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 Equations

Basic Transport Equation

The basic transport equation is

$$ u_t + c u_x = 0, \quad u(0, x) = f(x). \tag{a} $$

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

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

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 $\xi{(t,x)} = x - ct.$

Now, we'll let perform a change of variables (to a moving coordinate system that will create stations waves) and let $\xi{(t,x)} = x - ct.$ 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

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

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 found that $\frac{dv}{dt} = 0,$ 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 $u(t,x) = v(t, \xi),$ our solution is $u(t,x) = f(x - ct) = f(\xi)$ for any $f \in C^1.$ This meany any reasonable function of $\xi$ will solve our PDE, i.e. $\xi^2 +1$ or $\cos{\xi}$ will produce a corresponding solution such as $(x-ct)^2 + 1$ or $\cos{(x - ct)}.$

Transport with Decay

Let $a > 0$ be a positive constant, and $c$ an arbitrary constant. The homogeneous linear first-order partial differential equation

$$ \frac{\partial u}{\partial t} + c \frac{\partial u }{\partial x} + a u = 0 $$

models the transport of, for example, a radioactively decaying solute in a uniform fluid flow with wave speed $c,$ and the coefficient $a$ modeling the rate of decay.

We can reuse the same characteristic as we used in the basic transport equation (since, I believe, it is determined only by the differential terms of the equation.) Then, following what we did in (c) and (d) above we get

$$ \frac{dv}{dt} + av = 0. $$

This is a first order linear differential equation. We find the integrating factor to be $e^{at},$ and the solution to be

$$ e^{at} v(t, \xi) = C. \tag{e} $$

If we let $t = 0, $ we get $v(0, \xi) = f(\xi) = C.$ Then if we divide both sides by $e^{at}$ we get

$$ v(t, \xi) = f(\xi)e^{-at}, $$

and since $u(t,x) = v(t, \xi)$ we have $u(t,x) = f(x - ct)e^{-at}$ as our solution.

Non-Uniform Transport

The non-uniform transport problem is another generalization (still linear) where the wave speed $c(x)$ is now allowed to depend on the spatial position

$$ u_t + c(x) u_x = 0. $$

To use characteristics, we will parameterize as $h(t) = u(t, x(t)).$ Now,

$$ \frac{dh}{dt} = \frac{d}{dt} u(t, x(t)) = \frac{\partial}{\partial}u(t, x(t)) + \frac{\partial}{\partial x} u(t, x(t)) \frac{dx}{dt}. $$

So, if $\frac{dx}{dt} = c(x),$ then

$$ \frac{dh}{dt} = \frac{\partial}{\partial}u(t, x(t)) + c(x) \frac{\partial}{\partial x} u(t, x(t)) = 0. $$

Solving $\frac{dx}{dt} = c(x)$ via separation of variables gives:

$$ \begin{aligned} \int \frac{dx}{c(x)} & = \int dt \\ \beta(x) & := \int \frac{dx}{c(x)} = t + k, \end{aligned} $$

so $\beta(x) - t = k$ is constant and is our characteristic variable, i.e. $\xi(t,x) = \beta(x) - t,$ and hence

$$ u(t,x) = v(\beta{x} -t) $$

is our solution for any $v(\xi) \in C^1.$

Let's say we're given initial condition $u(0, x) = f(x).$ Then $v(\xi) = f(x).$ For $t=0,$ $\xi = \beta(x) - 0,$ so $x(t) = \beta^{-1}(\xi),$ and $v(0, \xi) = f(\beta^{-1}(\xi)).$ Now, $v(t, \xi)$ is constant, so $v(t, \xi) = f(\beta^{-1}(\xi)).$ Substituting $\xi = \beta{(x)} - t,$ we have $u(x,t) = v(t, \xi) = f(\beta^{-1}(\beta(x) - t)).$

For example, if we have $u_t + \frac{1}{x} u_x = 0,$ then $c(x) = 1/x.$ We solve $\frac{dx}{dt} = 1/x$ to get $\beta{x} = \frac{x^2}{2} = t + k,$ i.e. $\xi = \frac{x^2}{2} + t.$ Now, to find $\beta^{-1}, let t = \frac{x^2}{2}.$ Swapping $x$ and $y$ gives $x = \frac{y^2}{2},$ and solving for $y = x(t)$ gives $\beta^{-1} = x(t) = \sqrt{2t}.$ Now our solution is

$$ \begin{aligned} u(t,x) & = f(\sqrt{2 ( \beta{x} - t)}) \\ & = f(\sqrt{2 ( \frac{x^2}{2} -t )} \\ & = f(\sqrt{x^2 -2t}). \\ \end{aligned} $$