- #1
stanley.st
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Hello!
I recently tried to prove following theorem: Let \phi:B\to\mathbb{R}^2 be a diffeomorphism (regular, injective mapping). Then
\int_{\phi(B)}f(\mathbf{x})\,\mathrm{d}x=\int_{B}f(\phi(\mathbf{t}))\left|{\mathrm{det}}\mathbf{J}_{\phi}\right|\mathrm{d}t
With following I can't proof this theorem. Look, I start with integral sums
(*)\quad\sum_{i=1}^{n}f(x_i,y_j)(x_{i+1}-x_i)(y_{j+1}-y_{j})
According to the transformation phi, we have
x_i=\phi_x(r(x_i,y_j),t(x_i,y_j))y_i=\phi_y(r(x_i,y_j),t(x_i,y_j))
We can imagine phi as a polar coordinate system transformation, so I use notation with variables r,t. Then we have using Taylor formula
\begin{array}{ll}x_{i+1}-x_{i}&=\phi_x(r(x_{i+1},y_j),t(x_{i+1},y_j))-\phi_x(r(x_{i},y_j),t(x_i,y_j))\\&=\frac{\partial \phi_x}{\partial r}(\xi,\eta)(r(x_{i+1},y_j)-r(x_{i},y_j))+\frac{\partial \phi_x}{\partial r}(\xi,\eta)(t(x_{i+1},y_{j})-t(x_{i},y_j))\\&=\frac{\partial \phi_x}{\partial r}\delta r_{i+1,j}+\frac{\partial \phi_x}{\partial r}\delta t_{i+1,j}\quad(\textrm{short form})\end{array}
In the same way I can derive
y_{j+1}-y_j=\frac{\partial \phi_y}{\partial r}\delta r_{i,j+1}+\frac{\partial \phi_y}{\partial r}\delta t_{i,j+1}
If I put this into (*) I get
\sum_{i,j=1}^{n}f(\phi(r_{ij},t_{ij}))\left(\frac{\partial \phi_x}{\partial r}\delta r_{i+1,j}+\frac{\partial \phi_x}{\partial r}\delta t_{i+1,j}\right)\left(\frac{\partial \phi_y}{\partial r}\delta r_{i,j+1}+\frac{\partial \phi_y}{\partial r}\delta t_{i,j+1}\right)
But this is different than I expected. I expected it in the form like
\sum_{i,j=1}^{n}f(\phi(r_{ij},t_{ij}))\left(\frac{\partial \phi_x}{\partial r}\frac{\partial \phi_y}{\partial t}-\frac{\partial \phi_y}{\partial t}\frac{\partial \phi_x}{\partial r}\right)\delta r\delta t
Do I something wrong? Thanks a lot..
I recently tried to prove following theorem: Let \phi:B\to\mathbb{R}^2 be a diffeomorphism (regular, injective mapping). Then
\int_{\phi(B)}f(\mathbf{x})\,\mathrm{d}x=\int_{B}f(\phi(\mathbf{t}))\left|{\mathrm{det}}\mathbf{J}_{\phi}\right|\mathrm{d}t
With following I can't proof this theorem. Look, I start with integral sums
(*)\quad\sum_{i=1}^{n}f(x_i,y_j)(x_{i+1}-x_i)(y_{j+1}-y_{j})
According to the transformation phi, we have
x_i=\phi_x(r(x_i,y_j),t(x_i,y_j))y_i=\phi_y(r(x_i,y_j),t(x_i,y_j))
We can imagine phi as a polar coordinate system transformation, so I use notation with variables r,t. Then we have using Taylor formula
\begin{array}{ll}x_{i+1}-x_{i}&=\phi_x(r(x_{i+1},y_j),t(x_{i+1},y_j))-\phi_x(r(x_{i},y_j),t(x_i,y_j))\\&=\frac{\partial \phi_x}{\partial r}(\xi,\eta)(r(x_{i+1},y_j)-r(x_{i},y_j))+\frac{\partial \phi_x}{\partial r}(\xi,\eta)(t(x_{i+1},y_{j})-t(x_{i},y_j))\\&=\frac{\partial \phi_x}{\partial r}\delta r_{i+1,j}+\frac{\partial \phi_x}{\partial r}\delta t_{i+1,j}\quad(\textrm{short form})\end{array}
In the same way I can derive
y_{j+1}-y_j=\frac{\partial \phi_y}{\partial r}\delta r_{i,j+1}+\frac{\partial \phi_y}{\partial r}\delta t_{i,j+1}
If I put this into (*) I get
\sum_{i,j=1}^{n}f(\phi(r_{ij},t_{ij}))\left(\frac{\partial \phi_x}{\partial r}\delta r_{i+1,j}+\frac{\partial \phi_x}{\partial r}\delta t_{i+1,j}\right)\left(\frac{\partial \phi_y}{\partial r}\delta r_{i,j+1}+\frac{\partial \phi_y}{\partial r}\delta t_{i,j+1}\right)
But this is different than I expected. I expected it in the form like
\sum_{i,j=1}^{n}f(\phi(r_{ij},t_{ij}))\left(\frac{\partial \phi_x}{\partial r}\frac{\partial \phi_y}{\partial t}-\frac{\partial \phi_y}{\partial t}\frac{\partial \phi_x}{\partial r}\right)\delta r\delta t
Do I something wrong? Thanks a lot..
