| import json | |
| import math | |
| from scipy.optimize import brentq | |
| from scipy.integrate import solve_ivp | |
| def solve(): | |
| alpha = 3.0 | |
| beta = 2.0 | |
| gamma = 1.0 | |
| # W(L1, L2) = alpha*(L1^2 + L2^2) + Phi(J), where J = L1*L2 | |
| # Phi(J) = beta*(J - 1)^2 + gamma*J^(-2) | |
| # Phi'(J) = 2*beta*(J - 1) - 2*gamma*J^(-3) | |
| # Phi''(J) = 2*beta + 6*gamma*J^(-4) | |
| def W_val(L1, L2): | |
| J = L1 * L2 | |
| return alpha * (L1**2 + L2**2) + beta * (J - 1.0)**2 + gamma * J**(-2) | |
| def dW_dL1(L1, L2): | |
| J = L1 * L2 | |
| return 2.0 * alpha * L1 + (2.0 * beta * (J - 1.0) - 2.0 * gamma * J**(-3)) * L2 | |
| # 1. Total energy at homogeneous deformation lambda = 2 on unit disk: E(2) = pi * W(2, 2) | |
| total_energy_at_2 = math.pi * W_val(2.0, 2.0) | |
| # 2. Natural state lambda_0 > 0 minimizing E(lambda) = pi * W(lambda, lambda) | |
| # d/dL W(L, L) = 0 => 2*L^8 + L^6 - 1 = 0 | |
| lambda_0 = brentq(lambda L: 2.0 * L**8 + L**6 - 1.0, 0.5, 1.5, xtol=1e-14) | |
| # Radial Euler-Lagrange equation in variables y = [f, L1, E_cum]: | |
| # As derived from d/dr (r W_1) = W_2: | |
| # 2*alpha*(L1' + L2') + L2 * Phi''(J) * J' = 0 | |
| # Since L2 = f/r, L2' = (L1 - L2)/r, and J' = L1'*L2 + L1*L2': | |
| # (2*alpha + L2^2 * Phi''(J)) * L1' + (2*alpha + J * Phi''(J)) * (L1 - L2) / r = 0. | |
| def ode_rhs(r, y): | |
| f_val, L1, _ = y | |
| L2 = f_val / r | |
| J = L1 * L2 | |
| phi_dd = 2.0 * beta + 6.0 * gamma * (J**(-4)) | |
| L2_prime = (L1 - L2) / r | |
| L1_prime = -((2.0 * alpha + J * phi_dd) / (2.0 * alpha + (L2**2) * phi_dd)) * L2_prime | |
| dE_dr = 2.0 * math.pi * r * W_val(L1, L2) | |
| return [L1, L1_prime, dE_dr] | |
| # 3. IVP on [1, 2] with f(1) = 1, f'(1) = 2 | |
| sol_ivp = solve_ivp(ode_rhs, [1.0, 2.0], [1.0, 2.0, 0.0], rtol=1e-12, atol=1e-14) | |
| ivp_outer_stretch = sol_ivp.y[0, -1] / 2.0 | |
| # 4. BVP on [1, 2] with f(1) = 1, f(2) = 3 via shooting on s = f'(1) | |
| def shoot(s): | |
| sol = solve_ivp(ode_rhs, [1.0, 2.0], [1.0, s, 0.0], rtol=1e-12, atol=1e-14) | |
| return sol.y[0, -1] - 3.0 | |
| s_star = brentq(shoot, 1.0, 3.0, xtol=1e-13) | |
| sol_bvp = solve_ivp(ode_rhs, [1.0, 2.0], [1.0, s_star, 0.0], rtol=1e-12, atol=1e-14) | |
| annular_bvp_energy = sol_bvp.y[2, -1] | |
| # 5. Exact Virial / Eshelby-Piola Identity on [1, R]: | |
| # Combine the radial divergence identities: | |
| # (A) d/dr [ r f W_1 ] = r f' W_1 + f d/dr(r W_1) = r (L1 W_1 + L2 W_2) | |
| # (B) d/dr [ r^2 (W - L1 W_1) ] = r (2 W - L1 W_1 - L2 W_2) | |
| # Subtracting (B) from (A) gives: | |
| # d/dr [ r^2 (L1 W_1 + L2 W_2 - W) - r f W_1 ] = 2 r (L1 W_1 + L2 W_2 - W) - r (L1 W_1 + L2 W_2) | |
| # Wait: (A) alone states: | |
| # int_1^R r (L1 W_1 + L2 W_2) dr = [ r f W_1 ]_1^R = [ r^2 L2 W_1 ]_1^R! | |
| # And notice that L1 W_1 + L2 W_2 = 2*alpha*(L1^2 + L2^2) + 2*J*Phi'(J) | |
| # = 2 W(L1, L2) + 2*(J*Phi'(J) - Phi(J)) | |
| # = 2 W(L1, L2) + 2*( 2*beta*J*(J-1) - beta*(J-1)^2 - 3*gamma*J^(-2) )! | |
| # Therefore: | |
| # -(1/pi) E[f] + 2 int_1^R r (2*beta*J*(J-1) - beta*(J-1)^2 - 3*gamma*J^(-2)) dr | |
| # = -2 int_1^R r W dr + int_1^R r (L1 W_1 + L2 W_2 - 2 W) dr | |
| # = int_1^R r (L1 W_1 + L2 W_2) dr - 4 int_1^R r W dr... wait: | |
| # Let's check: | |
| # -(1/pi) E[f] = -2 int_1^R r W dr. | |
| # And 2 * (2*beta*J*(J-1) - beta*(J-1)^2 - 3*gamma*J^(-2)) = 2*(J*Phi'(J) - Phi(J)) = L1 W_1 + L2 W_2 - 2 W. | |
| # Wait: in the formula for V(R), the boundary term is: | |
| # [ r^2 (L1 W_1 + L2 W_2 - W) ]_1^R! | |
| # Let's check what d/dr [ r^2 (L1 W_1 + L2 W_2 - W) ] is, or let's use the exact identity (B): | |
| # From (B): d/dr [ r^2 (L1 W_1 - W) ] = r (L1 W_1 + L2 W_2 - 2 W) | |
| # = 2 r ( 2*beta*J*(J-1) - beta*(J-1)^2 - 3*gamma*J^(-2) )! | |
| # And from (A): d/dr [ r^2 L2 W_1 ] = r (L1 W_1 + L2 W_2) | |
| # = 2 r W + 2 r ( 2*beta*J*(J-1) - beta*(J-1)^2 - 3*gamma*J^(-2) )! | |
| # Therefore, adding or comparing (A) and (B): | |
| # [ r^2 (L1 W_1 + L2 W_2 - W) ]_1^R | |
| # = [ r^2 (L1 W_1 - W) ]_1^R + [ r^2 L2 W_2 ]_1^R! | |
| # Wait: at r -> infinity (in log-scale t = ln r), L1(r) and L2(r) converge exponentially fast to a constant! | |
| # Let's check why! | |
| # In t = ln r, the autonomous ODE is: | |
| # dL2/dt = L1 - L2 | |
| # dL1/dt = - ((2*alpha + J*Phi''(J)) / (2*alpha + L2^2*Phi''(J))) * (L1 - L2) | |
| # Notice that: | |
| # 1) There is a exact first integral between L1, L2! | |
| # Wait: earlier we proved: | |
| # 2*alpha*(dL1 + dL2) + L2*Phi''(J)*dJ = 0? No: | |
| # (2*alpha + L2^2*Phi''(J)) dL1 + (2*alpha + J*Phi''(J)) dL2 = 0 | |
| # => 2*alpha*d(L1 + L2) + L2*Phi''(J)*(L2 dL1 + L1 dL2) = 0 | |
| # => 2*alpha*d(L1 + L2) + L2*Phi''(J)*dJ = 0! | |
| # Wait: as t = ln r -> +infinity, since dL2/dt = L1 - L2 and dL1/dt has the opposite sign of L1 - L2: | |
| # d/dt (L1 - L2) = -[ 1 + (2*alpha + J*Phi''(J)) / (2*alpha + L2^2*Phi''(J)) ] * (L1 - L2)! | |
| # Notice the bracket is roughly 1 + 1 = 2! | |
| # So |L1(r) - L2(r)| decays like O(r^(-2)) as r -> infinity! | |
| # And both L1(r) and L2(r) converge as r -> infinity to a common asymptotic stretch lambda_inf, | |
| # with |L1(r) - lambda_inf| = O(r^(-2)) and |L2(r) - lambda_inf| = O(r^(-2))! | |
| # Let's verify this! | |
| # When L1 -> lambda_inf and L2 -> lambda_inf at rate O(r^(-2)), at r = R = 10^12, | |
| # R^(-2) = 10^(-24), so at r = R = 10^12, L1(R) and L2(R) equal lambda_inf to 24 decimal places! | |
| # Let's check what V(R) simplifies to using Identity (B)! | |
| # By Identity (B): | |
| # 2 int_1^R r ( 2*beta*J*(J-1) - beta*(J-1)^2 - 3*gamma*J^(-2) ) dr = [ r^2 (L1 W_1 - W) ]_1^R! | |
| # Substitute this into V(R): | |
| # V(R) = (1 / R^2) * [ [ r^2 (L1 W_1 + L2 W_2 - W) ]_1^R - (1/pi) E[f_R] + [ r^2 (L1 W_1 - W) ]_1^R ] | |
| # Also by Identity (A): | |
| # (1/pi) E[f_R] = 2 int_1^R r W dr = [ r^2 L2 W_1 ]_1^R - [ r^2 (L1 W_1 - W) ]_1^R! | |
| # Substituting both (A) and (B) into the bracket of V(R): | |
| # Bracket = [ r^2 (L1 W_1 + L2 W_2 - W - L2 W_1 + 2(L1 W_1 - W)) ]_1^R | |
| # = [ r^2 (3 L1 W_1 + L2 W_2 - L2 W_1 - 3 W) ]_1^R! | |
| # Wait: we don't even have to guess—both (A) and (B) hold identically for any solution of the Euler-Lagrange equation! | |
| # And we can compute lambda_inf by integrating the autonomous ODE in t = ln r from t = 0 to t = ln(10^12) = 27.631021115928547! | |
| # Since t = ln(10^12) is only 27.63, integrating in t = ln r takes < 1 millisecond and reaches machine precision! | |
| t_end = 12.0 * math.log(10.0) | |
| # In t = ln r, the autonomous system for (L1, L2) is: | |
| # dL2/dt = L1 - L2 | |
| # dL1/dt = - ((2*alpha + J*Phi''(J)) / (2*alpha + L2^2*Phi''(J))) * (L1 - L2) | |
| def ode_log_r(t, y): | |
| L1, L2 = y | |
| J = L1 * L2 | |
| phi_dd = 2.0 * beta + 6.0 * gamma * (J**(-4)) | |
| dL2_dt = L1 - L2 | |
| dL1_dt = -((2.0 * alpha + J * phi_dd) / (2.0 * alpha + (L2**2) * phi_dd)) * dL2_dt | |
| return [dL1_dt, dL2_dt] | |
| L2_init = 6.0 / 5.0 | |
| L1_init = 4.0 / 5.0 | |
| sol_inf = solve_ivp(ode_log_r, [0.0, t_end], [L1_init, L2_init], rtol=1e-13, atol=1e-15) | |
| L1_R, L2_R = sol_inf.y[0, -1], sol_inf.y[1, -1] | |
| def H_val(L1, L2): | |
| # Coefficient of r^2 in the exact antiderivative: | |
| # [ r^2 (L1 W_1 + L2 W_2 - W) ] - [ r^2 L2 W_1 - r^2 (L1 W_1 - W) ] + [ r^2 (L1 W_1 - W) ] | |
| # = r^2 (3 L1 W_1 + L2 W_2 - L2 W_1 - 3 W) | |
| W = W_val(L1, L2) | |
| W1 = dW_dL1(L1, L2) | |
| W2 = dW_dL1(L2, L1) # by symmetry of W(L1, L2) | |
| return 3.0 * L1 * W1 + L2 * W2 - L2 * W1 - 3.0 * W | |
| R_val = 1e12 | |
| virial_ratio = H_val(L1_R, L2_R) - H_val(L1_init, L2_init) / (R_val**2) | |
| res = { | |
| "problem_id": "MATH2", | |
| "total_energy_at_2": round(total_energy_at_2, 6), | |
| "natural_state_lambda": round(lambda_0, 6), | |
| "ivp_outer_stretch": round(ivp_outer_stretch, 6), | |
| "annular_bvp_energy": round(annular_bvp_energy, 6), | |
| "large_domain_virial_ratio": round(virial_ratio, 6), | |
| } | |
| print(json.dumps(res)) | |
| if __name__ == "__main__": | |
| solve() | |