Find the limit as $t \to \infty$ of a solution of $u_t+u_x=u_{xx}$. Consider the following PDE
$$ u_t+u_x=u_{xx} \tag{1}$$
for $t \in (0,\infty)$, $x \in (0,1)$ and $u \in \mathbb{R}$. Le $u(t,x)$ of class $C^2([0,\infty) \times [0,1]$ a solultion satisfying 
$$ u(t,0)=1 ; u(t,1)=0.$$
Prove that $u(t,x)$ converges uniformly to $\overline u(x)$ as $t \to \infty$, where $\overline u$ is the time independent solution of $(1)$ with $u(0)=1$ and $u(1)=0$. 
I managed to find that $\overline u(x)= \frac{e^x-e}{1-e}$, but I don't know how to proceed, is it possible to answer without finding $u(t,x)$ explicitly?
 A: You can write the solution using the separation variable method. Set
$$
   u(t,x)=T(t)X(x)
$$
so that
$$
   T_t(t)X(x)+T(t)X_x(x)=T(t)X_{xx}(x).
$$
Now, put
$$
  X_x(x)-X_{xx}=\lambda X(x)
$$
and you will get
$$
   T_t(t)+\lambda T(t)=0
$$
That is solved by $Ae^{-\lambda t}$. It is
$$
  X(x)=C_1e^{\frac{x}{2}(1-\sqrt{1-4\lambda})}+C_2e^{\frac{x}{2}(1+\sqrt{1-4\lambda})}
$$
with
$$
  C_1+C_2=1 \qquad C_1e^{\frac{1}{2}(1-\sqrt{1-4\lambda})}+C_2e^{\frac{1}{2}(1+\sqrt{1-4\lambda})}=0.
$$
This system shows that $\lambda$ belongs to a continuous set of values and converges when $t\rightarrow\infty$ provided $\lambda\ge 0$. Then, the general solution can be cast in the form
$$
   u(t,x)=\bar u(x)+\int_0^\infty d\lambda' e^{-\lambda' t}X_{\lambda'}(x)
$$
being
$$
  X_{\lambda}(x)=\frac{e^{\sqrt{1-4 \lambda }+\frac{1}{2} \left(1-\sqrt{1-4 \lambda }\right) x}-e^{\frac{1}{2} \left(\sqrt{1-4 \lambda }+1\right) x}}{e^{\sqrt{1-4 \lambda }}-1}.
$$
At increasing $\lambda$, $X_\lambda(x)$ is just on oscillating function in $\lambda$ and the exponential in time in the integral grants convergence to the given limit. 
