Sign of the solution to a PDE for all time Consider the following equation and initial contition $$\left\{ \begin{array}{ll} u_t-\frac{1}{2} u_{xx}+2au_x=0, & x\in \mathbb{R},t>0 \\ u(x,0)=u_0(x), & x\in \mathbb{R} \end{array} \right.$$ where $u_0(x)$ is odd, monotone increasing and bounded over $\mathbb{R}$.
It is easy to check that $u_t-\frac{1}{2}u_{xx}&lt0$. Is it possible to deduce from this the sign of the solution $u(0,t)$ for all time?
Thanks in advance for any insight.
 A: Let
$$
v(x,t)=u\bigl(\frac{x}{\sqrt2}+a\,t,t\bigr).
$$
Then $v_t-v_{xx}=0$ and $v(x,0)=u_0(x/\sqrt2)$. Since $u_0$ is odd
$$\begin{align*}
v(x,t)&=\frac{1}{\sqrt{4\,\pi\,t}}\int_{-\infty}^\infty e^{-\tfrac{(x-y)^2}{4t}}u_0\bigl(\frac{y}{\sqrt2}\bigr)\,dy\\
&=\frac{1}{\sqrt{4\,\pi\,t}}\int_{0}^\infty \Bigl(e^{-\tfrac{(x-y)^2}{4t}}-e^{-\tfrac{(x+y)^2}{4t}}\Bigr)u_0\bigl(\frac{y}{\sqrt2}\bigr)\,dy\\
&=\frac{1}{\sqrt{4\,\pi\,t}}\int_{0}^\infty e^{-\tfrac{(x-y)^2}{4t}}\Bigl(1-e^{-\tfrac{xy}{t}}\Bigr)u_0\bigl(\frac{y}{\sqrt2}\bigr)\,dy
\end{align*}$$
Since moreover $u_0$ is iscreasing, $u_0(x)\ge0$ if $x\ge0$. Thus we see that $v(x,t)\ge0$ if $x\ge0$, while $v(x,t)\le0$ if $x&lt0$. Since $u(0,t)=v(-a\,\sqrt2\,t,t)$, $u(0,t)$ will have the same sign as $-a$.
Observe that we don't need $u_0$ to be increasing. It is enough that $u_0$ has constant sign on $[0,\infty)$.
A: If $f=u_t-\frac{1}{2}u_{xx}&lt0\,$ then $u$ is a solution of the Chauchy problem
$$
\left\{ \begin{array}{ll} u_t-\frac{1}{2} u_{xx}=f, \\ 
u(x,0)=u_0(x).
\end{array} \right.
$$
It can be represented as a sum $u_1+u_2$, where $u_1$ is solution of the problem with rhs $f$ and zero initial condition and $u_2$ is solution for the zero rhs and intial condition $u_0$. Since $u_0$ is odd $u_2(0,t)=0$ and $u_1(x,t)&lt0\,$ for $t>0$ due to the maximum principle. 
