This may be a really stupid question, but hopefully someone will point out what i've been missing:
I've just started studying PDE and came across the classification of second order equations, for example in this pdf. It states that given second order equation $au_{xx}+2bu_{xy}+cu_{yy}+du_x+eu_y+fu=0$ if $b^2-4ac=0$ then given equation is parabolic, then goes on to state that the heat equation $u_t-u_{xx}=0$ is the most common example of a parabolic equation. I've also seen the one dimensional heat equation as $u_t-ku_{xx}=0$ with a non-negative constant k.

My question is: I don't see how this satisfies the parabolic condition. If $u(x,t)$ is the solution, and we take $t$ to be $x$ in the general equation above and $x$ to be $y$, then the coefficient 1 of $u_t$ is $e$ and $-1$ (or $-k$) is c. So $b^2-4ac=4$ or $4k$ which is zero only in the case that k is zero. How then is the heat equation parabolic?

  • 2
    $\begingroup$ One of a or c is always zero however you label it. $\endgroup$
    – Thompson
    Sep 13, 2016 at 1:56
  • $\begingroup$ D'oh! I knew it would be so simple. Thanks for answering. $\endgroup$ Sep 13, 2016 at 1:57

1 Answer 1


In this case $x=x$ and $y=t$. So $a=k$ and $b=c=0$. So $b^2-4ac=0^2-4k0=0$


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