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diff --git a/content/know/concept/sturm-liouville-theory/index.pdc b/content/know/concept/sturm-liouville-theory/index.pdc new file mode 100644 index 0000000..7ccd625 --- /dev/null +++ b/content/know/concept/sturm-liouville-theory/index.pdc @@ -0,0 +1,346 @@ +--- +title: "Sturm-Liouville theory" +firstLetter: "S" +publishDate: 2021-02-23 +categories: +- Mathematics +- Physics + +date: 2021-02-23T08:52:28+01:00 +draft: false +markup: pandoc +--- + +# Sturm-Liouville theory + +**Sturm-Liouville theory** defines the analogue of Hermitian matrix +eigenvalue problems for linear second-order ODEs. + +It states that, given suitable boundary conditions, any linear +second-order ODE can be rewritten using the **Sturm-Liouville operator**, +and that the corresponding eigenvalue problem, known as a +**Sturm-Liouville problem**, will give real eigenvalues and a complete set +of eigenfunctions. + + +## General operator + +Consider the most general form of a second-order linear +differential operator $\hat{L}$, where $p_0(x)$, $p_1(x)$, and $p_2(x)$ +are real functions of $x \in [a,b]$ which are non-zero for all $x \in ]a, b[$: + +$$\begin{aligned} + \hat{L} \{u(x)\} = p_0(x) u''(x) + p_1(x) u'(x) + p_2(x) u(x) +\end{aligned}$$ + +We now define the **adjoint** or **Hermitian** operator +$\hat{L}^\dagger$ analogously to matrices: + +$$\begin{aligned} + \braket*{f}{\hat{L} g} + = \braket*{\hat{L}^\dagger f}{g} +\end{aligned}$$ + +What is $\hat{L}^\dagger$, given the above definition of $\hat{L}$? +We start from the inner product $\braket*{f}{\hat{L} g}$: + +$$\begin{aligned} + \braket*{f}{\hat{L} g} + &= \int_a^b f^*(x) \hat{L}\{g(x)\} \dd{x} + = \int_a^b (f^* p_0) g'' + (f^* p_1) g' + (f^* p_2) g \dd{x} + \\ + &= \big[ (f^* p_0) g' + (f^* p_1) g \big]_a^b - \int_a^b (f^* p_0)' g' + (f^* p_1)' g - (f^* p_2) g \dd{x} + \\ + &= \big[ f^* \big( p_0 g' \!+\! p_1 g \big) \!-\! (f^* p_0)' g \big]_a^b + \int_a^b \! \big( (f p_0)'' - (f p_1)' + (f p_2) \big)^* g \dd{x} + \\ + &= \big[ f^* \big( p_0 g' + (p_1 - p_0') g \big) - (f^*)' p_0 g \big]_a^b + \int_a^b \big( \hat{L}^\dagger\{f\} \big)^* g \dd{x} +\end{aligned}$$ + +We now have an expression for $\hat{L}^\dagger$, but are left with an +annoying boundary term: + +$$\begin{aligned} + \braket*{f}{\hat{L} g} + &= \big[ f^* \big( p_0 g' + (p_1 - p_0') g \big) - (f^*)' p_0 g \big]_a^b + \braket*{\hat{L}^\dagger f}{g} +\end{aligned}$$ + +To fix this, +let us demand that $p_1(x) = p_0'(x)$ and that +$[p_0(f^* g' - (f^*)' g)]_a^b = 0$, leaving: + +$$\begin{aligned} + \braket*{f}{\hat{L} g} + &= \big[ p_0 \big( f^* g' - (f^*)' g \big) \big]_a^b + \braket{\hat{L}^\dagger f}{g} + = \braket*{\hat{L}^\dagger f}{g} +\end{aligned}$$ + +Using the aforementioned restriction $p_1(x) = p_0'(x)$, +we then take a look at the definition of $\hat{L}^\dagger$: + +$$\begin{aligned} + \hat{L}^\dagger \{f\} + &= (p_0 f)'' - (p_1 f)' + (p_2 f) + \\ + &= p_0 f'' + (2 p_0' - p_1) f' + (p_0'' - p_1' + p_2) f + \\ + &= p_0 f'' + p_0' f' + p_2 f + \\ + &= (p_0 f')' + p_2 f +\end{aligned}$$ + +The original operator $\hat{L}$ reduces to the same form, +so it is **self-adjoint**: + +$$\begin{aligned} + \hat{L} \{f\} + &= p_0 f'' + p_0' f' + p_2 f + = (p_0 f')' + p_2 f + = \hat{L}^\dagger \{f\} +\end{aligned}$$ + +Consequently, every such second-order linear operator $\hat{L}$ is self-adjoint, +as long as it satisfies the constraints $p_1(x) = p_0'(x)$ and $[p_0 (f^* g' - (f^*)' g)]_a^b = 0$. + +Let us ignore the latter constraint for now (it will return later), +and focus on the former: what if $\hat{L}$ does not satisfy $p_0' \neq p_1$? +We multiply it by an unknown $p(x) \neq 0$, and divide by $p_0(x) \neq 0$: + +$$\begin{aligned} + \frac{p(x)}{p_0(x)} \hat{L} \{u\} = p(x) u'' + p(x) \frac{p_1(x)}{p_0(x)} u' + p(x) \frac{p_2(x)}{p_0(x)} u +\end{aligned}$$ + +We now define $q(x)$, +and demand that the derivative $p'(x)$ of the unknown $p(x)$ satisfies: + +$$\begin{aligned} + q(x) = p(x) \frac{p_2(x)}{p_0(x)} + \qquad + p'(x) = p(x) \frac{p_1(x)}{p_0(x)} +\end{aligned}$$ + +The latter is a differential equation for $p(x)$, which we solve by integration: + +$$\begin{gathered} + \frac{p_1(x)}{p_0(x)} = \frac{1}{p(x)} \dv{p}{x} + \quad \implies \quad + \frac{p_1(x)}{p_0(x)} \dd{x} = \frac{1}{p(x)} \dd{p} + \\ + \implies \quad + \int_a^x \frac{p_1(\xi)}{p_0(\xi)} \dd{\xi} = \int_{p(a)}^{p(x)} \frac{1}{f} \dd{f} + = \ln\Big( \frac{p(x)}{p(a)} \Big) + \\ + \implies \quad + p(x) = p(a) \exp\!\Big( \int_a^x \frac{p_1(\xi)}{p_0(\xi)} \dd{\xi} \Big) +\end{gathered}$$ + +Now that we have $p(x)$ and $q(x)$, we can define a new operator $\hat{L}_p$ as follows: + +$$\begin{aligned} + \hat{L}_p \{u\} + = \frac{p}{p_0} \hat{L} \{u\} + = p u'' + p' u' + q u + = (p u')' + q u +\end{aligned}$$ + +This is the self-adjoint form from earlier! +So even if $p_0' \neq p_1$, any second-order linear operator with $p_0(x) \neq 0$ +can easily be put in self-adjoint form. + +This general form is known as the **Sturm-Liouville operator** $\hat{L}_{SL}$, +where $p(x)$ and $q(x)$ are non-zero real functions of the variable $x \in [a,b]$: + +$$\begin{aligned} + \boxed{ + \hat{L}_{SL} \{u(x)\} + = \frac{d}{dx}\Big( p(x) \frac{du}{dx} \Big) + q(x) u(x) + = \hat{L}_{SL}^\dagger \{u(x)\} + } +\end{aligned}$$ + + +## Eigenvalue problem + +A **Sturm-Liouville problem** (SLP) is analogous to a matrix eigenvalue problem, +where $w(x)$ is a real weight function, $\lambda$ is the **eigenvalue**, +and $u(x)$ is the corresponding **eigenfunction**: + +$$\begin{aligned} + \boxed{ + \hat{L}_{SL}\{u(x)\} = - \lambda w(x) u(x) + } +\end{aligned}$$ + +Necessarily, $w(x) > 0$ except in isolated points, where $w(x) = 0$ is allowed; +the point is that any inner product $\braket{f}{w g}$ may never be zero due to $w$'s fault. +Furthermore, the convention is that $u(x)$ cannot be trivially zero. + +In our derivation of $\hat{L}_{SL}$, +we removed a boundary term to get self-adjointness. +Consequently, to have a valid SLP, the boundary conditions for +$u(x)$ must be as follows, otherwise the operator cannot be self-adjoint: + +$$\begin{aligned} + \Big[ p(x) \big( u^*(x) u'(x) - (u'(x))^* u(x) \big) \Big]_a^b = 0 +\end{aligned}$$ + +There are many boundary conditions (BCs) which satisfy this requirement. +Some notable ones are listed here non-exhaustively: + ++ **Dirichlet BCs**: $u(a) = u(b) = 0$ ++ **Neumann BCs**: $u'(a) = u'(b) = 0$ ++ **Robin BCs**: $\alpha_1 u(a) + \beta_1 u'(a) = \alpha_2 u(b) + \beta_2 u'(b) = 0$ with $\alpha_{1,2}, \beta_{1,2} \in \mathbb{R}$ ++ **Periodic BCs**: $p(a) = p(b)$, $u(a) = u(b)$, and $u'(a) = u'(b)$ ++ **Legendre "BCs"**: $p(a) = p(b) = 0$ + +Once this requirement is satisfied, Sturm-Liouville theory gives us +some very useful information about $\lambda$ and $u(x)$. +From the definition of an SLP, we know that, given two arbitrary (and possibly identical) +eigenfunctions $u_n$ and $u_m$, the following must be satisfied: + +$$\begin{aligned} + 0 = \hat{L}_{SL}\{u_n\} + \lambda_n w u_n = \hat{L}_{SL}\{u_m^*\} + \lambda_m^* w u_m^* +\end{aligned}$$ + +We subtract these expressions, multiply by the eigenfunctions, and integrate: + +$$\begin{aligned} + 0 + &= \int_a^b u_m^* \big(\hat{L}_{SL}\{u_n\} + \lambda_n w u_n\big) - u_n \big(\hat{L}_{SL}\{u_m^*\} + \lambda_m^* w u_m^*\big) \:dx + \\ + &= \int_a^b u_m^* \hat{L}_{SL}\{u_n\} - u_n \hat{L}_{SL}\{u_m^*\} + u_n u_m^* w (\lambda_n - \lambda_m^*) \:dx +\end{aligned}$$ + +Rearranging this a bit reveals that these are in fact three inner products: + +$$\begin{aligned} + \int_a^b u_m^* \hat{L}_{SL}\{u_n\} - u_n \hat{L}_{SL}\{u_m^*\} \:dx + &= (\lambda_m^* - \lambda_n) \int_a^b u_n u_m^* w \:dx + \\ + \braket*{u_m}{\hat{L}_{SL} u_n} - \braket*{\hat{L}_{SL} u_m}{u_n} + &= (\lambda_m^* - \lambda_n) \braket{u_m}{w u_n} +\end{aligned}$$ + +The operator $\hat{L}_{SL}$ is self-adjoint by definition, +so the left-hand side vanishes, leaving us with: + +$$\begin{aligned} + 0 + &= (\lambda_m^* - \lambda_n) \braket{u_m}{w u_n} +\end{aligned}$$ + +When $m = n$, the inner product $\braket{u_n}{w u_n}$ is real and positive +(assuming $u_n$ is not trivially zero, in which case it would be disqualified anyway). +In this case we thus know that $\lambda_n^* = \lambda_n$, +i.e. the eigenvalue $\lambda_n$ is real for any $n$. + +When $m \neq n$, then $\lambda_m^* - \lambda_n$ may or may not be zero, +depending on the degeneracy. If there is no degeneracy, we +see that $\braket{u_m}{w u_n} = 0$, i.e. the eigenfunctions are orthogonal. + +In case of degeneracy, manual orthogonalization is needed, but as it turns out, +this is guaranteed to be doable, using e.g. the [Gram-Schmidt method](/know/concept/gram-schmidt-method/). + +In conclusion, **a Sturm-Liouville problem has real eigenvalues $\lambda$, +and all the corresponding eigenfunctions $u(x)$ are mutually orthogonal**: + +$$\begin{aligned} + \boxed{ + \braket{u_m(x)}{w(x) u_n(x)} + = \braket{u_n}{w u_n} \delta_{nm} + = A_n \delta_{nm} + } +\end{aligned}$$ + +When you're solving a differential eigenvalue problem, +knowing that all eigenvalues are real is a *huge* simplification, +so it is always worth checking whether you're dealing with an SLP. + +Another useful fact of SLPs is that they always +have an infinite number of discrete eigenvalues. +Furthermore, the eigenvalues always ascend to $+\infty$; +in other words, there always exists a *lowest* eigenvalue $\lambda_0 > -\infty$, +known as the **ground state**. + + +## Completeness + +Not only are the eigenfunctions $u_n(x)$ of an SLP orthogonal, they +also form a **complete basis**, meaning that any well-behaved function $f(x)$ can be +expanded as a **generalized Fourier series** with coefficients $a_n$: + +$$\begin{aligned} + \boxed{ + f(x) + = \sum_{n = 0}^\infty a_n u_n(x) + \quad \mathrm{for}\: x \in ]a, b[ + } +\end{aligned}$$ + +This series will converge significantly faster if $f(x)$ +satisfies the same BCs as $u_n(x)$. In that case the +expansion will even be valid for the inclusive interval $x \in [a, b]$. + +To find an expression for the coefficients $a_n$, +we multiply the above generalized Fourier series by $w(x) u_m^*(x)$ for an arbitrary $m$: + +$$\begin{aligned} + f(x) w(x) u_m^*(x) + &= \sum_{n = 0}^\infty a_n u_n(x) w(x) u_m^*(x) +\end{aligned}$$ + +By integrating we get inner products on both the left and the right: + +$$\begin{aligned} + \int_a^b f(x) w(x) u_m^*(x) \dd{x} + &= \int_a^b \Big(\sum_{n = 0}^\infty a_n u_n(x) w(x) u_m^*(x)\Big) \dd{x} + \\ + \braket{u_m}{w f} + &= \sum_{n = 0}^\infty a_n \braket{u_m}{w u_n} +\end{aligned}$$ + +Because the eigenfunctions of an SLP are mutually orthogonal, +the summation disappears: + +$$\begin{aligned} + \braket{u_m}{w f} + &= \sum_{n = 0}^\infty a_n \braket{u_m}{w u_n} + = \sum_{n = 0}^\infty a_n A_n \delta_{nm} + = a_m A_m +\end{aligned}$$ + +After isolating this for $a_n$, we see that +the coefficients are given by the projection of the target +function $f(x)$ onto the normalized eigenfunctions $u_n(x) / A_n$: + +$$\begin{aligned} + \boxed{ + a_n + = \frac{\braket{u_n}{w f}}{A_n} + = \frac{\braket{u_n}{w f}}{\braket{u_n}{w u_n}} + } +\end{aligned}$$ + +As a final remark, we can see something interesting +by rearranging the generalized Fourier series +after inserting the expression for $a_n$: + +$$\begin{aligned} + f(x) + &= \sum_{n = 0}^\infty \frac{1}{A_n} \braket{u_n}{w f} u_n(x) + = \int_a^b \Big(\sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) f(\xi) u_n(x) \Big) \dd{\xi} + \\ + &= \int_a^b f(\xi) \Big(\sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) u_n(x) \Big) \dd{\xi} + %= \int_a^b f(\xi) \delta(x - \xi) \dd{\xi} +\end{aligned}$$ + +Upon closer inspection, the parenthesized summation +must be the [Dirac delta function](/know/concept/dirac-delta-function/) $\delta(x)$ +for the integral to work out. +This is in fact the underlying requirement for completeness: + +$$\begin{aligned} + \boxed{ + \sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) u_n(x) = \delta(x - \xi) + } +\end{aligned}$$ + |