PDE's: Analysis without solving explicitly the equation I have the following PDE'S
$$\frac{\partial}{\partial t}u(t,r)=\frac{\partial}{\partial r}\big((r-1/2)u(t,r)\big)$$
with $u(0, r)=u_0(r)$ compact supported in the interval $[0,1]$ and such that $\int_0^1u_0(r)dr=1$.
I know the solution explicitly, Through the characteristic method we know that
that $$u(t,r)=u_0\big((r-1/2)e^t+1/2\big)e^t$$.
I am wondering about the $\lim_{t\to +\infty}\int_0^1u(t,r)G(r) dr$, with $G\in C([0,1], \mathbb R)$ a test function.
Since I know the explicit solution it is easy to understand that 
$$\lim_{t\to +\infty}\int_0^1u(t,r)G(r) dr=G(1/2)$$.
My question is, could I deduce something about the asymptotic behaviour without knowning the explicit solution but just looking at the expression of the PDE?
 A: This is really closely related to the method of characteristics, but doesn't explicitly use the solution.
Consider $$J = J(a(t), b(t), t) = \int_{a(t)}^{b(t)} u(t,r)\; dr $$ 
where $a(t)$ and $b(t)$ are continuously differentiable functions.  By the Fundamental Theorem of Calculus we have
$$ \eqalign{\dfrac{dJ}{dt} &= b'(t) u(t,b(t)) - a'(t) u(t,a(t)) + \int_{a(t)}^{b(t)} \dfrac{\partial u}{\partial t}(t,r) \; dr \cr
&= b'(t) u(t,b(t)) - a'(t) u(t,a(t)) + \int_{a(t)}^{b(t)} \dfrac{\partial}{\partial t} \left((r-1/2) u(t,r)\right)\; dr \cr
&= (b'(t) + b(t) - 1/2) u(t, b(t)) - (a'(t) + a(t) - 1/2) u(t, a(t))\cr} $$
In particular, this is $0$ if $b(t)$ and $a(t)$ are solutions of the differential equation $x' + x - 1/2 = 0$.  Those solutions all converge to $1/2$ as $t \to \infty$.  Thus if $[a(0), b(0)]$ contains the support of $u_0$, for large $t$
the support of $u(t,\cdot)$ is contained in the small interval $[a(t), b(t)]$ near
$1/2$, and $\int_{a(t)}^{b(t)} u(t,r)\; dr = 1$. Take $\epsilon > 0$. If the interval is small enough that $|G(r) - G(1/2)|< \epsilon$,
we have 
$$\left|\int_0^1 u(t,r) G(r)\; dr - G(1/2)\right| = \left|\int_{a(t)}^{b(t)} u(t,r) (G(r) - G(1/2))\; dr \right| < \epsilon \int_{a(0)}^{b(0)} |u(t,r)|\; dr $$ 
Thus we get $$ \int_0^1 u(t,r) G(r)\; dr \to G(1/2)\ \text{as}\ t \to \infty$$
