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.



Heat Equation with an integral boundary condition
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 $v(x,t)=u(x-1,t)+Z(x,t)$.


