# Deformable mesh in 1D elastic body

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 November 23, 2013, 08:40 Deformable mesh in 1D elastic body #1 New Member   Alexander Kazakov Join Date: Nov 2013 Posts: 1 Rep Power: 0 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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