The Fubini solution solves the lossless nonlinear approxiamte evolution equation (which is accurate to quadtaric order) as a Fourier sine series from 0 ≤ σ < 1, where σ is distance nondimensionalized by the shock-formation distance. The expansion coefficients *B _{n}* involve integrating over θ from θ = 0 to θ = π. The phase Φ of the Fubini solution is given by the dimensionless phase of the lossless nonlinear approximate solution, Φ = θ + σ sin(Φ). The question is, "Why does Φ take on the same limits as θ in the integration?"

To answer this quesiton, first consider the lower limit, θ = 0, of the integral for the expansion coefficients *B _{n}*. Then, the transcendental equation being solved is Φ = σ sin(Φ). In the plot below, σ is being swept from 0 to 1. You can see that for 0 ≤ σ < 1, the only solution to the transcendental equation is Φ = 0. So when θ = 0, Φ = 0 also.

Next consider the upper limit of the integral for the expansion coefficients *B _{n}*, θ = π. The transcendental equation now being solved is Φ = π + σ sin(Φ). As before, σ is being swept from 0 to 1. You can see that for 0 ≤ σ < 1, the only solution to the transcendental equation is Φ = π. So when θ = π, Φ = π also.