Grading Breakdown

  • HW: 40%
    • Assigned on Monday weekly. Due next monday
  • Exams: 20% each
  • Textbook: Haberman: Applied Partial Differential Equations

Heat Equation

  • Basic physical principle: Convervation of energy
    • The change in energy over time = flow of heat energy (flux) across the boundaries
    • The total heat energy at time t is the integral of the temperature over the domain: $E = \int_{a}^{b} u(y,t) dy$
    • The time derivative w.r.t. E then equals the heat flux as per Fourier’s law $\epsilon(x,t) = -K(x)\frac{\partial u}{\partial x}$
      • $K(x)$ is the thermal conductivity
    • Therefore, heat equation becomes: $\frac{\partial u}{\partial t} = -K(x) \frac{\partial^{2} u}{\partial x^{2}}$
  • To get a complete solution you need initial conditions and boundary conditions
    • Initial conditions: $u(x,t=t{0}) = f(x)$
    • Boundary conditions:
      • Dirichlet: Temperature is prescribed
      • Neumann: Derivative of temperature is perscribed

Steady state solutions

  • Assuming that the steady state exists, just drop all time derivatives to reduce 1D heat equation to ODE.
  • Dirichlet: Solution becomes a linear interpolation between ends of rod
  • Neumann: Need to use conservation of heat energy to relate initial energy (derived from integrating initial condition) to final energy (derived from integrating steady state solution)
  • To get something physical, we also demand the following:
    • Temperature is continuous ($lim_{x\rightarrow x_{0}-} u = lim_{x\rightarrow x_{0}+} u$)
    • Heat flow is continuous ($lim_{x\rightarrow x_{0}-} K(x) \frac{\partial u}{\partial x} = lim_{x\rightarrow x_{0}+} K(x) \frac{\partial u}{\partial x}$)

Time Dependent Solution

Seperation of Variables

  • Assume that the tempurature can be split into a time dependent part and a space dependent part: $u(x,t) = X(x)T(t)$
  • Suppose that we have the boundary conditions $u(0,t) = u(L,t) = 0$. This implies that $X(0) = X(L) = 0$
  • Plugging in potential solution results in two ODEs that equal to each other. This implies that each individually equals a constant. rearranging yields:
    • $X^{’’} = c X$
      • If c<0, then solutions of the form $X(x) = A exp(i\lambda x)+ B exp(-i\lambda x)$ where $-\lambda^{2} = c$
      • If c>0, then solutions of the form $X(x) = A exp(\lambda x)+B exp(-\lambda x)$ where $\lambda^{2} = c$
    • $T^{’} = cKT$
      • Solutions of the form $T(t) = T(0)exp(ckt)$
  • For the boundary conditions that we have, only the oscillating solutions are physical. Can solve for A,B and c via boundary conditions
  • General solution comes from taking a linear combination of these solutions:
    • $u(x,t) = \Sigma_{n=0}^{n=\infty} c_{n} T(t)X(x)$
    • $c_{n}$ can be computed using the orthogonality of sin and cos:
      • $\int_{-L}^{L} \sin(\frac{n\pi x}{L})\cos(\frac{m\pi x}{L}) dx = 0$
      • $\int_{-L}^{L} \cos(\frac{n\pi x}{L})\cos(\frac{m\pi x}{L}) dx = \frac{L}{2}\delta_{nm}$
      • $\int_{-L}^{L} \sin(\frac{n\pi x}{L})\sin(\frac{m\pi x}{L}) dx = \frac{L}{2}\delta_{nm}$

Laplace Equation

  • $\nabla^{2} u = \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial t^{2}} = 0$
    • Suppose that the following Dirchlet BC’s hold:
      • $u(0,y) = f_{1}(y)$
      • $u(L,y) = f_{2}(y)$
      • $u(x,0) = g_{1}(x)$
      • $u(0,L) = g_{2}(x)$
  • Using seperation of variables, we know from the heat equation that the following solution works:
    • $u = v(x)w(y)$
    • $w(y) = \sin(\frac{n\pi y}{L})$
    • $v(x) = C_{1}exp(cx)+C_{2}exp(-cx)$
      • $c = (\frac{n\pi}{L})^{2}$
    • General solution is then $u(x,y) = \Sigma A_{n} \sinh(\frac{n\pi x}{L}) \sin(\frac{n\pi y}{L})$

Polar Coordinates

  • $\nabla^{2} u = \frac{1}{r}\frac{\partial}{\partial r}(r \frac{\partial u}{\partial r})+ \frac{1}{r^{2}}\frac{\partial^{2} u }{\partial \theta^{2}} = 0$
  • BCs:
    • $u(r,\pi) = u(r,-\pi)$
    • $\frac{\partial}{\partial \theta}u(r,\pi) = \frac{\partial}{\partial \theta}u(r,-\pi)$
  • Using separation of variables: $u = R(r)\Theta(\theta)$
    • $u(r,\theta) = \Sigma_{n=0}^{\infty} A_{n} r^{n} \cos(n\theta)+ \Sigma_{n=0}^{\infty} B_{n} r^{n} \sin(n\theta)$

Mean Value Property

  • Assuming Dirchlet BCs, the value at center of disc is equal to the average value on the edge of the disk: $u(0,\theta) = \frac{1}{2\pi}\int_{-\pi}^{\pi} u(\alpha,\theta) d\theta$
    • This holds for all sub disks centered on the disk (mean value property)
  • Let $U(w) = \frac{1}{2\pi}\int u(x_{0}+w \cos (\theta),y_{0}+w \sin (\theta)) d\theta$
    • This is the average value on a circle of radius 2 centered at $x_{0},y_{0}$
  • Taking $\frac{d}{dw}$ of the above, applying chain rule, and rewriting as a dot product
    • Making the substitution $\vec{n} = (\cos\theta,\sin\theta)$ and applying the divergence theorem, and applying Laplace’s equation $\nabla^{2} u = 0$, we see that $\frac{dU}{dw} = 0$. Hence, the mean value property holds on any disk on the region

Maximum Principle

  • Laplace’s equation can only have an extrema on the boundary of the domain (unless it is constant everywhere)

Fourier Series

  • Assuming that a function is bounded and piecewise continuous, then it can be approximated as a Fourier series.
    • If we bound the function to the domain [0,L], then the function can be written as
    • $f(x) = a_{0}+\Sigma a_{n}\cos(\frac{n\pi x}{L})+\Sigma b_{n} \sin(\frac{n\pi x}{L})$
    • the coefficients can be found via Fourier trick
    • In the case of jump discontinuities, then the Fourier series converges to the average of the left and the right limits
  • We can extend the function to the domain [-L,L] in two ways: in an even manner (cosine series) or an odd manner (sine series)
  • If f(x) is bounded and is smooth on [-L,L]
    • The fourier series is continuous and converges to f(x) iff f(x) is continuous and $f(-L) = f(L)$
    • The Fourier cosine series is continuous and converges to f(x) on the interval [0,L] iff f(x) is continuous on interval (assuming even extension)
    • The Fourier sine series is continuous and converges to f(x) on the interval [0,L] iff f(x) is continuous on interval and f(0) = F(L) = 0 (assuming odd extension)
  • If f(x) solves a PDE, then it’s Fourier series also solves the PDE (and vice versa)

Convergence

  • The Fourier series converges to f(x) iff it is bounded and piecewise smooth
    • Bounded means there is a constant $M\geq 0$ such that $|f(x)|<M$ on the entire domain of f
    • piecewise smooth means that the interval can be broken up into a finite number of pieces, were on each subinterval, f(x) and f’(x) are continuous
  • On an interval from -L to L, the Fourier series converges if f is continous, or it converges to the average of the left and right limits of f has a jump discontinuity
    • This extends to the periodic extension of f(x)

Term by Term Differentiation

  • Useful for solving inhomogeneous PDEs
  • The following are necessary conditions for term by term differentiation to be valid
    • f(x) continuous on interval $[-L,L]$
    • f(L) = f(-L)

Solving Inhomogeneous Equations

  • Suppose that we have a nonhomogenous heat equation with source g(x)
  • $\frac{\partial u}{\partial t} = \kappa\frac{\partial^{2} u}{\partial x^{2}}-\alpha u +g(x)$
    • IC $u(x,0) = f(x)$
    • BC $\frac{\partial u}{\partial t}(0,t) = 0 = \frac{\partial u}{\partial y}(L,t)$
  • Due to the Neumann BCs, assume a consine series solution exists for f(x), g(x) and u(x)
  • Use term by term differentiation to solve for coefficients of $u(t)$

Wave Equation

  • $\frac{\partial^{2} u}{\partial t^{2}} = c^{2} \frac{\partial^{2} u}{\partial x^{2}}$
    • c defines the wave speed
  • Suppose BCs where u is fixed to 0 at x=0 and x=L
  • Use separation of variables to solve problem
    • $u(x,t) = \Sigma_{n=1}^{\infty} [A_{n}\cos(\frac{n\pi c t}{L})+B_{n}\sin(\frac{n\pi c t}{L})] \sin(\frac{n\pi x}{L})$
      • $A_{n}$ and $B_{n}$ come from IC via Fourier trick
  • Can rewrite above in terms of normal modes (ie. standing wavees)
    • $u(x,t) = \Sigma_{n=1}^{\infty} [\sqrt{A_{n}+B_{n}}\sin(\omega t+\theta)] \sin(\frac{n\pi x}{L})$
      • $\omega = \frac{n\pi c}{L}$
      • $\theta = tan^{-1}(\frac{A_{n}}{B_{n}})$
    • Each standing wave can be rewritten as two travelling waves
    • $\sin(\frac{n\pi x}{L})\sin(\frac{n\pi c t}{L}) = \frac{1}{2}[\cos(\frac{n\pi}{L}(x-ct))-\cos(\frac{n\pi}{L}(x+ct))]$
  • Total energy is conserved:
    • $E(t) = \frac{1}{2}\int|\frac{\partial u}{\partial t}|^{2} dx+\frac{c^{2}}{2}\int |\frac{\partial u}{\partial x}|^{2} dx$

Strum Liouville Theory

  • Strum Liouville problem defined as
    • $\mathbb{L}[\phi](x) = \lambda \sigma(x) \phi(x)$
      • $\mathbb{L}[\phi](x) = -\frac{d}{dx} (p(x) \frac{d\phi}{dx})-q(x)\phi(x)$
        • p,q, $\sigma$ are real functions on interval, including boundary
        • p(x) >0 and $\sigma(x)$ >0 for all x in open interval
        • Additional constraint of $q(x) \geq 0$ defines a regular SL problem
      • BCs
        • $\beta_{1} \phi(a) +\beta_{2} \frac{d\phi}{dx}(a) = 0$
        • $\beta_{3} \phi(b) +\beta_{4} \frac{d\phi}{dx}(b) = 0$
          • $|\beta_{1}|^{2}+|\beta_{2}|^{2} \neq 0$
          • $|\beta_{1}|^{3}+|\beta_{1}|^{4} \neq 0$
  • One can transform a general 2nd order ODE to Sl problem
    • Let $R(x)\frac{d^{2}\phi}{dx^{2}}+P(x) \frac{d\phi}{dx}+Q(x) \phi = \lambda \delta(x) \phi$
    • The above becomes an SL problem, where
      • $p(x) = exp(\int \frac{P(x)}{R(x)})$
      • $q(x) = p(x) \frac{Q(x)}{R(x)}$
      • $\sigma(x) = p(x) \frac{\delta (x)}{R(x)}$

SL properties

  • All eigenvalues $\lambda$ are real
  • There are infinitely many eigenvalues with some hierarchy such that $\lambda_{n} < \lambda_{n+1}$
    • Hence, there is some smallest eigenvalue $\lambda_{1}$
    • $\lambda_{n}$ goes to infinity as n goes to infinity
  • For each $\lambda_{n}$, there exists some $\phi_{n}(x)$ as it’s associated eigenfunction with exactly $n-1$ zeroes on the interval
  • The eigenfunctions for a complete basis, so that any function on the interval can be written as $f(x) = \Sigma_{n=1}^{\infty} a_{n} \phi_{n}(x)$
  • The eigenfunctions follow an orthogonality relationship
    • $\int_{a}^{b} \phi_{n} \phi_{m} \sigma dx = A \delta_{mn}$
  • The Rayleigh quotient relates the eigenfunction to it’s eigenvalue:
    • $\lambda = \frac{-p \phi \frac{d\phi }{d x} \bigg |_{a}^{b} +\int _{a}^{b} [p(\frac{d\phi}{dx})^{2}-q\phi^2]}{\int _{a}^{b} \phi^{2} \sigma dx}$

Green’s Identity

  • $\mathbb{L}[\phi](x) = -\frac{d}{dx} (p(x) \frac{d\phi}{dx})-q(x)\phi(x)$
  • The following holds for SL problems for any two functions u,v
    • $u \mathbb{L}[v]-v\mathbb{L}[u] = -\frac{d}{dx}(p(x) (u\frac{dv}{dx}- v\frac{du}{dx}))$
  • If $\int _{a}^{b} u \mathbb{L}[v]-v\mathbb{L}[u] = 0$, then $\mathbb{L}$ is said to be self-adjoint
  • With regular SL BCs, $\mathbb{L}$ is always self adjoint

Minimization Principle

  • For $n >1$ for regular SL problems, we have that
    • $\lambda_{r} = min RQ[v] = min_{v} \frac{\int _{a}^{b} v\mathbb{L}[v]dx}{\int _{a}^{b} v^{2} \sigma dx}$
      • where we are minimizing over all functions v which satisfy the BCs
    • Hence, if we find some v which satisfies the BCs, we can set an upper bound on $\lambda$
    • For reasonable upper bounds, v should be greater than 0 over entire domain

Parseval’s Identity

  • $\int_{a}^{b} |f(x)|^{2} \sigma(x) dx = \Sigma_{n=1}^{\infty} a_{n}^{2} \int_{a}^{b} |\phi_{n}|^{2} \sigma(x) dx$
    • $f(x) = \Sigma_{n=1}^{\infty} a_{n} \phi_{n} (x)$

Bessel’s Identity

  • $\int_{a}^{b} |f(x)|^{2} \sigma(x) dx \geq \Sigma_{n=1}^{M} a_{n} \int_{a}^{b} |\phi_{n}|^{2} \sigma dx$

Multi-Dimensional PDEs

  • $\frac{\partial y}{\partial t} = \kappa \nabla^{2} u$ (heat equation)
  • $\frac{\partial^{2} u}{\partial t^{2}} = c^{2} \nabla^{2} u$ (wave equation)
  • $\nabla^{2} = 0$ (Laplace equation)
  • Some homogeneous problems with seperation of variables just like before
  • For homogeneous problems with source terms, expand the source term w.r.t the eigenfunctions of the normal homogeneous case, expand the ICs the same way, then solve for the coefficients using term by term differentiation

Green’s Identity (Higher Dimensions)

  • let $\mathbb{L}[u] = \nabla^{2} u $
  • Recall that $<x,y> = \int x y dV$
  • $<u, \Delta v> -<\Delta u, v> = \int [u(\Delta v \cdot \vec{n})-v(\Delta u\cdot\vec{n})] dS$
  • The associated Rayleigh Quotient is then $RQ[u] = \frac{<u, \mathbb{L[u]}>}{<u,u>_{\sigma}}$

Polar Coordinates

  • Suppose we have the eigenproblem $-\nabla^{2} u = \lambda u$ in polar coordinates
  • Using seperation of variables, we must solve the following ODEs
    • $\frac{r}{f} \frac{d}{dr}(r \frac{df}{dr}) + \lambda r^{2} = \frac{-1}{g} \frac{d^{2}g}{d\theta^{2}} = \mu$
  • We recognize the r diff eq. as an SL problem, with $p = r$, $q = \frac{-\mu^{2}}{r}$ and $\sigma(r) = r$

Bessel’s equation

  • Make the change of variables $f(r) = h(\sqrt{\lambda} r)$ to arrive at the equation
    • $z^{2} \frac{d^{2} h}{dz^{2}} + z \frac{dh}{dz} + (z^{2}-\mu^{2})h = 0$ (called Bessel’s equation)
    • The general solution to Bessel’s equation is $h(z) = c_{1} J_{m}(z) +c_{2}Y_{m}(z)$
  • $J_{m}$ are called Bessel functions of the first kind, and have well defined asymptotic behavior as z approaches 0:
    • $J_{m}(z) \approx 1$ if $m=0$ as z approaches 0
    • $J_{m}(z) \approx \frac{z^{m}}{2^{m}m!}$ if $m >0$ as z approaches 0
    • We denote the roots of $J_{mn}(z)$ as $\rho_{mn}$
  • $Y_{m}$ is unbounded at z = 0
    • $Y_{m}(z) \approx \frac{2}{\phi} ln(z)$ if $m=0$ as z approaches 0
    • $Y_{m}(z) \approx \frac{-2^{m} (m-1)!}{\pi z^{m}}$ if $m >0$ as z approaches 0

Fourier Transforms

  • We define the Fourier transform (in this class) as follows:

    • $F(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x) exp(i\omega x) dx$
      • Can derive the above by thinking of Fourier sine series, and extending period to infinity
      • The $2\pi$ is associated with the integral from f(t) to $F(\omega)$ and not the other way around (kind of arbitrary, but whatever)
    • A result of the above is that $\frac{f(x+)+f(x-)}{2} = \int_{-\infty}^{\infty} F(\omega) e^(i\omega x)$

  • Using the above, one can write solutions to heat equation $\frac{\partial u}{\partial t} = k \frac{\partial^{2} u}{\partial x^{2}}$ as

    • $u(x,t) = \int_{-\infty}^{\infty} c(\omega) exp(-i\omega x) exp(-k\omega^{2} t) d\omega$
    • $c(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty} f(x) exp(i\omega x) dx$
      • where $f(x) = u(x,0)$

  • Alternatively, can write in terms of Green’s function:

    • $u(x,t) = \int_{-infty}^{\infty} f(x) \frac{1}{\sqrt{4\pi k t}} exp(\frac{-(x-\bar{x})^{2}}{4kt}) d\bar{x}$
      • $G(x,t,\bar{x},0) = \frac{1}{\sqrt{4\pi k t}} exp(\frac{-(x-\bar{x})^{2}}{4kt})$
        • Fun fact: $lim_{t\rightarrow 0} \frac{1}{\sqrt{4\pi k t}} exp(\frac{-(x-\bar{x})^{2}}{4kt}) = \delta(x-\bar{x})$

  • Parseval’s identity relates the energy content of the function to it’s Fourier Transform

    • $\frac{1}{2\pi}\int_{-\infty}^{\infty} |g(x)|^{2} dx = \int_{-\infty}^{\infty} |G(\omega)|^{2} d\omega$

  • Table of useful Fourier Transform Identities

    Function Fourier Transform
    $exp(-\alpha x^{2})$ $\frac{1}{\sqrt{4\pi a}} exp(-\frac{-\omega^{2}}{4\alpha})$
    $\sqrt{\frac{\pi}{\beta}}exp(\frac{-x^{2}}{4\beta})$ $exp(-\beta \omega^{2})$
    $\frac{\partial f}{\partial t}$ $\frac{\partial F}{\partial t}$
    $\frac{\partial f}{\partial x}$ $-i\omega F(\omega)$
    $\frac{\partial^{2} f}{\partial x^{2}}$ $(-i\omega)^{2} F(\omega)$
    $\frac{1}{2\pi}\int_{-infty}^{\infty} f(\bar{x})g(x-\bar{x}) d\bar{x}$ $F(\omega) G(\omega)$
    $\delta(x-x_0)$ $\frac{1}{2\pi} exp(i\omega x_{0})$
    $f(x-\beta)$ $exp(i\omega \beta) F(\omega)$
    $xf(x)$ $-i \frac{dF}{d\omega}$
    $\frac{2\alpha}{x^{2}\alpha^{2}}$ $exp(-\alpha|\omega|)$

Green’s Functions

Heat Equation Case Study

  • Suppose that we look at the heat equation with sources. The solution is
  • $u(x,t) = \Sigma_{n=1}^{\infty} a_{n}(t) \sin(\frac{n\pi x}{L}) e^{-k(\frac{n\pi}{L})^{2}}$
    • Solve for $a_n$ via Fourier Trick and coefficient matching to create 1st order ODE (expand Q(x,t) via eigendecomposition)
    • swap order of sum and integration,and relabel some things to get the following Green’s function:
  • $G(x,t,x_0,t_0) = \Sigma_{n=1}^{\infty} \frac{2}{L} \sin (\frac{n\pi x}{L})\sin (\frac{n\pi x_0}{L}) e^{-k(\frac{n\pi}{L})^{2}(t-t_0)}$
    • Can think of Green’s function as the system’s response to a delta function input
    • Final solution is then $u(x,t) = \int_{0}^{L} g(x_{0}) G(x,t,x_{0},0) + \int_{0}^{L}\int_{0}^{t} Q(x_0,t_0) G(x,t,x_{0},t_0) dt_{0} dx_{0}$
  • Maxwell Reciprocity: $G(x_1,x_2) = G(x_2,x_1)$

Dirac Delta

  • $\delta(x-x_i) = 0$ everywhere except at $x_i$, where it is infinite
  • $f(x) = \int f(x_{i}) \delta(x-x_{i}) dx_{i}$
  • Dirac delta function has unit area
  • Dirac delta function is even
  • Heaviside function $H(x-x_{i})$ = 0 $x<x_{i}$ and 1 otherwise
    • $H(x-x_i) = \int_{-\infty}^{x} \delta(x_{0}-x_{i}) dx_{0}$
  • Delta function scales as follows:
    • $\delta(c(x-x_i)) = \frac{1}{|c|} \delta (x-x_i)$

Green’s Function via Differential Equation

  • Defining ODE is $L(G) = \delta(x-x_0)$
    • Has homogeneous BCs
  • Integrate, and solve directly

Nonhomogeneous BCs

  • Use same Green’s function as homogeneous case. Utilize Green’s formula with v=G to get
  • $u(x) = \int_{0}^{L} f(x_0)G(x,x_0) dx_0 + u(L) \frac{dG}{dx_{0}}(x=L) - u(0) \frac{dG}{dx_0}(x=0)$

Green’s Function via Eigenvalue Expansion

  • Suppose that $L(u) = f(x)$. There is an associated eigenvalue problem $L(\phi) = -\lambda \sigma \phi$ subject to the same homogenous boundary conditions.
  • The above is solved via a eigenvalue expansion $u(x) = \Sigma_{n=1}^{\infty} a_{n}\phi_{n}(x)$. Plugging into eigenvalue problem, and abusing orthogonality yields
    • $-a_{n}\lambda_{n} = \frac{\int_{a}^{b} f(x) \phi_{n} dx}{\int_{a}^{b}\phi_{n}^{2}\sigma dx}$
    • Can substitute this formula for coefficient to back substitute into ansatz. Swapping sum and integral yields Green’s function
      • $G(x,x_{0}) = \Sigma_{n=1}^{\infty} \frac{\phi_{n}(x_{0})}{-\lambda_{n}\int_{a}^{b} \phi_{n}^{2}\sigma dx}$
      • Green’s function does NOT exist is $\lambda_{n} = 0$

Fredholm Alternative

  • When is $\lambda=0$ an eigenvalue of an differential equation? What can that tell us about possible solutions of a PDE?
  • Fredholm’s alternative states how this question can be answered for non-homogeneous problems.
  • From the eigenvalue
  • Suppose that $L(u) = f(x)$ is subject to homogeneous boundary conditions instead of non-homogeneous ones
  • Either: $u=0$ is the only homogenous solution, which means nonhomogeneous problem has unique solution
  • Or: there are nontrivial homogenous solutions ($\lambda=0$), which means that either no solutions exists, or an infinite number of solutions exist
    • This branch happens when $\int_{a}^{b} f(x)\phi_{n}(x) dx = 0$
    • Another way of saying the above is that the nonhomogeneous problem has a solution iff the forcing function is orthogonal to the eigenstates
  • A table summarizing the Fredholm Alternative
    Eigenpairs # $\int_{a}^{b} f(x)\phi_{n}(x) dx$
    $\phi_{n}=0 (\lambda\neq 0)$ 1 0
    $\phi_{n}\neq 0 (\lambda = 0)$ $\infty$ 0
    $\phi_{n}\neq 0 (\lambda = 0)$ = $\neq 0$