Suppose $v:[0,\infty)\times\mathbb R\to\mathbb R$ is a solution of $$\frac{\partial v}{\partial t}=\frac\partial{\partial y}\left(\frac{\partial v}{\partial y}+h'v\right)\;,\;\;\;v(0,y)=v_0(y)$$ for some $h\in C^1(\mathbb R)$ and $v_0\in C^0(\mathbb R)$ with $$\int v_0(y)\:{\rm d}y=1.\tag1$$
Why are we able to conclude $$\int v(t,y)\:{\rm d}y=1\tag2$$ for all $t\ge0$?
Compute using the fundamental theorem of calculus: $$ \frac{d}{dt} \int v(t,y) dy = \int \partial_t v(t,y) dy = \int \partial_y( \partial_y v(t,y) + h'(y) v(t,y))dy = \lim_{z \to \infty} \left[\partial_y v(t,z) + h'(z) v(t,z) - \partial_y v(t,-z) + h'(-z) v(t,-z) \right] =0, $$ assuming that $\partial_y v$ and $v$ decay sufficiently rapidly at infinity. This can be justified, for instance, if the solution lives in an appropriate Sobolev space. Then since the time derivative vanishes, we have $$ \int v(t,y) dy = \int v(0,y) dy = \int v_0(y) dy = 1 $$ for all $t \ge 0$.