The PDE $u_t = u_{xx}$ follows the path defined by $\dfrac{dx}{dt} = \pm \infty$ I have the PDE $u_t = u_{xx}$ (heat equation).
I am then told that, by writing the equation as $(\partial_x + (0)\partial_t)^2 u = u_t$, we see that its characteristics would follow the path defined by $\dfrac{dx}{dt} = \pm \infty$.
I wonder how they came to this conclusion? Please kindly explain. :)
 A: The rewriting of the PDE collects the leading order (second order) derivatives on the right-hand side, as the characteristics should be determined by the leading order terms.
If you have a vector $v=(a,b)$ in the $(x,t)$-plane, then the characteristics of the second order operator $(v\cdot\nabla)^2=(a\partial_x+b\partial_t)^2$ are lines in the direction of the vector $v$.
In this particular case $a=1$ and $b=0$, so the line points in the $x$ direction with no $t$ component.
This conclusion is then expressed — in my opinion clumsily — by saying that $\frac{dx}{dt}=\pm\infty$.
Since you gave no source in your question, I can't interpret everything.
But there is truth to the statement: the heat equation (not only in $1+1$ dimensions) has infinite speed of propagation, meaning that solutions travel infinitely fast in all directions.
This is described in the question as $\frac{dx}{dt}=\pm\infty$, but I do hope the source you are using also gives a description in words.
A: For a second order PDE in two variables $x$ and $t$,
$$
a(x,t) u_{xx} + 2b(x,t) u_{xt} + c(x,t) u_{tt} + \dots = 0
,
$$
there is an associated quadratic form at each point $(x,t)$,
$$
Q(h,k) = a(x,t) h^2 + 2b(x,t) hk + c(x,t) k^2
,
$$
and a characteristic curve is a curve whose normal vector $(h,k)$ satisfies $Q(h,k)=0$ at each point of the curve.
For the heat equation we have simply
$$
Q(h,k)=h^2
$$
so the normal vector should be $(h,k)=(0,1)$ at each point, which forces the characteristics to be lines of constant $t$.
