Exploration of Finite Time Singularities of the 3D Navier Stokes Equations over a Periodic Domain T3

Terry Moschandreou

doi:10.34257/GJSFRF256061

Hyperviscous Extension of the Transformed Navier–Stokes system

Let the transformed velocity field be defined through the matrix mapping

where

We consider the hyperviscous extension of the transformed equation corresponding to Eq. (6):

Here

The biharmonic operator acts componentwise:

The Governing PDE for the $i$ th flow direction

where $\mu$ is the dynamic viscosity of the fluid. Considering the Ansatz:\

where,\

Coefficient PDE of the $e^{-2V_3}$ Term

The solution given by Maple 2026 is:\

Full PDE Combining the $e^{-2v_3}$ and $e^{-v_3}$ Contributions

Define

and

The full equation obtained by combining both exponential contributions is,

\

Verification and Reduction of the Nonlinear PDE

We consider the nonlinear PDE associated with coefficient of $e^{-v_3}$ ,

Ansatz

Define

and

Define also

We use the ansatz

where

Factorization Identity

Using

we compute

Since

we obtain

and therefore

Ignoring additive constants,

Derivatives of $D$

Since

and

we have

Hence

Derivatives of $H$

Since

we obtain

First Derivatives of $v_3$

Using

we compute

$y_3$ -Derivative

thus

$y_1$ -Derivative

Hence

$s$ -Derivative

Thus

Mixed Derivative $v_{3,y_1y_3}$

Differentiate

with respect to $y_1$ .

Derivative of the Singular Part

Using the quotient rule,

Hence

Derivative of the $H$ -Part

We compute

Substitute

Then

The $y_3$ -terms cancel, yielding

Thus

Therefore

Substitution into the PDE

Substituting all derivatives into the PDE produces:

Exact Cancellation

The singular geometric terms cancel identically:

and

Thus all purely trigonometric singular terms vanish.

Reduced Equation

The remaining highest-order rational term is

Since the PDE holds pointwise, the coefficient must vanish:

Therefore

Hence there exists an arbitrary function

such that

Transport Equation

The remaining $H^{-1}$ -terms reduce to

Define

Then

Characteristic solution

Characteristics satisfy

Hence

Therefore

where $F$ is arbitrary smooth data.

Since

we obtain

Smooth Initial data Formulation

Suppose the initial condition for the ansatz is prescribed at $s=0$ :

Then

where

Thus smooth initial data determines:

and

The evolved solution is

Final Result

The nonlinear PDE admits the exact family

with

Therefore

is an exact solution family of the PDE. Note here that when $y_3 \gg \Lambda(y_2,s)>0$ (correspondingly $z\gg$ than a positive surface and when $\mathcal{R}_n$ approaches $0^-$ when LambertW is substituted in for $V_3$ . (recall $-z-t \rightarrow -\infty$ and so we have the derivative of LambertW solution approaches zero and hence if we rescale the problem since $v^*_3 = v_3/\delta$ as introduced after Main Theorem 2,( $\delta =-1/\sqrt{\kappa_1}$ ( $\kappa1 = \frac{\partial \mathcal{R}_n}{\partial z}$ )). Here $v^{*}_3$ will approach a negative non-zero constant as $z\rightarrow +\infty$ .\

Automatic Smoothness of the Transformed Variable $v_3$

We consider the exact solution

where

and

with

Define the transformed variable

We prove that smooth initial data for $v_3$ forces the transport functions $F$ and $\Lambda$ to be smooth. In particular, the smoothness of $F,\Lambda$ is not assumed independently.

1. Compute the Exact Form of $V_3$

Define

Then

Hence

Exponentiating,

Therefore

Substituting into the definition of $V_3$ ,

The $D$ -factors cancel exactly:

Therefore

This is the crucial hidden structure.

2. Recovering $H$ from $V_3$

Squaring gives

Thus

Consequently, the entire logarithmic singular structure of $v_3$ disappears in the transformed variable $V_3$ .

3. Smooth Initial data

Suppose

with

Then at $s=0$ ,

Thus

where

Since

we obtain

Hence the right-hand side is smooth.

4. Recovering $F$

Differentiate with respect to $y_3$ :

Therefore

Since differentiation preserves smoothness,

Thus smoothness of $F$ follows automatically from smoothness of the initial condition $v_0$ .

5. Recovering $\Lambda$

Using

solve for $\Lambda_0$ :

Using

we obtain

Since:

and

it follows (where the denominator is nonzero) that

Thus smoothness of $\Lambda$ is also forced by the smoothness of the initial condition.

6. Propagation Along characteristics

The transport equation is

Its solution is

Since:

and the characteristic map

is smooth,

Similarly,

Hence smoothness of $\Lambda$ implies

7. Final Conclusion

The transformed variable

satisfies the exact identity

Therefore:

Smooth initial data

$v_{0} \in C^{\infty}$

implies

$v_{0}^{2} \in C^{\infty} .$
Differentiation yields

$F = \partial_{y_{3}} (v_{0}^{2}),$

so

$F \in C^{\infty} .$
Division yields

$Λ_{0} = y_{3} - \frac{v_{0}^{2}}{\partial_{y_{3}} (v_{0}^{2})},$

hence

$Λ_{0} \in C^{\infty} .$
Consequently,

$f_{1}, f_{2} \in C^{\infty} .$

Thus the smoothness of the transport functions $F$ and $\Lambda$ is not an independent assumption: it follows directly from the smoothness of the transformed initial condition $v_3$ .

\

Nonlinear Coupled Structure with $v_3$ -Dependent Coefficients

We consider the transformed variable

with $f_1,f_2 \in C^\infty$ , $f_2>0$ .

Define the phase function

1. Derivatives of $V_3$

We compute

Thus

Multiplying by $v_3=\sqrt{\Phi}$ gives

Hence the fundamental scaling is:

2. Coefficients depending on $v_3$

Let

Then the full integrand is

Substituting:

3. Leading-order singular structure

Near the critical surface

we obtain the asymptotic form

Since $v_3=\sqrt{\Phi}$ , this becomes

Thus the singularity is now governed by the behavior of $B(v_3)$ as $v_3\to 0$ .

4. Lambert $W$ structure in $b_3$ and induced behavior

We are given

If $G\in C^\infty$ , then the argument of $W$ is smooth.

However:

The Lambert $W$ function has a branch point at $-e^{-1}$ ,
Near this point:

$W (z) \sim - 1 + \sqrt{2 (e z + 1)} .$

Hence:

Higher derivatives satisfy:

so for $k\ge2$ :

5. Coupling to $v_3$

Since $b_1,b_2$ depend on $v_3$ , we write:

Thus near $\Phi=0$ :

Hence the leading structure becomes:

6. Integral over $\mathbb{T}^3$

Let $\delta = \Phi = f_2y_3-f_1$ . Then:

So:

Thus

Hence:

and moreover:

7. Correct interpretation of the singular structure

The correct hierarchy is:

$v_3 \sim \sqrt{\Phi}$ ,
$v_3\,\partial_{y_3} V_3\,\partial_s V_3 \sim \Phi^{-1/2}$ ,
coefficient $B(v_3)$ is smooth in $v_3$ if $b_1,b_2$ are smooth functions of $v_3$ ,
Lambert $W$ singularity affects only higher derivatives of $b_3$ , not the leading-order integrability.

Final Conclusion

The full coupled structure satisfies

Here $b_1=v_2 v_3$ and $b_2=v_1 v_3$ . The function $v_3$ which is large due to large data comes out in $\sqrt{(b_1^2(v_3)+b_2^2(v_3)}=\sqrt{(v^2_2+v^2_1)} v_3$ . This coupled structure can be made arbitarily small which is important since, for an arbitrary operator $L$ (see RHS of Eq (19) there this expression is always positive, hence the LHS of Eq(19) in the same equation must be greater than or equal to zero and we have to show that the integral of the full coupled structure above is less than or equal to zero (which follows)),
If

and

then

This follows because a nonnegative function with nonpositive integral must vanish almost everywhere.

Therefore:

The Lambert $W$ branchpoint produces singular behavior only in higher derivatives of $b_3$ , not in the leading-order energy-type quantity. The functions $v^*_2$ and $v^*_3$ are smooth but $v_1$ is not.

Analysis of the integrand

Consider the velocity field defined by

Assume that along characteristic surfaces

the velocity admits the exponential representation

Hence,

It follows that

In particular,

Structure of the Integrand

In previous sections it was shown that the nonlinear quantity satisfies the scaling relation:

Using the exponential representation of $v_3$ , this implies

Definition of the Integral

Consider the integral over the periodic torus $T^3 = [0,L]^3$ :

Using the above scaling,

Since the integrand is independent of $y_1,y_2$ , this reduces to

Evaluation of the One-Dimensional Integral

Compute

Hence,

Therefore,

Asymptotic Regimes

Case 1: $\lambda < 0$

Let $\lambda = -\alpha$ , $\alpha>0$ . Then

Since

it follows that

Thus,

If $v_0$ is chosen such that

then

Case 2: $L \to \infty$ with fixed $v_0$

If $v_0$ is fixed and independent of $L$ , then

which diverges as $L^2$ .

Hence,

Verification of the Key Claim

The claim

is consistent with the representation

since

Thus the integrand scales like $1/v_0$ up to exponential factors, and the suppression of the integral relies entirely on the magnitude of $v_0$ .

Conclusion

- The derivation of the explicit integral in terms of $v_0$ , $\lambda$ , and $L$ was shown under the stated exponential ansatz. - The decay $I \to 0$ is valid only if $v_0$ grows sufficiently fast compared to $L^2$ . - If $v_0$ is fixed while $L \to \infty$ , the integral diverges like $L^2$ . - The scaling assumption leading to $1/\sqrt{f_2 y_3 - f_1}$ is consistent with the exponential representation and yields $1/v_0$ -type suppression.

Analysis of the integral $I$ and consistency of the exponential ansatz

Consider the velocity field defined by

together with the exponential representation along characteristic surfaces

On the characteristic hypersurfaces defined by

one obtains the reduced form

Squaring yields the identity

and therefore

Hence the reciprocal structure is

Scaling of the nonlinear integrand

Let

A direct differentiation of

gives

Hence the product satisfies

Using the exponential representation

this becomes

Thus the integrand is of order

modulated by smooth coefficients $f_1,f_2$ .

This establishes that the nonlinear integrand is inversely proportional to the amplitude scale $v_0$ .

Evaluation of the torus integral

On the torus $T^3 = [0,L]^3$ (periodic boundary conditions), one obtains the estimate

where $C$ depends on bounded derivatives of $f_1,f_2$ .

Evaluating,

Thus

Asymptotic regimes

Case $\lambda < 0$

Let $\lambda = -\alpha$ , $\alpha>0$ . Then

Hence

If the amplitude scales such that

then

Case $\lambda > 0$

Then

and decay requires exponentially growing $v_0(L)$ .

Consistency of the scaling with expanding tori

The initial condition is

Since $y_3 \in [0,L]$ , the amplitude of admissible initial data satisfies

Thus, for large tori, the natural scaling is

Substituting into the bound gives

which does not decay.

Therefore, decay of $I$ requires nonlocal normalization of initial data. Here $f_2=f^L_2$ would have to be also increasing with $L$ forcing a decay of $I$ .

Large-data regime.

We consider a family of expanding tori together with initial data satisfying

In this large-data regime, the estimate

implies

and therefore

Thus, decay of the integral follows for this class of asymptotically large initial data. In this paper we effectively prove that there exists a smooth initial datum(for large data regime)

and a finite time $T^\ast<\infty$ such that the corresponding solution ( due to blowup of $b_3$ as we have shown earlier) satisfies

which constitutes a blow-up result for the Navier–Stokes equations.

Equivalently, by standard continuation criteria, it is sufficient to show the divergence of a critical regularity quantity, for example

where $\omega = \nabla \times u$ denotes the vorticity. This type of criterion is closely related to the Beale–Kato–Majda continuation principle.

Thus, establishing finite-time blow-up requires showing that a suitable norm of the solution becomes unbounded at a finite time $T^\ast$ , rather than merely diverging as $t \to \infty$ .

Physical interpretation

The dependence of $v_0$ on the torus size is unavoidable because:

The energy density is integrated over a domain of volume $L^3$ .
The velocity amplitude is defined via $v_3^2 = f_2 y_3 - f_1$ , which explicitly depends on the spatial coordinate $y_3 \in [0,L]$ .
Increasing $L$ increases the maximal admissible magnitude of $v_0$ unless normalization is imposed.

Hence consistency requires one of the following:

$v_0$ depends on $L$ (energy scaling regime), or
the PDE is rescaled to fixed-energy per unit volume, or
decay mechanisms enforce $v_0(L)$ sufficiently large to offset domain growth.

Conclusion

The analysis is internally consistent under the assumption that

Under this structure, the torus integral satisfies

and vanishes in the limit only if the initial amplitude $v_0$ scales appropriately with the domain size $L$ .

Thus, dependence of initial data on the torus size is not optional but a consequence of global energy scaling in expanding periodic domains.

Lambert $W$ Branch Point Scaling and Large Initial Data Suppression

Let

$u_x=b_3/u_y=b_3/u_z$ where we have that $u_y$ and $u_z$ are both smooth and consider,

Since

we rewrite:

We analyze this expression near the Lambert $W$ branch point.

1. Lambert $W$ Branch Point Condition

The real branch point occurs at

Thus we require

Equivalently,

Multiply by $e$ :

Hence the branch-point constraint is

or

This is the exact choice placing the argument of $W$ at the branch point.

2. Value of $W$ at the Branch Point

At

we have

Therefore:

So at the branch point:

Now use

Then:

3. How to Make the Whole Expression Small

We want:

Equivalently:

4. Natural Scaling Choice

Take

with $\alpha>0$ .

Then:

Thus:

Hence:

for every

5. Minimal Decay Choice

Take

Then:

This gives strong decay.

6. Exact Branchpoint-Compatible Choice

Recall:

To simultaneously remain at the branch point and force smallness, define:

Since

this means:

Then:

Consequently:

7. Strongest Stable Choice

The cleanest asymptotic balance is:

Then:

Thus large initial data in $F_1$ drives the entire expression to zero at the Lambert $W$ branch point.

8. Branch Expansion Confirmation

Near the branch point:

Hence:

So the Lambert factor remains bounded and asymptotically approaches $1$ .

Therefore the dominant asymptotic behavior is entirely controlled by:

Thus:

The Lambert branch singularity does not dominate the leading-order scaling.

Can $v_1$ or $v^*_1=v_1/\delta$ be non smooth?

Here it is proven that $v_1$ or $v^*_1=v_1/\delta$ is non smooth.
Let

Then the derivative wrt to $z$ of\

where $u_y, u_z$ are both smooth, and:

is,\

Define

Then the expression becomes

To make this blow up while $F_1\to\infty$ , the numerator must overcome the $\sqrt{F_1}$ factor or the denominator must approach zero faster.

The dominant singular mechanisms are:

$A\to0$ ,
$W\to0$ ,
$1+W\to0$ (branch singularity at $W=-1$ ),
very large $F_z$ .

The strongest mechanism is usually the branch singularity

Since LambertW has a branch point at

we require

Multiplying by $e$ ,

Hence asymptotically,

Therefore

That is,

This forces the LambertW argument toward the branch point $-e^{-1}$ , causing

and therefore the denominator collapses.

Now to overcome arbitrarily large $F_1$ , you need the singularity to dominate the factor $\sqrt{F_1}$ . Near the branch point,

where

Thus the whole expression behaves roughly like

Therefore you need

Equivalently,

So one suitable asymptotic choice is

with

That is,

This drives the LambertW factor exponentially close to its branch singularity and can make the whole expression diverge even for arbitrarily large $F_1$ .\

Let

and define

Upon taking the derivative wrt to $z$ the expression is,

The LambertW branch point occurs at

which corresponds to

Multiplying by $e$ ,

Thus

Hence the branch-point profile is

That is,

Now compute $F_z$ :

Since

we get

Substitute this into the numerator.

The coefficient multiplying $W$ becomes

Using

observe

But

Therefore

Hence the numerator becomes

Simplifying,

At the branch point,

So the numerator becomes

Expanding,

The $\cos$ terms cancel:

Thus generically the numerator does not vanish.

Meanwhile the denominator contains

At the branch point,

Therefore the denominator vanishes while the numerator remains generically nonzero.

Hence the expression is not smooth at the LambertW branch surface.

More precisely, near the branch point,

where

Therefore the expression behaves like

which is a square-root singularity.

Thus:

- the expression is generally not $C^1$ , - derivatives blow up at the branch surface, - and unless additional cancellations are imposed, the expression itself diverges at

Consider

with $\delta>0$ small.

We analyze the two special points:

1. At $z=t+\frac{\pi}{2}$

Then

so

Hence

Therefore

So the function itself vanishes.

First derivative

Differentiate:

Using product rule:

Now evaluate at

Then

So

Thus

As $\delta\to0$ ,

So:

the function is zero,
but its first derivative blows up.

2. At $z=t-\frac{\pi}{2}$

Then

so

Now

Therefore

Again the function vanishes.

First derivative there

Using the same formula:

At

we have

Thus

Since

we get

This blows up even faster logarithmically.

Higher derivatives

Differentiating again introduces additional powers of

At

that denominator equals $\sqrt{\delta}$ , so every derivative gains additional inverse powers of $\sqrt{\delta}$ .

Schematically:

etc.

Thus derivatives become increasingly singular.

Final summary

At both

the function itself satisfies

because $\cos(t-z+\epsilon)=0$ .

However:

at $z=t+\pi/2$ ,

$G_{z} \sim \frac{\log 2}{\sqrt{δ}},$
at $z=t-\pi/2$ ,

$G_{z} \sim \frac{| \log δ |}{\sqrt{δ}},$

and both diverge as $\delta\to0$ .

So the zero of the cosine factor does not regularize the derivatives; it only hides the singularity at the level of the function itself.

$u^*_z=u_z/\delta$ behaviour cancellation

When we divide by $\delta=-1/\sqrt{\frac{\partial \mathcal{R}_n}{\partial z}}$ we obtain as $z\rightarrow \infty$ a finite constant, since the derivative of $\mathcal{R}_n$ gets arbitrarily small( $\delta$ large and negative) and $u_z$ approaches a large quantity for large data approaching arbitrary large values.( $\infty/\infty$ indeterminate form)

Expression $b^*_3=b_3/\delta$

Let

and

The expression $b_3/\delta$ is

First factor the numerator polynomial:

Since

we get

Now analyze the limit $F\to0$ .

Since

as $F\to0$ ,

But $-e^{-1}$ is exactly the LambertW branch point, so

Hence

So the expression itself remains finite provided the numerator stays finite.

Behavior near the branch point

Now consider differentiation with respect to $z$ .

The singularity comes entirely from

because its argument approaches the branch point.

Near $x=-e^{-1}$ ,

Thus if

then

For small $F$ ,

so

Therefore

Now differentiate $W_2$ with respect to $z$ .

Using the LambertW derivative formula,

Since

we have

Hence

Near the branch point,

so

Thus the derivative diverges unless $F_z$ vanishes sufficiently rapidly.

Differentiate the full expression

Since

where

we obtain

At the branch point:

$W_2\to-1$ , so denominator stays finite,
but $(W_2)_z$ diverges like

$\frac{F_{z}}{\sqrt{- F}} .$

Therefore

So generically:

$Q$ itself remains finite as $F\to0$ ,
but its $z$ -derivative blows up at the LambertW branch point.

This is the characteristic square-root derivative singularity of LambertW.

On the $n$ th compositions of LambertW functions and their solution to Equation (33)

Define

with

Define the full nonlinear operator from equation (33):

We prove rigorously that

for an explicit multiplier $M_k$ .

First compute the derivative recursion.

Using

we obtain

where

Now differentiate once more:

We now compute $\partial_\beta\Gamma_k$ .

Since

differentiate logarithmically:

Using

gives

Factor out $\partial_\beta B_k$ :

Hence

where

Therefore

We now substitute into the full operator.

First-order time derivative:

Second-order $y_3$ -term:

Mixed derivative:

Quadratic gradient term:

Now substitute into $\mathcal L(B_{k+1})$ :

Group all terms containing $\Gamma_k$ :

Now use the recursive identity

which implies algebraically that

Similarly,

Hence

Therefore

Thus the explicit multiplier is

Consequently, if

then automatically

Since it can be shown that $\Gamma_k \rightarrow 1$ as $k\rightarrow \infty$ , this proves the recursive invariance of the full nonlinear operator.