A How to Find Initial Conditions for a Fifth Order ODE in Visco-Elastic Models?

[Joa]

TL;DR Summary

I am struggling to find the initial conditions to a 5th order ODE describing visco-elastic material behavior

Hello,

I am trying to solve a differential equation corresponding to a visco-elastic material model consisting of 5 units of springs and dashpots connected in parallel as can be seen in the image below. I am able to come-up with a single fifth order ODE, however I am struggling to find the initial conditions for t=0. The differential equation is solved for epsilon(t) and is subject to a constant stress sigma_0. I am able to come-up with the first initial condition as can be seen in the image but I don't know how to find the remaining 4 conditions. Please help!

Joa

 

[Mark44]

Summary: I am struggling to find the initial conditions to a 5th order ODE describing visco-elastic material behavior

Hello,

I am trying to solve a differential equation corresponding to a visco-elastic material model consisting of 5 units of springs and dashpots connected in parallel as can be seen in the image below. I am able to come-up with a single fifth order ODE, however I am struggling to find the initial conditions for t=0. The differential equation is solved for epsilon(t) and is subject to a constant stress sigma_0. I am able to come-up with the first initial condition as can be seen in the image but I don't know how to find the remaining 4 conditions. Please help!

JoaView attachment 253342

I don't understand your question. There is a difference between solving a differential equation, and solving an initial value problem (which involves a differential equation and one or more initial conditions).

A simple example of a differential equation is y' - y = 0, for which the general solution is $y = Ae^t$. It is not possible to find the constant A nor is it possible to get a unique value for y(0), since no initial condition is given.

Now consider an initial value problem such as y' - y = 0, with y(0) = 2. Knowing the initial condition allows us to determine that A = 2, so the unique solution is $y = 2e^t$.

So in short, just being able to go from a differential equation to a general solution (or family of solutions) in no way let's us determine an initial condition.

 

[Joa]

I don't understand your question. There is a difference between solving a differential equation, and solving an initial value problem (which involves a differential equation and one or more initial conditions).

A simple example of a differential equation is y' - y = 0, for which the general solution is $y = Ae^t$. It is not possible to find the constant A nor is it possible to get a unique value for y(0), since no initial condition is given.

Now consider an initial value problem such as y' - y = 0, with y(0) = 2. Knowing the initial condition allows us to determine that A = 2, so the unique solution is $y = 2e^t$.

So in short, just being able to go from a differential equation to a general solution (or family of solutions) in no way let's us determine an initial condition.

Finding the general solution to the problem is not the issue here. I am able to formulate the general solution to the problem, but then i am left with 5 unknown constants which can be solved using the initial conditions. However formulating these initial conditions is the thing I am struggling with. So far I only found one initial conditions as can be seen in the image. Does this clarify my question?

 

[Mark44]

Does this clarify my question?

No. As I said in my previous post, being able to find the general solution to a differential equation doesn't enable one to find initial conditions.

Also, it's difficult for me to understand what you're doing. A typical problem involving a mass, spring, and damping dashpot is a second order differential equation. To find a unique solution you would need the initial position and initial velocity of the mass in this system.

Also, it's not clear to me what the symbols you're using represent. You have $\epsilon_1, \dot{\epsilon_1}$ etc., $\tau_1$ etc. $\sigma_1, \dot{\sigma_1}, \mu_1$, etc. and $E_1$ etc.
Presumably the $\mu$s are the masses attached to the springs, and possibly $\sigma_i$ is a position and $\dot{\sigma_i}$ is a velocity, but which of the symbols you're using is an acceleration?

 

[Joa]

No. As I said in my previous post, being able to find the general solution to a differential equation doesn't enable one to find initial conditions.

Also, it's difficult for me to understand what you're doing. A typical problem involving a mass, spring, and damping dashpot is a second order differential equation. To find a unique solution you would need the initial position and initial velocity of the mass in this system.

Also, it's not clear to me what the symbols you're using represent. You have $\epsilon_1, \dot{\epsilon_1}$ etc., $\tau_1$ etc. $\sigma_1, \dot{\sigma_1}, \mu_1$, etc. and $E_1$ etc.
Presumably the $\mu$s are the masses attached to the springs, and possibly $\sigma_i$ is a position and $\dot{\sigma_i}$ is a velocity, but which of the symbols you're using is an acceleration?

The following document provides some background to what I am trying to model. Specifically on page 20-22 a problem similar to mine is evaluated. There two Maxwell units in parallel are used. I am trying to use five units.
http://homepages.engineering.auckla...scoelasticity/10_Viscoelasticity_Complete.pdf

 

[BvU]

I don't see a fifth order ODE at all, but instead five first order ODEs -- assuming $\sigma$ is the input¹ and $\varepsilon$ is to be found². They can be added up to one first order ODE. Please enlighten me if I am wrong ? (And I think I am... )

1 I derive from

2 You found already

With your

(Gee, it's a drag to quote from a picture -- $\LaTeX$ is soooo much better... )

Anyway with that you have established ${\dot\epsilon}_0 = 0$ and also that $\dot\sigma_0 = 0$ -- I think.

Specifically on page 20-22 a problem similar to mine is evaluated

Strange: page 20 (302) is already further:10.3b Retardation and Relaxation spectra.

But 10.3.6 Generalized Models on page 300 has a picture like the one in your post #1:

with a warning that

​