広島大学 先進理工系科学研究科 情報科学プログラム 2022年1月実施 専門科目I 問題2
Author
samparker, 祭音Myyura
Description
(1) 関数 \(z = z(u, v)\) は、実変数 \((u, v) \in \mathbb{R}^2\) に関して、\(C^2\) 級であると仮定する。 また、\((x, y) \in \mathbb{R}^2\) に対して、写像 \((u, v) = (x + y, x - y)\) を定義する。
\[
z_{xx} = \frac{\partial^2 z}{\partial x^2}, \quad z_{xy} = \frac{\partial^2 z}{\partial x \partial y}, \quad z_{yy} = \frac{\partial^2 z}{\partial y^2}
\]
を
\[
z_{uu} = \frac{\partial^2 z}{\partial u^2}, \quad z_{uv} = \frac{\partial^2 z}{\partial u \partial v}, \quad z_{vv} = \frac{\partial^2 z}{\partial v^2}
\]
を用いて表せ。
(2) 点 \(O(0, 0)\) 以外で定義された関数
\[
z = (x + y) \ln\{2(x^2 + y^2)\}
\]
の全ての極値およびそのときの \((x, y)\) を求めよ。
(1) Suppose that the function \(z = z(u, v)\) is of class \(C^2\) in real variables \((u, v) \in \mathbb{R}^2\). Define the mapping by \((u, v) = (x + y, x - y)\) for \((x, y) \in \mathbb{R}^2\). Express
\[
z_{xx} = \frac{\partial^2 z}{\partial x^2}, \quad z_{xy} = \frac{\partial^2 z}{\partial x \partial y}, \quad z_{yy} = \frac{\partial^2 z}{\partial y^2}
\]
in terms of
\[
z_{uu} = \frac{\partial^2 z}{\partial u^2}, \quad z_{uv} = \frac{\partial^2 z}{\partial u \partial v}, \quad z_{vv} = \frac{\partial^2 z}{\partial v^2}.
\]
(2) Find all local extrema and their extremum points of the function
\[
z = (x + y) \ln\{2(x^2 + y^2)\}
\]
defined for \((x,y) \neq (0,0)\).
Kai
(1)
\[
\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial x} = \frac{\partial z}{\partial u} + \frac{\partial z}{\partial v}
\]
\[
\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial y} = \frac{\partial z}{\partial u} - \frac{\partial z}{\partial v}
\]
\[
z_{xx} = \frac{\partial^2 z}{\partial x^2} = \frac{\partial^2 z}{\partial u^2} + \frac{\partial^2 z}{\partial u \partial v} + \frac{\partial^2 z}{\partial v \partial u} + \frac{\partial^2 z}{\partial v^2} = \frac{\partial^2 z}{\partial u^2} + 2\frac{\partial^2 z}{\partial u \partial v} + \frac{\partial^2 z}{\partial v^2}
\]
Similarly,
\[
z_{yy} = \frac{\partial^2 z}{\partial u^2} - 2\frac{\partial^2 z}{\partial u \partial v} + \frac{\partial^2 z}{\partial v^2}
\]
\[
z_{xy} = \frac{\partial^2 z}{\partial u^2} - \frac{\partial^2 z}{\partial v^2}
\]
(2)
Let
\[
x = r \cos \theta, y = r \sin \theta
\]
Then we have
\[
z = r(\cos \theta + \sin \theta) \ln(2r^2)
\]
\[
\frac{\partial z}{\partial r} = (\sin \theta + \cos \theta) \left(\ln (2r^2) + 2\right)
\]
\[
\frac{\partial z}{\partial \theta} = r(\cos \theta - \sin \theta) \ln(2r^2)
\]
Solve the following equations
\[
\begin{cases}
\frac{\partial z}{\partial r} = (\sin \theta + \cos \theta) \left(\ln (2r^2) + 2\right) = 0 \\
\frac{\partial z}{\partial \theta} = r(\cos \theta - \sin \theta) \ln(2r^2) = 0
\end{cases}
\]
we get extremum points
\[
(-\frac{1}{\sqrt{2}e}, \frac{\pi}{4} + 2k\pi), (\frac{1}{\sqrt{2}e}, \frac{\pi}{4} + 2k\pi), (\frac{1}{\sqrt{2}e}, \frac{3\pi}{4} + 2k\pi), (-\frac{1}{\sqrt{2}e}, \frac{3\pi}{4} + 2k\pi)
\]
where \(k\) is an integer, and extrema are
\[
\frac{2}{e}, -\frac{2}{e}.
\]