Title: Part IB / Methods
Tags: Fourier_Series Fourier_Transform Sturm_Liouville Linear_PDEs Greens_Functions

Q: Definition. Inner product and norm. Consider a vector space of 'nice' L-periodic function from the reals to complex numbers, what are the norm and inner product?
A: \[<f,g> = \int_0^L f(\theta)\overline{g(\theta)} \mathrm{d} \theta\]
\[||f|| = \sqrt{<f,f>}\]

Q: Definition. Complex Fourier Series. For L-periodic function f define its complex Fourier series by [...]
A: \[\sum_n \hat{f}_n e^{2\pi i n \theta / L}, \; \hat{f}_n = \frac{1}{L}\int_0^L f(\theta) e^{-2 \pi i n \theta / L} \mathrm{d} \theta\]
Back Extra: Where the coefficients are called complex Fourier coefficients.
Customary to write \[f(\theta) \sim \sum_n \hat{f}_n e^{2 \pi i n \theta ' L}\]
to mean infinite series on right is complex Fourier series for thing on left

Q: Definition. Fourier Series. For L-periodic function f define its Fourier series by [...]
A: \[ \frac{1}{2} a_0 + \sum_{n=1}^{\infty} a_n \cos \left ( \frac{2 \pi n \theta}{L} \right ) + b_n \sin \left( \frac{2 \pi n \theta}{L} \right )\]
Back Extra: Where \(a_n = \frac{2}{L} \int_0^L f(\theta) \cos (\frac{2 \pi n \theta}{L}) d \theta, b_n = \frac{2}{L} \int_0^L f(\theta) \sin (\frac{2 \pi n \theta}{L}) d \theta\).
So if f is odd, \(a_n = 0\) and if f is even, \(b_n = 0\)

Q: Proposition.  Let \(f : \mathbb{R} \to \mathbb{C}\), L-periodic with
- f' exists away from a countable collection of points in \([0,L)\)
- \(\int_0^L |f'(\theta)| \mathrm{d}\theta < \infty\)
then [...]
for each \(\theta \in [0,L)\) where \[\theta_{\pm} = \lim_{\epsilon \to 0} \theta \pm \epsilon\]
A: \[(S_N f)(\theta) \to \frac{f(\theta_+) + f(\theta_-)}{2}\]
Back Extra: So tends to function at continuous parts, and average of function at discontinuous parts.

Q: Definition. Dirichlet Kernel. Define it to be \[\sum_{|n| \leq N} e^{i n\theta}\].
[...]
[...]
A: i) It is continuous, even and periodic with period \(2 \pi\).; ii) \(\int_{- \pi}^{\pi} D_{N} (\theta) \mathrm{d}\theta = 2 \pi\) for every N.

Q: Proposition. Assume that \(f: \mathbb{R} \to \mathbb{C}\) is smooth and \(2\pi\) periodic. Fix \(\theta_0 \in [0,2 \pi]\). What is \((S_N f)(\theta_0) - f(\theta_0)\)?
A: Using definition we get \[(S_N f)(\theta_0) - f(\theta_0) = \frac{1}{2\pi}\int_{\pi}^{-\pi} D_N(\theta - \theta_0)[f(\theta) - f(\theta_0)] d\theta\]
Replace variables and shift domain over which we integrate, use the fact that \(D_N(\theta)\) is a geometric series, and definition of integral to end up with \[\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{1}{N+1/2} \cos [(N+ 1/2)\theta] F'(\theta ; \theta_0 ) d\theta  = 0(1/N) \to 0\]

Q: Definition. For \(f : [0,L) \to \mathbb{C}\) define the sine and cosine series for f by [...]
A: \[\sum_{n=1}^{\infty} b_n \sin (\frac{n \pi \theta}{L}) , \frac{1}{2} a_0 + \sum_{n=1}^{\infty} a_n \cos (\frac{n \pi \theta}{L}) \]
Back Extra: With \[b_n = \frac{2}{L} \int_0^L f(\theta) \sin (\frac{\pi n \theta}{L}) d\theta \] \[a_n = \frac{2}{L} \int_0^L f(\theta) \cos (\frac{\pi n \theta}{L}) d\theta\]

Q: Proposition. If \(f : \mathbb{R} \to \mathbb{C}\) is k-times continuously differentiable and L-periodic then \[\hat{f}_n = o(1/n^k)\]
A: Integration by parts.
\(\hat{f}_n = \frac{1}{L} \int_0^L f(\theta) e^{-2 \pi i  n \theta / L} d\theta = \frac{-L}{2\pi i n} \left[ f(\theta) e^{2\pi i n \theta / L} \right ]^L_0 + \frac{L}{2\pi i n} \frac{1}{L} \int_0^L f'(\theta) e^{-2\pi i n \theta / L} d\theta = \dots = \left (\frac{L}{2\pi i n} \right)^k \frac{1}{L} \int_0^L f^{(k)}(\theta) e^{-2\pi i n \theta / L} d\theta = o(n^{-k})\)
As the final integral is \(o(1)\).

Q: Theorem. If f is L-periodic and f' exists throughout (0,L), and if f is continuous. The Fourier series of the derivative is [...]
A: the term wise differentiation of the original Fourier series.

Q: Theorem. Parseval's Theorem. Suppose f has Fourier coefficients \(a_i , b_i\). Then \[\int_0^{2L} |f(x)|^2 dx = L \left [ \frac{1}{2} a_0^2 + \sum_{n=1}^{\infty} (a_n^2 + b_n^2) \right ]\]
A: Prove from definition of norm and using the fact that \(< e_n, e_m > = \delta_{nm}\)

Q: Definition. We say a linear differential operator L is self-adjoint on \((V, <\cdot, \cdot>_w )\) if [...]. We say that \((y, \lambda) \in V \times \mathbb{C}\) is eigenfunction-eigenvalue pair if [...].
A: \(<Ly_1, y_2>_w = <y_1, Ly_2>_w \; \; \forall y_1, y_2 \in V\); \(y \neq 0\) and \(Ly = \lambda y\)

Q: Proposition. If L is self-adjoint on \((V, ,<\cdot , \cdot>_w)\)
i) eigenvalues are real
ii) eigenfunctions with different eigenvalues are orthogonal
iii) the countable collection of eigenfunctions form a complete orthogonal set on V
A: If \(L y = \lambda y\)
i) \((\lambda - \overline{\lambda}) ||y||^2_w = (\lambda - \overline{\lambda}) <y,y>_w = <\lambda y, y>_w - <y, \lambda y>_w = <Ly , y>_w - <y, Ly>_w = 0\)
ii) \(L y_1 = \lambda_1 y_1, L y_2 = \lambda_2 y_2, \lambda_1 \neq \lambda_2\).
  \( (\lambda_1 - \lambda_2)< y_1 , y_2>_w = <\lambda_1 y_1 , y_2>_w - <y_1, \lambda_2 y_2>_w = <L y_1, y_2>_w = <y_1, L y_2>_w = 0 \)

Q: Definition.  For Sturm-Liouville problem on \((a,b)\), say endpoint \(c \in \{a,b\}\) is singular if [...]. It is non-singular otherwise.
A: \(p(c) = 0\)
Back Extra: We only specify boundary conditions at non-singular endpoints.

Q: Proposition. If \(y_1 , y_2 \in C^2 [a,b]\) and L is a Sturm-Liouville operator \(L = \frac{1}{W}[-\frac{d}{dx}(p\frac{dy}{dx}) + q]\) then \[<Ly_1, y_2>_w - <y_1, L y_2>_w = p(x)W(y_1, \overline{y_2})(x)|^b_{x=a}\] where \(W(y_1, \overline{y_2}) = y_1 \overline{y_2}' - y_1' \overline{y_2}\) is the Wronskian. In particular, if \(y_1 , y_2 \in V\) then \(<L y_1, y_2>_w = <y_1, L y_2>_w\), i.e. L is self-adjoint on \((V , <\cdot, \cdot>_w)\).
A: Write out integrals and simplify loads and you end up at right answer using FTC. Can check if c is singular end point, we get the final expression being 0. If non-singular then \[\alpha_c y_1(c) + \beta_c y_1'(c) = 0, \alpha \overline{y_2}(c) + \beta_c \overline{y_2}'(c) = 0\] with \(\alpha_c^2 + \beta_c^2 \neq 0\) which implies the Wronskian is 0.

Q: Proposition. Given a generic eigenvalues problem for \(x \in (a,b)\) \[\alpha(x)\frac{d^2y}{dx^2} + \beta(x) \frac{dy}{dx} + \gamma(x)y + \lambda y = 0\] where \(\alpha > 0\). We can convert this to a Sturm-Liouville problem by [...].
A: dividing through by \(\alpha (x)\) and multiply by integrating factor \(I(x) = \exp \int^x \frac{\beta (t)}{\alpha (t)}\)
Back Extra: The equation is equivalent to \[-\frac{d}{dx} (p \frac{dy}{dx}) + ay = \lambda w y\] where \(p(x) = I(x), q(x) = -\frac{\gamma (x) I (x)}{\alpha (x)}, w(x) = \frac{I(x)}{\alpha(x)}\). Proof is fairly simple.

Q: Proposition. Let L be a general differential operator of the form \(\alpha \frac{d^2}{dx^2} + \beta \frac{d}{dx} + \gamma\), and let \(y_1, y_2\) be two linearly independent solutions to \(Ly = 0\) such that \(y_1(a) = y_2(b) = 0\). For each fixed \(\xi \in (a,b)\) define \[G(x, \xi) = \frac{1}{\alpha(\xi) W(y_1 m y_2)(\xi)} \times \{y_1(x)y_2(\xi) \text{ if a < x < } \xi, y_1(\xi)y_2(x) \text{if } \xi < x < b\}\] where W is the Wronskian. Then \[L_x[G(x, \xi)] = \delta(x - \xi)\] and \[G(a,\xi) = G(b , \xi) = 0\]
A: Clearly, \(L_x G = 0\) on \(x \neq \xi\) since \(Ly_1 = Ly_2 = 0\) and G is continuous at \(x = \xi\). Also \[\left [\frac{dG}{dx} \right ]_{x = \xi^-}^{x = \xi^+}\] as we know Wronskian is non-zero since the solutions are linearly independent. So this satisfies all the necessary properties and so it is the Dirichlet Green's function for L.

Q: Definition. The Fourier transform of a function f is [...]
A: \[\hat{f}(\lambda) = \mathcal{F}[f](\lambda) = \int_{-\infty}^{\infty} f(x) e^{-i \lambda x} dx\]
Back Extra: The inverse transform is \[f(x) = \frac{1}{2 \pi} \int e^{i\lambda x} \hat{f}(\lambda) d\lambda\]

Q: Definition. Convolution of f and g is defined as [...]
A: \[(f * g)(x) = \int f(x-y) g(y) dy\]
Back Extra: We have \[\mathcal{F}[ f*g(x)] = \hat{f}(\lambda) \hat{g}(\lambda)\]

Q: Definition. Fourier Transform in \(\mathbb{R}^n\). Define \[\hat{f}(\mathbf{\lambda}) = [...]\] \[f(\mathbf{x}) = [...]\]
A: \int e^{-i \mathbf{\lambda \cdot x} } f(\mathbf{x}) d^n \mathbf{x}; \frac{1}{(2\pi)^n} \int e^{i \mathbf{\lambda \cdot x} } \hat{f}(\mathbf{\lambda}) d^n \mathbf{\lambda}
Back Extra: Same results still follow like convolution theorem, differentiation stuff etc.

Q: Proposition. The solution to \[\psi_{tt} - c^2 \Delta \psi = 0, \text{ in } \Omega \times (0,\infty) \]
\[\psi = f, \psi_t = g \text{ on } \Omega \times \{t = 0\} \text{ and } \psi = 0 \text{ on } \partial \Omega \times [0,\infty)\]is unique.
A: If \(\psi_1\) and \(\psi_2\) were two solutions to this then by linearity \(\psi = \psi_1 - \psi_2\) would satisfy this with \(f = g = 0\). This means that the energy \[E(t) = \frac{1}{2} \int_{\Omega} (\psi_t^2 + c^2 | \nabla \psi|^2 ) dV) = \frac{1}{2} \int_{\Omega} (|g|^2 + c^2 |\nabla f|^2 ) dV = 0\]so \(\psi_t = |\nabla \psi | = 0\) on \(\Omega \times (0,\infty)\) meaning \(\psi\) is constant, but since it is 0 on the boundary it must be zero everywhere so \(\psi_1 = \psi_2\)

Q: Proposition. Suppose we have an inhomogenenous problem of the form \[L_xy =f, x \in (a,b) \text{ with } y(a) = y(b) = 0\]where \(L_x\) is a second order linear differential operator, which problem would we like our Green's function to solve and why would this help solve our inital problem and what condition should it satisfy?
A: If we have some \(G = G(x, \xi)\) for each \(\xi \in (a,b)\), that satisfies \[L_x G = \delta(x - \xi), x\in(a,b) \text{ with } G(a,\xi) = G(b ,\xi) = 0\]Then setting \(y(x) = \int_a^b G(x,\xi)f(\xi) d\xi\) this will solve our problem.
We will need G to be continuous at \(x = \xi\) and dG/dx has a jump discontinuity at \(x = \xi\) defined by \[\left [ \frac{dG}{dx}(x, \xi) \right ]_{x=\xi^-}^{x = \xi^+} = \frac{1}{\alpha(\xi)}\]

Q: Proposition. If \(f = f(x)\) has Fourier transform \(\hat{f} = \hat{f}(\lambda)\) then we have the inversion formula \[f(x) = \frac{1}{2\pi} \int e^{i \lambda x} \hat{f}(\lambda) d\lambda\]
A: We do the integral from -n to n and change the order of integration, noting that \[\frac{1}{2\pi} \int_{-n}^n e^{i \lambda (x-y)} d\lambda = \frac{\sin [n(x-y)]}{\pi (x-y)}\]so we get that \[\frac{1}{2\pi} \int e^{i \lambda x} \hat{f}(\lambda) d\lambda -f(x) = \int \sin(ny) \left [ \frac{f(x+y) -f(x)}{\pi y} \right ] dy \to 0\]as \(n \to \infty\) so long as the term in square brackets is integrable, which it will be for nice functions

Q: Proposition. Parseval's Theorem. \[\int f(x) \overline{g(x)} dx = \frac{1}{2\pi} \int \hat{f}(\lambda) \overline{\hat{g}(\lambda)} d\lambda\] and so \[\int |f(x)| ^2 dx = \frac{1}{2\pi} \int |\hat{f}(\lambda)|^2 d\lambda\]
A: Using the definition of the Fourier transform we see the right hand side is \[\frac{1}{2\pi} \int \left [ \int e^{- i \lambda x} f(x) dx\right ] \left [ e^{i \lambda y} \overline{g(y)} dy\right ] d\lambda = \int \int f(x) \overline{g(y)} \left[ \frac{1}{2\pi} \int e^{i\lambda (x-y)} d\lambda\right ] dx \; dy\]and the term in the square brackets is \(\delta(x-y)\) so we get our result.

Q: Proposition. \[\mathcal{F}_{x \to \lambda} [(f * g) (x)] = \hat{f}(\lambda) \hat{g}(\lambda)\]
A: \[\int (f * g)(x)e^{-i \lambda x} dx = \int \left [\int f(x-y)g(y) e^{-i\lambda x} \right ] dy = \int \left[ \int f(x) g(y) e^{-i \lambda (x+y)} dx\right ]dy\]
The final integral on the right hand side can be written as \[\int f(x) e^{-i \lambda x} \left [ \int e^{-i \lambda y} g(y) dy \right ] dx = \hat{f}(\lambda) \hat{g}(\lambda)\]which is our desired result.

Proposition. Important Fourier Transforms.
\[f(x) = H(x)e^{-\sigma x}, \text{ Re }\sigma > 0\iff \hat{f}(\lambda) = \frac{1}{\sigma + i \lambda} \]
\[f(x) = \frac{1}{\sqrt{2 \pi} } e^{-x^2/2}  \iff \hat{f}(\lambda) = e^{- \lambda^2/2} \]
\[f(x) = \delta(x) \iff \hat{f}(\lambda) = 1\]
Back Extra: The last one tells us \[\delta(x) = \frac{1}{2\pi} \int e^{i \lambda x} dx\]

Q: Proposition. Poisson Summation Formula. \[\sum_n f(x+n) = \sum_n \hat{f} (2\pi n)e^{2 \pi i n x}\]
A: Both sides of the equation define a 1 periodic function so we can write both sides as a Fourier series. If F denotes the function on the left and G denotes the function on the right, then clearly \(\hat{G}_n = \hat{f}(2 \pi n)\). Also \[\hat{F}(n) = \int_0^1 \left ( \sum_m f(x+m) \right ) e^{- 2 \pi i n x} = \sum_m \int_0^1 f(x+m) e^{-2 \pi i n x}dx\]Making the substitution \(y = m +x\) we find \[\hat{F}_n = \sum_m \int_m^{m+1} f(y) e^{-2\pi i n y}dy = \int f(y) e^{-2 \pi i n y}dy = \hat{f}(2 \pi n)\]so \(\hat{F}(n) = \hat{G}(n)\) and the functions must be the same.

Q: Proposition. The inversion formula holds for discrete Fourier transforms \[X_n = \frac{1}{N} \sum_{k=0}^{N-1} \hat{X}_k e^{i (2\pi k / N)n}, \; 0 \leq n \leq N\]
A: By direct computation \[\sum_{k=0}^{N-1} \hat{X}_k e^{i (2 \pi k / N)n} = \sum_{k,m = 0}^{N-1} X_m e^{- i(2\pi m /N) k} e^{i (2 \pi k /N)n} = \sum_{k,m=0}^{N-1} X_m e^{2\pi i k(n-m)/N}\]Writing \(\omega_N = e^{2\pi i /N}\) we note that for \(n \neq m\) \[\sum_{k=0}^{N-1}\omega_N^{(n-m)k} = \frac{1 - \omega_N^{N(n-m)}}{1 - \omega_N^{n-m}} = 0\]and clearly if n=m then the same sum evaluates to N. So \[\sum_{k=0}^{N-1}\hat{X}_k e^{i (2 \pi k /N)n} = \sum_{m=0}^{N-1} X_m (N \delta_{mn}) = NX_n\]