General pathline of a particle x at point epsilon at time ta

[hqjb]

Homework Statement

Show that the path line of a particle at point x currently, and point ξ at time τ is given by

ξ(τ) = x + (τ-t)Lx

Homework Equations

Pathline is solution to
dx/dt = u
x(t)|t=τ = X

L is the velocity gradient and is a 2nd order tensor L_ij = du_i/dx_j

The Attempt at a Solution

I am not really sure how to start, any hints or leads on how to begin proving this?
I know how to obtain pathlines if the velocity field is given in terms of x,y,z,t (via integration) but how do I integrate the equations to obtain a general equation with a velocity gradient?

Thanks

 

[mark726]

for posting this question! Here is a possible approach to proving this equation:

1. Start by defining the pathline function ξ(τ) as the solution to the differential equation dx/dt = u, with the initial condition x(τ) = ξ. This means that at any given time τ, the particle's position x is given by ξ.

2. From the given information, we know that the velocity gradient L is a 2nd order tensor, which means it can be represented by a matrix. Let's denote this matrix as [L].

3. Using this notation, we can rewrite the differential equation as dx/dt = [L]x. This is a linear system of differential equations, which can be solved using the matrix exponential function.

4. The solution to this linear system is given by x(t) = e[t[L]]x(0), where x(0) is the initial position of the particle and e[t[L]] is the matrix exponential of [L].

5. Now, let's plug in the initial condition x(τ) = ξ. This gives us x(t) = e[t[L]]ξ.

6. Finally, we can rewrite this as ξ(t) = e[t[L]]ξ = ξ + (t-τ)Lξ, which is the same as the given equation ξ(τ) = x + (τ-t)Lx.

I hope this helps! Let me know if you have any further questions or need clarification.