This is a mighty dumb question but I can not seem to figure out how a 2-D heat equation is parabolic. $$ \frac{\partial u}{\partial t}=\Delta u $$ I know that the discriminant : $b^2-ac=0$ for a parabolic PDE.

Here, $b=0,a=c=-1$ . Then how is it parabolic ? I think I am missing out something.


$u$ is a function of $x$ and $t$. In the discriminant, $c$ is the co-efficient of $\frac{\partial^2 u}{\partial t^2}$, which is $0$, so the discriminant is indeed $0$.

  • $\begingroup$ Isn't $c$ the co-efficient of $\frac{\partial^2 u}{\partial y^2}$ and $a$ is the co-efficient of $\frac{\partial^2 u}{\partial x^2}$ ? ? $\endgroup$ – user1157 Oct 9 '18 at 11:14
  • $\begingroup$ See en.wikipedia.org/wiki/Parabolic_partial_differential_equation for the general case. The key point is that there is no term in $\frac{\partial^2 u}{\partial t^2}$ or in $\frac{\partial^2 u}{\partial x \partial t}$ or in $\frac{\partial^2 u}{\partial y \partial t}$. $\endgroup$ – gandalf61 Oct 9 '18 at 11:44
  • $\begingroup$ I think I misinterpreted the general case. The function is $u(x,y)$ in the general case but in this heat equation it is $u(x,t)$;hence, $c=0$. Your answer was clear. $\endgroup$ – user1157 Oct 9 '18 at 19:13

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