What is the $\lim_{t\to 0^+}u(1,t)$? Consider $u_t=u_{xx}$ $x\in \mathbb{R}, t&gt;0$

$
u(x,0) = 
     \begin{cases}
       \text{1,} &amp;0\leq x\leq1 \\
       \text{0,} &amp;\quad\text{otherwise.} \\ 
     \end{cases}
$

What is the $\lim_{t\to 0^+}u(1,t)$

Is the question complete? I mean no boundary conditions are given. 

I know how to solve the problem when the boundary conditions are given. Here is the source from which I am studying http://tutorial.math.lamar.edu/Classes/DE/SolvingHeatEquation.aspx

Can somebody help me in finding the solution to this problem.  

Thanks a lot.
Actually the boundary values are given via the function $u(x,0)$. So what you have is a standard heat equation whose solution is obtained by the standard methods. It is:

$$u(x,t)=\frac{1}{\sqrt{\pi}}\left(\int_0^{\frac{1-x}{2\sqrt{t}}}e^{-s^2}\,ds+\int_0^{\frac{x}{2\sqrt{t}}}e^{-s^2}\,ds\right)$$
Therefore
$$u(1,t)=\frac{1}{\sqrt{\pi}}\int_0^{\frac{1}{2\sqrt{t}}}e^{-s^2}\,ds$$
so that
$$\lim_{t\to 0^{+}}u(1,t)=\frac{1}{\sqrt{\pi}}\int_0^{\infty}e^{-s^2}\,ds=\frac{1}{2}$$
Actually the boundary values are given via the function $u(x,0)$. So what you have is a standard heat equation whose solution is obtained by the standard methods. It is:

$$u(x,t)=\frac{1}{\sqrt{\pi}}\left(\int_0^{\frac{1-x}{2\sqrt{t}}}e^{-s^2}\,ds+\int_0^{\frac{x}{2\sqrt{t}}}e^{-s^2}\,ds\right)$$
Therefore
$$u(1,t)=\frac{1}{\sqrt{\pi}}\int_0^{\frac{1}{2\sqrt{t}}}e^{-s^2}\,ds$$
so that
$$\lim_{t\to 0^{+}}u(1,t)=\frac{1}{\sqrt{\pi}}\int_0^{\infty}e^{-s^2}\,ds=\frac{1}{2}$$