What a partial differential equation is

From one variable to two

Lesson 6.1.1 met ordinary differential equations (ODEs): an unknown function of one variable, $y(t)$, and an equation linking it to its derivatives. For example, $\frac{dy}{dt} = -2y$ has the solutions $y = Ce^{-2t}$, one for each number $C$.

A partial differential equation (PDE) does the same for a function of two or more variables. The unknown is a function such as $u(x, t)$, where $x$ is a position and $t$ a time. The equation links $u$ to its partial derivatives, written as in lesson 4.1.2: $u_t = \frac{\partial u}{\partial t}$, $u_x = \frac{\partial u}{\partial x}$ and $u_{xx} = \frac{\partial^2 u}{\partial x^2}$.

A solution is a function whose partial derivatives make the equation true at every point of a region of the $(x, t)$ plane. The order of the equation is the order of its highest partial derivative, in any of the variables.

Units

Every term of an equation must have the same units. In the heat equation, $u_t$ is in units of $u$ per unit of time. But $u_{xx}$ is in units of $u$ per unit of length squared.

So a physical heat equation carries a constant: $u_t = k\,u_{xx}$, with $k > 0$ in length² per unit of time. This lesson takes $k = 1$, which amounts to choosing the unit of time. In the transport equation, $c$ is in length per unit of time.

The heat equation

The heat equation $u_t = u_{xx}$ describes heat spreading along a thin rod, with $u(x, t)$ the temperature at position $x$ and time $t$. It says: the temperature rises where the profile bends up ($u_{xx} > 0$) and falls where it bends down. Module 6.4 studies it in depth.

A candidate. Take $u = e^{-t}\sin x$. Holding $x$ fixed, $u_t = -e^{-t}\sin x$; holding $t$ fixed, $u_x = e^{-t}\cos x$ and $u_{xx} = -e^{-t}\sin x$. The two sides agree at every point, so it is a solution.

A candidate that fails. Try $u = e^{-t}\sin 2x$. Now $u_t = -e^{-t}\sin 2x$, but by the chain rule $u_{xx} = -4e^{-t}\sin 2x$. The residual, the left side minus the right side, is $u_t - u_{xx} = 3e^{-t}\sin 2x$: not zero, so this is not a solution.

The wiggle $\sin 2x$ bends four times as sharply as $\sin x$, so it must die four times as fast: $e^{-4t}\sin 2x$ is a solution. In general $e^{-k^2 t}\sin kx$ solves the equation for every number $k$. So does a polynomial such as $x^2 + 2t$, with $u_t = 2$ and $u_{xx} = 2$.

Checking by central differences

When $u$ is a program rather than a formula, estimate each partial by lesson 2.8.1's central differences, with a step $h$. For $u_t$, move $t$ by $\pm h$ and keep $x$ fixed; for $u_{xx}$, use the second difference in $x$.

With $h = 0.01$, at three points:

For the solution the residual is tiny, and it shrinks like $h^2$. At $(1, 0.5)$ it is $-1.3 \times 10^{-3}$, $-1.3 \times 10^{-5}$ and $-1.3 \times 10^{-7}$ for $h$ = 0.1, 0.01 and 0.001: truncation error, not a failure. For $e^{-t}\sin 2x$ the residual closes in on its true value, $3e^{-0.5}\sin 2 \approx 1.6546$, and stays there.

Where a candidate fails: a corner

The heat equation needs $u_{xx}$, so a candidate must be smooth enough to have it. The profile $u = |x|$, frozen in time, has $u_t = 0$ and $u_{xx} = 0$ everywhere except at $x = 0$, where it has a corner.

There the second difference is $\frac{h - 0 + h}{h^2} = \frac{2}{h}$: 20, 200 and 2,000 for $h$ = 0.1, 0.01 and 0.001. It grows without limit, so $|x|$ does not solve the heat equation at its corner. Module 6.4 shows that heat smooths such a corner at once.

The transport equation

The transport equation $u_t + c\,u_x = 0$ describes a shape that moves at speed $c$ without changing, such as dye carried along by a steady stream. Take any differentiable function $f$ of one variable, and set $u(x, t) = f(x - ct)$.

By the chain rule (lesson 2.3.3), with the inside $x - ct$ having slope $-c$ in $t$ and 1 in $x$:

So every $f(x - ct)$ is a solution. Its value depends only on $x - ct$, so it is the same all along each line $x = ct + \text{constant}$. Whatever is at $x = 0$ at $t = 0$ is at $x = c$ at $t = 1$.

An example by numbers. Take $c = 2$ and the bump $u = e^{-(x - 2t)^2}$, whose peak is at $x = 2t$.

Each row is the row above moved one place to the right. The bump moves 1 in half a unit of time: a speed of 2.

Both directions, and a corner

The sign of $c$ sets the direction. For $c > 0$ the shape moves right; for $c < 0$ it moves left; for $c = 0$ the equation is $u_t = 0$, and the shape stays put.

A shape written the other way round, $f(x + ct)$, moves left. For example $(x + 2t)^2$ gives $u_t + 2u_x = 8x + 16t$, not zero: it solves $u_t - 2u_x = 0$ instead.

The function $f$ may have a corner, such as $u = |x - 2t|$. Then the partial derivatives fail on the line $x = 2t$, and the equation holds everywhere else; the corner simply travels with the shape.

Laplace's equation

Laplace's equation $u_{xx} + u_{yy} = 0$ has no time in it. It describes a steady state, such as the temperature across a metal plate once it has stopped changing.

It says the bend of the slice in $x$ cancels the bend of the slice in $y$. Where one bends up, the other bends down as much.

The saddle $x^2 - y^2$ of lesson 4.1.1 is a solution: $u_{xx} = 2$ and $u_{yy} = -2$. The bowl $x^2 + 2y^2$ is not, since both its slices bend up: $2 + 4 = 6$. Two more solutions are $xy$, whose slices are straight lines, and $e^x\sin y$, with $u_{xx} = e^x\sin y$ and $u_{yy} = -e^x\sin y$.

Linear equations and adding solutions

An equation is linear if $u$ and its partials appear only on their own, each times a number or a function of the variables. They are never multiplied together, squared, or put inside another function. The heat, transport and Laplace equations are linear.

The equation $u_t + u\,u_x = 0$ is not: it multiplies $u$ by $u_x$. Neither is $u_t = (u_x)^2$.

For example, $e^{-t}\sin x + 3e^{-4t}\sin 2x$ solves the heat equation, because each part does. Module 6.4 builds solutions from many such parts.

The rule fails in two ways. For the nonlinear $u_t + u\,u_x = 0$, the function $u = \frac{x}{1 + t}$ is a solution, but $2u$ leaves the residual $\frac{2x}{(1 + t)^2}$.

It also fails for a term without $u$, such as the $+1$ in $u_t = u_{xx} + 1$. There $u = t$ is a solution, but $2t$ leaves a residual of 1: the $+1$ is not doubled.

Every equation in this lesson has many solutions: $e^{-k^2 t}\sin kx$ for every $k$, or $f(x - ct)$ for every $f$. Lesson 6.3.2 adds the conditions that pick out one of them. Case study 6.3.4 checks in the same way that the Black–Scholes formula solves the Black–Scholes equation.

Common mistakes

1. Hold the other variable fixed. In $u_t$ of $e^{-t}\sin x$, the factor $\sin x$ is a constant.
2. Check everywhere, not at one point. The residual $3e^{-t}\sin 2x$ of $e^{-t}\sin 2x$ is zero at $x = 0$, yet the candidate fails.
3. Keep the chain-rule factor. For $u = f(x - ct)$, $u_t = -c\,f'(x - ct)$, not $f'(x - ct)$. For $\sin 2x$, $u_{xx}$ carries a factor 4.
4. Mind the direction. $f(x - ct)$ moves right when $c > 0$; $f(x + ct)$ moves left and solves $u_t - c\,u_x = 0$.
5. The order is the highest partial in any variable. $u_t = u_{xx}$ has order 2, though it is first order in $t$.
6. Adding solutions needs a linear equation with no term free of $u$.
7. A small residual from central differences is a pass if it shrinks like $h^2$. A residual that settles on a nonzero value is a fail. At a corner, the second difference grows like $\frac{1}{h}$.