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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.

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    $\begingroup$ Is $T$ fixed or the integral condition should hold for any $T>0$? If the first, would it make any difference if to change the rhs to $C$? If the second how Laplace transform is supposed to be applied? And btw does $\Omega=(-\infty,1)$? $\endgroup$
    – Andrew
    Jul 6 '17 at 10:23
  • $\begingroup$ @Andrew, $T$'s condition should hold for all $T>0$. I have replaced the $\Omega$, sorry about that. I could be mistaken about the Laplace transform, what i was thinking was that if we can work out $v(1, t)$ and $v$'s partial derivative wrt to $x$ at $x = 1$, based on the other boundary conditions, then these are the pieces needed to apply the Laplace transform to the PDE wrt to $x$. no ? $\endgroup$
    – Comic Book Guy
    Jul 6 '17 at 14:44
  • $\begingroup$ The last formula holds at least if $\psi\in C^\gamma$, $\gamma>1/2$ and $\psi(0)=0$. $\endgroup$
    – Andrew
    Jul 6 '17 at 20:46
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    $\begingroup$ For $T< 1$ $\log T$ is negative, what about it? The problem can be reduced to the zero initial condition considering function $u(x,t)=v(x,t)-Z(x,t)$ where $Z$ is the fundamental solution for the heat equation. The rhs in the integral condition will change somewhat, the change can be written down explicitly via $\mathtt{erf}$. The solution $u$ can be sought in the form of a double layer potential, the density $\psi$ of which is equal up to factor 2$\alpha$ the boundary values of the solution: $u(1,t)=2\alpha \psi(t)$, and it's derivative $u_x(1,t)=D^{1/2}_t\psi(t)$. $\endgroup$
    – Andrew
    Jul 6 '17 at 20:49
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    $\begingroup$ Btw is $\alpha$ the same in the heat equation and the exponent in the rhs of the integral condition? In this form the integral condition very nicely go into (integrating $Z_x$ from the double layer potential) into $\int_{-\infty}^{1}v\left(x, 0\right) dx=C(\alpha)I^{1/2}_t\psi(t)$. Hence $\psi(t)=C(\alpha) D_t^(1/2)g(t)$. $\endgroup$
    – Andrew
    Jul 6 '17 at 21:03
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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 $v(x,t)=u(x-1,t)+Z(x,t)$.

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