Note that while solutions to systems of ODEs depend on arbitrary constants, solutions to systems of PDEs depend on arbitrary functions (why?)

Counting principle: we can expect the solution to an $n$th order PDE involving $m$ independent variables to depend on $n$ arbitrary functions of $m-1$ variables.

Simplest PDE

The simplest PDE, for a function $u(t,x)$ of two variables is

$$ \frac{\partial u}{\partial t} = 0. $$

This is a first-order, homogeneous, linear equation. We can solve it by integrating both from $0$ to $t:$

$$ 0 \ \int_{0}^{t} \frac{\partial u}{\partial t} (s,x) ds = u(t,x) - u(0,x). $$

The solution, then, takes the form

$$ u(t,x) = f(x), \quad \text{where} \quad f(x) = u(0,x). $$

Note that this solution is a function of the space variable $x$ alone. We only require that $f(x)$ be continuously differentiable (why?). This solution represents a stationary wave - it does not change in time. The initial profile stays frozen in place and the system remains in equilibrium (in the same meaning as equilibrium in dynamical systems - a fixed point?)

Transport Equation

The basic transport equation is

$$ u_t + c u_x = 0. \tag{a} $$

We'll find its characteristic curves. First, let's parameterize $u(x,t)$ to get $v(t) = u(x(t), t).$ Now,

$$ \frac{dv}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial t} \frac{dt}{dt} = u_t + \frac{dx}{dt} u_x. \tag{b} $$

Comparing (b) to (a), we see that if if $u_t + c u_x = 0,$ then $\frac{dx}{dt} = c.$ Now, we solve that ODE:

$$ \begin{aligned} \frac{dx}{dt} & = c \\ dx & = cdt \\ \int dx & = \int cdt \\ x & = ct + k \\ x - ct & = k. \end{aligned} $$

So, we see that $x - ct$ is constant, and if we let $t=0,$ we get $x_0 = k.$ So, our characteristic curve is $\xi{(x,t)} = x - ct.$

Now, we'll let perform a change of variables and let $\xi = x - ct.$ So we have

$$ u(t,x) = v(t, \xi) = v(t, x - ct). $$

Finding $u_t$ and $c u_x$ (the terms in our original PDE in (a)) we get

$$ \begin{aligned} u_t & = \frac{\partial}{\partial t} v(t, x - ct) + \frac{\partial}{\partial \xi} v(t, x - ct) (-c) \\ & = v_t + -c v_{\xi} \\ c u_x & = c \frac{\partial}{\partial t} v(t, x - ct) \frac{dt}{dx} + c \frac{\partial}{\partial \xi} v(t, x - ct) \frac{d \xi}{d \xi} \\ & = 0 + c v_{\xi} \\ \end{aligned}. \tag{c} $$

Substituting our terms from (c) into (a) gives us

$$ \begin{aligned} v_t - c v_{\xi} + c v_{\xi} & = 0 \\ v_t = \frac{dv}{dt} = 0. \end{aligned} $$

Note: we can go from $v_t to \frac{dv}{dt}$ because $v(t, x - ct)$ is only a function of $t,$ because $x - ct$ is constant.

Now we solve this ODE:

$$ \begin{aligned} \int_0^t \frac{dv}{ds}(s, \xi) ds & = 0 \\ v(t, \xi) - v(0, x_0) & = 0 \\ v(t, \xi) & = v(0, x_0). \\ \end{aligned} $$

We were given that $u(0, x) = f(x),$ and $\frac{dv}{dt} = 0,$ i.e. $v$ doesn't change with time so $v(t, \xi) = v(t, x_0) = v(0, x_0) = f(x_0) = f(\xi) = f(x - ct).$

Since we defined $u(t,x) = v(t, \xi),$ so our solution is $u(t,x) = f(x - ct).$