Background: I study a massive quantum field in 1 dimension space with length 2L. The wave functional of ground state is: $$\Psi[\phi(x)]\propto\int D[\phi]e^{-1/2\int\dot{\phi}^2+(\nabla \phi)^2+u^2 \phi^2 dx dt}$$
where boundary condition is: $$\phi(x,t=0)=\phi(x)$$ and the path integral is define in the area: $$t>0,\quad -L<x<L$$
To get the part which is related to $\phi(x)$, I divide the $\phi(x,t)$ into two part:
$$\phi(x,t)=\phi_{c}(x,t)+\phi'(x,t)$$
Then I specify part of the boundary condtion for $\phi_{c}(x,t)$: $$\phi_{c}(x,t=0)=\phi(x)$$
Thus the corresponding boundary condition for $\phi'$ is: $$\phi'(x,t=0)=0$$
The integral becomes:
$$\int dx dt[\dot{\phi_c}^2+(\nabla \phi_c)^2+u^2 \phi_c^2]+[\dot{\phi'}^2+(\nabla \phi')^2+u^2 \phi'^2]+[2\dot{\phi_c}\dot{\phi'}+2\nabla \phi_c\nabla \phi'+2u^2 \phi_c\phi']$$
To get rid of the terms in the last square bracket, I can use the freedom of divide the $\phi$ into 2 parts and impose the boundary condition properly.
After integration by part, the last square bracket becomes:
$$\int dx (2\dot{\phi_c}\phi')|^{+\infty}_{t=0}+\int dt (2\nabla \phi_c \phi')|^{L}_{x=-L}+\int dx dt [-2\ddot{\phi_c}-2\nabla^2\phi_c+2u^2\phi_c]\phi'$$
By demanding that:
$$\dot{\phi_c}(x,t=+\infty)=0, \quad \nabla \phi_c(x=\pm L)=0,\quad -2\ddot{\phi_c}-2\nabla^2\phi_c+2u^2\phi_c=0$$
we can get rid of the square bracket.
Then the path integral becomes:
$$\Psi[\phi(x)]\propto e^{-1/2\int[ \dot{\phi_c}^2+(\nabla \phi_c)^2+u^2 \phi_c^2 ]dx dt }\int D[\phi]e^{-1/2\int\dot{\phi}^2+(\nabla \phi)^2+u^2 \phi^2 dx dt}$$
The path integral part is totally a number independent of $\phi(x)$ thus we can drop it:
$$\Psi[\phi(x)]\propto e^{-1/2\int[ \dot{\phi_c}^2+(\nabla \phi_c)^2+u^2 \phi_c^2 ]dx dt }$$
with condition:
$$\quad -2\ddot{\phi_c}-2\nabla^2\phi_c+2u^2\phi_c=0,\quad \phi_c(x,t=0)=\phi(x)$$ $$\dot{\phi_c}(x,t=+\infty)=0, \quad \nabla \phi_c(x=\pm L)=0$$
Use the Equation of motion: $$\quad -2\ddot{\phi_c}-2\nabla^2\phi_c+2u^2\phi_c=0$$
By integrating by part, the integral becomes:
$$e^{1/2\int dx \phi_c(x)\dot{\phi_c(x)}}$$
The integration is done alone t=0.
MATH Part: Finally comes the pure math part, to solve the PDE:
$$\quad -2\ddot{\phi_c}-2\nabla^2\phi_c+2u^2\phi_c=0,\quad \phi_c(x,t=0)=\phi(x)$$ $$\dot{\phi_c}(x,t=+\infty)=0, \quad \nabla \phi_c(x=\pm L)=0$$
I use Green's function. $$(\partial_t^2 +\partial_x^2- u^2)G(x,t_1;y,t_2)=\delta(x-y,t_1-t_2)$$
The boundary condtion of $\phi_c$ demands the corresponding boundary condtion for $G$ to be:
$$\partial_tG(x,t_1=+\infty,y,t_2)=0,\quad G(x,t_1=0;y,t_2)=0$$ $$\partial_x G(x=\pm L,t_1;y,t_2)=0$$
I don't know how to solve it, wish some help me solve it.