# How do you explain what a hyperbolic PDE is…?

A friend of mine who just finished an introductory calculus class saw my books on fluid mechanics and asked me what a different PDEs look like. I explained elliptical and parabolic by giving examples of a two way and one way space coordinates and time coordinates.

How do I explain hyperbolic without involving too much detail about characteristic? I tried explaining it from the standpoint of the wave equation, but he wanted something a little more mathematical.

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See my answer here; or go back to Hadamard's Lectures on Cauchy's Problem for Linear Partial Differential Equations, where (near the beginning of the book) he classified second order partial differential equations using the polynomial symbols. –  Willie Wong May 18 '12 at 9:57

Sorry this answer looks so long and scary! It's actually very short and pretty, once you understand the stuff about polynomials. If you're intimidated by the giant mass of text, just skip to the end...

Before I explain what a hyperbolic PDE is, I have to tell you a little about polynomials.

You're probably familiar with the idea of plugging a real number, like $3$, into a polynomial, like $Q(X) = X^2 + 2$. You take a power of a real number by multiplying it by itself a bunch of times, and the "zeroth power" of a real number is just the number $1$, so

$$Q(3) = 3 \times 3 + 2 \times 1 \\ = 9 + 2 \\ = 11.$$

You may know also that you can plug a matrix, like $\left(\begin{array}{cc} 2 & 1 \\ 0 & 1 \end{array}\right)$, into a polynomial. You take a power of a matrix by multiplying it by itself a bunch of times, and the "zeroth power" of a matrix is just the identity matrix, so

$$Q\left(\begin{array}{cc} 2 & 1 \\ 0 & 1 \end{array}\right) = \left(\begin{array}{cc} 2 & 1 \\ 0 & 1 \end{array}\right)\left(\begin{array}{cc} 2 & 1 \\ 0 & 1 \end{array}\right) + 2\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) \\ = \left(\begin{array}{cc} 4 & 3 \\ 0 & 1 \end{array}\right) + \left(\begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array}\right) \\ = \left(\begin{array}{cc} 6 & 3 \\ 0 & 3 \end{array}\right).$$

An 2-by-2 matrix is just a fancy way of writing down a linear function from $\mathbb{R}^2$ to $\mathbb{R}^2$. For example, the matrix we used before represents the function $A(x, y) = (2x + y, y)$. Multiplying two matrices is the same as composing the associated functions, and the identity matrix represents the function $I(x, y) = (x, y)$. So, if we can plug a matrix into a polynomial, we can also plug a linear function into a polynomial:

$$Q(A) = A \circ A + 2I$$

$$Q(A)(x, y) = A(A(x, y)) + 2I(x, y) \\ = A(2x + y, y) + 2(x, y) \\ = (4x + 3y, y) + (2x, 2y) \\ = (6x + 3y, 3y).$$

Let's call the set of differentiable functions $\mathcal{F}$. You can think of differentiation as a linear function from $\mathcal{F}$ to $\mathcal{F}$. Since we know how to plug a linear function into a polynomial, we can plug the operation of differentiation into a polynomial!

$$Q(\tfrac{d}{dx}) = \tfrac{d}{dx} \circ \tfrac{d}{dx} + 2I$$

$$Q(\tfrac{d}{dx})\sin(x) = \tfrac{d}{dx}[\tfrac{d}{dx}\sin(x)] + 2\sin(x) \\ = \tfrac{d}{dx}\cos(x) + 2\sin(x) \\ = -\sin(x) + 2\sin(x) \\ = \sin(x).$$

Hold that thought.

Say you have a second-order polynomial in two variables,

$$P(X, Y) = \alpha_{20} X^2 + \alpha_{11} XY + \alpha_{02} Y^2 + \alpha_{10} X + \alpha_{01} Y + \alpha_{00}.$$

If you plot all the pairs of real numbers $x$ and $y$ for which $P(x, y) = 0$, you'll generally get a smooth curve---an ellipse, a parabola, or a hyperbola. It makes sense to call a polynomial "elliptic" if its zero-set is an ellipse, "parabolic" if its zero-set is a parabola, and "hyperbolic" if its zero-set is a hyperbola.

If you plug the operations $\tfrac{d}{dx}$ and $\tfrac{d}{dy}$ into $P$, you get a really simple way of writing down a differential equation...

$$P(\tfrac{d}{dx}, \tfrac{d}{dy})f = 0$$

...that would look really complicated in more familiar notation.

$$\alpha_{20}\tfrac{d^2f}{dx^2} + \alpha_{11} \tfrac{d^2f}{dx\,dy} + \alpha_{02} \tfrac{d^2f}{dy^2} + \alpha_{10} \tfrac{df}{dx} + \alpha_{01} \tfrac{df}{dy} + \alpha_{00} f = 0.$$

You might be tempted to call this differential equation "elliptic" if $P$ is an elliptic polynomial, "parabolic" if $P$ is a parabolic polynomial, and "hyperbolic" if $P$ is a hyperbolic polynomial. This actually turns out to be a really good way of classifying differential equations, so the names stuck!

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