# Non-minimal coupling ξ , Minkowski false vacuum decay case (shooting method) (An update to the previous question)

I asked a few days ago this question: Non-minimal coupling ξ - numerical bounce solutions (shooting method, false vacuum decay)

Alex Trounev helped me improve my code building based on this paper (Appendix A.6):https://arxiv.org/pdf/2205.03140v2.pdf?fbclid=IwAR0uOAZObsvf1BILxFgbuAZnfhe6L9e-hsO-jElhBJtEHShqBZ7_sQtIEB4

I improved the code a little bit more using some Manipulate. This is the code i used:

{fv, xi, a, b, c} = {0, 0.1, 1, .1, 0};
u = -1/4 a^2 (3 b - 1) phi[x]^2 + 1/2 a (b - 1) phi[x]^3 + 1/4 phi[x]^4 + a^4 c;

Manipulate[
R = k/(1 - k*xi*phi[x]^2)*(phi'[x]^2 + 4 u - 6 xi (phi'[x]^2 + phi[x] phi''[x] + 3 phi'[x]*phi[x]*r'[x]/r[x]));
due = D[u, phi[x]] /. phi[x] -> e;
eq1 = phi''[x] + (3 r'[x]/r[x] phi'[x] - xi*phi[x]*R - D[u, phi[x]]) == 0;
eq2 = r''[x] + k*r[x]/(3 (1 - k*xi*phi[x]^2)) (phi'[x]^2 + u - 3 xi (phi'[x]^2 + phi''[x] phi[x] + phi'[x] phi[x] r'[x]/r[x])) == 0;
ic = {phi[xmin] == e + 1/8 due xmin^2, phi'[xmin] == 1/4 due xmin, r[xmin] == xmin, r'[xmin] == 1};

sol = ParametricNDSolve[Flatten[{eq1, eq2, ic}], {phi, r}, {x, xmin, xmax}, {e}];
sol1 = FindRoot[phi[e][xmax] == fv /. sol, {e, e0}];

p1 = Plot[Evaluate[{phi[e][x]} /. sol1 /. sol], {x, xmin, xmax}, PlotLegends -> {"ξ=0.1"}];
q1 = Plot[Evaluate[{r[e][x]}/10 /. sol1 /. sol], {x, xmin, xmax}, PlotLegends -> {"ξ=0.1"}];

Show[p1, q1],
{{k, 1}, 0.1, 2, 0.01},
{{xmin, 0.01}, 0.0001, 0.01, 0.0001},
{{xmax, 35}, 15, 40, 0.1},
{{e0, 0.99}, 0.89, 1.5, 0.01}
]


Repeating the process for xi=0 and xi=0.2 I took the following results:  Where $$η$$ is $$τ$$ in their paper and x in my code (sorry for that, but I want $$\eta$$ in my plots to match with another paper heh)

Ok not bad a approach it is, but the field does not reach fv=0 as in the paper and $$ρ$$ lines are not in the same order for different values of ξ as in the paper. Then I read the paper I took the problem from carefully this time (https://arxiv.org/pdf/1701.05731.pdf) and they state that in the Nunmericals section:

In numerical calculations containing ξ we solve the field EOM (eq. of $$φ$$) with Ricci scalar expressed by the scalar field and the second Friedman equation (eq. of $$ρ$$) with the boundary conditions posted in the first question (for Minkowski false vacuum)

(approximating $$ρ(0)$$ as proportional to initial $$τ = ε$$ (this is our $$τ$$ is our x and $$ε$$ our xmin and $$\dotρ = 1$$. Our EOM is the equation of a particle in potential $$−V(φ)$$ with a time-dependent friction $$3 \dotρ/ρ$$ and we expand it around both vacua in $$ε$$ that can be arbitrarily small. We neglect higher orders of both expansions and we find iteratively their leading orders that fulfill the boundary condition. Last initial condition the field value $$φ_0$$ (our e) corresponding to CDL, can be found by a undershoot/overshoot method known from the flat setup (by this final statment I think that i means the process of Taylor expanding the $$\dot\phi$$ and $$\phi$$ as appendix A.6 does in the first question.

My main problem is that I don't understand the process with the Taylor expansion of the EOM, I chated a little with GPT-4. It gave me this


{fv, xi, a, b, c} = {0, 0.1, 1, .1, 0};
u = -1/4 a^2 (3 b - 1) phi[x]^2 + 1/2 a (b - 1) phi[x]^3 + 1/4 phi[x]^4 + a^4 c;
Manipulate[
R = k/(1 - k*xi*phi[x]^2)*(phi'[x]^2 + 4 u - 6 xi (phi'[x]^2 + phi[x] phi''[x] + 3 phi'[x]*phi[x]*r'[x]/r[x]));
due = D[u, phi[x]] /. phi[x] -> e;

eq1 = Normal[Series[phi''[x] + (3 r'[x]/r[x] phi'[x] - xi*phi[x]*R - D[u, phi[x]]), {x, 0, 2}]] == 0;
eq2 = r''[x] + k*r[x]/(3 (1 - k*xi*phi[x]^2)) (phi'[x]^2 + u - 3 xi (phi'[x]^2 + phi''[x] phi[x] + phi'[x] phi[x] r'[x]/r[x])) == 0;

ic = {phi[xmin] == e + 1/8 due xmin^2, phi'[xmin] == 1/4 due xmin, r[xmin] == xmin, r'[xmin] == 1, phi'[xmax] == 0,};

sol = ParametricNDSolve[Flatten[{eq1, eq2, ic}], {phi, r}, {x, xmin, xmax}, {e}];
sol1 = FindRoot[phi[e][xmax] == fv /. sol, {e, e0}];

p1 = Plot[Evaluate[{phi[e][x]} /. sol1 /. sol], {x, xmin, xmax}, PlotLegends -> {"ξ=0.1"}];
q1 = Plot[Evaluate[{r[e][x]}/10 /. sol1 /. sol], {x, xmin, xmax}, PlotLegends -> {"ξ=0.1"}];

Show[p1, q1],
{{k, 1}, 0.1, 2, 0.01},
{{xmin, 0.01}, 0.0001, 0.01, 0.0001},
{{xmax, 35}, 15, 40, 0.1},
{{e0, 0.99}, 0.89, 1.5, 0.01}]



Its results are full of errors. But this was my thought too, the Taylor Expansion of the EOM. Can anybody help me adjust the text of the paper in code to reproduce their results:

Last note: My code has worked better for the dS transition case ,which is most crucial in my work, where $$c=0.05$$ check my results:

Ok maybe $$ξ=0$$ does not match with the other curves but the shapes are fine. I am sorry for not updating in the first question, but i had a lot of information to share with you and I believe a new question is fine. I have no problem my first approach (first question code) it does work almost fine, but I would try to take the exact result if I could.

Ps: Check the first question for the full problem equations-b.c., codes etc etc.

• Do you try to test the numerical method given in the paper? Mar 21 at 4:29
• Yeap, I tried a little bit to transform the code with an EOM expansion, but i don't exactly understand the text description. Also in the φ0 initial condition they give wrong paper in the reference. I am sure they refer to this: sci-hub.se/https://journals.aps.org/prd/abstract/10.1103/… Furthermore, I am confused with their conditions. Why they approximate $\dot\rho=1$ , i am sure the right thing is $\dot\rho(x_{min})=1$ and this false statement is in a second edition of the paper with small differences (arxiv.org/pdf/1606.07808.pdf) Mar 21 at 11:55

Finally solved them. The c=0 case, with the following code.


{fv, xi, a, b, c} = {0, 0, 1, .1, 0};
u = -1/4 a^2 (3 b - 1) phi[x]^2 + 1/2 a (b - 1) phi[x]^3 +
1/4 phi[x]^4 + a^4 c;

Manipulate[
R = k/(1 - k*xi*phi[x]^2)*(phi'[x]^2 + 4 u -
6 xi (phi'[x]^2 + phi[x] phi''[x] + 3 phi'[x]*phi[x]*r'[x]/r[x]));
due = D[u, phi[x]] /. phi[x] -> e;
eq1 = phi''[
x] + (3 r'[x]/r[x] phi'[x] - xi*phi[x]*R - D[u, phi[x]]) == 0;
eq2 = r''[x] +
k*r[x]/(3 (1 - k*xi*phi[x]^2)) (phi'[x]^2 + u -
3 xi (phi'[x]^2 + phi''[x] phi[x] +
phi'[x] phi[x] r'[x]/r[x])) == 0;
ic = {phi[tmin] == e + 1/8 due tmin^2, phi'[tmin] == 1/4 due tmin,
r[tmin] == tmin, r'[tmin] == 1};
sol = ParametricNDSolve[
Flatten[{eq1, eq2, ic}], {phi, r}, {x, tmin, tmax}, {e}];
sol1 = FindRoot[phi[e][tmax] == fv /. sol, {e, e0}];
p1 = Plot[Evaluate[{phi[e][x]} /. sol1 /. sol], {x, tmin, tmax},
PlotLegends -> {"ξ=0.1"}];
q1 = Plot[Evaluate[{r[e][x]}/10 /. sol1 /. sol], {x, tmin, tmax},
PlotLegends -> {"ξ=0.1"}];
Show[p1, q1], {{k, 1.9, "κ"}, 1, 3, 0.1}, {{tmin, 0.01, "τ_min"},
0.0001, 0.01, 0.0001}, {{tmax, 30, "τ_max"}, 15, 35,
0.1}, {{e0, 0.999, "e0"}, 0.89, 1.1, 0.01}]


• The key point is to play with e0 to avoid overshoot, place the starting value at 3 decimals from the e solution Mar 27 at 20:12