# 2.6: Classification of Second Order PDEs

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We have studied several examples of partial differential equations, the heat equation, the wave equation, and Laplace’s equation. These equations are examples of parabolic, hyperbolic, and elliptic equations, respectively. Given a general second order linear partial differential equation, how can we tell what type it is? This is known as the classification of second order PDEs.

Let $$u = u(x, y)$$. Then, the general form of a linear second order partial differential equation is given by

$\label{eq:1}a(x,y)u_{xx}+2b(x,y)u_{xy}+c(x,y)u_{yy}+d(x,y)u_x+e(x,y)u_y+f(x,y)u=g(x,y).$

In this section we will show that this equation can be transformed into one of three types of second order partial differential equations.

Let $$x = x(ξ, η)$$ and $$y = y(ξ, η)$$ be an invertible transformation from coordinates $$(ξ, η)$$ to coordinates $$(x, y)$$. Furthermore, let $$u(x(ξ, η), y(ξ, η)) = \mathcal{U}(ξ, η)$$. How does the partial differential equation $$\eqref{eq:1}$$ transform?

We first need to transform the derivatives of $$u(x, t)$$. We have

\begin{align}u_x&=\mathcal{U}_\xi \xi_x+\mathcal{U}_\eta \eta_x,\nonumber \\ u_y&=\mathcal{U}_\xi\xi_y+\mathcal{U}_\eta\eta_y,\nonumber \\ u_{xx}&=\frac{\partial}{\partial x}(\mathcal{U}_\xi\xi_x+\mathcal{U}_\eta\eta_x),\nonumber \\ &=\mathcal{U}_{\xi\xi}\xi_x^2+2\mathcal{U}_{\xi\eta}\xi_x\eta_x+\mathcal{U}_{\eta\eta}\eta_x^2+\mathcal{U}_\xi\xi_{xx}+\mathcal{U}_\eta\eta_{xx},\nonumber \\ u_{yy}&=\frac{\partial}{\partial y}(\mathcal{U}_\xi\xi_y+\mathcal{U}_\eta\eta_y),\nonumber \\ &=\mathcal{U}_{\xi\xi}\xi_y^2+2\mathcal{U}_{\xi\eta}\xi_y\eta_y+\mathcal{U}_{\eta\eta}\eta_y^2+\mathcal{U}_\xi\xi_{yy}+\mathcal{U}_\eta\eta_{yy},\nonumber \\ u_{xy}&=\frac{\partial}{\partial y}(\mathcal{U}_\xi\xi_x+\mathcal{U}_\eta\eta_x),\nonumber \\ &=\mathcal{U}_{\xi\xi}\xi_x\xi_y+\mathcal{U}_{\xi\eta}\xi_x\eta_y+\mathcal{U}_{\eta\xi}\xi_y\eta_x+\mathcal{U}_{\eta\eta}\eta_x\eta_y+\mathcal{U}_\xi\xi_{xy}+\mathcal{U}_\eta\eta_{xy}.\label{eq:2} \end{align}

Inserting these derivatives into Equation $$\eqref{eq:1}$$, we have

\begin{align}g − f \mathcal{U} &= au_{xx} + 2bu_{xy} + cu_{yy} + du_x + eu_y\nonumber \\ &=a\left( \mathcal{U}_{ξξ} ξ_x^2 + 2\mathcal{U}_{ξη}ξ_xη_x + \mathcal{U}_{ηη}η_x^2+ \mathcal{U}_ξ ξ_{xx} + \mathcal{U}_ηη_{xx}\right)\nonumber \\ &\quad +2b( \mathcal{U}_{ξξ} ξ_xξ_y + \mathcal{U}_{ξη}ξ_xη_y + \mathcal{U}_{ξη}ξ_yη_x\nonumber \\ &\quad + \mathcal{U}_{ηη}η_xη+y + \mathcal{U}_ξ ξ_{xy} + \mathcal{U}_ηη_{xy})\nonumber \\ &\quad +c \left( \mathcal{U}_{ξξ} ξ_y^2 + 2\mathcal{U}_{ξη}ξ_yη_y + \mathcal{U}_{ηη}η_y^2 + \mathcal{U}_ξ ξ_{yy} + \mathcal{U}_ηη_{yy}\right)\nonumber \\ &\quad +d (\mathcal{U}_ξ ξ_x + \mathcal{U}_ηη_x)\nonumber \\ &\quad +e (\mathcal{U}_ξ ξ_y + \mathcal{U}_ηη_y)\nonumber \\ &= (aξ_x^2 + 2bξ_xξ_y + cξ_y^2)\mathcal{U}_{\xi\xi}\nonumber \\ &\quad +(2aξ_xη_x + 2bξ_xη_y + 2bξ_yη_x + 2cξ_yη_y)\mathcal{U}_{\xi\eta}\nonumber \\ &\quad +(aη_x^2 + 2bη_xη_y + cη_y^2)\mathcal{U}_{\eta\eta}\nonumber \\ &\quad +(aξ_{xx} + 2bξ_{xy} + cξ_{yy} + dξ_x + eξ_y)\mathcal{U}_\xi\nonumber \\ &\quad +(aη_{xx} + 2bη_{xy} + cη_{yy} + dη_x + eη_y)\mathcal{U}_\eta\nonumber \\ &=A\mathcal{U}_{\xi\xi}+2B\mathcal{U}_{\xi\eta}+C\mathcal{U}_{\eta\eta}+D\mathcal{U}_\xi +E\mathcal{U}_\eta .\label{eq:3}\end{align}

Picking the right transformation, we can eliminate some of the second order derivative terms depending on the type of differential equation. This leads to three types: elliptic, hyperbolic, or parabolic.

For example, if transformations can be found to make $$A ≡ 0$$ and $$C ≡ 0$$, then the equation reduces

$\mathcal{U}_{\xi\eta}=\text{lower order terms}.\nonumber$

. Such an equation is called hyperbolic. A generic example of a hyperbolic equation is the wave equation.

The conditions that $$A ≡ 0$$ and $$C ≡ 0$$ give the conditions

\begin{align}a\xi_x^2+2b\xi_x\xi_y+c\xi_y^2&=0.\nonumber \\ a\eta_x^2+2b\eta_x\eta_y+c\eta_y^2&=0.\label{eq:4}\end{align}

) We seek $$ξ$$ and $$η$$ satisfying these two equations, which are of the same form. Let’s assume that $$ξ = ξ(x, y)$$ is a constant curve in the $$xy$$-plane. Furthermore, if this curve is the graph of a function, $$y = y(x)$$, then

$\frac{d\xi}{dx}=\xi_x+\frac{dy}{dx}\xi_y=0.\nonumber$

Then

$\frac{dy}{dx}=-\frac{\xi_x}{\xi_y}.\nonumber$

Inserting this expression in $$A = 0$$, we have

\begin{align}A&=a\xi_x^2+2b\xi_x\xi_y+c\xi_y^2\nonumber \\ &=\xi_y^2\left(a\left(\frac{\xi_x}{\xi_y}\right)^2+2b\frac{\xi_x}{\xi_y}+c\right)\nonumber \\ &=\xi_y^2\left(a\left(\frac{dy}{dx}\right)^2-2b\frac{dy}{dx}+c\right)=0.\label{eq:5}\end{align}

This equation is satisfied if $$y(x)$$ satisfies the differential equation

$\frac{dy}{dx}=\frac{b\pm\sqrt{b^2-ac}}{a}.\nonumber$

So, for $$A = 0$$, we choose $$ξ$$ and $$η$$ to be constant on these characteristic curves.

## Example $$\PageIndex{1}$$

Show that $$u_{xx} − u_{yy} = 0$$ is hyperbolic.

###### Solution

In this case we have $$a = 1 = −c$$ and $$b = 0$$. Then,

$\frac{dy}{dx}=\pm 1.\nonumber$

This gives $$y(x) = ±x + c$$. So, we choose $$ξ$$ and $$η$$ constant on these characteristic curves. Therefore, we let $$ξ = x − y,$$ $$η = x + y$$.

Let’s see if this transformation transforms the differential equation into a canonical form. Let $$u(x, y) = \mathcal{U}(ξ, η)$$. Then, the needed derivatives become

\begin{align} u_x &= \mathcal{U}_ξ ξ_x + \mathcal{U}_ηη_x = \mathcal{U}_ξ + \mathcal{U}_η.\nonumber \\ u_y &= \mathcal{U}_ξ ξ_y + \mathcal{U}_ηη_y = −\mathcal{U}_ξ + \mathcal{U}_η.\nonumber \\ u_{xx}&=\frac{\partial}{\partial x}(\mathcal{U}_\xi +\mathcal{U}_\eta )\nonumber \\ &= \mathcal{U}_{ξξ} ξ_x + \mathcal{U}_{ξη}η_x + \mathcal{U}_{ηξ} ξ_x + \mathcal{U}_{ηη}η_x\nonumber \\ &= \mathcal{U}_{ξξ} + 2\mathcal{U}_{ξη} + \mathcal{U}_{ηη}.\nonumber \\ u_{yy}&=\frac{\partial}{\partial y}(-\mathcal{U}_\xi +\mathcal{U}_\eta )\nonumber \\ &= −\mathcal{U}_{ξξ} ξ_y − \mathcal{U}_{ξη}η_y + \mathcal{U}_{ηξ} ξ_y + \mathcal{U}_{ηη}η_y\nonumber \\ &= \mathcal{U}_{ξξ} − 2\mathcal{U}_{ξη} + \mathcal{U}_{ηη}.\label{eq:6}\end{align}

Inserting these derivatives into the differential equation, we have

$0=u_{xx}-u_{yy}=4\mathcal{U}_{\xi\eta}.\nonumber$

Thus, the transformed equation is $$\mathcal{U}_{ξη} = 0$$. Thus, showing it is a hyperbolic equation.

We have seen that $$A$$ and $$C$$ vanish for $$ξ(x, y)$$ and $$η(x, y)$$ constant along the characteristics

$\frac{dy}{dx}=\frac{b\pm\sqrt{b^2-ac}}{a}\nonumber$

for second order hyperbolic equations. This is possible when $$b^2 − ac > 0$$ since this leads to two characteristics.

In general, if we consider the second order operator

$L[u] = a(x, y)u_{xx} + 2b(x, y)u_{xy} + c(x, y)u_{yy},\nonumber$

then this operator can be transformed to the new form

$L'[\mathcal{U}]=B\mathcal{U}_{\xi\eta}\nonumber$

if $$b^2 − ac > 0$$. An example of a hyperbolic equation is the wave equation, $$u_{tt} = u_{xx}$$.

When $$b^2 − ac = 0$$, then there is only one characteristic solution, $$\frac{dy}{dx}=\frac{b}{a}$$. This is the parabolic case. But, $$\frac{dy}{dx}=-\frac{\xi_x}{\xi_y}$$. So,

$\frac{b}{a}=-\frac{\xi_x}{\xi_y},\nonumber$

or

$a\xi_x+b\xi_y=0.\nonumber$

Also, $$b^2-ac=0$$ implies that $$c=b^2/a$$.

Inserting these expression into coefficient $$B$$, we have

\begin{align} B &= 2aξ_xη_x + 2bξ_xη_y + 2bξ_yη_x + 2cξ_yη_y\nonumber \\ &= 2(aξ_x + bξ_y)η_x + 2(bξ_x + cξ_y)η_y\nonumber \\ &= 2 \frac{b}{a} (aξ_x + bξ_y)η_y = 0.\label{eq:7}\end{align}

Therefore, in the parabolic case, $$A = 0$$ and $$B = 0$$, and $$L[u]$$ transforms to

$L'[\mathcal{U}]=C\mathcal{U}_{\eta\eta}\nonumber$

when $$b^2 − ac = 0$$. This is the canonical form for a parabolic operator. An example of a parabolic equation is the heat equation, $$u_t = u_{xx}$$.

Finally, when $$b^2 − ac < 0$$, we have the elliptic case. In this case we Elliptic case. cannot force $$A = 0$$ or $$C = 0$$. However, in this case we can force $$B = 0$$. As we just showed, we can write

$B = 2(aξ_x + bξ_y)η_x + 2(bξ_x + cξ_y)η_y.\nonumber$

Letting $$η_x = 0$$, we can choose $$ξ$$ to satisfy $$bξ_x + cξ_y = 0$$.

$A=a\xi_x^2+2b\xi_x\xi_y+c\xi_y^2=a\xi_x^2-c\xi_y^2=\frac{ac-b^2}{c}\xi_x^2\nonumber$

$C=a\eta_x^2+2b\eta_x\eta_y+c\eta_y^2=c\eta_y^2\nonumber$

Furthermore, setting $$\frac{ac-b^2}{c}\xi_x^2=c\eta_y^2$$, we can make $$A=c$$ and $$L[u]$$ transforms to

$L'[\mathcal{U}]=A[\mathcal{U}_{\xi\xi}+\mathcal{U}_{\eta\eta}]\nonumber$

when $$b^2 − ac < 0$$. This is the canonical form for an elliptic operator. An example of an elliptic equation is Laplace’s equation, $$u_{xx} + u_{yy} = 0$$.

## Classification of Second Order PDEs

The second order differential operator

$L[u] = a(x, y)u_{xx} + 2b(x, y)u_{xy} + c(x, y)u_{yy},\nonumber$

can be transformed to one of the following forms:

• $$b^2-ac>0$$. Hyperbolic: $$L[u]=B(x,y)u_{xy}$$
• $$b^2-ac=0$$. Parabolic: $$L[u]=C(x,y)u_{yy}$$
• $$b^2-ac<0$$. Elliptic: $$L[u]=A(x,y)[u_{xx}+u_{yy}]$$

As a final note, the terminology used in this classification is borrowed from the general theory of quadratic equations which are the equations for translated and rotated conics. Recall that the general quadratic equation in two variable takes the form

$\label{eq:8}ax^2+2bxy+cy^2+dx+ey+f=0.$

One can complete the squares in $$x$$ and $$y$$ to obtain the new form

$a(x-h)^2+2bxy+c(y-k)^2+f'=0.\nonumber$

So, translating points $$(x, y)$$ using the transformations $$x' = x − h$$ and $$y' = y − k$$, we find the simpler form

$ax^2+2bxy+cy^2+f=0.\nonumber$

Here we dropped all primes.

We can also introduce transformations to simplify the quadratic terms. Consider a rotation of the coordinate axes by $$θ$$,

\begin{align}x'&=x\cos\theta+y\sin\theta\nonumber \\ y'&=-x\sin\theta+y\cos\theta ,\label{eq:9}\end{align}

or

\begin{align}x&=x'\cos\theta -y'\sin\theta\nonumber \\ y&=x'\sin\theta +y'\cos\theta .\label{eq:10}\end{align}

The resulting equation takes the form

$Ax'^2+2Bx'y'+Cy'^2+D=0,\nonumber$

where

\begin{align} A&=a\cos^2\theta +2b\sin\theta\cos\theta +c\sin^2\theta .\nonumber \\ B&=(c-a)\sin\theta\cos\theta +b(\cos^2\theta -\sin^\theta ).\nonumber \\ C&=a\sin^2\theta -2b\sin\theta\cos\theta +c\cos^2\theta .\label{eq:11}\end{align}

Furthermore, one can show that $$b^2 − ac = B^2 − AC$$. From the form $$Ax'^2 + 2Bx'y' + Cy'^2 + D = 0$$, the resulting quadratic equation takes one of the following forms:

• $$b^2-ac>0$$. Hyperbolic: $$Ax^2-Cy^2+D=0$$.
• $$b^2-ac=0$$. Parabolic: $$Ax^2+By+D=0$$.
• $$b^2-ac<0$$. Elliptic: $$Ax^2+Cy^2+D=0$$.

Thus, one can see the connection between the classification of quadratic equations and second order partial differential equations in two independent variables.

This page titled 2.6: Classification of Second Order PDEs is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Russell Herman via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.