数学物理方程第2次作业

Chasse_neige

(a)

\frac{\partial^{2} u}{\partial x^{2}} - 2 \frac{\partial^{2} u}{\partial x \partial y} - 3 \frac{\partial^{2} u}{\partial y^{2}} = 0

(D_{x}^{2} - 2 D_{x} D_{y} - 3 D_{y}^{2}) u = 0

(D_{x} + D_{y}) (D_{x} - 3 D_{y}) u = 0

u = f (x - y) + g (3 x + y)

(b)

\frac{\partial^{2} u}{\partial x^{2}} - \frac{\partial^{2} u}{\partial x \partial y} = 0

(D_{x}^{2} - D_{x} D_{y}) u = 0

D_{x} (D_{x} - D_{y}) u = 0

u = f (y) g (x + y)

(c)

\frac{\partial^{2} u}{\partial t^{2}} - \frac{c^{2}}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial u}{\partial r}) = 0, c \neq 0

(提示: $\frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial u}{\partial r}) = \frac{1}{r} \frac{\partial^{2}}{\partial r^{2}} (r u)$ )

(D_{t}^{2} - c^{2} D_{r}^{2}) (r u) = 0

(D_{t} - c D_{r}) (D_{t} + c D_{r}) (r u) = 0

u = \frac{1}{r} (f (r + c t) + g (r - c t))

热传导方程表示为

\frac{\partial u}{\partial t} = κ \frac{\partial^{2} u}{\partial x^{2}}

分离变量

u (x, t) = X (x) T (t)

\frac{1}{T} T^{'} = \frac{κ}{x} X^{″} = - λ

解得

T (t) = e^{- λ t}

X (x) = A (ω) \sin (ω x) + B (ω) \cos (ω x)

其中

ω = \sqrt{\frac{λ}{κ}}

由于 $u |_{t = 0} = a x (l - x)$

所以 $B (ω) = 0$

且 $ω$ 满足

ω = \frac{n π}{l}

所以对应参数 $ω$ 的分量系数大小为

A (ω) = \frac{2}{l} \int_{0}^{l} a x (l - x) \sin (ω x) d x = \frac{8 a l^{2}}{(2 m - 1)^{3} π^{3}}

其中 $n = 2 m - 1$

所以解为

u (x, t) = \sum_{m = 1}^{\infty} (\frac{8 a l^{2}}{(2 m - 1)^{3} π^{3}} \sin (\frac{(2 m - 1) π}{l} x)) e^{- κ \frac{(2 m - 1)^{2} π^{2}}{l^{2}} t}

数学物理方程 第2次作业 ​