# Navier Stokes EquationQuestion about the Diff EQ

1. Nov 7, 2009

Hello! I am going over an example in my fluid mechanics text and I am confused about a few lines. My question is more about the math then the fluid mechanics. In fact, I doubt you need to understand the FM at all; if you understand Diff eqs, you can probably answer my question.

I am given the components of the velocity field:

$$u = a(x^2-y^2)\,\,\,\,\,\,\,\,\,\,\,v = -2axy\,\,\,\,\,\,\,\,\,\,w=0$$

and I am also given that gravity only acts in the downward z direction $g_z = -g$.

The example question asks to first determine under what conditions the given velocity field is a solution to the Navier-Stokes equation. Then, assuming that these conditions are met, determine the resulting pressure distribution p(x,y,z).

The Navier-Stokes EQs are given by:

Their approach is as follows. They say
1st question: what do they mean by exact? What would it mean to not be exact?

Alright, now after a whole bunch a solving that you don't need to see, we end up with the following pressure gradients:

$$\frac{\partial{p}}{\partial{x}}=-2a^2\rho(x^3+xy^2)$$

$$\frac{\partial{p}}{\partial{y}}=-2a^2\rho(x^2y+y^3)$$

$$\frac{\partial{p}}{\partial{z}}=\rho g$$

The book then says:
2nd question Why do I need to do this? What do they mean compatible? I know that this is reminisce of an 'exact diff EQ,' but I don't really know what that means ... I just know how to solve one by going through the motions.

So we then go ahead and cross-differentiate and find that

$$\frac{\partial{}}{\partial{y}}(\frac{\partial{p}}{\partial{x}}) = \frac{\partial{}}{\partial{x}}(\frac{\partial{p}}{\partial{y}})$$

But I still do not know why we care. And secondly, why don't we cross-differentiate

$$\frac{\partial{p}}{\partial{z}}=\rho g$$

with anything?

Any insight would be greatly appreciated.

~Casey

Last edited: Nov 8, 2009
2. Nov 8, 2009

### arildno

Okay:

1. Remember that the curl of the gradient of a scalar function is zero!
2. The converse is also true:
IF the curl of a vector field is zero, then that vector field MUST be the gradient of a scalar function!
Thus, by making the "cross-differentiation", what you are actually doing is to compute the z-component of the curl, and finding out it is zero.
The x-and y-components of the curl are trivially zero, since differentiation of the vector field's x-and y-components with respect to z yields 0, AND the differentiation of -pg with respect to x-and y- also yields zero.

YOu can also solve this explicitly as follows:
$$\frac{\partial{p}}{\partial{z}}=-\rho{g}\to{p}(x,y,z)=-\rho{g}z+H(x,y) (1)$$
where H is some function of x and y.
$$\frac{\partial{p}}{\partial{x}}=-2a^{2}\rho(x^{3}+x^{2}y)\to{p}(x,y,z)=}}=-2a^{2}\rho(\frac{1}{4}x^{4}+\frac{1}{2}x^{2}y^{2})+F(y,z)(2)$$
$$\frac{\partial{p}}{\partial{y}}=-2a^{2}\rho(y^{3}+y^{2}x)\to{p}(x,y,z)=}}=-2a^{2}\rho(\frac{1}{4}y^{4}+\frac{1}{2}x^{2}y^{2})+G(x,z)(3)$$

Now, these three expressions, (1),(2) and (3) for p must specify the SAME function!
We can do this by setting:
$$H(x,y)=-2a^{2}\rho(\frac{1}{4}x^{4}+\frac{1}{2}x^{2}y^{2}+\frac{1}{4}y^{4})+K$$
K being a constant.
$$F(y,z)=-2a^{2}\rho(\frac{1}{4}y^{4})-\rho{gz}+K$$
$$G(x,z)=-2a^{2}\rho(\frac{1}{4}x^{4})-\rho{gz}+K$$

Thus, (1), (2) and (3) represents the SAME equation, and we get the final pressure distribution:
$$p(x,y,z)=)=-2a^{2}\rho(\frac{1}{4}x^{4}+\frac{1}{2}x^{2}y^{2}+\frac{1}{4}y^{4})-\rho{gz}+K=-\frac{a^{2}\rho}{2}(x^{2}+y^{2})^{2}-\rho{gz}+K$$

Last edited: Nov 8, 2009
3. Nov 8, 2009

Dude. My mind just got blown. I think I have another question, but I have no idea how to voice it.

Let me ask you this arildno, just to satisfy my curiosity: Did you solve this by taking a fluid mechanics approach or just from a mathematical stance? (or both?)

I only ask because the word curl never even crossed my mind, nor has it been mentioned up to this point. So I get the feeling it was the latter.

Thank you so much for your help thus far!

4. Nov 8, 2009

### arildno

Since I have a Master's in F.M., I am no longer sure how I learnt it once back in the stone age.

But, just to reassure myself:

You DO know what the curl of a vector field is, right?

5. Nov 8, 2009

My multivariate skills are rubbish I will admit; and I am working on that. But yes, I do know that the curl of a vector field tells us something about its rate of rotation. And I do know how to compute the curl of a vector field using the notation abusive 'determinant method.'

I am just not sure how you even decided to invoke the use of curl? It seems purely mathematical (not that that is a bad thing, I just am not there yet).

6. Nov 8, 2009

### Andy Resnick

I don't know what the author's intentions are by 'exact' and 'compatible', but my answers would be:

1) If a function P(x,y,z) can be determined, the velocity function then satisfies the N-S equations, and so can be considered an analytic solution, rather than an approximate solution.

2) The gradient of the function P(x,y,z) is given, and from that, we guess that P(x,y,z) can be separated into a function that goes like P(x,y)*Q(z). The business of cross-derivatives comes from some mathematical property that ensures the function itself is "nice" (I forget the formal logic... 'continuous', maybe?)

http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Clairaut.27s_theorem

Anyhow, because the x and y dependence is separated from z, the second derivative rule is automatically satisfied for those pairs that you note are missing.

hope this helps...

7. Nov 9, 2009

### arildno

OKay:

1. You were given a velocity field.

2. You were asked to verify that this can, indeed, be a valid solution to the Navier-Stokes equations.

3. If it is to be a valid solution, then it must be possible to find a SCALAR FUNCTION "p" that can play the role of pressure.

4. By taking the curl of the given vectorial expression, we find that this is 0, i.e, that N-S has been reduced to a GRADIENT expression for the pressure.

5. Therefore, the given velocity field CAN BE a valid solution to the N-S equation.

6. If 4. had given something else than 0, then your velocity field is unphysical, in the sense it cannot be a solution to the N-S equation.