Total variation of a continuous, two variable function is defined by:
$$ TV(\phi) = \lim_{\epsilon\to 0} \sup \int\int |\phi(x + \epsilon,y) - \phi(x, y)| dx dy + \lim_{\epsilon\to 0} \sup \int\int |\phi(x,y + \epsilon) - \phi(x, y)| dx dy $$
Consider solutions to $ \frac{\partial \phi(x, t)}{\partial t} + \frac{\partial }{\partial x}(u \phi(x, t)) + \frac{\partial }{\partial y}(v\phi(x, t)) = 0 $ in solid body rotation with $ u = -y $ and $ v = x $ then:
$$ (*) \frac{\partial \phi}{\partial t} -y \frac{\partial \phi}{\partial x} + x\frac{\partial \phi}{\partial y} = 0 $$ with
$ \phi(x, y, 0) = 1 $ if $ |x| \le 1 $ and $ |y| \le 1 $ and $ \phi(x, y, 0) = 0 $ otherwise
Show that after distribution of $ \phi $ rotates through an angle of $ \pi/4 $ then TV increases by $ \sqrt2 $.
My solution so far:
At time $ t = \bar t $ apply transformation of rotation through an angle of $ \pi/4 $.
$$ (x, y) \rightarrow ((x + y)/\sqrt2, (y - x)/\sqrt2) $$
with
$$ \bar x = (x + y)/\sqrt2 $$ $$ \bar y = (y - x)/\sqrt2 $$
thus, pde (*), becomes:
$$ \frac{\partial \phi}{\partial t} + \frac{\partial \phi}{\partial \bar x}\frac{\bar x - \bar y}{\sqrt 2} + \frac{\partial \phi}{\partial \bar y}\frac{\bar x + \bar y}{\sqrt 2} = 0 $$
with initial condition
$ \phi(\bar x, \bar y, 0) = 1 $ if $ |\bar x| \le 1 $ and $ |\bar y| \le 1 $ and $ \phi(\bar x, \bar y, 0) = 0 $ otherwise
How can I use this to show that the TV has increased by $ \sqrt 2 $ after the transformation?