Casey Chu

- 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