東京大学 工学系研究科 機械工学専攻 機械工学第1部 2025年8月実施 問題 2 Author 祭音Myyura (Based on zzy's answer refined with GPT 6 Astra)
Description 図2に示すように，密度 ρ \rho ρ ν \nu ν x x x z z z z z z z z z t = 0 t=0 t = 0 x x x u 0 u_0 u 0 x x x
(1) 流体の x x x
∂ u ∂ t + u ∂ u ∂ x + w ∂ u ∂ z = − 1 ρ ∂ p ∂ x + ν ( ∂ 2 u ∂ x 2 + ∂ 2 u ∂ z 2 ) . \frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+w\frac{\partial u}{\partial z}
=-\frac1\rho\frac{\partial p}{\partial x}
+\nu\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial z^2}\right). ∂ t ∂ u + u ∂ x ∂ u + w ∂ z ∂ u = − ρ 1 ∂ x ∂ p + ν ( ∂ x 2 ∂ 2 u + ∂ z 2 ∂ 2 u ) . ここで，u , w u,w u , w x , z x,z x , z p p p u u u
(2) 設問(1)で求めた偏微分方程式を解くために，以下の式で定義される変数 η \eta η
η = z 2 ν t . \eta=\frac{z}{2\sqrt{\nu t}}. η = 2 ν t z . 流速 u u u η \eta η u = f ( η ) u=f(\eta) u = f ( η ) f ( η ) f(\eta) f ( η )
(3) 流速 u u u u 0 , ν , z , t u_0,\nu,z,t u 0 , ν , z , t
erf ( α ) = 2 π ∫ 0 α exp ( − s 2 ) d s . \operatorname{erf}(\alpha)=\frac2{\sqrt\pi}\int_0^\alpha\exp(-s^2)\,ds. erf ( α ) = π 2 ∫ 0 α exp ( − s 2 ) d s . なお，lim α → ∞ erf ( α ) = 1 \displaystyle\lim_{\alpha\to\infty}\operatorname{erf}(\alpha)=1 α → ∞ lim erf ( α ) = 1
(4) 壁が動き出してから十分時間が経ったときの流速 u u u z = 0 z=0 z = 0 z → ∞ z\to\infty z → ∞
题目描述
如图2所示，密度为 ρ \rho ρ ν \nu ν x x x z z z z z z z z z t = 0 t=0 t = 0 x x x u 0 u_0 u 0 x x x
(1) 流体沿 x x x
∂ u ∂ t + u ∂ u ∂ x + w ∂ u ∂ z = − 1 ρ ∂ p ∂ x + ν ( ∂ 2 u ∂ x 2 + ∂ 2 u ∂ z 2 ) . \frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+w\frac{\partial u}{\partial z}
=-\frac1\rho\frac{\partial p}{\partial x}
+\nu\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial z^2}\right). ∂ t ∂ u + u ∂ x ∂ u + w ∂ z ∂ u = − ρ 1 ∂ x ∂ p + ν ( ∂ x 2 ∂ 2 u + ∂ z 2 ∂ 2 u ) . 其中 u , w u,w u , w x , z x,z x , z p p p u u u
(2) 为求解第(1)问得到的偏微分方程，引入变量
η = z 2 ν t . \eta=\frac{z}{2\sqrt{\nu t}}. η = 2 ν t z . 流速 u u u η \eta η u = f ( η ) u=f(\eta) u = f ( η ) f ( η ) f(\eta) f ( η )
(3) 用 u 0 , ν , z , t u_0,\nu,z,t u 0 , ν , z , t u u u
erf ( α ) = 2 π ∫ 0 α exp ( − s 2 ) d s , \operatorname{erf}(\alpha)=\frac2{\sqrt\pi}\int_0^\alpha\exp(-s^2)\,ds, erf ( α ) = π 2 ∫ 0 α exp ( − s 2 ) d s , 其中 lim α → ∞ erf ( α ) = 1 \displaystyle\lim_{\alpha\to\infty}\operatorname{erf}(\alpha)=1 α → ∞ lim erf ( α ) = 1
(4) 画出壁面开始运动后经过足够长时间时的流速 u u u z = 0 z=0 z = 0 z → ∞ z\to\infty z → ∞
Kai (1)
w = 0 w=0 w = 0 u x = 0 u_x=0 u x = 0 p x = 0 p_x=0 p x = 0
∂ u ∂ t = ν ∂ 2 u ∂ z 2 . \boxed{\frac{\partial u}{\partial t}=\nu\frac{\partial^2u}{\partial z^2}}. ∂ t ∂ u = ν ∂ z 2 ∂ 2 u . (2)
∂ η ∂ t = − η 2 t , \frac{\partial\eta}{\partial t}=-\frac\eta{2t}, ∂ t ∂ η = − 2 t η , ∂ η ∂ z = 1 2 ν t
\frac{\partial\eta}{\partial z}=\frac1{2\sqrt{\nu t}} ∂ z ∂ η = 2 ν t 1 より、u t = − η f ′ / ( 2 t ) u_t=-\eta f'/(2t) u t = − η f ′ / ( 2 t ) u z z = f ′ ′ / ( 4 ν t ) u_{zz}=f''/(4\nu t) u zz = f ′′ / ( 4 ν t )
f ′ ′ + 2 η f ′ = 0 . \boxed{f''+2\eta f'=0}. f ′′ + 2 η f ′ = 0 . (3)
積分して f ′ = C 1 e − η 2 f'=C_1e^{-\eta^2} f ′ = C 1 e − η 2 f = C 1 ∫ 0 η e − s 2 d s + C 2 f=C_1\int_0^\eta e^{-s^2}ds+C_2 f = C 1 ∫ 0 η e − s 2 d s + C 2 f ( 0 ) = u 0 f(0)=u_0 f ( 0 ) = u 0 f ( ∞ ) = 0 f(\infty)=0 f ( ∞ ) = 0 C 2 = u 0 C_2=u_0 C 2 = u 0 C 1 = − 2 u 0 / π C_1=-2u_0/\sqrt\pi C 1 = − 2 u 0 / π
u ( z , t ) = u 0 [ 1 − erf ( z 2 ν t ) ] . \boxed{u(z,t)=u_0\left[1-\operatorname{erf}\left(\frac{z}{2\sqrt{\nu t}}\right)\right]}. u ( z , t ) = u 0 [ 1 − erf ( 2 ν t z ) ] . z > 0 z>0 z > 0 t ↓ 0 t\downarrow0 t ↓ 0 u → 0 u\to0 u → 0
(4)
壁面では u = u 0 u=u_0 u = u 0 u → 0 u\to0 u → 0 ν t \sqrt{\nu t} ν t
Reference 
