I would like to show that no new maxima or minima can develop in smooth solutions to
$ \frac{\partial \phi(x, t)}{\partial t} + \frac{\partial }{\partial x}(f(\phi(x, t)) = 0 $
My solution:
Since solutions are smooth, they can be differentiated. Thus equation becomes:
$ \frac{\partial \phi}{\partial t} + \frac{\partial \phi}{\partial x} f'(\phi) = 0 $
We can also impose an initial condition $ \phi(x,0) = g(x) $ where g(x) has a fixed amount of extrema, say M. Our goal is then to show that $ \phi(x,t) $ has no further extrema for all $ t $. In other words, for all $ t_0 > 0 $, there are no more than M points such that $ \frac{\partial \phi(x, t_0)}{\partial x} = 0 $.
I have then attempted to solve this by the method of characteristics:
Let $ x $ and $ t $ be parameterised by $ s $ then:
$ z(s) = \phi(x(s), t(s)) $ so that:
$$ \frac{d z}{d s} = \frac{d x}{d s}\frac{\partial \phi}{\partial x} + \frac{d t}{d s}\frac{\partial \phi}{\partial t} $$
comparison with the above equation gives the following charactristic equations:
$$ \frac{d z}{d s} = 0 $$ $$ \frac{d x}{d s} = f'(z) $$ $$ \frac{d t}{d s} = 1 $$
with initial conditions $ t(0) = 0 $ and $ x(0) = x_0 $
giving solution:
$$ t = s $$ $$ z = g(x_0) $$ $$ x = f'(g(x_0))t + x_0 $$
Now, if we can invert the last equation such that $ x_0 = h(x, t) $ then solution is given by:
$$ \phi = g(h(x, t)) $$
And so $ \phi $ will not generate any new maxima since it is determined by g.
I don't feel confident in this argument and would appreciate any hints or alternative solutions.
A related question is:
Show solutions conserve amplitude of all local maxima and minima in the initial data.