Jtdcftul

I am using TransformedField to convert a system of ODEs from Cartesian to polar coordinates:

TransformedField[

  "Cartesian" -> "Polar",

  {μ x1 - x2 -  σ x1 (x1^2 + x2^2), x1 + μ x2 - σ x2 (x1^2 + x2^2)}, 

  {x1, x2} -> {r, θ}

] // Simplify

and I get the result

{r μ - r^3 σ, r}

but I am pretty sure that the right answer should be

{r μ - r^3 σ, 1}

Where is the error?

Mathematica's answer is correct and consistent with your expectations, but you are not accounting for the basis of the vector field.

TransformedField transforms a vector field between two coordinate systems and bases. In this case, it is converting from $f(x,y)hat x+g(x,y)hat y$ to the same geometrical vector field expressed as $u(r,theta)hat r + v(r,theta) hat theta$. Mathematica's answer can therefore be interpreted as saying
$$left(μ x_1 - x_2 - σ x_1 (x_1^2 + x_2^2)right)hat x + left( x_1 + μ x_2 - σ x_2 (x_1^2 + x_2^2)right)hat y = left(r μ - r^3 σright)hat r + r hattheta$$

Notice that the expressions $r'$ and $theta'$ don't appear anyhwere. Those are dynamical quantities, not geometrical ones (unless working in the jet bundle, but let's not go there). Also notice the hats! As stated in the documentation, TransformedField assumes inputs are in an orthonormal basis, and returns outputs in the same basis. That will be important for later on.

Now, you are dealing with a differential equation, and based on your expected answer I'll assume what you have is a first-order system and you are transforming the associated vector field (AKA the "right-hand side"). Finding solutions means find the integral curves of the vector field. This gives as a nice relationship between the geometric variables the dynamical ones, except that this relationship is of necessity expressed in the so called coordinate basis, written $(r',theta') = a frac{partial}{partial r} + b frac{partial}{partial theta}$. So to get the answer expressed in your desired basis, we need the relationship between the coordinate and orthonormal basis vectors. As is covered in books on vector calculus (and elsewhere), the relationsip is $hat r = frac{partial}{partial r}$ and $hat theta = frac{1}{r}frac{partial}{partial theta}$. Substituting this into the answer Mathematica gave above, we get
$$left(r μ - r^3 σright)hat r + r hattheta = left(r μ - r^3 σright) frac{partial}{partial r} + (1) frac{partial}{partial theta},$$
which is the answer you expected.

We can test the built-in TransformedField by defining our own functions.

From $x',y'$ to $r',theta'$, we derive:
$$
r' = left(sqrt{x^2 +y^2} right)'
= frac{(x^2 +y^2)'}{2
sqrt{x^2 +y^2}}=frac{xx' +yy'}{r}
$$
and
$$
theta' = left(arctan frac{y}{x} right)'
= frac{(y/x)'}{1+(y/x)^2} = frac{y' x -x' y}{r^2}.
$$

First, we define

rdot[x1_, x2_] := (x1 (μ x1 - x2 - σ x1 (x1^2 + x2^2)) + x2 (x1 + μ x2 - σ x2 (x1^2 + x2^2)))/r

We now make the substitution and simplify

rdot[r Cos[t], r Sin[t]] // FullSimplify

This yields (matches Mathematica)

$$r' = mu r-r^3 sigma$$

We now do the same for the other

thetadot[x1_,x2_]:=(x1 (x1+μ x2-σ x2 (x1^2+x2^2)) - x2(μ x1-x2-σ x1 (x1^2+x2^2)))/r^2

We now make the substitution and simplify

thetadot[r Cos[t], r Sin[t]] // FullSimplify

This yields (does not match Mathematica - it is as though they are forgetting to divide by $r$)

$$theta'= 1$$

I have asked this question before on this site in two different ways and have never gotten an answer that resolves the matter, but that could just be my denseness!