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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

标签：

Author​

samparker, 祭音Myyura

Description​

(1) 関数 z=z(u,v)z = z(u, v) は、実変数 (u,v)∈R2(u, v) \in \mathbb{R}^2 に関して、C2C^2 級であると仮定する。 また、(x,y)∈R2(x, y) \in \mathbb{R}^2 に対して、写像 (u,v)=(x+y,x−y)(u, v) = (x + y, x - y) を定義する。

zxx=∂2z∂x2,zxy=∂2z∂x∂y,zyy=∂2z∂y2z_{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

zuu=∂2z∂u2,zuv=∂2z∂u∂v,zvv=∂2z∂v2z_{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)O(0, 0) 以外で定義された関数

z=(x+y)ln⁡{2(x2+y2)}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)(x, y) を求めよ。

(1) Suppose that the function z=z(u,v)z = z(u, v) is of class C2C^2 in real variables (u,v)∈R2(u, v) \in \mathbb{R}^2. Define the mapping by (u,v)=(x+y,x−y)(u, v) = (x + y, x - y) for (x,y)∈R2(x, y) \in \mathbb{R}^2. Express

zxx=∂2z∂x2,zxy=∂2z∂x∂y,zyy=∂2z∂y2z_{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

zuu=∂2z∂u2,zuv=∂2z∂u∂v,zvv=∂2z∂v2.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(x2+y2)}z = (x + y) \ln\{2(x^2 + y^2)\}

defined for (x,y)≠(0,0)(x,y) \neq (0,0).

Kai​

(1)​

∂z∂x=∂z∂u∂u∂x+∂z∂v∂v∂x=∂z∂u+∂z∂v\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}
∂z∂y=∂z∂u∂u∂y+∂z∂v∂v∂y=∂z∂u−∂z∂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}
zxx=∂2z∂x2=∂2z∂u2+∂2z∂u∂v+∂2z∂v∂u+∂2z∂v2=∂2z∂u2+2∂2z∂u∂v+∂2z∂v2z_{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,

zyy=∂2z∂u2−2∂2z∂u∂v+∂2z∂v2z_{yy} = \frac{\partial^2 z}{\partial u^2} - 2\frac{\partial^2 z}{\partial u \partial v} + \frac{\partial^2 z}{\partial v^2}
zxy=∂2z∂u2−∂2z∂v2z_{xy} = \frac{\partial^2 z}{\partial u^2} - \frac{\partial^2 z}{\partial v^2}

(2)​

Let

x=rcos⁡θ,y=rsin⁡θx = r \cos \theta, y = r \sin \theta

Then we have

z=r(cos⁡θ+sin⁡θ)ln⁡(2r2)z = r(\cos \theta + \sin \theta) \ln(2r^2)
∂z∂r=(sin⁡θ+cos⁡θ)(ln⁡(2r2)+2)\frac{\partial z}{\partial r} = (\sin \theta + \cos \theta) \left(\ln (2r^2) + 2\right)
∂z∂θ=r(cos⁡θ−sin⁡θ)ln⁡(2r2)\frac{\partial z}{\partial \theta} = r(\cos \theta - \sin \theta) \ln(2r^2)

Solve the following equations

{∂z∂r=(sin⁡θ+cos⁡θ)(ln⁡(2r2)+2)=0∂z∂θ=r(cos⁡θ−sin⁡θ)ln⁡(2r2)=0\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

(−12e,π4+2kπ),(12e,π4+2kπ),(12e,3π4+2kπ),(−12e,3π4+2kπ)(-\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 kk is an integer, and extrema are

2e,−2e.\frac{2}{e}, -\frac{2}{e}.
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