Begin a new journey !!

Today, I went back to blogging and I decided to recreate my personal website to begin a new journey. My upcoming posts will be fascinating topics in Math, Physics and some interesting stories of daily life around us. Thank for following and supporting me !!

Change of Variable Formula

We will start the series of my posts with pleasant topics. In our earlier study of the integral in the one-dimensional case, we have already used an important formula for change of variable to compute some complicated integral. Such formula has the form

    \displaystyle \int_{\alpha(a)}^{\alpha(b)}f \mathrm{d}\alpha =\int_{a}^{b}f(\alpha(x))\alpha'\mathrm{d}x
    In this post, I would like to provide readers with a similar formula in the general case and its proof. The main result we will obtain is

    Theorem 1. If \varphi : D_t \rightarrow D_x is a diffeomorphism of a bounded open set D_t \subset \mathbb{R}^n onto a set D_x = \varphi(D_t) \subset \mathbb{R}^n, f is a function integrable over D_x and is supported in D_x , then f \circ \varphi |\mathrm{det} \varphi'| is integrable over D_t. The following formula holds

    \displaystyle \int_{D_x=\varphi(D_t)} f \mathrm{d}x= \int_{D_t}f \circ \varphi |\mathrm{det}\varphi'| \mathrm{d}t (1)

    We now divide the proof of this result into a few steps

    Step 1: We now prove this result in the one-dimensional case. Firstly, we proceed with the proof for the case of D_x, D_t are closed intervals in \mathbb{R}^1.

    Indeed, let P_x=\left\{x_0<x_1<...<t_n\right\} of D_t through the mapping \varphi^{-1}. We have \mathrm{diam}(P_x) \to 0 \Leftrightarrow \mathrm{diam}(P_t) \to 0 from the uniform continuity of \varphi, \varphi^{-1}. Since Lagrange ‘s Theorem, |x_i-x_{i-1}|=|\varphi(t_i)-\varphi(t_{i-1}|=|\varphi'(\tau_i)|.|(t_i-t_{i-1}|.
    We now write the Riemann sums for partitions P_x,P_t with chosen points \xi_i=\varphi(\tau_i)

    \displaystyle \sum_{i}f(\xi_i)|x_i-x_{i-1}|= \sum_{i} f(\varphi(\tau_i))|\varphi'(\tau_i)||t_i-t_{i-1}|

    Send \mathrm{diam}(P_x),  \mathrm{diam}(P_t) \to 0 , we obtain the formula (1).

    Next, We point out that (1) holds for D_x, D_t are arbitrary bounded open sets in \mathbb{R}^1. Let K= \mathrm{supp}f \in D_x, It follows the boundedness and closedness of K that K is compact. Then, K can be covered by a finite collection of closed intervals I_i in D_x such that two of which have no common interior points. Therefore,

    \displaystyle \int_{D_x} f \mathrm{d}x= \int_{K} f \mathrm{d}x = \sum_{i} \int_{I_i} f \mathrm{d}x = \sum_{i} \int_{\varphi^{-1}(I_i)} f  \circ \varphi |det \varphi'| \mathrm{d}t =\int_{\varphi^{-1}(K)} f  \circ \varphi |det \varphi'| \mathrm{d}t=\int_{D_t} f  \circ \varphi |det \varphi'| \mathrm{d}t

    Step 2: Now we prove (1) for the class of diffeomorphisms have the relatively “simple” form, that is element diffeomorphism. We recall the following definition.

    Definition. The diffeomorphism \varphi: D_t \rightarrow D_x is elementary if its coordinate representation has the form

    x^1=\varphi^1(t^1,t^2,\dots , t^n)=t^1
    x^2=\varphi^2(t^1,t^2,\dots , t^n)=t^2
    x^{k-1}=\varphi^{k-1}(t^1,t^2,\dots , t^n)=t^{k-1}
    x^{k+1}=\varphi^{k+1}(t^1,t^2,\dots , t^n)=t^{k+1}
    x^{n}=\varphi^n(t^1,t^2,\dots , t^n)=t^{n}

    We need to prove (1) hold for \varphi is elementary.

    Indeed, assume without loss of generality that \varphi changes only the nth coordinate. Denote

    D_{x^n}(\tilde{x_0}):= \left\{(\tilde{x},x^n) \in D_x | \tilde{x}=\tilde{x_0} \in \mathbb{R}^{n-1} \right\}
    D_{t^n}(\tilde{t_0}):= \left\{(\tilde{t},t^n) \in D_t | \tilde{t}=\tilde{t_0} \in \mathbb{R}^{n-1} \right\}
    Let D_x \subset I_{\tilde{x}} \times I_{x^n} with I_{\tilde{x}} is a closed (n-1)-dimensional interval and I_{x^n} is a closed interval of nth coordinate.
    Respectively, we can choose an interval I_{\tilde{t}} \times I_{t^n} contains D_t, taking account of the coincidence of first (n-1) coordinates of D_t,D_x we can choose I_{\tilde{t}} \equiv I_{\tilde{x}}. By Fubini ‘s theorem,

    \displaystyle \int_{D_x}f \mathrm{d}x=\int_{ I_{\tilde{x}} \times I_{x^n}}f \chi_{D_x}\mathrm{d}x=\int_{I_{\tilde{x}}}\mathrm{d}\tilde{x} \int_{I_{x^n}} f\chi_{D_{x^n}(\tilde{x})}\mathrm{d}x^n=\int_{I_{\tilde{x}}}\mathrm{d}\tilde{x} \int_{D_{x^n}(\tilde{x})} f\mathrm{d}x^n

    \displaystyle = \int_{I_{\tilde{t}}}\mathrm{d}\tilde{t} \int_{D_{t^n}(\tilde{t})} f(\tilde{t},\varphi^n(\tilde{t},t^n))|\partial_{t^n}\varphi^n|\mathrm{d}t^n=\int_{I_t}(f \circ \varphi|\mathrm{det} \varphi'| \chi_{D_t})\mathrm{d}t

    \displaystyle =\int_{D_t}(f \circ \varphi|\mathrm{det} \varphi'|)\mathrm{d}t

    Where \chi_A is indicator function of the set A . In this computation, we have already used the fact \partial_{t^n} \varphi^n = \det( \varphi').

    Step 3: Now we admit without proof the following result about the local resolution of the diffeomorphism. The proof will be provided in my later post.

    Lemma 1. If g: G_{\mbox{open}} \subset \mathbb{R}^n \rightarrow \mathbb{R}^n is a diffeomorphism, then for any x_0 \in G there is a neighborhood of the point in which the representation g=f_1 \circ f_2 \cdots \circ f_m holds, where f_1,f_2, \dots, f_m are elementary diffeomorphisms.

    We need another result as a bridge to use the lemma 1, the result states as follows:
    Lemma 2. If \psi: D_{\tau} \rightarrow D_t, \varphi: D_t \rightarrow D_x are diffeomorphisms for each of which formula (1) holds, then it holds also for the composition \psi \circ \varphi

    Because the proof of Lemma 2 is quite simple, so it is left as an exercise for readers.

    Complete the proof. In the final step, using results that we got above, we now provide a complete proof for the general case.

    For each point of p of the compact set K_t = \mbox{supp}((f \circ \varphi) |\det( \varphi')|) \subset D_t, there exist a ball B(p,\delta_p) \subset D_t in which \varphi decomposes into a composition of elementary diffeomorphisms. Since the compactness of K_t we can choose a finite covering B(p_1, \delta_1), \dots , B(p_k,\delta_k) of K_t. Let \delta = \frac{1}{2} \min\{\delta_1, \dots, \delta_2 \}. Then any set whose diameter is smaller than \delta and its closure intersects K_t must be contained in at least one of the neighborhoods B(p_1,\delta_1), B(p_2,\delta_2), \dots, B(p_k,\delta_k).

    Let \{I_i\} is a collection of disjoint closed interval cover D_t such that \mbox{diam}(I_i) < \delta. Then intervals \{I_j\} that have nonempty intersection with K_t are contained in some U_i=B(p_i,\delta_i) \subset D_t and the integral over I_i is equal to 0 iff it has an empty intersection with K_t . Then,

    \displaystyle \int_{D_t} (f \circ \varphi) |\det(\varphi')| \mathrm{d}t = \sum_{j} \int_{I_j}(f \circ \varphi) |\det(\varphi')| \mathrm{d}t (2)


    \displaystyle \int_{D_x} f(x) \mathrm{d}x = \sum_{j} \int_{\varphi(I_j)} f(x) \mathrm{d}x (3)

    Finally, Since I_j \subset U_i, by Lemma 1, 2 and Step 2, we obtain that

    \int_{\varphi(I_j)} f(x) \mathrm{d}x = \int_{I_j} (f \circ \varphi)|\det(\varphi')| \mathrm{d}t (4)

    Combining (2),(3), (4) , we get a complete proof for Theorem 1.

    A straightforward corollary that we have often used is

    Corollary . Let g: A \subset \mathbb{R}^n \rightarrow \mathbb{R}^n an injective, continuously differentiable such that the set \{ x \in A| \det g'(x) = 0 \} has the measure zero. Then formula (1) holds for all integrable function f: g(A) \rightarrow \mathbb{R} .