Discriminant (in the context of PDE classification): $b^2 - 4ac$ or $b^2 - ac$?

Question

I'm reading two textbooks on partial differential equations. In their respective sections on classification of PDEs (hyperbolic, parabolic, elliptical), they differ in what they describe as being the discriminant. One textbook says that the discriminant is $b^2 - 4ac$, while the other describes it being $b^2 - ac$. Are these both correct, or is one correct and the other incorrect?

EDIT: What are the $a$, $b$, and $c$? The second-order linear PDE in two independent variables is $Au_{xx} + Bu_{xy} + Cu_{yy} + Du_x + Eu_y + Fu = G$, where $A$, $B$, $C$, $D$, $E$, $F$, $G$ are functions of $x$ and $y$ and could be constants.

Answer 1

In short, either can be correct, depending on how the original equation is expressed.

Assuming that the mixed partials of $u$ are equal, we can write your differential equation in operator form: $$\left( A\partial_x^2+B\partial_x\partial_y+C\partial_y^2+D\partial_x+E\partial_y+F \right)u = 0.$$ The parenthesized operator is reminiscent of the left-hand side of the general conic equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, which can be written in matrix form as $$\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix} A & \frac B2 \\ \frac B2 & C \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} + \begin{bmatrix}D&E\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+F = 0.$$ The discriminant of this equation, which indicates the type of conic that the equation represents, is the negative determinant of the $2\times2$ matrix of the quadratic part of this equation, i.e., $B^2/4-AC$. This expression conventionally gets multiplied by $4$ for convenience in writing, if nothing else: $B^2-4AC$.† On the other hand, it can also be convenient not to have those stray factors of two floating around. They can be eliminated by using $2B$ instead of $B$ as the coefficient of the mixed term, with corresponding matrix $$\begin{bmatrix}A&B\\B&C\end{bmatrix},$$ in which case the discriminant becomes $B^2-AC$. If you remember that it’s related to the determinant of the quadratic part of the equation, you’ll get the right version.

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† In fact, multiplying by $4$ gives you the negative determinant of the Hessian of the left-hand side of the conic equation, which is another way to arrive at the discriminant.

Answer 2

Some authors/texts like to write the middle term as $2Bu_{xy}$ as opposed to $Bu_{xy}.$ In that case, it is $B^2-AC$ and in your case, it should be $B^2-4AC.$

Answer 3

A few points -

*) If we are discussing the second order linear PDE of two independent variables, which say is of the form - $Au_{xx} + Bu_{xy} + Cu_{yy} + Du_x + Eu_y + Fu = G$, where $A$, $B$, $C$, $D$, $E$, $F$, $G$ are functions of $x$ and $y$ and could be constants, then we must be careful about the coefficients of the PDE, they are $A,B,C,D,E,F,G$ and not $a,b,c,d,e,f,g$, (very simple yet much important).

*) If we are considering the PDE $Au_{xx} + Bu_{xy} + Cu_{yy} + Du_x + Eu_y + Fu = G$, where $A$, $B$, $C$, $D$, $E$, $F$, $G$ are functions of $x$ and $y$ and could be constants, then their classification is as follows -

We check the Discriminant which is $d = B^2(x_{0},y_{0}) - 4 A(x_{0},y_{0})C(x_{0},y_{0})$.

At $(x_{0},y_{0})$, the equation is said to be -

1) Elliptic if $d <0$

2) Parabolic if $d = 0$

3) Hyperbolic if $d>0$

If this is true for all points $(x_{0},y_{0}) \in $ domain $\Omega$, then the equation is said to be Elliptic, Parabolic or Hyperbolic in that domain.

Extra point -

Now if there is the case of $n$ independent variables like $x_{1},x_{2},...,x_{n}$and second order linear PDE (of the form $\sum \sum a_{i,j} u_{x_{i},x_{j}}+ $ lower order terms $= 0$ then the classification depends on the signature of the eigenvalues of the coefficient matrix.

*) Elliptic if the eigenvalues are all positive or all negative.

*) Parabolic - The eigenvalues are all positive or all negative, save one which is zero

*) Hyperbolic - There is only one negative eigenvalue and all the rest are positive, or there is only one positive eigenvalue and all the rest are negative.

Few references similar to this question -

*) How to determine where a non-linear PDE is elliptic, hyperbolic, or parabolic?

*) Characterizing 2nd order partial differential equations

*)Classification of a system of two second order PDEs with two dependent and two independent variables

Perhaps by checking the second reference above and the answers to your questions above you will know the difference of usage of $B^2 -4AC$ and $B^2 - AC$!