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Why are the Partial Differential Equations so named? i.e, elliptical, hyperbolic, and parabolic. I do know the condition at which a general second order partial differential equation becomes these, but I don't understand why they are so named?

Does it has anything to do with the ellipse, hyperbolas and parabolas?

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3 Answers 3

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A general 2nd order linear PDE in two variables is written

$$Au_{xx} + 2Bu_{xy} + Cu_{yy} + Du_x + Eu_y + F = 0$$

and $A,B,C,D,E,F$ can be functions depending on $x$ and $y$. We say a PDE is elliptic, hyperbolic or parabolic if \begin{align} B^2 - AC &= 0, &\text{parabolic} \\ B^2 - AC &>0, &\text{hyperbolic} \\ B^2 - AC &<0, &\text{elliptic} \end{align} Note that if $A,B,C,D,E,F$ depend on $x$ or $y$, there can be regions where the PDE is elliptic, hyperbolic or parabolic and different techniques are used to solve each type. If the coefficients are constant the naming comes form considering the polynomial equation $$Ax^2 + 2Bxy + Cy^2 + Dx + Ey + F = 0$$ depending on the sign of $B^2 - AC$, this forms an ellipse, hyperbola or parabola in $\mathbb{R}^2$. This can be extended to higher dimensions as well with hyperboloids, paraboloids, or ellipsoids.

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    $\begingroup$ Thank You :) , But I think the solutions of the corresponding differential equations have nothing to do with their name? Or do they have?. $\endgroup$ Commented Apr 30, 2020 at 8:50
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    $\begingroup$ @VivekanandMohapatra actually, the solutions to simple elliptical PDEs around a small pertubation tend to come out as “blobs”, ellipse-ish, to parabolic PDEs they disperse ever slower like the arms of a parabola, and for hyperbolic they wander off asymptotically straight towards infinity like a hyperbola. But I'm not sure if there's really a rigorous mathematical aspect to this. $\endgroup$ Commented Apr 30, 2020 at 15:32
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    $\begingroup$ @leftaroundabout could you give an explicit example of some of these? I don't quite understand what you could consider a hyperbola-like blob in a solution of the hyperbolic wave equation, for example. $\endgroup$
    – Ruslan
    Commented May 1, 2020 at 5:06
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    $\begingroup$ @Ruslan if you start with $u_0(x) = 2\cdot e^{-x^2}$, then $u(x,t) = e^{-(x-t)^2} + e^{-(x+t)^2}$. On a colour plot of this in the $t\geq0$ half-plane, you see something V-shaped with straight arms and rounded near 0, which is also what a $t^2 - x^2 = c$ hyperbola looks like. $\endgroup$ Commented May 1, 2020 at 9:11
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    $\begingroup$ @leftaroundabout There is a large rigorous mathematical theory of that kind of things: microlocal analysis. It allows to generalize and understand the concept of ellipticity (or lack thereof) much further. $\endgroup$ Commented May 1, 2020 at 11:54
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I'd like to address this comment of the OP to one of the answers:

But I think the solutions of the corresponding differential equations have nothing to do with their name? Or do they have?.

Actually, we can relate the names to the corresponding geometrical curves. To do this, consider how the following homogeneous PDE will look in the Fourier-transformed form.

Original PDE (with $u^{(n,m)}(x,y)$ denoting $n$th partial derivative of $u$ in $x$ and $m$th in $y$):

\begin{multline} Au^{(2,0)}(x,y) + 2Bu^{(1,1)}(x,y) + Cu^{(0,2)}(x,y) +\\ + Du^{(1,0)}(x,y) + Eu^{(0,1)}(x,y) + Fu(x,y) = 0.\tag1 \end{multline}

Fourier-transformed one (with $\hat u(k_x,k_y)$ denoting the Fourier transform of $u(x,y)$):

$$\mathcal L\hat u(k_x,k_y) = 0,\tag2$$

where

$$\mathcal L=A\kappa_x^2 + 2B\kappa_x\kappa_y + C\kappa_y^2 + D\kappa_x + E\kappa_y + F\tag3$$

with $\boldsymbol{\kappa}=i\mathbf{k}.$ This multiplication by $\mathcal L$ is the Fourier-space version of the differential operator from $(1)$. Notice that $\mathcal L$ is just a second-degree polynomial in $\kappa_x$, $\kappa_y$. Now consider the nodes of $\mathcal L$ in the $(\kappa_x,\kappa_y)$ plane: they will be defined by the equation

$\mathcal L=0.\tag4$

Geometrically, these nodes will be exactly the kinds of curves with the names corresponding to the name of the kind of PDE, i.e. ellipses, hyperbolas or parabolas.

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    $\begingroup$ This is a very nice answer, and I think is aimed more towards what the OP was asking. Further, @leftaroundabout's response to that comment in that small perturbations also seem to geometrically represent such equations, is in line with describing the perturbations as proportional to the fourier transform of the equation. I'm curious how to say this in a rigorous way, let me know if this makes some sense. $\endgroup$
    – Jon Snow
    Commented Dec 2, 2020 at 4:16
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    $\begingroup$ I think the main idea is good, but if you think at Fourier transform in space, space derivatives transform as $\partial / \partial x_i \rightarrow i k_i$, or w/o using components $\nabla \rightarrow i \mathbf{k}$. Here I don't see any imaginary unit. It looks like "your $\mathbf{k}$" is Fourier $i \mathbf{k}$ $\endgroup$
    – basics
    Commented Sep 17, 2022 at 7:44
  • $\begingroup$ @basics you're right, I always forget about this imaginary unit. Will edit. $\endgroup$
    – Ruslan
    Commented Sep 17, 2022 at 10:36
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All quadratic curves can be studied using the equation $Ax^2+2Bxy+Cy^2 + Dx + Ey + F=0$ the discriminant of which is $B^2-AC$ and the solution curve will be a ellipse, hyperbola, or parabola depending on whether the discriminant is positive, negative, or zero.

Partial second-order differential equations take a much similar form with $Au_{xx}+2Bu_{xy}+Cu_{yy} + Du_x + Eu_y + F = 0$ and the discriminant plays a similar role in classifying the solutions so they are named after the algebraic curves that resemble this expression.

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    $\begingroup$ I thought using 2Bxy rather than Bxy made the discriminant B^2-AC, not B^2-4AC? $\endgroup$
    – No Name
    Commented Apr 30, 2020 at 2:23

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