Let $u(x,t)$ satisfy the IVP $u_{t}=u_{xx}$,$x\in \mathbb{R},t>0$ and $$u(x,0)=\begin{cases} 1, & x\in[0,1]\\ 0, & \text{otherwise} \end{cases}$$ Then the value of $$\lim_{t\to 0^+} u(1,t)$$ is
$e$
$\pi$
$\frac{1}{2}$
$1$
My efforts: I have three solution of the given problem
$$u(x,t)=(ax+b)$$
$$u(x,t)=(ae^{kx}+be^{-kx})ce^{k^2t}$$
$$u(x,t)=(a\cos(kx)+b\sin(kx))ce^{-k^2t}$$
Now, how can I apply boundary conditions to get the answer? Answer is 3.
Hint
The general answer of the heat equation is as follows:$$u(x,t)=\int_0^\infty e^{-\omega^2t}\Big(A(\omega)\cos\omega x+B(\omega)\sin\omega x\Big)$$with$$A(\omega)={1\over \pi}\int_{-\infty}^\infty u(x,0)\cos\omega xdx\\B(\omega)={1\over \pi}\int_{-\infty}^\infty u(x,0)\sin\omega xdx$$now if we assume an even expansion of $u(x,0)$ around $x=0$ we obtain$$B(\omega)=0\\A(\omega)={2\over\pi}\int_0^1\cos\omega xdx={2\sin\omega\over \pi\omega}$$therefore by substitution$$u(x,t)={2\over \pi}\int_0^\infty e^{-\omega^2 t}{\sin\omega\over\omega}\cos\omega xdx$$