Let $F: \mathbb{R}^n \to \mathbb{R}$ be an integrable function, and $\phi: \mathbb{R}^n \to \mathbb{R}^n$ be a smooth function with compact support (i.e., vanishing outside a certain ball).
Justify that the transformation $$\Phi_t: \mathbb{R}^n \to \mathbb{R}^n \ , \ \Phi_t(x) = x + t \cdot \phi(x)$$ is a diffeomorphism for $t$ in some interval $(-t_0, t_0)$. Determine the derivative of $\det D\Phi_t(x)$ with respect to $t$ at $t = 0$.
Show that the derivative of the function $$f: \mathbb{R} \to \mathbb{R} \ , \ f(t) := \int_{\mathbb{R}^n} F(x + t \cdot \phi(x)) dx$$ at $t = 0$ is $$f'(0) = -\int_{R^n} F(x) \cdot \left(\frac{\partial \phi_1}{\partial x_1}(x) + \cdots + \frac{\partial \phi_n}{\partial x_n}(x)\right) dx$$.
Hint: You can assume the formula $\det(I + tA) = 1 + t \cdot \operatorname{tr} A + o(t)$ is known.
SOLUTION:
(1)
To prove that $\Phi_t$ is a diffeomorphism for $t$ in some interval $(-t_0, t_0)$, we need to show that:
- $\Phi_t$ is differentiable and invertible,
- inverse of $\Phi_t$ is also differentiable.
Differentiability of $\Phi_t$:
$\Phi_t(x) = x + t \cdot \phi(x)$ is differentiable because it is the sum of differentiable functions ($x$ and $t \cdot \phi(x)$).
Invertibility of $\Phi_t$:
We know that $\Phi_0(x) = x$.
To show that $\Phi_t$ is invertible for $t$ in some interval $(-t_0, t_0)$, it is enough to show that $\det D\Phi_t(x) \neq 0$ for $t$ in this interval.
We have:
$$ D\Phi_t(x) = I + t \cdot D\phi(x), $$
where $I$ is the identity matrix. Now let's compute the derivative of $\det D\Phi_t(x)$ with respect to $t$ at $t = 0$:
$$ \frac{d}{dt} \det D\Phi_t(x) \Big|_{t=0} = \frac{d}{dt} \det (I + t \cdot D\phi(x)) \Big|_{t=0} $$
Using Jacobi's theorem on determinant derivatives, we get:
$$ \frac{d}{dt} \det (I + t \cdot D\phi(x)) \Big|_{t=0} = \text{tr}(D\phi(x)), $$
where $\text{tr}$ denotes the trace of the matrix.
Since $\text{tr}(D\phi(x))$ does not depend on $t$, we can choose a sufficiently small $t_0 > 0$ such that $\det D\Phi_t(x) \neq 0$ for $t \in (-t_0, t_0)$.
Differentiability of the inverse $\Phi_t^{-1}$:
Since $\Phi_t$ is differentiable and invertible, its inverse $\Phi_t^{-1}$ is also differentiable.
Therefore, $\Phi_t$ is a diffeomorphism for $t$ in some interval $(-t_0, t_0)$.
(2)
To show that $\Phi_t$ is a flow, we need to prove two properties:
- $\Phi_0 = \text{id}$, where $\text{id}$ is the identity.
- $\Phi_{s+t} = \Phi_s \circ \Phi_t$ for any $s, t \in \mathbb{R}$.
Property (1) is obvious, since $\Phi_0(x) = x + 0 \cdot \phi(x) = x$.
Property (2) can be proven using the differentiability of $\Phi_t$ and the fact that $\Phi_t$ is a diffeomorphism for $t$ in some interval $(-t_0, t_0)$.
Is my thinking correct?