Finite difference method for fourth order PDE

BoxQwerty · February 14, 2021, 13:26

How can I solve this initial boundary problem?
WhatsApp Image 2021-02-14 at 20.58.08.jpeg

Are my approximations right?
WhatsApp Image 2021-02-14 at 23.18.58.jpeg

it's the nonlinear mathematical model of the dynamics of a rotating drill string compressed at both ends by the longitudinal force N(x3, t)

u2 - displacement of the center of bending of the cross-section of the bar along the x2 axe due to bending
E = 2.1e11 # Pa - Young's modulus
rho = 7800 # kg/м3 - density
v = 0.28 # Poisson's ratio
d1 = 0.12 # м - drill string inner diameter
d2 = 0.2 # м - external diameter
A = 2.01e-2 # Column cross-sectional area
Ix1 = PI / 64. * (d2 ** 4 - d1 ** 4) # Ring section moment of inertia
Omega = 0.33 # drill string angular velocity
C1 = 0.01 # constant determining the rate of deviation of the drill string axis from the initial position in the Ox2x3 plane at the initial moment
time.

Thank you in advance

FMDenaro · February 14, 2021, 15:23

I don't know the physics described by this PDE, the problem is not only in a consistent discretization but also on the mathematical character of the equation. A second order in time PDE could be hyperbolic but I see also fourth order spatial terms. Central discretization in time and space is consistent but I don't know if it is also stable.
Then, also the BCs for this problem should be analysed.
Have you some specific reference?

BoxQwerty · February 15, 2021, 06:48

Material I have is all in Russian.

In short, it's the nonlinear mathematical model of the dynamics of a rotating drill string compressed at both ends by the longitudinal force N(x3, t)

u2 - displacement of the center of bending of the cross-section of the bar along the x2 axe due to bending
E = 2.1e11 # Pa - Young's modulus
rho = 7800 # kg/м3 - density
v = 0.28 # Poisson's ratio
d1 = 0.12 # м - drill string inner diameter
d2 = 0.2 # м - external diameter
A = 2.01e-2 # Column cross-sectional area
Ix1 = PI / 64. * (d2 ** 4 - d1 ** 4) # Ring section moment of inertia
Omega = 0.33 # drill string angular velocity
C1 = 0.01 # constant determining the rate of deviation of the drill string axis from the initial position in the Ox2x3 plane at the initial moment
time.

BoxQwerty · February 15, 2021, 06:49

Material I have is all in Russian.
In short it's the nonlinear mathematical model of the dynamics of a rotating drill string compressed at both ends by the longitudinal force N(x3, t)
u2 - displacement of the center of bending of the cross-section of the bar along the x2 axe due to bending
E = 2.1e11 # Pa - Young's modulus
rho = 7800 # kg/м3 - density
v = 0.28 # Poisson's ratio
d1 = 0.12 # м - drill string inner diameter
d2 = 0.2 # м - external diameter
A = 2.01e-2 # Column cross-sectional area
Ix1 = PI / 64. * (d2 ** 4 - d1 ** 4) # Ring section moment of inertia
Omega = 0.33 # drill string angular velocity
C1 = 0.01 # constant determining the rate of deviation of the drill string axis from the initial position in the Ox2x3 plane at the initial moment
time.

February 14, 2021, 13:26	Finite difference method for fourth order PDE	#1
BoxQwerty New Member Box Join Date: Feb 2021 Posts: 4 Rep Power: 6	How can I solve this initial boundary problem? WhatsApp Image 2021-02-14 at 20.58.08.jpeg Are my approximations right? WhatsApp Image 2021-02-14 at 23.18.58.jpeg it's the nonlinear mathematical model of the dynamics of a rotating drill string compressed at both ends by the longitudinal force N(x3, t) u2 - displacement of the center of bending of the cross-section of the bar along the x2 axe due to bending E = 2.1e11 # Pa - Young's modulus rho = 7800 # kg/м3 - density v = 0.28 # Poisson's ratio d1 = 0.12 # м - drill string inner diameter d2 = 0.2 # м - external diameter A = 2.01e-2 # Column cross-sectional area Ix1 = PI / 64. * (d2 4 - d1 4) # Ring section moment of inertia Omega = 0.33 # drill string angular velocity C1 = 0.01 # constant determining the rate of deviation of the drill string axis from the initial position in the Ox2x3 plane at the initial moment time. Thank you in advance Last edited by BoxQwerty; February 16, 2021 at 00:37.

February 14, 2021, 15:23		#2
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,942 Rep Power: 73	I don't know the physics described by this PDE, the problem is not only in a consistent discretization but also on the mathematical character of the equation. A second order in time PDE could be hyperbolic but I see also fourth order spatial terms. Central discretization in time and space is consistent but I don't know if it is also stable. Then, also the BCs for this problem should be analysed. Have you some specific reference? aero_head and BoxQwerty like this.

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February 15, 2021, 06:48		#3
BoxQwerty New Member Box Join Date: Feb 2021 Posts: 4 Rep Power: 6	Material I have is all in Russian. In short, it's the nonlinear mathematical model of the dynamics of a rotating drill string compressed at both ends by the longitudinal force N(x3, t) u2 - displacement of the center of bending of the cross-section of the bar along the x2 axe due to bending E = 2.1e11 # Pa - Young's modulus rho = 7800 # kg/м3 - density v = 0.28 # Poisson's ratio d1 = 0.12 # м - drill string inner diameter d2 = 0.2 # м - external diameter A = 2.01e-2 # Column cross-sectional area Ix1 = PI / 64. * (d2 4 - d1 4) # Ring section moment of inertia Omega = 0.33 # drill string angular velocity C1 = 0.01 # constant determining the rate of deviation of the drill string axis from the initial position in the Ox2x3 plane at the initial moment time.

February 15, 2021, 06:49		#4
BoxQwerty New Member Box Join Date: Feb 2021 Posts: 4 Rep Power: 6	Material I have is all in Russian. In short it's the nonlinear mathematical model of the dynamics of a rotating drill string compressed at both ends by the longitudinal force N(x3, t) u2 - displacement of the center of bending of the cross-section of the bar along the x2 axe due to bending E = 2.1e11 # Pa - Young's modulus rho = 7800 # kg/м3 - density v = 0.28 # Poisson's ratio d1 = 0.12 # м - drill string inner diameter d2 = 0.2 # м - external diameter A = 2.01e-2 # Column cross-sectional area Ix1 = PI / 64. * (d2 4 - d1 4) # Ring section moment of inertia Omega = 0.33 # drill string angular velocity C1 = 0.01 # constant determining the rate of deviation of the drill string axis from the initial position in the Ox2x3 plane at the initial moment time.