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Understanding Davis artificial viscosity

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Old   August 6, 2017, 04:24
Default Understanding Davis artificial viscosity
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Oleg Sutyrin
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I'm solving 2D Euler's equations in Cartesian coordinates:

\frac{\partial U}{\partial t} + \frac{\partial F}{\partial x} + \frac{\partial G}{\partial y} = 0, \qquad 
U = \left( \begin{array}{c} \rho \\ \rho u \\ \rho v \\ e \end{array} \right), \quad
F = \left( \begin{array}{c} \rho u \\ p+\rho u^2 \\ \rho uv \\ (e+p)u \end{array} \right), \quad
G = \left( \begin{array}{c} \rho v \\ \rho uv \\ p+\rho v^2 \\ (e+p)v \end{array} \right)

using finite difference MacCormack method with Davis artificial viscosity which dampens non-physical oscillations. The MacCormack method is pretty simple and not shown here. Davis viscosity works good in my simulations, but is quite difficult; I'm trying to fully understand it in order to apply it cylindrical coordinates later. I hope that some of you may help me with unclear parts of it (shown in bold font below).

Viscous terms are added after 2nd MacCormack step as follows (formulated in 1D for simplicity):

U_j^{n+1}=U_j^{(m)} + \left( D_{j+\frac{1}{2}}-D_{j-\frac{1}{2}} \right),

where U_j^{(m)} denote field values (at j-th coordinate point and (n+1)-th time layer) calculated by MacCormack scheme as usual, and D_{j+\frac{1}{2}}, D_{j-\frac{1}{2}}, are forward and backward differences of U with non-linear coefficients:

D_{j+\frac{1}{2}} := \frac{1}{2} C(\nu_j) \left( 2 - \phi(r_j^+) - \phi(r_{j+1}^-) \right) \Delta U_{j+\frac{1}{2}}, \quad \Delta U_{j+\frac{1}{2}} = U_{j+1} - U_j
D_{j-\frac{1}{2}} := \frac{1}{2} C(\nu_{j-1}) \left( 2 - \phi(r_{j-1}^+) - \phi(r_{j}^-) \right) \Delta U_{j-\frac{1}{2}}, \quad \Delta U_{j-\frac{1}{2}} = U_{j} - U_{j-1}

Where C(\nu) is a limited coefficient based on Courant number:

C(\nu) := \begin{cases} \nu(1-\nu) & \;\; \text{if} \;\; \nu \le 0.5 \\ 0.25 & \;\; \text{otherwise} \end{cases}, \qquad \nu = \rho(A(U_j)) \frac{\Delta t}{\Delta x},

A(U) is the Jacobian matrix of vector F, \rho(A(U)) is it's spectral radius (for Euler's equations it equals \text{max}(|u|+a, |u|-a), where a is the speed of sound).

\phi(r) is simple limiting function: \begin{cases} 0, & r < 0, \\ 2r, & 0 \le r \le 0.5, \\ 1, & r >0.5\\ \end{cases}

And finally r_j^+,r_j^- are slope ratios:

r_j^+ := \frac{\left( \Delta U_{j-\frac{1}{2}}, \Delta U_{j+\frac{1}{2}}\right)}{\left( \Delta U_{j+\frac{1}{2}}, \Delta U_{j+\frac{1}{2}}\right)}, \qquad
r_j^- := \frac{\left( \Delta U_{j-\frac{1}{2}}, \Delta U_{j+\frac{1}{2}}\right)}{\left( \Delta U_{j-\frac{1}{2}}, \Delta U_{j-\frac{1}{2}}\right)}

Where \left( \cdot , \cdot \right) is scalar product. These ratios are positive in monotone areas and negative in non-monotone areas, so that in general coefficients like \left( 2 - \phi(r_j^+) - \phi(r_{j+1}^-) \right) are equal 2 in non-monotone areas; about 1 in monotone, but "curvy" areas; and equal 0 in monotone (and rectilinear) areas.

If both these (2-\phi-\phi) coefficients are equal 2, and both C(\nu) are equal 0.25 we have an effective second derivative of the field value:

D_{j+\frac{1}{2}}-D_{j-\frac{1}{2}} = 0.25 \left( U_{j+1} - 2U_j + U_{j-1} \right).

But where is the division by \Delta x^2? If the viscosity would be simply the 2nd derivative, the scheme would look like

U_j^{n+1}=U_j^{(m)} + \frac{\Delta t}{\Delta x^2} \left( U_{j+1} - 2U_j + U_{j-1} \right).

Is C(\nu) effectively equals \frac{\Delta t}{\Delta x^2}?

And, what if D_{j+\frac{1}{2}} is non-zero and D_{j-\frac{1}{2}} is zero? Does it still yield an effective 2nd derivative?
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Old   August 6, 2017, 05:10
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I think you should see the artificial viscosity as a term that disappears as dx->0. Therefore, multiply and divide by dx^2 and you get a second order derivative multiplied by dx^2.
You wrote dt/dx^2 but the dimensions are no longer congruent
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Old   August 6, 2017, 06:00
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Thanks for such a fast answer, FMDenaro!

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Originally Posted by FMDenaro View Post
I think you should see the artificial viscosity as a term that disappears as dx->0. Therefore, multiply and divide by dx^2 and you get a second order derivative multiplied by dx^2.
So, you mean that artificial viscosity is not like physical viscosity (which effectively is a second derivative with coefficient that doesn't depend on grid), but more like physical viscosity multiplied by dx^2?
But spurious oscillations do not disappear as dx \to 0. If artificial viscosity disappears as dx \to 0, how would it dampen the oscillations then?

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You wrote dt/dx^2 but the dimensions are no longer congruent
I can't say that I understand that completely. Could you elaborate?
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Old   August 6, 2017, 06:10
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The artificial viscosity has nothing to do with the physical viscosity. It is just an added terms that has the aim of dumping oscillations. But it has to be consistent with the original PDE equation as h->0.
I suggest a reading in the fundemental textbooks about the numerical solution of Euler equations.
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Old   August 6, 2017, 06:20
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Yeah, it seems that deep diving into the theory is inevitable :-)
Thanks again!
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Old   August 7, 2017, 14:54
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As a general statement, artificial viscosity is a consequence of numerical differentiation. It is not a physical term. It can be determined, in a hand waving way, from the wave equations. In a nutshell, central difference is a result of taking the limit when coming from the right and left.

In a sense, dF/dx is a function of lim (+ and -)->0 [dF/dx(+) + dF/dx(-)] + lim (+ and -)->0 [dF/dx(+) - dF/dx(-)]. If everything was perfect, and assuming no discontinuities, [dF/dx(+) - dF/dx(-)] would be zero. But numerically it is not. Thus the C*(U(i+1)-2*U(i)+U(i-1)) where C is a knob.
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Old   August 12, 2017, 05:34
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Quote:
Originally Posted by Martin Hegedus View Post
If everything was perfect, and assuming no discontinuities, [dF/dx(+) - dF/dx(-)] would be zero. But numerically it is not.
Thanks for the explanation! As far as I know, ENO/WENO methods try to alleviate this very problem by choosing most smooth stencil over some region.

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Originally Posted by Martin Hegedus View Post
Thus the C*(U(i+1)-2*U(i)+U(i-1)) where C is a knob.
"Knob" has too many meanings in English language, so I can't understand the meaning of this phrase :-)
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Old   August 12, 2017, 07:21
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I suggest having a reading to the texbook of Leveque (for example pag.71-72) to start with the concept of artificial viscosity explicitly added to the discretization in order to dump oscillations and numerical viscosity present in the local truncation error that is implicitly induced by the discretization.
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Old   August 12, 2017, 07:33
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Quote:
Originally Posted by FMDenaro View Post
I suggest having a reading to the texbook of Leveque (for example pag.71-72) to start with the concept of artificial viscosity explicitly added to the discretization in order to dump oscillations and numerical viscosity present in the local truncation error that is implicitly induced by the discretization.
Could you provide the name of the book? There are several books by Randall J. LeVeque in the web, but I'm not sure which one do you mean (pages 71-72 of them contain something different).
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Old   August 12, 2017, 07:37
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http://www.cambridge.org/catalogue/c...=9780521009249
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Old   August 12, 2017, 07:57
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Quote:
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Ah, I missed this one because I was looking only on finite difference methods :-) Thanks!
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