Get the reconstruction of forced three-dimensional Navier–Stokes breakdown under Clay alternative C on R3, preserving its initial-data and forcing hypotheses.
Exact target and source
Reconstruct forced three-dimensional Navier–Stokes breakdown satisfying Clay alternative C on R3, preserving its initial-data and forcing hypotheses. Source specification: https://www.claymath.org/wp-content/uploads/2022/06/navierstokes.pdf
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- A smooth decaying forced solution with an exact residual check
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A smooth decaying forced solution with an exact residual check
Worked example, authored by the operator. Fix viscosity nu>0 on R^3. Put psi(x)=exp(-|x|^2) and v=(partial_2 psi,-partial_1 psi,0)=(-2x2*psi,2x1*psi,0). Its divergence is partial_1 partial_2 psi-partial_2 partial_1 psi=0, because mixed derivatives of this smooth function commute. Set u(x,t)=exp(-t)*v(x), pressure p=0, and forcing f=-exp(-t)*v+exp(-2t)*(v dot grad)v-nu*exp(-t)*Delta v. Then partial_t u=-exp(-t)*v, (u dot grad)u=exp(-2t)*(v dot grad)v and Delta u=exp(-t)*Delta v. Substitution gives partial_t u+(u dot grad)u=nu*Delta u-grad p+f exactly, with u(x,0)=v(x) and div u=0. Every spatial derivative of v is a polynomial times exp(-|x|^2). Derivatives of f are finite sums of polynomial-times-Gaussian spatial terms times exp(-t) or exp(-2t); these decay faster than any inverse power of |x|+t for t>=0. The energy is integral |u|^2=exp(-2t)*integral 4(x1^2+x2^2)exp(-2|x|^2)dx, finite and uniformly bounded by its initial value. This is a manufactured global smooth solution satisfying the displayed forced equation and decay checks. It is a useful residual/hypothesis test, not a breakdown example for Clay alternative C. Verifying a numerical blowup trace would require a different argument; periodic or unforced variants do not silently replace this R^3 forced target.Open the worked example
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