九州大学 システム情報科学府 情報理工学専攻・電気電子工学専攻 2024年8月実施 解析学・微積分

标签：

• Kyushu-University
• Double-Integral
• Laplace-Transform
• Second-Order-Linear-Ordinary-Differential-Equation
• Cauchy-Riemann-Equations

Author

祭音Myyura (assisted by ChatGPT 5.5 Thinking)

Description

(1) Calculate the following integral, where $a$ and $b$ are positive constants and $\mathbb{R}$ denotes the set of all real numbers.

$$\int_D (x^2+y^2)\,dxdy, \qquad D= \left\{ (x,y)\in\mathbb{R}^2 \mid \frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1 \right\}.$$

(2) Find the solution to the following differential equation using Laplace transforms. Here, $y'$ denotes the first-order derivative of a function $y(x)$ with respect to $x$.

$$y''+4y=\cosh x-\sinh x, \qquad y(0)=2, \qquad y'(0)=3.$$

(3) Let $z=x+iy$, where $x$ and $y$ are real numbers, and $i=\sqrt{-1}$. Answer the following questions.

• (a) Find a real number $a$ for which the function $e^{2x}(\cos ay+i\sin ay)$ is holomorphic.
• (b) Consider a holomorphic function $w(z)$. Suppose the real part of the function $w(z)$ is given as $\operatorname{Re}(w(z))=\cos x\cosh y,$ where $\operatorname{Re}(w(z))$ denotes the real part of the function $w(z)$. Then, find a formula for the function $w(z)$.

题目描述

1. 设 $a,b>0$，$\mathbb R$ 为实数集。对椭圆区域
   $$D=\left\{(x,y)\in\mathbb R^2\ \middle| \frac{x^2}{a^2}+\frac{y^2}{b^2}\le1\right\},$$
   计算
   $$\iint_D(x^2+y^2)\,dx\,dy.$$
2. 使用拉普拉斯变换求解初值问题
   $$y''+4y=\cosh x-\sinh x,\qquad y(0)=2,\quad y'(0)=3,$$
   其中 $y'$ 表示 $y(x)$ 对 $x$ 的一阶导数。
3. 令 $z=x+iy$，其中 $x,y$ 为实数、$i=\sqrt{-1}$。
   • 求使 $e^{2x}(\cos ay+i\sin ay)$ 为全纯函数的实数 $a$；
   • 设 $w(z)$ 全纯，且 $\operatorname{Re}w(z)=\cos x\cosh y$，求 $w(z)$ 的一般表达式。

考点

• 二重积分与椭圆变量代换：用缩放极坐标参数化椭圆，计算 Jacobian 后积分二次函数。
• 拉普拉斯变换求微分方程：将二阶初值问题变为代数方程，作部分分式分解并逆变换。
• 柯西–黎曼方程：由实部、虚部偏导关系确定参数 $a$，并从给定实部求调和共轭函数。

Kai

(1)

We calculate

$$\int_D (x^2+y^2)\,dxdy,$$

where

$$D= \left\{ (x,y)\in\mathbb{R}^2 \mid \frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1 \right\}.$$

Use the change of variables

$$x=ar\cos\theta, \qquad y=br\sin\theta,$$

where

$$0\leq r\leq 1, \qquad 0\leq \theta\leq 2\pi.$$

The Jacobian is

$$\left| \frac{\partial(x,y)}{\partial(r,\theta)} \right| =abr.$$

Also,

$$x^2+y^2 = a^2r^2\cos^2\theta+b^2r^2\sin^2\theta.$$

Therefore,

$$\begin{aligned} \int_D (x^2+y^2)\,dxdy &= \int_0^{2\pi}\int_0^1 \left( a^2r^2\cos^2\theta+b^2r^2\sin^2\theta \right)abr\,drd\theta \\ &= ab \int_0^1 r^3\,dr \int_0^{2\pi} \left( a^2\cos^2\theta+b^2\sin^2\theta \right) d\theta. \end{aligned}$$

Since

$$\int_0^1 r^3\,dr=\frac14,$$

and

$$\int_0^{2\pi}\cos^2\theta\,d\theta = \int_0^{2\pi}\sin^2\theta\,d\theta = \pi,$$

we get

$$\begin{aligned} \int_D (x^2+y^2)\,dxdy &= ab\cdot \frac14 \cdot \pi(a^2+b^2). \end{aligned}$$

Hence,

$$\boxed{ \int_D (x^2+y^2)\,dxdy = \frac{\pi ab(a^2+b^2)}{4} }.$$

(2)

Let

$$Y(s)=\mathcal{L}\{y(x)\}.$$

Taking the Laplace transform of both sides of

$$y''+4y=\cosh x-\sinh x,$$

we have

$$\mathcal{L}\{y''\}+4\mathcal{L}\{y\} = \mathcal{L}\{\cosh x-\sinh x\}.$$

Using

$$\mathcal{L}\{y''\}=s^2Y(s)-sy(0)-y'(0),$$

and the initial conditions

$$y(0)=2, \qquad y'(0)=3,$$

we obtain

$$s^2Y(s)-2s-3+4Y(s) = \mathcal{L}\{\cosh x-\sinh x\}.$$

Since

$$\cosh x-\sinh x=e^{-x},$$

we have

$$\mathcal{L}\{\cosh x-\sinh x\} = \mathcal{L}\{e^{-x}\} = \frac{1}{s+1}.$$

Thus,

$$(s^2+4)Y(s)-2s-3 = \frac{1}{s+1}.$$

Hence,

$$Y(s) = \frac{2s+3}{s^2+4} + \frac{1}{(s+1)(s^2+4)}.$$

Now decompose

$$\frac{1}{(s+1)(s^2+4)} = \frac{A}{s+1} + \frac{Bs+C}{s^2+4}.$$

Then

$$1 = A(s^2+4)+(Bs+C)(s+1).$$

Comparing coefficients gives

$$A=\frac15, \qquad B=-\frac15, \qquad C=\frac15.$$

Therefore,

$$\frac{1}{(s+1)(s^2+4)} = \frac{1}{5}\cdot\frac{1}{s+1} - \frac{1}{5}\cdot\frac{s}{s^2+4} + \frac{1}{5}\cdot\frac{1}{s^2+4}.$$

Thus,

$$Y(s) = \frac{2s}{s^2+4} + \frac{3}{s^2+4} + \frac{1}{5}\cdot\frac{1}{s+1} - \frac{1}{5}\cdot\frac{s}{s^2+4} + \frac{1}{5}\cdot\frac{1}{s^2+4}.$$

Taking the inverse Laplace transform, we get

$$y(x) = 2\cos 2x + \frac32\sin 2x + \frac15 e^{-x} - \frac15\cos 2x + \frac{1}{10}\sin 2x.$$

Therefore,

$$y(x) = \frac95\cos 2x + \frac85\sin 2x + \frac15 e^{-x}.$$

Hence,

$$\boxed{ y(x)=\frac95\cos 2x+\frac85\sin 2x+\frac15 e^{-x} }.$$

(3)

(a)

Let

$$f(z)=e^{2x}(\cos ay+i\sin ay).$$

Write

$$f(z)=u(x,y)+iv(x,y),$$

where

$$u(x,y)=e^{2x}\cos ay, \qquad v(x,y)=e^{2x}\sin ay.$$

For $f$ to be holomorphic, the Cauchy-Riemann equations must hold:

$$u_x=v_y, \qquad u_y=-v_x.$$

Compute each derivative:

$$u_x=2e^{2x}\cos ay,$$

$$v_y=ae^{2x}\cos ay,$$

$$u_y=-ae^{2x}\sin ay,$$

$$v_x=2e^{2x}\sin ay.$$

The first Cauchy-Riemann equation gives

$$2e^{2x}\cos ay = ae^{2x}\cos ay.$$

Thus,

$$a=2.$$

The second Cauchy-Riemann equation also gives the same condition:

$$-ae^{2x}\sin ay = -2e^{2x}\sin ay.$$

Hence,

$$\boxed{a=2}.$$

Indeed, when $a=2$,

$$e^{2x}(\cos 2y+i\sin 2y) = e^{2x}e^{2iy} = e^{2z},$$

which is holomorphic.

(b)

We are given

$$\operatorname{Re}(w(z))=\cos x\cosh y.$$

Recall that

$$\cos z = \cos(x+iy) = \cos x\cosh y - i\sin x\sinh y.$$

Therefore, the real part of $\cos z$ is

$$\operatorname{Re}(\cos z)=\cos x\cosh y.$$

Hence, one such holomorphic function is

$$w(z)=\cos z.$$

Since adding a purely imaginary constant does not change the real part, the general formula is

$$\boxed{ w(z)=\cos z+iC }$$

where $C$ is a real constant.