Breakdown of Navier-Stokes Equations

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

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© 2026 G.M. Jackson