TITLE: Proving solution to a PDE remains nonnegative.
QUESTION [0 upvotes]: The following problem is taken from Question 3, Chapter 5 of:
D.R. Durran, Numerical Methods for Fluid Dynamics: With Applications to Geophysics,. 1. Texts in Applied Mathematics 32, DOI 10.1007/978-1-4419-6412-0 1
It states the following:
Show that if $ \phi(x, 0) \ge 0 $
, the solution to 
$ \frac{\partial \phi(x, t)}{\partial t} + \frac{\partial }{\partial x}(c(x)\phi(x, t)) = 0 $ remains nonnegative for all $ t \ge 0 $. Assume $ c(x) $ and $ \phi(x,t) $ have continuous derivatives to simplify argument. Does the result generalise to 2 and 3 spatial dimenions?
Hint:
For $ \phi(x,t) \lt 0 $, must have a first time  $ t_{0} $ and some point $ x_{0} $ s.t. $ \phi(x_{0}, t_{0}) = 0 $ and $ \frac{\partial \phi(x_{0}, t_{0})}{\partial t} \lt 0 $. Show that this is not possible.
My solution:
$$ \frac{\partial \phi(x, t)}{\partial t} = - c^´\phi(x, t) - c(x)\frac{\partial }{\partial x}\phi(x, t) $$
Evalutating above at $ (x_{0}, t_{0}) $ gives
$ \frac{\partial \phi(x_{0}, t_{0})}{\partial t} = - c(x_{0})\frac{\partial }{\partial x}\phi(x_{0}, t_{0}) $
Thus if $ \frac{\partial \phi(x_{0}, t_{0})}{\partial t} \lt 0 $ must have $ c(x_{0})\frac{\partial }{\partial x}\phi(x_{0}, t_{0}) \gt 0 $
At this point I do not know how to show this inequality. I haven`t used the relation $ \phi(x, 0) \ge 0 $ yet. But I cannot see how this can be used since we are looking for information at time $ t_{0} $, not $ t = 0 $. Furthermore, we know nothing about $ c(x) $.  Presumably I need to prove that if
$ c(x_{0}) \gt 0 $ 
then 
$ \frac{\partial }{\partial x}\phi(x_{0}, t_{0}) \lt 0 $ 
and if 
$ c(x_{0}) \lt 0 $ 
then 
$ \frac{\partial }{\partial x}\phi(x_{0}, t_{0}) \gt 0 $ thus showing that it is impossible to obtain $ \frac{\partial \phi(x_{0}, t_{0})}{\partial t} \lt 0 $.
I have proved that indeed $ \phi(x, t) $ remains nonnegative by the method of characteristics but I wanted to try proving it following the hint. As for the question of whether this result extends to 2 and 3 spatial dimensions; I believe it does but I am struggling to come up with a convincing explanation.

REPLY [0 votes]: If  $ \phi_{x}(x,t) \lt 0 $ then there exists an $ \epsilon \gt 0 $ with $ x_{1} + \epsilon $ such that $ \phi(x_{1},t_{0}) \lt 0 $. However, from the initial condition, $ \phi(x_{1},0) \gt 0 $ thus there must exist a $ t_{1} \in [0, t_{0}) $ s.t. $ \phi(x_{1},t_{1}) = 0 $ which is a contradiction against $ t_{0} $ being the first time s.t. $ \phi(x_{0}, t_{0}) = 0 $. 
Similar argument applies for  $ \phi_{x}(x,t) \gt 0 $.
However, I still have not convinved myself for the case of 2 or 3 spatial dimensions where the governing equation is extended to:
$ \frac{\partial \phi(x, t)}{\partial t} + \nabla . (\textbf{c}(x)\phi(x, t)) = 0 $
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