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 !!

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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
    \cdots
    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}
    \cdots
    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)

    Similarly,

    \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} .