# 2-D heat equation being parabolic?

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 $$0$$.
• Isn't $c$ the co-efficient of $\frac{\partial^2 u}{\partial y^2}$ and $a$ is the co-efficient of $\frac{\partial^2 u}{\partial x^2}$ ? ? – user1157 Oct 9 '18 at 11:14
• 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 $\frac{\partial^2 u}{\partial y \partial t}$. – gandalf61 Oct 9 '18 at 11:44
• 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. – user1157 Oct 9 '18 at 19:13