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alexanderkazak November 23, 2013 08:40

Deformable mesh in 1D elastic body
Good afternoon!
I'm solving the pde's of 1D deformable solid body (just an elastic case without viscosity)

du/dt + A*du/dx = 0,

where u is a 2D vector u = (velocity, tension) = (v, t)
As I understand, there are two methods: Euler's and Lagrange's
In Euler's coordinates, "x", which are not moving in the space, the matrix A is

( v -1/d
-E v ), d is density, E is Young's modulus

In Lagrange's coordinates, "y", which are moving with points of body

A = ( 0 -1/d
-E 0 )
Here equation is the same, du/dt + A * du/dx = 0, (not du/dt + A * du/dy = 0 !), but
but the coordinate of the particle is changed: dx = v*dt

1) Is it correct?

I use a grid-characteristic method to solve it.
In Euler's method, I diagonalize A and transfer the values of two invariants along the corresponding characteristics.
In Lagrange's method, I do the same with Lagrange's matrix (and formulas for recalculation invariants from (v, t) is the same!) and after that I do dx = v*dt for each node of my mesh.
This methods are equal, to my mind.

2) Is it correct?

3) What's the exact solution? It's not like a "f(x - at) + g(x + at)", where a = sqrt(E/d), because of "dx = v*dt" in Lagrange's method and diagonal "v" in Euler's method!

I'm looking forward for any answers! :)

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