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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 $u_t-\frac{1}{2}u_{xx}<0$. Is it possible to deduce from this the sign of the solution $u(0,t)$ for all time?

Thanks in advance for any insight.

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Note that $U(x,t) = \exp(2 a x - 2 a^2 t) u(x,t)$ satisfies the heat equation $U_t - \frac{1}{2} U_{xx} = 0$, and has the same sign as $u(x,t)$. –  Robert Israel Apr 15 '12 at 6:02

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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 $x<0$. 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 $[0,\infty)$.

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If $f=u_t-\frac{1}{2}u_{xx}<0\,$ 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 $u_1(x,t)<0\,$ for $t>0$ due to the maximum principle.

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