j\right\}\\
&=\sum_k\omega_{ik}\wedge\theta_k.
\end{align*}

From $\omega=dA\cdot{}^tA$,
$$\omega_{ij}=\sum_kda_{ik}a_{jk}.\ \ \ \ \ \mbox{(3)}$$
The exterior derivative of (3) is
\begin{align*}
d\omega_{ij}&=\sum_k da_{jk}\wedge d_{ik}\\
&=-\sum_k da_{ik}\wedge da_{jk},
\end{align*}
i.e.
\begin{align*}
d\omega&=-dA\wedge{}^t(dA)\\
&=-(\omega\cdot A)\cdot({}^tA\cdot{}^t\omega)\\
&=-\omega\cdot (A\cdot{}^tA)\cdot{}^t\omega\\
&=-\omega\cdot{}^t\omega\ \ \ (A\cdot{}^tA=I)\\
&=\omega\cdot\omega.\ \ \ (\mbox{$\omega$ is skew-symmetric.})
\end{align*}
This is equivalent to the second structural equations.

Example. [Structural Equations for the Spherical Frame Field] Let us first calculate the dual forms and connection forms.

From the spherical coordinates
\begin{align*}
x_1&=\rho\cos\varphi\cos\theta,\\
x_2&=\rho\cos\varphi\sin\theta,\\
x_3&=\rho\sin\varphi,
\end{align*}
we obtain differentials
\begin{align*}
dx_1&=\cos\varphi\cos\theta d\rho-\rho\sin\varphi\cos\theta d\varphi-\rho\cos\varphi\sin\theta d\theta,\\
dx_2&=\cos\varphi\sin\theta d\rho-\rho\sin\varphi\sin\theta d\varphi+\rho\cos\varphi\cos\theta d\theta,\\
dx_3&=\sin\varphi d\rho+\rho\cos\varphi d\varphi.
\end{align*}
From the spherical frame field $F_1$, $F_2$, $F_3$ discussed here, we find its attitude matrix
$$A=\begin{pmatrix} \cos\varphi\cos\theta & \cos\varphi\sin\theta & \sin\varphi\\ -\sin\theta & \cos\theta & 0\\ -\sin\varphi\cos\theta & -\sin\varphi\sin\theta & \cos\varphi \end{pmatrix}.$$
Thus by (2) we find the dual 1-forms
\begin{align*}
\begin{pmatrix}
\theta_1\\
\theta_2\\
\theta_3
\end{pmatrix}&=\begin{pmatrix}
\cos\varphi\cos\theta & \cos\varphi\sin\theta & \sin\varphi\\
-\sin\theta & \cos\theta & 0\\
-\sin\varphi\cos\theta & -\sin\varphi\sin\theta & \cos\varphi
\end{pmatrix}\begin{pmatrix}
dx_1\\
dx_2\\
dx_3
\end{pmatrix}\\
&=\begin{pmatrix}
d\rho\\
\rho\cos\theta d\theta\\
\rho d\varphi
\end{pmatrix}.
\end{align*}
\begin{align*}
&dA=\\
&\begin{bmatrix}
-\sin\varphi\cos\theta d\varphi-\cos\varphi\sin\theta d\theta & -\sin\varphi\sin\theta d\varphi+\cos\varphi\cos\theta d\theta & \cos\varphi d\varphi\\
-\cos\theta d\theta & -\sin\theta d\theta & 0\\
-\cos\varphi\cos\theta d\varphi+\sin\varphi\sin\theta d\theta & -\cos\varphi\sin\theta d\varphi-\sin\varphi\sin\theta d\theta & -\sin\varphi d\varphi
\end{bmatrix}\end{align*}
and so,
\begin{align*}
\omega&=\begin{pmatrix}
0 & \omega_{12} & \omega_{13}\\
-\omega_{12} & 0 & \omega_{23}\\
-\omega_{13} & -\omega_{23} & 0
\end{pmatrix}\\
&=dA\cdot{}^tA\\
&=\begin{pmatrix}
0 & \cos\varphi d\theta & d\varphi\\
-\cos\varphi d\theta & 0 & \sin\varphi d\theta\\
-d\varphi & -\sin\varphi d\theta & 0
\end{pmatrix}.
\end{align*}
From these dual 1-forms and connections forms one can immediately verify the first and the second structural equations.

# MAT 101 Online Lecture Notes: 3.1 Polynomial Functions and Models

A polynomial is a function of the form
$$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0.$$ The number $n$ is called the degree of the polynomial $P(x)$. The term $a_nx^n$ is called the leading term and $a_n$ is called the leading coefficient. The number $a_0$ is called the constant term. $P(x)$ is called linear if $n=1$, quadratic if $n=2$, cubic if $n=3$, quartic if $n=4$, quintic if $n=5$, sextic if $n=6$, septic if $n=7$, and so on so forth. You don’t really have to worry about memorizing these jargons. Names are less important. However you need to remember at least what the degree is and what the leading coefficient is.

There is a pattern for the long term behavior of a polynomial, i.e. the behavior of a polynomial when $x\to\infty$ or $x\to -\infty$. The behavior can be characterized as follows.

• $n=\mbox{even}$ and $a_n>0$:

Example. $f(x)=3x^4-2x^3+3$

• $n=\mbox{even}$ and $a_n<0$:

Example. $f(x)=-x^6+x^5-4x^3$

• $n=\mbox{odd}$ and $a_n>0$:

Example. $f(x)=x^5+\frac{1}{4}x+1$

• $n=\mbox{odd}$ and $a_n<0$:

Example. $f(x)=-5x^3-x^2+4x+2$

Finding zeros of a polynomial $P(x)$

By factoring, solve the equation $P(x)=0$. The solutions are the zeros of $P(x)$.

Example. Find the zeros of $P(x)=x^3+2x^2-5x-6$.

Solution. \begin{align*}
P(x)&=x^3+2x^2-5x-6\\
&=(x^3+x^2)+(x^2-5x-6)\ \mbox{(grouping)}\\
&=x^2(x+1)+(x-6)(x+1)\\
&=(x+1)(x^2+x-6)\\
&=(x+1)(x+3)(x-2).
\end{align*}
Hence, $P(x)$ has zeros $x=-3,-1,2$.

How do we determine whether $x=a$ is a zero of a polynomial $P(x)$?

To only check whether $x=a$ is a zero of $P(x)$, you don’t really have to factor $P(x)$. This is what you need to know. If $P(a)=0$, then $x=a$ is a zero of the polynomial $P(x)$.

Example. Consider $P(x)=x^3+x^2-17x+15$. Determine whether each of numbers 2 and $-5$ is a zero of $P(x)$.

Solution. $P(2)=(2)^3+(2)^2-17(2)+15=-7$, so $x=2$ is not a zero. $P(-5)=(-5)^3+(-5)^2-17(-5)+15=0$, so $x=-5$ is a zero of $P(x)$.

Even and Odd Multiplicity

Even and odd multiplicity is an important property for sketching the graph of a polynomial function. Suppose that $k$ is the largest integer such that $(x-c)^k$ is a factor of $P(x)$. The number $k$ is called the multiplicity of the factor $x-c$.

1. If $k$ is odd, the graph of $P(x)$ crosses the $x$-axis at $(c,0)$.
2. If $k$ is even, then the graph of $P(x)$ is tangent to the $x$-axis, i.e. touches the $x$-axis without crossing at $(c,0)$.

Example. Consider $f(x)=x^2(x+3)^2(x-4)(x+1)^4$. The factors $x$ and $x+3$ have multiplicity 2 and the factor $x+1$ has multiplicity 4. Hence the graph of $f(x)$ touches the $x$-axis without crossing at $x=0$, $x=-3$ and $x=-1$. The factor $x-4$ has multiplicity 1, so the graph crosses the $x$-axis at $x=4$. This is also shown in the following figure.

# Legendre Functions II: Recurrence Relations and Special Properties

In this lecture, we derive some important recurrence relations of Legendre functions and use them to show that Legendre functions are indeed solutions of a differential equation, called Legendre’s differential equation.

Differentiating the generating function
$$g(x,t)=(1-2xt+t^2)^{-1/2}=\sum_{n=0}^\infty P_n(x)t^n,\ |t|<1\ \ \ \ \ \mbox{(1)}$$
with respect to $t$, we get
\begin{align*}
\frac{\partial g(x,t)}{\partial t}&=\frac{x-t}{(1-2xt+t^2)^{3/2}}\ \ \ \ \ \mbox{(2)}\\&=\sum_{n=0}^\infty nP_n(x)t^{n-1}.\ \ \ \ \ \mbox{(3)}\end{align*}
(2) can be written as
$$\frac{x-t}{(1-2xt+t^2)(1-2xt+t^2)^{1/2}}=\frac{(x-t)(1-2xt+t^2)^{-1/2}}{1-2xt+t^2}.$$
By (1) and (3), we obtain
$$(x-t)\sum_{n=0}^\infty P_n(x)t^n=(1-2xt+t^2)\sum_{n=0}^\infty nP_n(x) t^{n-1}$$ or
$$(1-2xt+t^2)\sum_{n=0}^\infty nP_n(x) t^{n-1}+(t-x)\sum_{n=0}^\infty P_n(x)t^n=0$$
which can be written out as
\begin{align*}
\sum_{n=0}^\infty nP_n(x)t^{n-1}-\sum_{n=0}^\infty &2xnP_n(x)t^n+\sum_{n=0}^\infty nP_n(x)t^{n+1}\\&+\sum_{n=0}^\infty P_n(x)t^{n+1}-\sum_{n=0}^\infty xP_n(x)t^n=0.\ \ \ \ \ \mbox{(4)}\end{align*}
In (4) replace $n$ by $n+1$ in the first term, and replace $n$ by $n-1$ in the third and fourth term. Then (4) becomes
\begin{align*}
\sum_{n=0}^\infty (n+1)P_{n+1}(x)t^n-\sum_{n=0}^\infty &2xnP_n(x)t^n+\sum_{n=0}^\infty (n-1)P_{n-1}(x)t^n\\&+\sum_{n=0}^\infty P_{n-1}(x)t^n-\sum_{n=0}^\infty xP_n(x)t^n=0.
\end{align*}
This can be simplified to
$$\sum_{n=0}^\infty[(n+1)P_{n+1}(x)-(2n+1)xP_n(x)+nP_{n-1}(x)]t^n=0$$
which implies that
$$(2n+1)xP_n(x)=(n+1)P_{n+1}(x)+nP_{n-1}(x).\ \ \ \ \ \mbox{(5)}$$
The recurrence relation (5) can be used to calculate Legendre polynomials. For example, we found $P_0(x)=1$ and $P_1(x)=x$ here. For $n=1$, (5) is
$$3xP_1(x)=2P_2(x)+P_0(x)$$
i.e.
$$P_2(x)=\frac{1}{2}(3x^2-1).$$
Continuing this using the recurrence relation (5), we obtain
\begin{align*}
P_3(x)&=\frac{1}{2}(5x^3-3x),\\
P_4(x)&=\frac{1}{8}(35x^4-30x^2+3),\\
P_5(x)&=\frac{1}{8}(63x^5-70x^3+15x),\\
\cdots.
\end{align*}
A great advantage of having the recurrence relation (5) is that one can easily calculate Legendre polynomials using a computer with a simple programming. This can be easily done for instance in Maxima.

Let us load the following simple program to run the recurrence relation (5).

(%i1) Legendre(n,x):=block ([],
if n = 0 then 1
else
if n = 1 then x
else  ((2*n – 1)*x*Legendre(n – 1, x)-(n – 1)*Legendre(n – 2,x))/n);

(%o1) Legendre(n, x) := block([], if n = 0 then 1
else (if n = 1 then x else ((2 n – 1) x Legendre(n – 1, x)
- (n – 1) Legendre(n – 2, x))/n))

Now we are ready to calculate Legendre polynomials. For example, let us calculate $P_3(x)$.

(%i2) Legendre(3,x);

The output is not exactly what we may like because it is not simplified.

In Maxima, simplification can be done by the command ratsimp.

(%i3) ratsimp(Legendre(3,x));

The output is

That looks better. Let us calculate one more, say $P_4(x)$.

Now we differentiate $g(x,t)$ with respect to $x$.
$$\frac{\partial g(x,t)}{\partial x}=\frac{t}{(1-2xt+t^2)^{3/2}}=\sum_{n=0}^\infty P_n’(x)t^n.$$
From this we obtain
$$(1-2xt+t^2)\sum_{n=0}^\infty P_n’(x)t^n-t\sum_{n=0}^\infty P_n(x)t^n=0$$
$$P_{n+1}’(x)+P_{n-1}’(x)=2xP_n’(x)+P_n(x).\ \ \ \ \ \mbox{(6)}$$
Add 2 times $\frac{d}{dx}(5)$ to $2n+1$ times (6). Then we get
$$(2n+1)P_n=P_{n+1}’(x)-P_{n-1}’(x).\ \ \ \ \ \mbox{(7)}$$
$\frac{1}{2}[(6)+(7)]$ results
$$P_{n+1}’(x)=(n+1)P_n(x)+xP_n’(x).\ \ \ \ \ \mbox{(8)}$$
$\frac{1}{2}[(6)-(7)]$ results
$$P_{n-1}’(x)=-nP_n(x)+xP_n’(x).\ \ \ \ \ \mbox{(9)}$$
Replace $n$ by $n-1$ in (7) and add the result to $x$ times (9):
$$(1-x^2)P_n’(x)=nP_{n-1}(x)-nxP_n(x).\ \ \ \ \ \mbox{(10)}$$
Differentiate (10) with respect to $x$ and add the result to $n$ times (9):
$$(1-x^2)P_n^{\prime\prime}(x)-2xP_n’(x)+n(n+1)P_n(x)=0.\ \ \ \ \ \mbox{(11)}$$
The linear second-order differential equation (11) is called Legendre’s differential equation and as seen $P_n(x)$ satisfies (11). This is why $P_n(x)$ is called a Legendre polynomial.

In physics (11) is often expressed in terms of differentiation with respect to $\theta$. Let $x=\cos\theta$. Then by the chain rule,
\begin{align*}
\frac{dP_n(\cos\theta)}{d\theta}&=-\sin\theta\frac{dP_n(x)}{dx},\ \ \ \ \ \mbox{(12)}\\ \frac{d^2P_n(\cos\theta)}{d\theta^2}&=-x\frac{dP_n(x)}{dx}+(1-x^2)\frac{d^2P_n(x)}{dx^2}.\ \ \ \ \ \mbox{(13)}
\end{align*}
Using (12) and (13), Legendre’s differential equation (11) can be written as
$$\frac{1}{\sin\theta}\frac{d}{d\theta}\left[\sin\theta\frac{dP_n(\cos\theta)}{d\theta}\right]+n(n+1)P_n(\cos\theta)=0.$$

# Tensors I

Tensors may be considered as a generalization of vectors and covectors. They are extremely important quantities for studying differential geometry and physics.

Let $M^n$ be an $n$-dimensional differentiable manifold. For each $x\in M^n$, let $E_x=T_xM^n$, i.e. the tangent space to $M^n$ at $x$. We denote the canonical basis of $E$ by $\partial=\left(\frac{\partial}{\partial x^1},\cdots,\frac{\partial}{\partial x^n}\right)$ and its dual basis by $\sigma=dx=(dx^1,\cdots,dx^n)$, where $x^1,\cdots,x^n$ are local coordinates. The canonical basis $\frac{\partial}{\partial x^1},\cdots,\frac{\partial}{\partial x^1}$ also simply denoted by $\partial_1,\cdots,\partial_n$.

Covariant Tensors

Definition. A covariant tensor of rank $r$ is a multilinear real-valued function
$$Q:E\times E\times\cdots\times E\longrightarrow\mathbb{R}$$
of $r$-tuples of vectors. A covariant tensor of rank $r$ is also called a tensor of type $(0,r)$ or shortly $(0,r)$-tensor. Note that the values of $Q$ must be independent of the basis in which the components of the vectors are expressed. A covariant vector (also called covector or a 1-form) is a covariant tensor of rank 1. An important of example of covariant tensor of rank 2 is the metric tensor $G$:
$$G(v,w)=\langle v,w\rangle=\sum_{i,j}g_{ij}v^iw^j.$$

In componenents, by multilinearity
\begin{align*}
Q(v_1\cdots,v_r)&=Q\left(\sum_{i_1}v_1^{i_1}\partial_{i_1},\cdots,\sum_{i_r}v_r^{i_r}\partial_{i_r}\right)\\
&=\sum_{i_1,\cdots,i_r}v_1^{i_1}\cdots v_r^{i_r}Q(\partial_{i_1},\cdots,\partial_{i_r}).
\end{align*}
Denote $Q(\partial_{i_1},\cdots,\partial_{i_r})$ by $Q_{i_1,\cdots,i_r}$. Then
$$Q(v_1\cdots,v_r)=\sum_{i_1,\cdots,i_r}Q_{i_1,\cdots,i_r}v_1^{i_1}\cdots v_r^{i_r}.\ \ \ \ \ \mbox{(1)}$$
Using the Einstein’s convention, (1) can be shortly written as
$$Q(v_1\cdots,v_r)=Q_{i_1,\cdots,i_r}v_1^{i_1}\cdots v_r^{i_r}.$$
The set of all covariant tensors of rank $r$ forms a vector space over $\mathbb{R}$. The number of components in such a tensor is $n^r$. The vector space of all covariant $r$-th rank tensors is denoted by
$$E^\ast\otimes E^\ast\otimes\cdots\otimes E^\ast=\otimes^r E^\ast.$$

If $\alpha,\beta\in E^\ast$, i.e. covectors, we can form the 2nd rank covariant tensor, the tensor product $\alpha\otimes\beta$ of $\alpha$ and $\beta$: Define $\alpha\otimes\beta: E\times E\longrightarrow\mathbb{R}$ by
$$\alpha\otimes\beta(v,w)=\alpha(v)\beta(w).$$
If we write $\alpha=a_idx^i$ and $\beta=b_jdx^j$, then
$$(\alpha\otimes\beta)_{ij}=\alpha\otimes\beta(\partial_i,\partial_j)=\alpha(\partial_i)\beta(\partial_j)=a_ib_j.$$

Contravariant Tensors

A contravariant vector, i.e. an element of $E$ can be considered as a linear functional $v: E^\ast\longrightarrow\mathbb{R}$ defined by
$$v(\alpha)=\alpha(v)=a_iv^i,\ \alpha=a_idx^i\in E^\ast.$$

Definition. A contravariant tensor of rank $s$ is a multilinear real-valued function $T$ on $s$-tuples of covectors
$$T:E^\ast\times E^\ast\times\cdots\times E^\ast\longrightarrow\mathbb{R}.$$ A contravariant tensor of rank $s$ is also called a tensor of type $(s,0)$ or shortly $(s,0)$-tensor.
For 1-forms $\alpha_1,\cdots,\alpha_s$
$$T(\alpha_1,\cdots,\alpha_s)=a_{1_{i_1}}\cdots a_{s_{i_s}}T^{i_1\cdots i_s}$$
where
$$T^{i_1\cdots i_s}:=T(dx^{i_1},\cdots,dx^{i_s}).$$
The space of all contravariant tensors of rank $s$ is denoted by
$$E\otimes E\otimes\cdots\otimes E:=\otimes^s E.$$
Contravariant vectors are contravariant tensors of rank 1. An example of a contravariant tensor of rank 2 is the inverse of the metric tensor $G^{-1}=(g^{ij})$:
$$G^{-1}(\alpha,\beta)=g^{ij}a_ib_j.$$

Given a pair $v,w$ of contravariant vectors, we can form the tensor product $v\otimes w$ in the same manner as we did for covariant vectors. It is the 2nd rank contravariant tensor with components $(v\otimes w)^{ij}=v^jw^j$. The metric tensor $G$ and its inverse $G^{-1}$ may be written as
$$G=g_{ij}dx^i\otimes dx^j\ \mbox{and}\ G^{-1}=g^{ij}\partial_i\otimes\partial_j.$$

Mixed Tensors

Definition. A mixed tensor, $r$ times covariant and $s$ times contravariant, is a real multilinear function $W$
$$W: E^\ast\times E^\ast\times\cdots\times E^\ast\times E\times E\times\cdots\times E\longrightarrow\mathbb{R}$$
on $s$-tuples of covectors and $r$-tuples of vectors. It is also called a tensor of type $(s,r)$ or simply $(s,r)$-tensor. By multilinearity
$$W(\alpha_1,\cdots,\alpha_s, v_1,\cdots, v_r)=a_{1_{i_1}}\cdots a_{s_{i_s}}W^{i_1\cdots i_s}{}_{j_1\cdots j_r}v_1^{j_1}\cdots v_r^{j_r}$$
where
$$W^{i_1\cdots i_s}{}_{j_1\cdots j_r}:=W(dx^{i_1},\cdots,dx^{i_s},\partial_{j_1},\cdots,\partial_{j_r}).$$

A 2nd rank mixed tensor may arise from a linear operator $A: E\longrightarrow E$. Define $W_A: E^\ast\times E\longrightarrow\mathbb{R}$ by $W_A(\alpha,v)=\alpha(Av)$. Let $A=(A^i{}_j)$ be the matrix associated with $A$, i.e. $A(\partial_j)=\partial_i A^i{}_j$. Let us calculate the component of $W_A$:
$$W_A^i{}_j=W_A(dx^i,\partial_j)=dx^i(A(\partial_j))=dx^i(\partial_kA^k{}_j)=\delta^i_kA^k{}_j=A^i{}_j.$$
So the matrix of the mixed tensor $W_A$ is just the matrix associated with $A$. Conversely, given a mixed tensotr $W$, once convariant and once contravariant, we can define a linear transformation $A$ such that $W(\alpha,v)=\alpha(A,v)$. We do not distinguish between a linear transformation $A$ and its associated mixed tensor $W_A$. In components, $W(\alpha,v)$ is written as
$$W(\alpha,v)=a_iA^i{}_jv^j=aAv.$$

The tensor product $w\otimes\beta$ of a vector and a covector is the mixed tensor defined by
$$(w\otimes\beta)(\alpha,v)=\alpha(w)\beta(v).$$ The associated transformation is can be written as
$$A=A^i{}_j\partial_i\otimes dx^j=\partial_i\otimes A^i{}_jdx^j.$$

For math undergraduates, different ways of writing indices (raising, lowering, and mixed) in tensor notations can be very confusing. Main reason is that in standard math courses such as linear algebra or elementary differential geometry (classical differential geometry of curves and surfaces in $\mathbb{E}^3$) the matrix of a linear transformation is usually written as $A_{ij}$. Physics undergraduates don’t usually get a chance to learn tensors in undergraduate physics courses. In order to study more advanced differential geometry or physics such as theory of special and general relativity, and field theory one must be able to distinguish three different ways of writing matrices $A_{ij}$, $A^{ij}$, and $A^i{}_j$. To summarize, $A_{ij}$ and $A^{ij}$ are bilinear forms on $E$ and $E^\ast$, respectively that are defined by
$$A_{ij}v^iv^j\ \mbox{and}\ A^{ij}a_ib_j\ (\mbox{respectively}).$$ $A^i{}_j$ is the matrix of a linear transformation $A: E\longrightarrow E$.

Let $(E,\langle\ ,\ \rangle)$ be an inner product space. Given a linear transformation $A: E\longrightarrow E$ (i.e. a mixed tensor), one can associate a bilinear covariant bilinear form $A’$ by
$$A’(v,w):=\langle v,Aw\rangle=v^ig_{ij}A^j{}_k w^k.$$ So we see that the matrix of $A’$ is
$$A’_{ik}=g_{ij}A^j{}_k.$$ The process can be said as “we lower the index $j$, making it a $k$, by mans of the metric tensor $g_{ij}$.” In tensor analysis one uses the same letter, i.e. instead of $A’$, one writes
$$A_{ik}:=g_{ij}A^j{}_k.$$ This is clearly a covariant tensor. In general, the components of the associated covariant tensor $A_{ik}$ differ from those of the mixed tensor $A^i{}_j$. But if the basis is orthonormal, i.e. $g_{ij}=\delta^i_j$ then they coincide. That is the reason why we simply write $A_{ij}$ without making any distiction in linear algebra or in elementary differential geometry.

Similarly, one may associate to the linear transformation $A$ a contravariant bilinear form
$$\bar A(\alpha,\beta)=a_iA^i{}_jg^{jk}b_k$$ whose matrix components can be written as
$$A^{ik}=A^i{}_jg^{jk}.$$

Note that the metric tensor $g_{ij}$ represents a linear map from $E$ to $E^\ast$, sending the vector with components $v^j$ into the covector with components $g_{ij}v^j$. In quantum mechanics, the covector $g_{ij}v^j$ is denoted by $\langle v|$ and called a bra vector, while the vector $v^j$ is denoted by $|v\rangle$ and called a ket vector. Usually the inner product on $E$
$$\langle\ ,\ \rangle:E\times E\longrightarrow\mathbb{R};\ \langle v,w\rangle=g_{ij}v^iw^j$$ is considered as a covariant tensor of rank 2. But in quantum mechanics $\langle v,w\rangle$ is not considered as a covariant tensor $g_{ij}$ of rank 2 acting on a pair of vectors $(v,w)$, rather it is regarded as the braket $\langle v|w\rangle$, a bra vector $\langle v|$ acting on a ket vector $|w\rangle$.

# Helmholtz Equation

Helmholtz equation
$$\nabla^2\psi+k^2\psi=0\ \ \ \ \ \mbox{(1)}$$
is extremely important in physics. Solving many physically important partial differential equations such as heat equation, wave equation (Klein-Gordon equation), Maxwell’s equations, and Schrödinger equation, etc. often require solving Helmholtz equation (1).

In this notes, we discuss how to solve Helmholtz equation using separation of variables in rectangular, cylindrical, and spherical coordinate systems. The solutions we discuss here will be used when you solve boundary value problems associated with Helmholtz equation.

Helmholtz Equation in Rectangular Coordinates

Assume that $\psi(x,y,z)=X(x)Y(y)Z(z)$. Then the equation (1) becomes
$$YZ\frac{d^2X}{dx^2}+XZ\frac{d^2Y}{dy^2}+XY\frac{d^2Z}{dz^2}+k^2XYZ=0.\ \ \ \ \ \mbox{(2)}$$
Dividing (2) by $XYZ$, we obtain
$$\frac{1}{X}\frac{d^2X}{dx^2}+\frac{1}{Y}\frac{d^2Y}{dy^2}+\frac{1}{Z}\frac{d^2Z}{dz^2}+k^2=0.\ \ \ \ \mbox{(3)}$$
Let us write (3) as
$$\frac{1}{X}\frac{d^2X}{dx^2}=-\frac{1}{Y}\frac{d^2Y}{dy^2}-\frac{1}{Z}\frac{d^2Z}{dz^2}-k^2.\ \ \ \ \ \mbox{(4)}$$
Now we have a paradox. The LHS of (4) depends only on the $x$-variable while the RHS of (4) depends on $y$ and $z$-variables. One way to to avoid this paradox is to assume that the LHS and the RHS of (4) is a constant, say $-l^2$. If you are wondering why we choose a negative constant, the reason comes from physics. For a physical reason, we need an oscillating solution which can be obtained by choosing a negative separation constant. Often boundary conditions for Helmholtz equation lead to a trivial solution for a positive separation constant. Continuing a similar process, we separate Helmholtz equation into three ordinary differential equations:
\begin{align*}
\frac{1}{X}\frac{d^2 X}{dx^2}&=-l^2,\\
\frac{1}{Y}\frac{d^2Y}{dy^2}&=-m^2,\\
\frac{1}{Z}\frac{d^2Z}{dz^2}&=-n^2,
\end{align*}
where $k^2=l^2+m^2+n^2$.

Each mode is given by
$$\psi_{lmn}(x,y,z)=X_l(x)Y_m(y)Z_n(z)$$ and the most general solution is given by the linear combination of the modes
$$\psi(x,y,z)=\sum_{i,m,n}a_{lmn}\psi_{lmn}(x,y,z).$$

Helmholtz Equation in Cylindrical Coordinates

In cylindrical coordinate system $(\rho,\varphi,z)$, Helmholtz equation (1) is written as
$$\frac{1}{\rho}\frac{\partial}{\partial\rho}\left(\rho\frac{\partial\psi}{\partial\rho}\right)+\frac{1}{\rho^2}\frac{\partial^2\psi}{\partial\varphi^2}+\frac{\partial^2\psi}{\partial z^2}+k^2\psi=0.\ \ \ \ \ \mbox{(5)}$$

We assume that $\psi(\rho,\varphi,z)=P(\rho)\Phi(\varphi)Z(z)$. Then (5) can be written as
$$\frac{\Phi Z}{\rho}\frac{\partial}{\partial\rho}\left(\rho\frac{\partial\psi}{\partial\rho}\right)+\frac{PZ}{\rho^2}\frac{\partial^2\psi}{\partial\varphi^2}+P\Phi\frac{\partial^2\psi}{\partial z^2}+k^2=0.\ \ \ \ \ \mbox{(6)}$$
As we have done in rectangular coordinate system, by introducing the separation constants we can separate (6) into three ordinary differential equations
\begin{align*}
\frac{d^2Z}{dz^2}=l^2z,\\
\frac{d^2\Phi}{d\phi^2}=-m^2\Phi,\\
\rho\frac{d}{d\rho}\left(\rho\frac{dP}{d\rho}\right)+(n^2\rho^2-m^2)P=0,\ \ \ \ \ \mbox{(7)}
\end{align*}
where $n^2=k^2+l^2$. The last equation (7) is Bessel’s differential equation.

The general solution of Helmholtz equation in cylindrical coordinates is given by
$$\psi(\rho,\varphi,z)=\sum_{m,n}a_{mn}P_{mn}(\rho)\Phi_m(\varphi)Z_n(z).$$

Helmholtz Equation in Spherical Coordinates

In spherical coordinates $(r,\theta,\varphi)$, Helmholtz equation (1) is written as
$$\frac{1}{r^2\sin\theta}\left[\sin\theta\frac{\partial}{\partial r}\left(r^2\frac{\partial\psi}{\partial r}\right)+\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right)+\frac{1}{\sin\theta}\frac{\partial^2\psi}{\partial\varphi^2}\right]=-k^2\psi.\ \ \ \ \ \mbox{(8)}$$
Assume that $\psi(r,\theta,\varphi)=R(r)\Theta(\theta)\phi(\varphi)$. Then (8) can be written as
$$\frac{1}{Rr^2}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\frac{1}{\Theta r^2\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)+\frac{1}{\Phi r^2\sin^2\theta}\frac{d^2\Phi^2}{d\varphi^2}=-k^2.\ \ \ \ \ \mbox{(9)}$$
By introducing separation constants, (9) is separated into three ordinary differential equations
\begin{align*}
\frac{1}{\Phi}\frac{d^2\Phi}{d\varphi^2}=-m^2,\\
\frac{1}{\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)+\left(Q-\frac{m^2}{\sin^2\theta}\right)\Theta=0,\ \ \ \ \ \mbox{(10)}\\
\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\left(k^2-\frac{Q}{r^2}\right)R=0.\ \ \ \ \ \mbox{(11)}
\end{align*}
The second equation (10) is the associated Legendre equation with $Q=l(l+1)$. The third equation (11) is spherical Bessel equation with $k^2>0$.

The general solution of Helmholtz equation (8) is then given by
$$\psi(r,\theta,\varphi)=\sum_{Q,m}R_Q(r)\Theta_{Qm}(\theta)\Phi_m(\varphi).$$

The restriction that $k^2$ be a constant is unnecessary. For instance the separation process will still be possible for $k^2=f(r)$. If $k^2=f(r)$, (11) is the associated Laguerre equation. The associated Laguerre equation is appeared in the hydrogen atom problem in quantum mechanics.

# Connection Forms

Let $E_1, E_2, E_3$ be an arbitrary frame field on $\mathbb{E}^3$. At each $v\in T_p\mathbb{E}^3$, $\nabla_v E_i\in T_p\mathbb{E}^3$, $i=1,2,3$. So, there exists uniquely 1-forms $\omega_{ij}:T_p\mathbb{E}^3\longrightarrow\mathbb{R}$, $i,j=1,2,3$ such that
\begin{align*}
\nabla_vE_1&=\omega_{11}(v)E_1(p)+\omega_{12}(v)E_2(p)+\omega_{13}(v)E_3(p),\\
\nabla_vE_2&=\omega_{21}(v)E_1(p)+\omega_{22}(v)E_2(p)+\omega_{23}(v)E_3(p),\\
\nabla_vE_3&=\omega_{31}(v)E_1(p)+\omega_{32}(v)E_2(p)+\omega_{33}(v)E_3(p)
\end{align*}
for each $v\in T_p\mathbb{E}^3$. These equations are called the connection equations of the frame field $E_1$, $E_2$, $E_3$. One can clearly see that $\omega_{ij}$ is determined by
$$\omega_{ij}(v)=\nabla_v E_i\cdot E_j(p).\ \ \ \ \ \mbox{(1)}$$ The 1-forms $\omega_{ij}$ are called the connection forms of the frame field $E_1,E_2,E_3$. Often the matrix $\omega=(\omega_{ij})$ is called the connection 1-form of the frame field $E_1,E_2,E_3$. The linearity of $\omega_{ij}$ is due to the linearity of the covariant derivative $\nabla E_i$.

Proposition. The matrix $\omega$ is a skew symmetric matrix, i.e. $\omega+{}^t\omega=0$.

Proof. Since $E_i\cdot E_j=0$, the directional derivative $v[E_i\cdot E_j]=0$. On the other hand, by Leibniz rule,
\begin{align*}
v[E_i\cdot E_j]&=\nabla_vE_i\cdot E_j(p)+E_i(p)\cdot \nabla_vE_j\\
&=\omega_{ij}(v)+\omega_{ji}(v).
\end{align*}
Hence,
$$\omega_{ij}+\omega_{ji}=0.\ \ \ \ \ \mbox{(2)}$$

If $i=j$ in (2), we get $\omega_{ii}=0$. So, the connection 1-form $\omega$ is written as
$$\omega=\begin{pmatrix} 0 & \omega_{12} & \omega_{13}\\ -\omega_{12} & 0 &\omega_{23}\\ -\omega_{13} & -\omega_{23} & 0 \end{pmatrix}.\ \ \ \ \ \mbox{(3)}$$

Remark. The set of all $3\times 3$ skew symmetric matrices is denoted by $\mathfrak{o}(3)$. It is the Lie algebra of the orthogonal group $\mathrm{O}(3)$. The orthogonal group $\mathrm{O}(3)$ is the set of all $3\times 3$ orthogonal matrices and it is a Lie group. Recall that a square matrix $A$ is orthogonal if and only if $A\cdot{}^tA=I$, i.e. $A^{-1}={}^tA$.

The connection equations of the frame field $E_1$, $E_2$, $E_3$
$$\nabla_VE_i=\sum_i\omega_{ij}(V)E_j,\ i=1,2,3\ \ \ \ \ \mbox{(4)}$$
where $V$ is a vector field on $\mathbb{E}^3$ become
$$\begin{array}{ccccccc} \nabla_VE_1&=&&&\omega_{12}(V)E_2&+&\omega_{13}(V)E_3,\\ \nabla_VE_2&=&-\omega_{12}(V)E_1& & &+&\omega_{23}(V)E_3,\\ \nabla_VE_3&=&-\omega_{13}(V)E_1&-&\omega_{23}(V)E_2. \end{array}$$
The connections equations are in fact a generalization of the Frenet-Serret formulas.

Let $Y$ be a vector field defined on a region containing a curve $\alpha(t)$. Then $Y_\alpha(t):=Y(\alpha(t))$ defined a vector field on the curve $\alpha(t)$. Then one can easily see that
$$\nabla_{\dot\alpha(t)}Y=\frac{d}{dt}Y_\alpha(t).$$
Let $\alpha(t)$ be a curve with unit speed. Let $E_1=T$, $E_2=N$, $E_3=B$. Then
\begin{align*}
\omega_{12}&=\nabla_{\dot\alpha_(t)}E_1\cdot E_2=\dot T\cdot N=(\kappa N)\cdot N=\kappa,\\
\omega_{13}&=\nabla_{\dot\alpha_(t)}E_1\cdot E_3=\dot T\cdot B=0,\\
\omega_{23}&=\nabla_{\dot\alpha_(t)}E_2\cdot E_3=\dot N\cdot B=(-\kappa T+\tau B)=\tau.
\end{align*}
The connection equations (4) are then nothing but the Frenet-Serret formulas
$$\begin{array}{ccccccc} \dot T&=&&&\kappa N&&\\ \dot N&=&-\kappa T& & &+&\tau B\\ \dot B&=&&-&\tau N. \end{array}$$

The frame $E_1,E_2,E_3$ can be written in terms of the natural frame $U_1,U_2,U_3$ as
\begin{align*}
E_1&=a_{11}U_1+a_{12}U_2+a_{13}U_3,\\
E_2&=a_{21}U_1+a_{22}U_2+a_{23}U_3,\\
E_3&=a_{31}U_1+a_{32}U_2+a_{33}U_3.
\end{align*}
Each real-valued function $a_{ij}:\mathbb{E}^3\longrightarrow\mathbb{R}$ is uniquely determined by $a_{ij}=E_i\cdot U_j$. The matrix $A=(a_{ij})$ is called the attitude matrix (also called rotation matrix or orientation matrix) of the frame field $E_1,E_2,E_3$. One can clearly see that the attitude matrix $A$ is an orthogonal matrix. In the above remark, I mentioned that the set of all $3\times$ skew symmetric matrices is the Lie algebra $\mathfrak{o}(3)$. The Lie algebra $\mathfrak{g}$ of a Lie group $G$ is defined to be the tangent space $T_e G$ to $G$ at the identity element $e$. (A Lie group is a differentiable manifold, so it make sense to talk about tangent spaces to $G$.)

Let us define a curve $\gamma: \mathbb{R}\longrightarrow\mathrm{O}(3)$ by
$$\gamma(t)=A(t)\cdot{}^tA(0).$$
Then $\gamma(0)=I$.
Hence $\dot{\gamma}(0)=\frac{dA(t)}{dt}|_{t=0}\cdot{}^tA(0)$ is a tangent vector to $\mathrm{O}(3)$ at the identity matrix $I$. That is, $\dot{\gamma}(0)\in\mathfrak{o}(3)$. Hence one can easily expect that the following theorem holds.

Theorem. If $A=(a_{ij})$ is the attitude matrix and $\omega=(\omega_{ij})$ the connection 1-form of a frame field $E_1, E_2, E_3$, then
$$\omega=dA\cdot{}^tA\ \ \ \ \ \mbox{(4)}$$
or equivalently
$$\omega_{ij}=\sum_k da_{ik} \cdot a_{jk}\ \mbox{for}\ i,j=1,2,3.$$

Proof. For each $v\in T_p\mathbb{E}^3$,
$$\omega_{ij}(v)=\nabla_vE_i\cdot E_j(p).$$
In terms of the natural field $U_i$, $i=1,2,3$,
$$E_i=\sum_ka_{ik}U_k,\ i=1,2,3.$$
So,
\begin{align*}
\nabla_vE_i&=\sum_k v[a_{ik}]U_k(p)\\
&=\sum_k da_{ik} U_k(p).
\end{align*}
Hence,
$$\omega_{ij}=\sum_k da_{ik}a_{jk},$$
i.e.
$$\omega=dA\cdot{}^tA.$$

Remark. In general, if $G$ is a Lie group then its Lie algebra $\mathfrak{g}$ is given by the set of differential $1$-forms
$$\mathfrak{g}=\{g^{-1}dg:\ g\in G\}=\{(dg^{-1})g:\ g\in G\}.$$

Example. Let us compute the connection forms of the cylindrical frame field. The attitude matrix is
$$A=\begin{pmatrix} \cos\theta & \sin\theta & 0\\ -\sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{pmatrix}.$$ Thus
$$dA=\begin{pmatrix} -\sin\theta d\theta & \cos\theta d\theta & 0\\ -\cos\theta d\theta & -\sin\theta d\theta & 0\\ 0 & 0 & 0 \end{pmatrix}.$$
Hence,
\begin{align*}
\omega&=dA\cdot{}^tA\\
&=\begin{pmatrix}
-\sin\theta d\theta & \cos\theta d\theta & 0\\
-\cos\theta d\theta & -\sin\theta d\theta & 0\\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix}
\cos\theta & -\sin\theta & 0\\
\sin\theta & \cos\theta & 0\\
0 & 0 & 1\end{pmatrix}\\
&=\begin{pmatrix}
0 & d\theta & 0\\
-d\theta & 0 & 0\\
0 & 0 & 0
\end{pmatrix}.
\end{align*}
The connection equations of the cylindrical frame field are then
\begin{align*}
\nabla_VE_1&=d\theta(V)E_2=V[\theta]E_2,\\
\nabla_VE_2&=-d\theta(V)E_1=-V[\theta]E_1,\\
\nabla_VE_3&=0
\end{align*}
for all vector fields $V$. As expected the vector field $E_3$ is parallel.