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Consider material traveling in a two-dimensional space with its velocity at any given location in space (capital letters correspond to Eulerian, lower case to Lagrangian) given by $\overrightarrow{V}(\overrightarrow{X},t) =\begin{pmatrix}X^1+t\\(X^2)^2 \end{pmatrix}$. (Vectors are written using Cartesian coordinates,$ \overrightarrow{X}= \overrightarrow{R}$,and $\overrightarrow{x}=\overrightarrow{r}$.)

Find the position of a parcel/particle that starts at the location$\overrightarrow{x}=\begin{pmatrix} x^1\\x^2\end{pmatrix}.$

I solved the system of ordinary differential equations $\frac{d}{dt}(X^1) =X^1+t; \frac{d}{dt}(X^2) = (X^2)^2$, but I'm having trouble using the initial conditions to plug stuff in? I feel like I need to find $\overrightarrow{x}(\overrightarrow{X},t)$ first, but I also have no clue how to do that. Do I need to convert from Eulerian to Lagrangian and then plug in the initial conditions? Given what I have, I'm not really sure how to do that? Thanks for any help!

Here is what I have:

$X^1=Ke^t-t-1$

$X^2=-\frac{1}{t+C}$

sadsloth_96
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  • Mathematics is case-sensitive. Is your $X^1=x^1?$ If so, please edit your post so they're all lower-case (the more usual case for position variables). – Adrian Keister Jan 31 '19 at 20:50
  • Nope! As sated above in parenthesis, the capital ones are Eulerian coordinates while the lowercase are Lagrangian. – sadsloth_96 Jan 31 '19 at 20:51

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I think I figured it out!

Since we are given that the initial coordinates are $\overrightarrow{x}$, we know that $X^1(0)=x^1$ and $X^2(0)=x^2$. So, we can solve for the constants in terms of $x^1$ and $x^2$. So, the end product looks like this:

$X^1=(x^1+1)e^t-t+1$

$X^2=\frac{-1}{t-\frac{1}{x^2}}$

(Note that $x^1, x^2, X^1, X^2$ are not powers; they are like x and y)

Then, finding $\overrightarrow{x}(\overrightarrow{X},t)$ is pretty simple!

Long story short, I made things way too hard.

sadsloth_96
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