[미적분학] Inverse of Jacobian determinant

미적분학을 배웠다면 Jacobian은 알고 있을 것이다.

하지만 얄팍한 지식으론 다음과 같은 문제를 만나면 난관에 봉착한다.

 

Problem.

 

Solve

$$ I=\int_0^\infty\int_0^\infty \frac{x^2+y^2}{1+(x^2-y^2)^2}e^{-2xy}dxdy$$

 

Hint : make the change of variables
$$u=x^2-y^2 \\ v=2xy$$

 

풀이야 간단하다. Jacobian determiant $J=J(u,v)$ 만 구하면 끝난다.

근데 막상 $\partial x/\partial u$를 구하려니 만만치가 않다.

 

바로 이때 유용하게 쓰이는 것이 Inverse of Jacobian determinant 이다.

 

Theorem.

 

 

Let $J$ be the jacobian determiant corresponding to the coordinate transmation from $x_i$ to $u_i$.
And let $J^{-1}$ be the jacobian determiant corresponding to the coordinate transmation from $u_i$ to $x_i$.
Then
$$J=\frac{1}{J^{-1}}$$

 

Proof.

 

 

Let $\mathbf{J}$ be the jacobian matrix corresponding to the coordinate transmation from $x_i$ to $u_i$.
And let $\mathbf{J^{-1}}$ be the jacobian determiant corresponding to the coordinate transmation from $u_i$ to $x_i$.

 

Then $$\mathbf{J}_{ij}=\frac{\partial x_i}{\partial u_j} \quad\text{and}\quad \mathbf{J^{-1}}_{ij}=\frac{\partial u_i}{\partial x_j}$$

 

Consider the matrix product $\mathbf{J}\mathbf{J^{-1}}$.

$$[\mathbf{J}\mathbf{J^{-1}}]_{ij}=\sum_k \frac{\partial x_i}{\partial u_k}\frac{\partial u_k}{\partial x_j}=\frac{\partial x_i}{\partial x_j}=\delta_{ij}$$

 

Thus $\mathbf{J}\mathbf{J^{-1}}=\mathbf{I}$.
And so $\det(\mathbf{J})\det(\mathbf{J^{-1}})=1$.

 

Therefore
$$\det(\mathbf{J})=\frac{1}{\det(\mathbf{J^{-1}})}.$$

 

따라서 위의 문제는 간단하게 $J=J(x,y)$를 구함으로써 풀 수 있다.

 

Solve

 

\begin{align*}J(x,y)&= \left|\begin{matrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{matrix}\right| = 4(x^2+y^2).\\ \\I&=\int_0^\infty \int_{-\infty}^{\infty}\frac{x^2+y^2}{1+(x^2-y^2)^2}e^{-2xy}|J(u,v)|dudv \\ \\ &=\int_0^\infty \int_{-\infty}^{\infty}\frac{x^2+y^2}{1+(x^2-y^2)^2}e^{-2xy}\frac{1}{|J(x,y)|}dudv \\ \\ &=\frac{1}{4}\int_0^\infty \int_{-\infty}^{\infty}\frac{1}{1+u^2}e^{-v}dudv \\ \\ &=\frac{\pi}{4}.\end{align*}