P != NP

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Abstract:

When clues are absent, a deterministic Turing machine $M$ cannot deterministically solve an NP-complete problem in polynomial time. Doing so would violate the fundamental definition of a mathematical function and/or the second law of thermodynamics.

1. Introduction:

Is it possible to solve an NP-complete problem in polynomial time? Using the haystack analogy, is it possible to find the needle without using brute force—that is, sifting through every piece of straw? If one has a magnet, the needle could be drawn out instantly. However, that would require a powerful magnetic field capable of covering the entire haystack at light speed, which is analogous to parallel processing where multiple cores evaluate every element simultaneously. Because the P vs. NP question applies strictly to deterministic Turing machines that process a problem sequentially, parallel processing and the metaphorical magnet are ruled out. To avoid pure brute force, an algorithm analogous to a metal detector remains a viable option. Gradient descent mimics this approach, utilizing local feedback clues to efficiently navigate and narrow down the search space. Clues appear to be essential to solving a hard problem; although, they may not always be present or sufficient. In the sections below, we explore whether they are truly essential and whether a function or algorithm exists that can take the output of a Boolean satisfiability problem and return the satisfying assignment in polynomial time.

The Inverse Function

Cook-Levin Theorem. The Boolean satisfiability problem is NP-complete.

Let $\phi$ be a Boolean satisfiability instance in conjunctive normal form with $n$ variables. Here, $n \in \mathbb{N}$. The maximum runtime $t$ needed to solve $\phi$ via brute force is $O(2^n)$. There are $2^n$ possible truth-value assignments, but not all of them will satisfy $\phi$. Below is a list of the possible assignments:

\[\phi_{\text{assignments}} = \begin{pmatrix} x_1 & x_2 & \dots & x_I \\ 0 & 0 & \dots & 0 \\ 0 & 0 & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & 1 \end{pmatrix} \hspace{1 cm} (1)\]

Assume that a satisfying assignment has been found using brute force:

\[ \phi(x_1, x_2, \dots, x_I) = \phi(0, 1, \dots, 0) = 1 \quad (1 \equiv \text{true}, \; 0 \equiv \text{false}) \hspace{1 cm} (2) \]

Given this specific solution to $\phi$, assume it is possible to construct and verify an inverse function or algorithm denoted as $\phi^{-1}$:

\[ \phi^{-1}(1) = (0, 1, \dots, 0) \hspace{1 cm} (3) \]

The inverse function $\phi^{-1}$ successfully returns the correct satisfying assignment in polynomial time when given the input $1$. The ultimate objective is to determine whether such an inverse mapping $\phi^{-1}$ can be generalized to efficiently find satisfying assignments for all Boolean satisfiability instances in polynomial time.

Let $\psi$ be a Boolean satisfiability instance distinct from $\phi$, such that $\psi$ has a unique satisfying assignment of $(1, 1, \dots, 1)$. If one attempts to use the previously defined inverse function $\phi^{-1}$ to find this assignment in polynomial time, a contradiction arises:

\[ \phi^{-1}(1) = (0, 1, \dots, 0) \neq (1, 1, \dots, 1) \hspace{1 cm} (4) \]

Because $\psi(1, 1, \dots, 1) = 1$, evaluating $\phi^{-1}(1)$ yields an assignment that fails to satisfy $\psi$.

By the fundamental definition of a mathematical function, a single input cannot map to multiple distinct outputs. Consequently, the distinct instance $\psi$ requires its own unique inverse mapping, which can only be constructed and verified if the solution to $\psi$ is already known via brute force or an alternative search method. This requirement of prior knowledge undermines the utility of using a predetermined inverse function to solve new instances. It can be inferred that this limitation generalizes to all instances of the Boolean satisfiability problem and, by extension, any NP-complete problems to which they reduce.

Entropy

The Shannon entropy $H(X)$ is as follows:

\[ H(X) = - \sum_{i=1}^{N}p(x_i)\log_2(p(x_i)) \hspace{1 cm} (5) \]

Where $X$ is the overall random event being analyzed; $p(x_i)$ is the probability of a specific individual outcome; $N \in \mathbb{N}$, is the number of specific individual outcomes. Assume $\phi$ is an instance reduced from the hardest NP-complete problem. We can express its maximum search space size in terms of Shannon entropy:

\[ H(X) = - \sum_{i=1}^{2^n}\frac{1}{2^n}\log_2(\frac{1}{2^n}) = -log_2(\frac{1}{2^n}) \hspace{1 cm} (6) \]

Using the power rule of logarithms $(\mu \log(\nu) = \log(\nu^\mu))$:

\[ H = \log_2(2^n) \hspace{1 cm} (7) \]

According to the second law of thermodynamics (SLOT), entropy stays the same or increases over time. To reduce the maximum runtime $O(2^n)$ to $O(n^k)$ requires a reduction of entropy:

\[ \ H < log_2(n^k) \hspace{1 cm} (8) \]

where $k \in \mathbb{N}$ is a constant, where a search space of size $n^k$ can be traversed in polynomial time. The inequality at (8) illustrates why, given insufficient clues, no deterministic algorithm has been found that can solve NP-complete problems in polynomial time. Such an algorithm is prohibited from violating the SLOT.

Now, it is possible to increase the amount of order of a system, i.e., reduce its entropy by doing work on it, but disorder or entropy will increase elsewhere. We model this in the following manner:

\[ H = \log_2(n^k) + \log_2(2^n) - \log_2(n^k) + \epsilon \hspace{.5 cm}(\epsilon\geq 0) \hspace{1 cm} (9) \]

Where $\epsilon$ is any increase in the overall entropy. The first term on the right side of equation (9) is reduced entropy that corresponds to a reduced search space of $n^k$. The remaining terms on the right side represent where entropy increases.

It is work that causes this redistribution of entropy. If equation (7) is the initial state of the system, and brute force (work) is used, the next state looks like this:

\[ H = \log_2(2^n-1) + \log_2(2^n) - \log_2(2^n-1) + \epsilon \hspace{.5 cm}(\epsilon\geq 0) \hspace{1 cm} (10) \]

If $2^n-n^k$ steps of work are performed, the search space shrinks to $n^k$ and can be searched in polynomial time (see equation (9)). The $P=NP?$ question could be reframed as follows: Is there a way to reduce the search-space entropy without doing the astronomical amount of brute-force work required?

The Value of Clues

To avoid violating the SLOT, let $A$ represent the active search-space entropy, which is initially $\log_2(2^n)$. Let $B$ represent the discard-space entropy corresponding to assignments that have either been bypassed (i.e., not assigned) or assigned but failed to satisfy $\phi$. Initially, $B = 0$. The initial entropy equation is expressed as follows:

\[ H = \log_2(2^n) + 0 \hspace{.5 cm}(A= \log_2(2^n);\hspace{.1 cm}B=0) \hspace{1 cm} (11) \]

To solve an NP-complete instance $\phi$ in polynomial time, a Turing machine $M$ requires an algorithm $\alpha$ capable of deciding which assignments should be tested ($A$), and which can be safely bypassed ($B$). Without clues, there is no deterministic basis for this partition; there is only a random or arbitrary basis that may cause the solution(s) to be missed. Clues encompass, but are not limited to, patterns, extrinsic or intrinsic evidence, and any helpful structural information. Thus, the framework for $\alpha$ can be structured as follows:

1. Evaluate Clues: Execute conditional statements that evaluate available clues to determine the largest number of potential assignments that can be safely bypassed, making entropy $A$ smaller and entropy $B$ larger.

2. Search Remainder: Perform a brute-force search over the remaining potential assignments. Each tested assignment reduces entropy A and increases entropy B.

3. Iterate: Repeat steps 1 and 2 until $A = log_2(\gamma)$; where $\gamma$ is the smallest number that includes the satisfying assignment(s).

To demonstrate this mechanism, suppose $M$ is given a maximal clue: the exact solution to $\phi$. The entropy equation is as follows:

\[ H = \log_2(1) + \log_2(2^n) - \log_2(1) + \epsilon \hspace{.5 cm}(A = \log(1); B = \log_2(2^n) - \log_2(1); \hspace{.1 cm}\epsilon \geq 0) \hspace{1 cm} (12) \]

$M$ executes step 1 by evaluating the clue and isolating the solution, i.e., reducing $A$ while increasing $B$. During step 2, $M$ tests the remaining single assignment. The total runtime is $O(1)$, bypassing step 3 entirely.

Now, suppose $M$ is provided with minor clues. The entropy equation can be modeled in the following manner:

\[ H = \log_2(n^k) + \log_2(2^n) - \log_2(n^k) + \epsilon \hspace{.5 cm}(A = \log(n^k); B = \log_2(2^n) - \log_2(n^k); \hspace{.1 cm}\epsilon \geq 0) \hspace{1 cm} (13) \]

where $M$ can solve the instance in polynomial time using the framework above. However, if $M$ has no clues to leverage, step 1 is bypassed, and only steps 2 and 3 are executed. The entropy equation reverts to:

\[ H = \log_2(2^n) \hspace{1 cm} (14) \]

Absent any clues, $M$ cannot solve $\phi$ deterministically in polynomial time. A groundless reduction in search space size would violate the SLOT:

\[ H \neq \log_2(n^k) \hspace{1 cm} (15) \]

Conclusion

Based on the fundamental definition of a mathematical function, we conclude that a general, deterministic inverse function is untenable. Based on the SLOT and the necessity of clues to deterministically partition a combinatorial search space, we conclude that a universal polynomial-time algorithm for unguided instances cannot exist. Therefore, we state that:

\[ P \neq NP. \hspace{1 cm} (16) \]

References

1. Cook, Stephen. \textit{The P versus NP Problem}. Clay Mathematics Institute.

2. Cook, Stephen. 1971. \textit{The complexity of theorem proving procedures}. Proceedings of the Third Annual ACM Symposium on Theory of Computing. pp. 151–158. doi:10.1145/800157.805047. ISBN 9781450374644. S2CID 7573663.

3. Herrmann, Paul Peter. 1973. \textit{On the Reducibility Among Combinatorial Problems}. Massachusetts Institute of Technology.

4. Karp, Richard M. 1972. \textit{Reducibility Among Combinatorial Problems}. In Miller, Raymond E.; Thatcher, James W. (eds.). \textit{Complexity of Computer Computations}. New York: Plenum. pp. 85–103. ISBN 0-306-30707-3.

© 2026 G.M. Jackson

Breakdown of Navier-Stokes Equations

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Abstract:

Given limited energy and a small mass density or large kinematic viscosity, this work shows why blowups of velocity $u$ can occur within a short time period. Additionally, by examining the Eulerian and Lagrangian perspectives and exposing inequalities, this work demonstrates that, given an arbitrary existing vector field, there is no solution of $u$ via the Navier-Stokes momentum equation that is always physically reasonable and/or mathematically consistent and can successfully model the vector field. Because a valid calculation of pressure $p$ via the said equation depends on a valid solution of $u$, there is no solution of $p$.

1. Introduction:

The Navier Stokes equations (NSE) were created during the 19th century to model viscous fluids. It was a time before atoms were established, a time before special relativity placed a speed limit on how fast a fluid can move, so velocities approaching infinity did not raise eyebrows. Fluids were believed to be infinitely divisible, and could be incompressible. So why should we be shocked or surprised that the NSE sometimes yield non-physical results by today’s standards? There are many exact solutions to the NSE such as Couette or Poiseuille solutions which result from simple textbook problems consisting of steady, fully developed flows where many of the troublesome NSE terms can be cancelled or eliminated. The discussion below seeks to keep the "troublesome" terms in play for the purpose of demonstrating how they lead to a lack of a solution.

2. The Challenge

Within his paper (“Existence and Smoothness of the Navier-Stokes Equation”), posted at the Clay Mathematics Institute (CMI) website, Charles L. Fefferman states in pertinent part:

“The Navier-Stokes equations are to be solved for an unknown velocity vector $u(x,t)$ … and pressure $p(x,t)$… defined for position $x \in \mathbb{R}^n$ and time $t \geq0$ .”

Below are the Navier-Stokes (NS) equations, the conditions imposed by the CMI along with the statement to be proved: (C).

\[ \frac{\partial u_i}{\partial t} + \sum_{j=1}^{n} u_j \frac{\partial u_i}{\partial x_j} = \nu\Delta u_i - \frac{\partial p}{\partial x_i} + f_i(x,t) \hspace{1 cm} (1) \] \[ div\hspace{.1 cm} u = \sum_{i=1}^{n} \frac{\partial u_i}{\partial x_i} \hspace{1 cm} (2) \] We assume mass density $\rho$ is set to $1$. \[ u(x,0) = u^\circ(x) \hspace{1 cm} (3) \] \[ |\partial^\alpha_x u^\circ(x)| \leq C_{\alpha K}(1+|x|)^{-K} \hspace{.3 cm} on \hspace{.1 cm} \mathbb{R}^n \hspace{.1 cm} for \hspace{.1 cm} any \hspace{.1 cm} \alpha,K \hspace{1 cm} (4) \] \[ |\partial^\alpha_x \partial^m_t f(x,t)| \leq C_{\alpha m K}(1+|x|+t)^{-K} \hspace{.3 cm} on \hspace{.1 cm} \mathbb{R}^n \times [0,\infty) \hspace{.1 cm} for \hspace{.1 cm} any \hspace{.1 cm} \alpha,m,K \hspace{1 cm} (5) \] \[ p,u \in C^\infty \hspace{.1 cm}(\mathbb{R}^n \times [0,\infty)) \hspace{1 cm} (6) \] \[ \int_K |u(x,t)| dx \leq C \hspace{1 cm} (7) \]

(C) Take $\nu >0$ and $n = 3$ . Then there exist a smooth, divergence-free vector field $u^\circ(x)$ on $\mathbb{R}^3$ and a smooth $f(x,t)$ on $\mathbb{R}^3 \times [0,\infty)$, satisfying (4), (5), for which there exist no solutions of $(p, u)$ of (1), (2), (3), (6), (7) on $\mathbb{R}^3 \times [0,\infty)$.

3. Definitions and Authorities

We define functions, components, indices, and list authorities below: \[ x=\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} \hspace{1 cm} (8) \] \[ u=\begin{bmatrix} u_1\\ u_2\\ u_3 \end{bmatrix} \hspace{1 cm} (9) \]

Index $i = 1,2,$ or $3$; $j = 1,2$ or $3$.

Conservation of Energy: Energy can neither be created nor destroyed, but only transformed from one kind to another.

Newton's Second Law: If a net or resultant force acts upon an object of non-zero mass, the object accelerates in the direction of the force. The acceleration is proportional to the force and inversely proportional to the mass of the object.

Identity Principle: $u=u$.

Constant Rule: The derivative of a constant is zero.

Constant Multiple Rule: $\psi \delta u = \delta \psi u$, where $\delta$ is a constant and $\psi$ is an operator.

4. Preliminary Proofs

4.1. Divergence-free

Vector component $u_i$ is an instantaneous velocity: \[ u_i = \frac{dx_i}{dt}= \lim_{h\to0}\frac{x_i(t +h)-x_i(t)}{h} \hspace{1 cm} (10) \] Take the partial derivative with respect to $x_i$: \[ \frac{\partial u_i}{\partial x_i}= \frac{\partial}{\partial x_i}\frac{x_i(t +h)-x_i(t)}{h} = \frac{1-1}{h}= 0 \hspace{1 cm} (11) \] The result is zero. Since $i = 1,2,3$, then \[ \frac{\partial u_1}{\partial x_1} + \frac{\partial u_2}{\partial x_2} + \frac{\partial u_3}{\partial x_3} = 0 \hspace{1 cm} (12) \]

Equation (12) is also true if $x_i$ is constant, and $u_1=u_2=u_3 = 0$.

4.2. The Material Derivative

Take the total derivative of $u_i$: \[ \frac{d}{dt}u_i(x,t) = \frac{\partial u_i}{\partial t} + \sum_{j=1}^3\frac{\partial u_i}{\partial x_j}\frac{dx_j}{d t} \hspace{1 cm} (13) \] \[ u_j = \frac{dx_j}{dt} \hspace{1 cm} (14) \] Substitute, then take the derivative using the product rule: \[ \frac{\partial}{\partial x_j}u_i u_j = \frac{\partial u_i}{\partial x_j}u_j + u_i\frac{\partial u_j}{\partial x_j} \hspace{1 cm} (15) \] \[ \frac{\partial u_j}{\partial x_j} = 0 \hspace{.5 cm}(see \hspace{.1 cm}4.1) \hspace{1 cm} (16) \] The last term on the right of (15) vanishes. Then we have the symmetric property: \[ \frac{\partial u_i}{\partial x_j}u_j = u_j\frac{\partial u_i}{\partial x_j} \hspace{1 cm} (17) \] Therefore, we have the following: \[ \frac{d}{dt}u_i(x,t) = \frac{\partial u_i}{\partial t} + \sum_{j=1}^3 u_j\frac{\partial u_i}{\partial x_j} \hspace{1 cm} (18) \] The material derivative is as follows: \[ \frac{du_i}{dt} =\frac{Du_i}{Dt} = \frac{\partial u_i}{\partial t} + \sum_{j=1}^3 u_j\frac{\partial u_i}{\partial x_j} \hspace{1 cm} (19) \]

5. In Search of u(x,t)

We select an existing arbitrary, smooth, divergence-free vector field $u^\circ(x)$ on $\mathbb{R}^n$ and a smooth $f(x,t)$ on $\mathbb{R}^n \times [0,\infty)$, satisfying all CMI requirements including (4), (5), where $\nu >0$, $n = 3$, where if $t>0$, $u \neq 0$, and $u$ changes at each position $x$ in response to changes in $p$ and $f$.

5.1. The Bounded Energy Blowup

First, let us examine what can go terribly wrong and why. Assume the NS momentum equation is always an equation, that the material-derivative proof is valid, and assume the vector field at all times satisfies (7), i.e., always has limited energy $E$, where $E= Vx(\rho f-\nabla p)$ and $V$ is the fluid volume.

5.1.1. Infinitesmal Mass Density

Assume mass density $\rho$ is constant but also infinitesmal, that $\rho = 1 \hspace{.1 cm}\rho_{unit}$. Suppose that at $t=0$, energy $E = 0$: \[ V\rho_{unit}\frac{Du}{Dt}x= E + Vx\nu\Delta u = 0 \hspace{1 cm} (20) \] If $E=0$, assume that $u$ and all its derivatives $=0$, and (4) is satisfied. At an arbitrary finite time $t >0$ we add a limited amount of energy to the system: \[ V\rho_{unit}\frac{Du}{Dt}x= E +Vx\nu\Delta u \neq 0 \hspace{1 cm} (21) \] The material derivative is derived from and equal to the total derivative (see 4.2): \[ \frac{du}{dt} = \frac{Du}{Dt} \hspace{1 cm} (22) \] Make a substitution: \[ V\rho_{unit}\frac{du}{dt}x= E + Vx\nu\Delta u \hspace{1 cm} (23) \] Solve for acceleration: \[ \frac{du}{dt}= \lim_{\rho_{unit} \to 0}\frac{E + Vx\nu\Delta u }{x V\rho_{unit}} \sim \pm\infty \hspace{1 cm} (24) \] Integrate both sides, assume the integration constant = 0: \[u = \lim_{\rho_{unit} \to 0} \int_{dt}^t \frac{E + Vx\nu\Delta u}{xV\rho_{unit}}dt \sim \pm \infty \hspace{.3 cm}(0 < dt < t < \infty) \hspace{1 cm} (25)\] If the mass density $\rho$ is constant but infinitesimal (or set to 1 infinitesimal unit) and $x$ is constant, then to conserve energy, the acceleration at (24) and velocity at (25) must have a $\pm$ infinite upper limit at any arbitrary finite time $t > 0$. In Newton's universe, velocity is not limited to light speed: infinite speed in n-dimensional space is perfectly fine, where n = 1, 2, or 3. Ironically, any solution of $u$ that does not allow $u$ to blow up to $\pm$ infinity if the mass density is infinitesmal, violates the energy conservation principle. Granted, this is a physically unreasonable result by today's standards, but a plausible result when using Newtonian-based equations including the NSE.

5.1.2. Unlimited Viscosity

Here is what results if there is no upper bound on $\nu$ given the same finite conserved energy $E$ added to the system when $t>0$: \[ V\rho_{unit}\frac{Du}{Dt}x - \lim_{\nu\to\infty} Vx\nu\Delta u = E \hspace{1 cm} (26) \] To conserve energy $E$, $Du/Dt$ must blow up to $\pm$ infinity. It follows that a $\pm$ infinite acceleration will lead to a $\pm$ infinite velocity $u$ in a finite period of time $t$: \[ u= \lim_{\nu \to \infty} \int_{dt}^t\frac{E + Vx\nu\Delta u }{ xV\rho_{unit}}dt\sim \pm\infty \hspace{.3 cm}(0 < dt < t < \infty) \hspace{1 cm} (27) \] Equation (27) was derived from (20) through (26).

5.2. The Eulerian Perspective

If we focus on a single position in the vector field, i.e., if we focus on a fixed position $x$ and measure the velocity $u_j(x,t)$ located there, we find to our dismay that the math does not match the reality if $u_j \neq 0$. We restate (14): \[ \frac{d}{dt}x_j = u_j(x,t) \hspace{1 cm} (28) \] But coordinate $x_j$ of a fixed position $x$ is not a function of time $t$. In fact, it is constant for all time $t$. Thus, we are forced to conclude the following: \[ \frac{d}{dt}x_j = 0 \neq u_j(x,t) \hspace{.5 cm}(t>0) \hspace{1 cm} (29) \] The derivative of a constant is always zero, and satisfies (4) where $dx/dt = u^\circ(x) = 0$, but the velocities of our selected existing vector field are not always zero if $t>0$ (see section 5). From the Eulerian perspective, there is no solution of $u_j$ that models the vector field. Since index $j=1,2,3$, we deduce that there is no solution of $u$ as well.

5.3. The Lagrangian Perspective

If $x_j = x_j(t)$, then $dx_j(t)/dt \neq 0$. However, if $dx_j/dt = x_j/t$ where $t=0$, we have the following: \[ u_j(x,0) = \frac{x(t)}{0} \hspace{1 cm} (30) \] To avoid a singularity at $t=0$, we use the difference quotient, (10) and (14): \[ u_j(x,0) = \frac{dx_j}{dt} = \lim_{h\to0}\frac{x_j(0+h)-x_j(0)}{0 + h-0} \hspace{.3 cm}(h \neq 0) \hspace{1 cm} (31) \] The right side of equations (31) perfectly models a divergence-free velocity vector (see 4.1) and is perfectly consistent with the material derivative proof at 4.2 and equation (14). Since $j=1,2,3$, we deduce the following: \[ u(x,t) = \lim_{h\to0} \frac{x(t+h)-x(t)}{t+h-t} = \lim_{h\to0} \frac{x(t+h)-x(t)}{h} \hspace{1 cm} (32) \] The far right side of (32) shows that the input $t$ is inconsequential in the denominator. The $h$ is the valid time input there; however, its forever shrinking value cannot be pinned down exactly. We can replace it with the SI unit for time: 1 second or $s$. We can also replace $x(t+h)$ with $x_f$ which represents a future position of $x$ if the instantaneous velocity is sustained for a period of $s$. We then have the following: \[ \frac{x_f - x}{s} = \lim_{h\to0} \frac{x(t+h)-x(t)}{h} \hspace{1 cm} (33) \] Where $x = x(t)$ and \[ x_f = us + x \hspace{1 cm} (34) \] \[ u(x,t) = u(x,x_f) \hspace{1 cm} (35) \] Note that the function on the left side of (35) requires inputs $x,t$; whereas, the function on the right side requires inputs: $x,x_f$. If we plug in a value for $x$ on both sides of (35), we might have an equation, but if we proceed to plug in a value for $x_f$, the left side has no reason to change, but the right side could change, so we could have the following inequality: \[ u(x,t) \neq u(x,x_f) \hspace{1 cm} (36) \] An obvious solution to this inequality problem is to make $t$ and $x_f$ constants, then only an input of $x$ is required on both sides of (35). Unfortunately, each $x$ will have only one value of $u$ (see 33). This is inconsistent with the selected vector field where $u$ changes. Now, will the inequality problem be solved if we let $x_f = x_f(x)$? If so, once again, only the input of $x$ would be required for both sides of (35). The answer to the question posed is a resounding no. Equation (34) shows that $x_f$ is a function of $u$, not just $x$. Velocity $u$ is the unknown we are trying to solve. Thus, we need to know and input $x_f$ like we know and input $x$ to solve $u$. Equations (31) through (35) show that $dx/dt = u(x,x_f)$. Equation (36) shows the following can be true: \[ u(x,t) \neq \frac{dx}{dt} \hspace{1 cm} (37) \] At the material-derivative proof (4.2), it was assumed, not proven, that (14) is true. We have shown here that (14) may not always be true. Since $u(x,t)$ represents any solution of $u(x,t)$, the inequality at (36) can occur for any solution of $u(x,t)$ where the only input variables are $x$ and $t$. Thus, there is no solution of $u(x,t)$ that solves this inequality problem and successfully models the selected existing vector field.

5.4. The Navier-Stokes Inequality

Here is a compact version of the NS momentum equations: \[ \frac{\partial u}{\partial t} + (u \cdot \nabla) u - \nu\Delta u = f - \nabla p \hspace{1 cm} (38) \] Assume $u = u(x,t)$ and $F = F(x,t)=f(x,t)- \nabla p(x,t)$: \[ \frac{\partial u}{\partial t} + (u \cdot \nabla) u - \nu\Delta u = F \hspace{1 cm} (39) \] According to 4.2, the following is true: \[ \frac{D u}{Dt}=\frac{\partial u}{\partial t} + (u \cdot \nabla) u \hspace{1 cm} (40) \] We make a substitution: \[ \frac{D u}{D t} - \nu\Delta u = F \hspace{1 cm} (41) \] Multiply both sides of (41) by a constant $\delta$: \[ \delta\frac{Du}{D t} - \delta\nu\Delta u = \delta F \hspace{1 cm} (42) \] Apply the constant multiple rule in reverse: \[ \frac{D (\delta u)}{D t} - \nu\Delta (\delta u) = \delta F \hspace{1 cm} (43) \] Where $u(x',t')=\delta u(x,t)$, $F(x',t') = \delta F(x,t)$ and $t'>t$. We observe that $u \propto F$. Now, multiply both sides of (39) by $\delta$: \[ \delta\frac{\partial u}{\partial t} + \delta(u \cdot \nabla) u - \delta\nu\Delta u = \delta F \hspace{1 cm} (44) \] Once again, apply the constant multiple rule in reverse: \[ \frac{\partial (\delta u)}{\partial t} + ( u \cdot \nabla)(\delta u) - \nu\Delta (\delta u) = \delta F \hspace{1 cm} (45) \] At the (45) advection term, $u = u(x,t)$ and $u=\delta u(x,t) = u(x',t')$. This is a violation of the identity principle and we have ... \[ u \neq u \hspace{1 cm} (46) \] If we fix the identity problem and enforce the observation that $u \propto F$, we have ... \[ \frac{\partial (\delta u)}{\partial t} + (\delta u \cdot \nabla) \delta u - \nu\Delta (\delta u) \neq \delta F \hspace{1 cm} (47) \]

...the Navier-Stokes Inequality. This inequality occurs if $p,f,u$ change with time and space. Since $u$ represents any solution of $u$, there is no solution of $u$ that can fix this problem and model the selected existing vector field.

6. In Search of p

We calculate $p$ using (1): \[ \frac{\partial p}{\partial x_i}=\nu\Delta u_i -\frac{\partial u_i}{\partial t} - \sum_{j=1}^3 u_j\frac{\partial u_i}{\partial x_j} + f_i \hspace{1 cm} (48) \] \[ p= \int(\nu\Delta u_i -\frac{\partial u_i}{\partial t} - \sum_{j=1}^3 u_j\frac{\partial u_i}{\partial x_j} + f_i)dx_i- C_P \hspace{1 cm} (49) \]

Equation (49) reveals that a solution of $p$ depends on solutions of $u_i,u_j$. We have shown there is no solution of either $u_i$, $u_j$ or $u$ that is mathematically consistent and/or physically reasonable for all time and space. Thus, it follows that there is no solution of $p$ that is mathematically consistent and/or physically reasonable for all time and space.

Conclusion

Based on the foregoing, if $\nu >0$ and $n = 3$, then there exists a smooth, divergence-free vector field $u^\circ(x)$ on $\mathbb{R}^3$ and a smooth $f(x,t)$ on $\mathbb{R}^3 \times [0,\infty)$, satisfying (4), (5), for which there exist no solutions of $(p, u)$ of (1), (2), (3), (6), (7) on $\mathbb{R}^3 \times [0,\infty)$.

References

1. Fefferman, Charles L. Existence and Smoothness of the Navier-Stokes Equation. Clay Mathematics Institute.

2. Lo, Asane. October 28, 2018. Smooth Solutions of the Three Dimensional Navier-Stokes Problem. King Fahd University of Petroleum and Minerals.

3. Porton, Victor. Analyzing Navier-Stokes Equations Using A Linearly Extended Funcional. Orcid 0000-0001-7064-7975.

4. Tao, Terrence. Finite Blowup for an Averaged Three-Dimensional Navier-Stokes Equation. arxiv.org.

5. Ladyzhenskaya, O.A.. 1958. Solution 'in the Large' to the Bounary Value Problem for the Navier-Stokes Equations in Two Space Variables Doki Akad Nauk SSSR, Volume 123, No. 3, P. 427-429

6. Peng Shi. General Solution to 2D Steady Navier-Stokes Equation for Incompressible Flow without vorticity diffusion. Well Logging Key Laboratory, China National Logging Corporation,

© 2026 G.M. Jackson