I've just started learning Fluids and have been given an assignment of which i want to make sure i'm right before i hand it in. If it's possible, can you fine ladies and gents of math stack exchange please have a tertiary glance over it and if you see any glaring errors, perhaps make me aware of it. specifically, if there's any gaps in knoweledge i would appreciate the help immensely
Thanks for taking the time to read this.
Consider The 2D Eulerian Velocity Field given by $\mathbf{u}(\mathbf{r},t) = \left(\frac{2x}{1+t},0\right)$
- Find the Position, $\mathbf{r}(t;\mathbf{a})$ of a particle released in the flow from $\mathbf{a} = (a,0)$ at $t=0$ (ie. the particle path)
now from my understanding $\mathbf{r}(t;\mathbf{a})$ is just saying the function r (location vector) with initial conditions a,
My solution: We know that pathlines are given by the following equation. $$\frac{d \mathbf{r}}{dt} = \mathbf{u}(\mathbf{r},t)~\implies$$ $$\left\{\begin{matrix} \frac{dx}{dt} &= u(x,y,z,t) &= \frac{2x}{1+t} \\ \frac{dy}{dt} &= v(x,y,z,t) &= 0 \end{matrix}\right.$$ $$\implies$$ $$\left\{\begin{matrix} x(t) &= \int u(x,y,z,t) dt &= 2x \ln (1+t) + c_1 \\ y(t) &= \int v(x,y,z,t) dt &= c_2 \end{matrix}\right.$$
plugging in ititial conditions gives
$$\left\{\begin{matrix} x(0) &= 2x \ln (1) + c_1 = a\\ y(0) &= c_2 = 0 \end{matrix}\right. \implies$$ $$\mathbf{r}(t) = \left(\begin{matrix} x(t) \\ y(t) \end{matrix}\right) = \left(\begin{matrix} 2x \ln (1+t) + a \\ 0 \end{matrix}\right)$$
- Find the Lagrangian Velocity, $\mathbf{v}(t;a),$ of the particle and it's (Lagrangian) acceleration
We know that the Lagrangian Velocity is $v(t) = \frac{d \mathbf{r}}{dt}$
This seems somewhat odd to me given that we used $\frac{d \mathbf{r}}{dt}$ to find r in the first place, in which case would we not simply just use the definition of u to quote v? this seems somewhat confusing to me, i'm sure the question is well defined but i'm finding it difficult to quite understand what i'm being asked here
in anycase. $$\mathbf{v}(t) = \left(\frac{2x}{1+t},0\right)$$
Because Lagrangian Velocity and formulation of Fluid flow has us moving within the flow i imagine that the definition of Lagrangian Acceleration is just $$\frac{d \mathbf{v}}{dt} = \left(\frac{-2x}{(1+t)^2},0\right)$$
- Confirm that the relationship $\mathbf{u}(\mathbf{r}(t;a),t) = \mathbf{v}(t;a)$ between the eulerian and lagrangian velocities hold
I'm interpreting this as making sure that u and v are equal to each other and they remain equal at the initial conditions. in which case
$$\mathbf{u}(\mathbf{r}(t;\mathbf{a}),t) = \mathbf{v}(t;a) = \left(\frac{2x}{1+t},0\right)$$
and
$$\mathbf{u}(\mathbf{r}(0;\mathbf{a}),0) = \mathbf{v}(0;a) = \left(\frac{2a}{1},0\right)$$
and so the properties hold.
- Compute the quantity
$$\frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u} \cdot \mathbf{\nabla})\mathbf{u} $$
evaluated at the point $\mathbf{r}(t;\mathbf{a})$ and confirm it matches the lagrangian acceleration
$$\frac{\partial u}{\partial t} = -\frac{2x}{(1+t)^2}$$ $$(\mathbf{u \cdot \nabla})\mathbf{u} = \left(\frac{\partial}{\partial x} \frac{2x}{1+t} + \frac{\partial}{\partial y}0\right)\left(\frac{2x}{1+t},0\right) = \frac{4x}{(1+t)^2}$$ $$\frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u \cdot \nabla})\mathbf{u} = \frac{2x}{(1+t)^2}$$
where have i gone wrong? (for instance i can see that my version of the Lagrangian acceleration doesn't match up to the Lagrangian Derivative...)
cheers for any help given, very much appreciated.