This is an extension to my question here: Clay Institute Navier Stokes . Last time my solution was wrong because my velocity vector did not use bounded functions. Also someone said that I am only showing one example, but I'm trying to prove a statement that says "there exists" so I should only have to find one right? I have looked into smooth and bounded functions and it looks like Gaussian functions are good examples, so I have rewritten another example. If you could please tell me what is still wrong/not satisfactory.

We are trying to satisfy:

\begin{equation} \frac{\partial \textbf{u}}{\partial t} + (\textbf{u}\cdot\nabla)\textbf{u}=-\frac{\nabla P}{\rho} + \nu\nabla^{2}\textbf{u}+\textbf{f}, \end{equation}

\begin{equation}\label{incompr} \nabla\cdot\textbf{u}=0. \end{equation}

We get to pick a velocity (u) and pressure (P) function to satisfy these equations. We take $n=3$ by putting our velocity and pressure vectors in the 3D Cartesian plane with $x,y,z$. We let $\textbf{f}$ be 0 according to the problem description, and assume kinematic viscosity (nu) to be greater than 0. Let

\begin{equation} \textbf{u}(x,t)=\begin{bmatrix} e^{-t^2} \\ e^{-t^2} \\ e^{-t^2} \end{bmatrix} \end{equation}

Then $\textbf{u}(x,t)$ satisfies the divergence free condition because

\begin{equation} \nabla \cdot \textbf{u} = \frac{\partial u_1}{\partial x}+\frac{\partial u_2}{\partial y}+\frac{\partial u_3}{\partial z} \end{equation}

which is

\begin{equation} \nabla \cdot \textbf{u} = 0+0+0=0 \end{equation}


\begin{equation}\frac{\partial \textbf{u}}{\partial t}=\begin{bmatrix} -2te^{-t^2} \\ -2te^{-t^2} \\ -2te^{-t^2} \end{bmatrix} \end{equation}


\begin{equation} \textbf{u} \cdot \nabla = (e^{-t^2})\frac{\partial}{\partial x} + (e^{-t^2})\frac{\partial}{\partial y}+(e^{-t^2})\frac{\partial}{\partial z} \end{equation}


\begin{equation} \begin{split} (\textbf{u} \cdot \nabla)\textbf{u} &= (e^{-t^2})\frac{\partial \textbf{u}}{\partial x} + (e^{-t^2})\frac{\partial \textbf{u}}{\partial y}+(e^{-t^2})\frac{\partial \textbf{u}}{\partial z} \\ &=(e^{-t^2})\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}+(e^{-t^2})\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}+(e^{-t^2})\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ &= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \end{split} \end{equation}

Since laplacian of $\textbf{u}$ is 0 the whole kinematic velocity term goes to 0. And the final Navier-Stokes expression is:

\begin{equation} \begin{bmatrix} -2te^{-t^2} \\ -2te^{-t^2} \\ -2te^{-t^2} \end{bmatrix}+\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}=-\frac{\nabla P}{\rho} \end{equation}

So we set $\rho$ equal to 1 for easy calculation. Bring the negative to the left hand side and combine the left hand side.

\begin{equation} \begin{bmatrix} 2te^{-t^2} \\ 2te^{-t^2} \\ 2te^{-t^2} \end{bmatrix}=\nabla P \end{equation}

and we solve for a solution for P

\begin{equation} P=2 e^{-t^2} t x + 2 e^{-t^2} t y + 2 e^{-t^2} t z \end{equation}

So we now have infinitely differentiable functions for u and P. Where did I make a mistake/misunderstand the problem? http://www.claymath.org/sites/default/files/navierstokes.pdf . I'm trying to prove statement A

  • 3
    $\begingroup$ You have shown that there is a solution in the special case $u(x,0) = (1,1,1)^T$. The problem statement asks for a proof of the existence of a smooth solution for any smooth initial data $u(x,0) = u^0(x)$. $\endgroup$ – ekkilop Aug 15 '17 at 14:39
  • $\begingroup$ How can I get from (1,1,1)^T to u^0 (x) ? $\endgroup$ – Seth Kitchen Aug 15 '17 at 14:47
  • 17
    $\begingroup$ If I knew that, I would be very famous :) $\endgroup$ – ekkilop Aug 15 '17 at 14:52

I think you're misunderstanding the problem. The problem is not to show that functions $\{\mathbf{u}(\mathbf{x},t), P(\mathbf{x},t)\}$ exist which satisfy the Navier-Stokes equations; that much is relatively easy. The problem is to show that given any function $\mathbf{u}^\circ(\mathbf{x})$, there exists functions $\{\mathbf{u}(\mathbf{x},t), P(\mathbf{x},t)\}$ that satisfy the Navier-Stokes equations and satisfy $$ \mathbf{u}(\mathbf{x},0) = \mathbf{u}^\circ(\mathbf{x}). $$ This function $\mathbf{u}^\circ(\mathbf{x})$ is called the initial data; it basically tells you what the system is doing at the moment you start your stopwatch.

For your function, you have $\mathbf{u}(\mathbf{x},0) = (1,1,1)$. I haven't gone carefully through your math, but even if this were a solution of the Navier-Stokes equations, it would only be a drop in the bucket. You'd then have to find a solution with $\mathbf{u}(\mathbf{x},0) = (e^{-(x^2+y^2 + z^2)}, 0, 0)$. And then $\mathbf{u}(\mathbf{x},0) = e^{-(x^2+y^2 + z^2)}(\sin(z^2+x), x^2, \tanh(y))$. And then every other conceivable function that I could give you for the initial data.

Obviously, it's impossible to write down a solution for every conceivable function, for the simple reason that there are infinitely many such functions and you don't have an infinite amount of time. Rather, when mathematicians are trying to prove existence of solutions to PDEs, they try to find a general solution in the form of integrals involving the initial data, and then prove that these integrals are well-behaved no matter what the initial data is. Sometimes, they have to impose certain conditions on the initial data to make sure that their integrals are well-behaved; these are the conditions that you ran into in your last post.


As with many hard math problems, the trick is to prove something always exists. This was a good exercise, though - you found a certain function that satisfies a nonlinear PDE, and that's not always easy to do explicitly. Now can you find a solution that satisfies something like $$ {\bf u}(x,0) = {\bf u}^0(x) = \exp(-\|x\|^2)[y,-x,0]^T? $$

  • $\begingroup$ could you explain or link me to some information on that notation? What does a exp(−∥x∥2)[1,1,1]^T look like? $\endgroup$ – Seth Kitchen Aug 15 '17 at 15:07
  • $\begingroup$ It's a vector field with each component given by the function $\exp(-(x^2+y^2+z^2))$. $\endgroup$ – icurays1 Aug 15 '17 at 17:10
  • 1
    $\begingroup$ but p would be basically integral -2*xexp(-(x^2 + y^2 + z^2))^2 - 2*yexp(-(x^2 + y^2 + z^2))^2 - 2*zexp(-(x^2 + y^2 + z^2))^2 + 2*(exp(-(x^2 + y^2 + z^2)) - 2*x^2*exp(-(x^2 + y^2 + z^2))) dx + integral -2*xexp(-(x^2 + y^2 + z^2))^2 - 2*yexp(-(x^2 + y^2 + z^2))^2 - 2*zexp(-(x^2 + y^2 + z^2))^2 + 2*(exp(-(x^2 + y^2 + z^2)) - 2*x^2*exp(-(x^2 + y^2 + z^2))) dy + integral -2*xexp(-(x^2 + y^2 + z^2))^2 - 2*yexp(-(x^2 + y^2 + z^2))^2 - 2*z*exp(-(x^2 + y^2 + z^2))^2 + 2*(exp(-(x^2 + y^2 + z^2)) - 2*x^2*exp(-(x^2 + y^2 + z^2))) dz $\endgroup$ – Seth Kitchen Aug 15 '17 at 17:21
  • 2
    $\begingroup$ The point is, if you can find a general technique and prove that it works for every possible initial condition satisfying the constraints discussed in the Clay problem statement, you've solved the problem and I will personally write you a check for the prize. If you solve the equation for the particular example I provided, you've only improved your own personal understanding of the problem (which is good), but you haven't solved the problem in general. There is a big big difference. $\endgroup$ – icurays1 Aug 16 '17 at 0:15
  • 1
    $\begingroup$ Think analogously to cancer: it is possible (indeed very common and increasingly so) to cure one person's cancer, but yet we cannot say that we have cured cancer because our treatments don't always work. $\endgroup$ – icurays1 Aug 16 '17 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.