Lecture Notes

Recall the first time we dipped in Chapter 13, partial differential equations, we discussed about the 'method of separation of variables'. We covered Boas 13.1 and 13.2. Today, we plan to talk about 13.3 and 13.4, the heat equation, Schroedinger equation and the wave equation.

$$ \d_t u = \Delta u $$ This equation discribe the diffusion of heat.

Consider 1-dimensional example, on a circle. Then function to be solved is $u(t, \theta)$ with $\theta \in [0, 2\pi]$. Suppose we are given the intial conditino $u_0(\theta)$ at $t=0$, then we try to solve for the general solution.

The basic idea is to find the eigenvalue of the operator $\Delta$ acting on function on $S^1$. The eigenvalues are $$ \lambda = 0 \z{ and } -n^2, n=1,2,\cdots $$ The eigen-function (up to scale) for $\lambda=0$ is $1$, and the eigenfunctino for $\lambda = -n^2$ is $\sin(n\theta)$ and $\cos(n\theta)$.

Hence, we have the following solution to the heat equation (ignoring the initial condition for now) $$1, \z{ and } e^{-n^2 t} \sin(n \theta), e^{-n^2 t} \cos(n \theta) $$ Thus, if we decompose the initial condition $u_0$ as $$ u_0(\theta) = c + \sum_{n = 1}^\infty a_n \cos(n \theta) + b_n \sin(n \theta)$$ then we have $$ u(t, \theta) = c + \sum_{n = 1}^\infty a_n e^{-n^2 t} \cos(n \theta) + b_n e^{-n^2 t} \sin(n \theta).$$

Let $u = u(t,x,y)$, $x \in [0,1], y \in [0,1]$. $$ \d_t u = \d_x^2 u + \d_y^2 u$$ and suppose the temperature on the boundary of the interval is held at a constant temperature $T$, and we have initial condition $u_0(x,y)$ given, compatible with the boundary condition. Let's try to find the evolution.

Let's first remove the boundary condition. We may write $$ u(t,x,y) = T + v(t,x,y) $$ then $v$ satisfies the same equation as $u$, and with initial condition given $v(0,x,y) = u_0(x,y) - T.$ We may now write down the basis of general solution to the equation (ignoring the initial condition) $$ v_{n,m}(t,x,y) = e^{- \pi^2 (n^2 + m^2) t} (\sin (n \pi x) \sin(n \pi y)), \quad n, m \geq 1 $$ Then the general solution can be written as $$ v(x,y) = \sum_{n,m} a_{n,m} v_{n,m}(t,x,y)$$ To fix the coefficients $v_{n,m}$, we use the initial conditions $$ v(0,x,y) = \sum_{n,m} a_{n,m} v_{n,m}(0,x,y) $$ so multiply both sides by $v_{n,m}(0,x,y)$ and integrate, only one term on the RHS contribute, and we get $$a_{n,m} = \frac{\int_{[0,1]^2} v(0,x,y) v_{n,m}(0,x,y) dx dy}{\int_{[0,1]^2} v^2_{n,m}(0,x,y) dx dy}.$$

Remark: if the boundary temperature is not constant $T$ (but still time-independent), we may still find a special solution first, a function $U(x,y)$ that satisfies the boundary condition, and $\Delta U(x,y) = 0$. Such function $U(x,y)$ exists and is unique, it is called the harmonic extension of the boundary value to the interior. Then we can still get rid of the boundary condition by setting $$ u(t,x,y) = U(x,y) + v(t,x,y) $$ where $v(t,x,y)$ now has boundary condition $0$, and initial condition $v(0,x,y) = u(0,x,y) - U(x,y)$.

$$ i \d_t u = - \Delta u $$

We may reuse the analysis for the heat equation, except replacing $t$ in heat equation to $it$. Thus, exponential decay now become oscillation.

$$ \d_t^2 u = \Delta u$$ Suppose $u$ lives on a domain $D$ with boundary value zero, or $u$ lives on a space without boundary, e.g $S^1$ or a torus. We may then consider eigenvalue of $\Delta$, $\lambda_1 \leq \lambda_2 \leq \cdots, $ with $\lambda_n \geq 0$, (repeated with multiplicity), with eigenfunction $u_1(x), u_2(x), \cdots,$ then we may write the general solution $$ u(t,x) = \sum_{n=1}^\infty (a_n \cos(\sqrt{\lambda_n} t) + b_n \sin(\sqrt{\lambda_n} t) ) u_n(x) $$ (if $\lambda_n=0$, then we may set $b_n=0$.) To fix the coefficient, we use initial condition $u(0, x)$ and $\dot u(0, x)$: $$ \int_D u(0,x) u_n(x) dx = a_n \int u_n^2 dx $$ $$ \int_D \dot u(0,x) u_n(x) dx = \sqrt{\lambda_n} b_n \int u_n^2 dx $$

Example: 1-dim string vibration on an interval.