The amount of pollutant in the interval at time $t$ is $M=\int_0^bu(x,t)dx$ .At later time t+h,the same molecules of pollutant have moved to the right by $ch$ centimetres.
$M=\int_0^bu(x,t)dx=\int_{ch}^{b+ch}u(x,t+h)dx$.
Differentiating with respect to b, we get
$u(b,t)=u(b+ch,t+h)$.
Can some one please explain to me how the expression after differentiating with respect to b is obtained $u(b,t)=u(b+ch,t+h)$. ?
Let $U$ be the $x$ antiderivative of $u$, so $U_x=u$, by the fundamental theorem of calculus we have:
$U(b,t)-U(0,t)=U(b+ch,t+ch)-U(ch,t+ch)$, differentiate with respect to $b$ and we have:
$U_b(b,t)=U_b(b+ch,t)$, but $U_b=u$, so:
$u(b,t)=u(b+ch,t)$