Substitute variable in differential equation

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If I am right, you want to transform the differential expression $\frac{d^2\phi}{dr^2}+\frac{2}{r}\frac{d\phi}{dr},$ where $\phi(r)=\frac{m(r)}{r},$ by changing the independent variable to $V$ . The relation between $V$ and $r$ is $V=\frac{4\pi}{3} r^3.$ We can think of $m(r)$ as the result of composing two functions: one that yields $V$ in terms of $r$ and then one that yields $m$ in terms of $V$ . In other words, an abusing a bit of the notation, $m(r)=m(V(r)).$

We can then transform the given differential expression as follows:

var("r,V")
vol = 4*pi*r^3/3
m(V) = function("m")(V)    # m as function of V
M = m(V.subs(V=vol))       # this is m(r)=m(V(r))
phi = M/r
divG = (diff(phi,r,2) + 2/r*diff(phi,r))
divG = divG.subs(vol==V, r^3==3*V/(4*pi)).full_simplify()
show(divG)

In a Jupyter notebook, one gets

$12 \pi V \frac{\partial^{2}}{(\partial V)^{2}}m\left(V\right) + 8 \pi \frac{\partial}{\partial V}m\left(V\right)$

that is, in a more proper mathematical notation,

$4\pi\left( 3V\frac{d^2m}{dV^2}(V)+2\frac{dm}{dV}(V)\right)$

I hope this has some meaning for you.

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Comments

Thanks a lot for showing me. - I got the first term of the solution by hand, but missed the second. The result has a meaning, but what it is I have to figure out, it seems the bracket contains sort of a "density operator".

Emil Widmann ( 4 years ago )

Just 2 more questions: How do I insert latex code here in this forum? How are the symbols D0 and D0,0 defined (result from show(divG), divG in the first version as function from r)?

Emil Widmann ( 4 years ago )

1) This forum uses MathJax to render LaTeX code. You just have to write it as you would do in a TeX file. For example, if you write

This is inline math: $V(r)=4\pi r^3/3$. This is display math:
$$V(r)=\frac{4\pi}{3}r^3.$$

you get, as expected,

This is inline math: $V(r)=4\pi r^3/3$ . This is display math: $V(r)=\frac{4\pi}{3}r^3.$

Juanjo ( 4 years ago )

2) If you insert print(divG), you will see that those symbols come from D[0,0](m) and D[0](m). For a multivariate function $f$ , D[i,j,k...](f) means the partial derivative of $f$ respect to the $i$ -th variable, then the $j$ -th variable, then the $k$ -th variable and so on. For example, for $f(x,y,z)$ , D[2,0,1,2](f) corresponds to $\frac{\partial^4f}{\partial z\partial x\partial y\partial z},$ that is, $\frac{\partial^4f}{\partial x\partial y\partial z^2}$ if $f\in C^4$ . Take into account that, in Python, indices start counting in $0$ .

For a univariate function, D[0](f), D[0,0](f), etc. are just their successive derivatives.

Juanjo ( 4 years ago )

Vielen Dank! I would upvote 2 times more if I could ...

Emil Widmann ( 4 years ago )

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