Partial Differential Equations

Partial Differential Equations
I took these notes in fall 2016 for CME 303, taught by Professor Lenya Ryzhik.
\tableofcontents
Methods
Method of characteristics
Suppose we have a first-order linear PDE such as $au_x + bu_y = f(x,y),$ with the boundary condition $u(x,0) = g(x)$ . We should think of this as $\partial_su = (a\partial_x + b\partial_y)u = f(x,y),$
which we interpret as saying that the directional derivative $\partial_s$ of $u$ at $(x,y)$ is given by $f(x,y)$ . Let $z(s) = (x(s), y(s))$ be an integral curve (a characteristic) starting at $z(0) = (x,y)$ of the vector field $\partial_s$ , i.e. satisfying $z'(s) = (a, b)$ . Then along this integral curve, the PDE becomes the ODE $(u \circ z)' = f \circ z$ in the variable $s$ . Once we solve it, we have a solution to the PDE given by $u(x,y) = (u \circ z)(0)$ .
Separation of variables
Integral representations
Suppose we have the PDE $Lu = f$ on $\mathbb{R}^n$ . Then if we have the fundamental solution $\Phi(x; y)$ that satisfies $L\Phi(x; y) = \delta(x - y)$ , then we have a solution to the inhomogeneous PDE $u(x) = \int_{\mathbb{R}^n} \Phi(x; y)\,f(y) \,dy = \Phi * f,$ with the last equality assumes $L$ is translation-invariant, allowing us to define $\Phi(x - y) \equiv \Phi(x - y; 0) = \Phi(x; y)$ . We can think of it as the impulse response at $y$ . Note that $\Phi(x;y)$ also satisfies the homogeneous equation $Lu = 0$ everywhere except $x = y$ .
Suppose we have the boundary-value problem $Lu = f$ on $U$ with the boundary condition $Bu = g$ on $\partial U$ . Then we can construct a Green’s function $G(x;y)$ , which satisfies $LG(x;y) = \delta(x - y)$ for $x \in U$ and $BG(x;y) = 0$ for $x \in \partial U$ . We can write it as an interior term minus a boundary term correction: $G(x; y) = \Phi(x; y) - \phi(x; y),$ so that $L\phi(x; y) = 0$ for $x \in U$ and $B\phi(x; y) = B\Phi(x; y)$ for $x \in \partial U$ . The homogeneous PDE for $\phi$ is solvable by some other method. Thus we can think of it as the impulse response at $y$ minus some homogeneous solution to make it zero at the boundary. We can then often write $u(x) = \int_U G(x; y) f(y) \,dy + \text{boundary terms involving }G\text{ and }g,$ derived from a Green’s identity of the form $\int_U (u\,L G - G\, L u) \,dx = \text{boundary terms involving }G\text{ and }u$ and substituting $LG = \delta$ and $Lu = f$ on the left.
Relatedly, we can use Duhamel’s principle to convert an inhomogeneous initial value problem into an integral of solutions to the homogeneous initial value problem. Suppose we have the initial value problem $\partial_tu + Lu = f(t,x)$ with zero initial condition. Then we can decompose the inhomogeneous part as a convolution $\partial_t u + Lu = \int_0^t \delta(t-s) \,f(s,x)\, ds,$ solve the family of PDEs, indexed by $s$ , $\partial_t v^{(s)} + Lv^{(s)} = \delta(t-s)\,f(s,x),$ also solvable as the initial value problem $\partial_t v^{(s)} + Lv^{(s)} = 0, \qquad v^{(s)}(s, x) = f(s,x),$ and arrive at the solution $u(t,x) = \int_0^t v^{(s)}(t,x) \,ds.$
Fourier transform
The Fourier transform transform of a function $f \in L^1(\mathbb{R})$ is the function $\hat{f}(\xi) = \int_{-\infty}^\infty f(x)\,e^{-2\pi i\xi x} \,dx.$
It relates multiplication to derivatives: $\partial_x f \mapsto 2\pi i \xi \hat{f}, \qquad xf \mapsto -\frac{1}{2\pi i}\partial_\xi \hat{f}.$
It relates convolution to multiplication: $f * g \mapsto \hat{f} \hat{g}.$
It relates smoothness to decay: if $f \in C^n(\mathbb{R})$ , then $\xi^n \hat{f}(\xi) \to 0$ as $\xi \to \pm\infty$ .
The Schwartz space $S(\mathbb{R}) \subset C^\infty(\mathbb{R})$ is the space of smooth, rapidly-decreasing functions on $\mathbb{R}$ , i.e. those for which $|x^\alpha \partial^\beta f(x)| < \infty$ for all $\alpha$ and $\beta$ . It is closed under addition and pointwise multiplication. The Fourier transform becomes a linear isomorphism on $S(\mathbb{R})$ with inverse $f(x) = \int_{-\infty}^\infty \hat{f}(\xi)\,e^{2\pi i \xi x} \, d\xi.$
The Gaussian $e^{-\pi x^2}$ is its own Fourier transform.
The Plancherel theorem says that the Fourier transform is unitary on $L^2(\mathbb{R})$ : $\int_{-\infty}^\infty f(x)\,\overline{g(x)} \,dx = \int_{-\infty}^\infty \hat{f}(\xi)\,\overline{\hat{g}(\xi)}\,d\xi.$
Variational methods
If a function $u : U \to \mathbb{R}$ minimizes the functional $I(u)$ over some set $A$ , with $I(u) = \int_U L(u, \nabla u, x) \,dx,$ then it satisfies the PDE given by the Euler-Lagrange equation $\frac{\partial L}{\partial u} - \nabla \cdot \frac{\partial L}{\partial(\nabla u)}= 0.$
Noether’s theorem gives conserved quantities in certain cases, such as the energy $E(t) = \int_{\mathbb{R}^n} (L - \frac{\partial L}{\partial u_t} u_t) \,dx.$
Types of PDEs
Conservation laws
Conservation law: $\partial_t u + \nabla \cdot F(u) = f(t,x)$
Advection equation: $\partial_t u + \mathbf{v} \cdot \nabla u = 0$
Burgers’ equation: $\partial_t u + u \,\partial_x u = 0$
These come from the integral conservation law $\frac{d}{dt}\int_U u \,dx = -\int_{\partial U} F(u) \cdot \mathbf{\hat{n}}\,dy + \int_U f(t,x) \,dx.$
They can frequently be solved using the method of characteristics. They present two difficulties: if characteristics cross (a shock) and if characteristics fail to cover the domain (rarefaction).
In the first case, if there is a discontinuity at time $t$ , then the speed $\sigma$ at which the discontinuity propagates is given by the Rankine-Hugoniot jump condition: $\sigma = \frac{F(u_+) - F(u_-)}{u_+ - u_-}.$
In the second case, we seek an entropy solution, which means that we fill the missing part with characteristics that must not intersect backwards in time.
Elliptic PDEs: Laplace’s equation
Laplace’s equation is $\nabla^2 u = 0$ , where $\nabla^2 = \partial_1^2 + \cdots + \partial_n^2$ .
Poisson’s equation is $-\nabla^2 u = f$ .
A function $u$ is harmonic if it satisfies Laplace’s equation $\nabla^2 u = 0$ .
Intuitive characterizations
- It describes the potential of a charge: $\nabla \cdot \mathbf{E} = \delta$ and set $-\nabla \varphi = \mathbf{E}$ .
- It has a probabilistic interpretation: suppose we are given a boundary condition $g(x)$ ; then $u(x)$ is the expected value of Brownian motion starting at $x$ collecting a reward at the boundary given by $g(x)$ , plus any reward $f(x)$ gathered along the walk.
- It’s the steady-state solution to the heat equation.
Variational characterization
- The unique solution to Poisson’s equation with a prescribed boundary uniquely minimizes $I(u) = \int_U \Big( \frac{1}{2}|\nabla u|^2 - uf \Big) dx.$
Integral representations
- The fundamental solution for $1$ , $2$ , and $n \ge 3$ dimensions is respectively $\Phi(x) = -\frac{1}{2}|x|, \qquad \Phi(x) = -\frac{1}{2\pi}\log |x|, \qquad \Phi(x) = \frac{1}{n(n-2)V_n}\frac{1}{|x|^{n-2}},$ where $V_n$ is the volume of an $n$ -ball.
- For a Dirichlet boundary condition ( $u = g$ on $\partial U$ ), the unique solution is $u(x) = \int_U G(x; y) f(y) \,dy - \int_{\partial U} [\mathbf{\hat{n}}_y \cdot \nabla_y G(x; y)]\, g(y) \, dy,$ where we interpret the normal derivative as $-\mathbf{\hat{n}}_y \cdot \nabla_y G(x;y) \approx \frac{1}{\epsilon} [G(x; y - \epsilon \mathbf{\hat{n}}) - G(x;y)] = \frac{1}{\epsilon}G(x; y - \epsilon \mathbf{\hat{n}}) \to \delta(x - y).$
Properties of harmonic functions
- Mean value property: $u * h = u$ for any spherically symmetric kernel $h$ , i.e. an $h$ such that $\int_U h = 1$ and $h = h(r)$ . Concretely, $u(x)$ is the average of $u$ on a sphere around $x$ (for all radii contained in $U$ ) iff $u$ is harmonic.
  - \proof Define $\phi(r) = \frac{1}{|S^{n-1}|} \int_{S^{n-1}} dz\, u(x + rz).$ Clearly $\phi(r) = u(x)$ , and $\begin{aligned} \phi'(r) &= \frac{1}{|S^{n-1}|} \int_{S^{n-1}} dz\, z\cdot \nabla u(x + rz) \\ &= \frac{r}{|S^{n-1}|} \int_{B^{n-1}} dz\, \nabla \cdot \nabla u(x + rz) \\ &= 0, \end{aligned}$ by the divergence theorem and because $\nabla^2 u = 0$ . Then, $\begin{aligned} u * h &= \int_{B^{n-1}} dz\,u(x - z) h(z) \\ &= \int_0^1 dr\,h(r)\, \int_{S^{n-1}} dz\,u(x - rz) \\ &= u(x). \end{aligned}$
- Maximum principle: $u$ only attains its maximum and minimum on the boundary of a bounded domain $U$ unless $u$ is constant.
  - \proof Consider $u$ as a function of $U$ (the interior). Suppose $u$ attains its maximum $M$ , and let $S = u^{-1}(\{ M \})$ . Since $u$ is continuous, $S$ is closed. Now, for each $x \in S$ , consider an open ball around $x$ . The average of $u$ in that ball is $M$ by the mean-value principle, so $S$ is open. If $U$ is connected, then $S$ is either empty or the whole interior.
- Uniqueness: There is a unique solution to Poisson’s equation with a prescribed boundary conditions.
  - \proof The difference of two solutions is harmonic with zero boundary condition, so by the maximum principle it is $0$ .
- Regularity: $u$ is smooth (and in fact analytic).
  - This is related to the fact that holomorphic functions have harmonic real and imaginary parts!
  - \proof Write $u = u * h$ by the mean value property where we choose $h$ to be smooth. The convolution of a smooth function with another function is smooth.
- Liouville theorem: $u$ is unbounded in $\mathbb{R}^n$ unless $u$ is a constant.
- Harnack’s inequality: if you fix a connected neighborhood $V$ and $u$ is non-negative, then the sup and inf of $u$ are within a constant factor of each other.
Parabolic PDEs: the heat equation
The heat equation is $\partial_t u - \nabla^2 u = 0$ .
Variational characterization
- The energy $E(t) = \frac{1}{2}\int_U u^2 \,dx.$ decreases as a function of time.
Integral representations
- The fundamental solution, a.k.a. the heat kernel, is defined for $t > 0$ as $\Phi(t,x) = \frac{1}{(4\pi t)^{n/2}} e^{-|x|^2/4t}.$ We may obtain this from the Fourier transform.
- For the initial value problem with $u(0, x) = g(x)$ , the solution is $u(t,x) = \int_{\mathbb{R}^n} \Phi(t, x-y)\,g(y)\,dy = \Phi \,*_x\, g,$ since $\Phi(0, x) = \delta(x)$ .
Properties
- Instantaneous regularity: If $g$ is bounded and continuous, then $u$ is smooth for $t > 0$ .
  - \proof Write $u = \Phi \,*_x\, g$ , and notice that $\Phi$ is smooth for $t > 0$ .
- Maximum principle for a bounded domain: Define $U_T = U \times (0, T]$ and its parabolic boundary $\Gamma_T = \bar{U}_T \setminus U_T$ . Then $u$ only achieves its maximum and minimum over $\bar{U}_T$ on $\Gamma_T$ , unless $u$ is constant.
  - \proof Set $v = u - \epsilon t$ , so that it satisfies $v_t - \nabla^2 v = - \epsilon$ . The maximum of $v$ occurs on $\Gamma_T$ , because if the maximum is in $U_T$ , then $v_t = 0$ or ( $v_t \ge 0$ if the maximum is at $t = T$ ) and $\nabla^2 v \le 0$ . Then $\max_{\bar{U}_T} u \le \epsilon T + \max_{\bar{U}_T} v = \epsilon T + \max_{\Gamma_T} v \le \epsilon T + \max_{\Gamma_T} u.$
- Maximum principle: $u$ only achieves its maximum and minimum over $\mathbb{R}^n$ at $t = 0$ , assuming that $u(t,x) \le Ae^{a|x|^2}$ , unless $u$ is constant.
- Infinite speed of propagation: If $g \ge 0$ is not identically $0$ and is bounded and continuous, then $u(t,x) > 0$ for all $x$ and $t > 0$ .
  - \proof Maximum principle.
- Uniqueness: Given an initial condition (and a boundary condition if the domain is bounded), there is at most one solution to the inhomogeneous heat equation (assuming the growth estimate for $\mathbb{R}^n$ case).
  - \proof Apply the maximum principle to $u - v$ , implying that $u-v = 0$ .
  - \proof The energy of $u-v$ is identically $0$ , implying that $u-v = 0$ .
- Estimates: $\begin{aligned} |u(t,x)| &\le \frac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} |g(y)|\,dy \\ |\nabla u(t,x)| &\le \frac{C}{t^{(n+1)/2}} \int_{\mathbb{R}^n} |g(y)|\,dy \end{aligned}$
  - \proof Use the explicit formula for the solution (convolution with the heat kernel).
  - \proof Fourier transform.
- Self-similarity: As $t \to \infty$ , $u(t,x) \sim \frac{1}{(4\pi t)^{n/2}} e^{-|x|^2/ 4t} \int_{\mathbb{R}^n} |g(y)|\,dy.$
  - \proof From the formula $u(t,x) = \int_{\R^n} \hat{g}(\xi) e^{-4\pi^2 |\xi|^2 t} e^{2\pi i\xi x} \,d\xi,$ make the change of variable $k = \xi \sqrt{4\pi t}$ , so we have $\begin{aligned} u(t,x) &= \frac{1}{(4\pi t)^{n/2}}\int_{\R^n} \hat{g}\Big(\frac{k}{\sqrt{4\pi t}}\Big) e^{-\pi |k|^2 } e^{2\pi ik (x/\sqrt{4\pi t})} \,d\xi, \\ &\sim \frac{\hat{g}(0)}{(4\pi t)^{n/2}}\int_{\R^n} e^{-\pi |k|^2 } e^{2\pi ik (x/\sqrt{4\pi t})} \,d\xi, \\ &= \frac{\hat{g}(0)}{(4\pi t)^{n/2}} e^{- |x|^2/4t }. \end{aligned}$
Hyperbolic PDEs: the wave equation
The wave equation is $u_{tt} - c^2 \nabla u = 0$ .
Variational characterization
- Solutions to the wave equation minimize the functional $I(u) = \frac{1}{2}\int_0^\infty\int_{\mathbb{R}^n} (u_t^2 - c^2 |\nabla u|^2) \,dx\,dt.$
- The energy $E(t) = \frac{1}{2}\int_{\mathbb{R}^n} (u_t^2 + c^2 |\nabla u|^2) \,dx.$ is constant as a function of time.
Integral representations
- For the one-dimensional wave equation with Cauchy initial conditions $u(0, x) = g(x)$ and $u_t(0,x) = h(x)$ , we have the d’Alembert formula $u(t,x) = \frac{1}{2}\left( g(x-ct) + g(x+ct) \right) + \frac{1}{2c}\int_{x-ct}^{x+ct} h(y) \,dy,$ (for example, use Fourier transform).
Properties
- Finite speed of propagation: The value $u(t_0, x_0)$ only depends on the values of the initial conditions within a ball around $x_0$ of radius $ct_0$ .
  - \proof Suppose $u_1$ and $u_2$ have the same initial conditions inside the ball, and consider the energy of $u_1 - u_2$ within the ball. The energy is non-negative, and it is initially $0$ . Show that it decreases (not obvious); conclude that $u_1 = u_2$ in the ball.
- Huygens principle: In odd dimensions, the value $u(t_0, x_0)$ only depends on the values of the initial conditions on the sphere around $x_0$ of radius $ct_0$ .
- Uniqueness: There is at most one solution to the Cauchy initial value problem (with $u(0, x)$ and $u_t(0, x)$ specified).
  - \proof $u_1 - u_2$ satisfies the wave equation with zero initial condition, so its energy is zero.
References
https://en.wikibooks.org/wiki/Partial_Differential_Equations/Poisson%27s_equation
http://math.arizona.edu/~kglasner/math456/greens.pdf
http://www.maths.tcd.ie/~ormondca/stuff/fyp.pdf