Heat Equation with an integral boundary condition I have been struggling with following Heat equation IBVP,
\begin{equation}
\frac{\partial v\left(x, t\right)}{\partial t} = \alpha \frac{\partial^2 v\left(x, t\right)}{\partial x^2}, \quad t \in \left(0, T\right], x \in \left(- \infty, 1\right]. \label{heatEqn}
\end{equation}
with the following initial and boundary conditions
\begin{equation}
v\left(x, 0\right) = \delta\left(x\right), \quad v\left(-\infty, t\right) = 0, \quad \int_{-\infty}^{1}v\left(x, T\right) dx =
\beta T^{\alpha-1}
\end{equation}
where $\alpha \in \left(0, 1\right)$ and $\beta$ is some positive constant.
Some interesting observation about this problem, if we consider the semi-infinite plane $x \in \left(- \infty, 1\right]$ $\times$ $t \in \left[0, T\right]$, we know line integrals along three of the boundaries of this rectangle, as in $\int_{-\infty}^{1}v\left(x, 0\right) dx = 1$, $\int_{0}^{T}v\left(-\infty, t\right) dt = 0$ and the boundary condition with the integral equation above. 
Also if we can deduce $v\left(1, t\right)$ and $\frac{\partial v\left(x, t\right)}{\partial x}|_{x = 1}$ then we can attack the problem using laplace transform.
I am wondering if there is version of Greens or Divergence Theorem which can be applied to tackle this problem.
Any help or guidance will be greatly appreciated.
Thank you.
 A: Here is the answer to your modified question via potential approach but simpler than proposed in the comments.
Denote $Z(x,t)=(4\alpha\pi t)^{-1/2} e^{-x^2/(4\alpha t)}$ the fundamental solution for the heat equation. I'll write $t$ instead of $T$ in the question. Subtracting $Z(x,t)$  from $v$ reduces the initial condition to zero and the integral to 
$$
\beta t^{\alpha-1}-\int_{-\infty}^1Z(x,t)\,dx=
\beta t^{\alpha-1}-\frac{1}{2} \left(\text{erf}\left(\frac{1}{2 \sqrt{\alpha  t}}\right)+1\right):=g(t).
$$
Shifting the side boundary to $x=0$ (for convenience) the problem becomes
$$
\left\{
\begin{array}{rccl}
u_t&=&\alpha u_{xx},&t>0,\ x<0,\\
u|_{t=0}&=&0,\\
\int_{-\infty}^0u(x,t)\,dx&=&g(t),&t>0.
\end{array}
\right.
$$
Its solution can be written down explicitly as a potential:
$$
u(x,t)=2\alpha\int_{-\infty}^0Z_{xx}(x,t-\tau)g(\tau)\,d\tau.
$$
To see that consider the double layer potential
$$
Wg(x,t)=\int_0^tZ_x(x,t-\tau)g(\tau)\,d\tau.
$$
It satisfies the heat equation for $x<0$ and for continuous densities $g$ has a property
$$
\lim_{x\to0-}Wg(x,t)=\frac{g(t)}{2\alpha}.
$$
Actually it's the jump relation for the double layer potential in this particular case.
For $\varepsilon>0$ we have
$$
\int_{-\infty}^{-\varepsilon}u(x,t)\,dx=
2\alpha\int_0^t\int_{-\infty}^{-\varepsilon}Z_{xx}(x,t-\tau)g(\tau)\,dx d\tau=
$$
$$
2\alpha\int_0^tZ_{x}(-\varepsilon,t-\tau)g(\tau)\,d\tau=
2\alpha Wg(-\varepsilon,t).
$$
Taking $\varepsilon\to0+$ gives the result.
The solution of the initial problem is $v(x,t)=u(x-1,t)+Z(x,t)$.
