数学物理方程第11次作业

Chasse_neige

1.计算下列积分 (a)

\int_{0}^{1} \sqrt{1 - x} \sin (a \sqrt{x}) d x, a > 0

作代换 $\sqrt{x} = u$

\begin{matrix} \int_{0}^{1} \sqrt{1 - x} \sin (a \sqrt{x}) d x = \int_{0}^{1} 2 u \sqrt{1 - u^{2}} \sin (a u) d u \\ = \int_{- 1}^{1} u \sqrt{1 - u^{2}} \sin (a u) d u = \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} θ \sin θ \sin (a \sin θ) d θ \\ = \frac{1}{2 i} \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} θ \sin θ (e^{i a \sin θ} - e^{- i a \sin θ}) d θ \end{matrix}

考虑

\begin{matrix} \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} θ \sin θ e^{i a \sin θ} d θ = {\frac{1}{i a} \cos θ \sin θ e^{i a \sin θ} |}_{- \frac{π}{2}}^{\frac{π}{2}} - \frac{1}{i a} \int_{- \frac{π}{2}}^{\frac{π}{2}} (2 \cos^{2} θ - 1) e^{i a \sin θ} d θ \\ = \frac{i}{2 a} \int_{- π}^{π} \cos 2 θ e^{i a \sin θ} d θ \\ = \frac{i}{4 a} \int_{- π}^{π} e^{i a \sin θ + i 2 θ} + e^{i a \sin θ - i 2 θ} d θ \\ = \frac{i π}{a} J_{2} (a) \end{matrix}

同理

\begin{matrix} \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} θ \sin θ e^{- i a \sin θ} d θ = - {\frac{1}{i a} \cos θ \sin θ e^{- i a \sin θ} |}_{- \frac{π}{2}}^{\frac{π}{2}} + \frac{1}{i a} \int_{- \frac{π}{2}}^{\frac{π}{2}} (2 \cos^{2} θ - 1) e^{- i a \sin θ} d θ \\ = - \frac{i}{2 a} \int_{- π}^{π} \cos 2 θ e^{- i a \sin θ} d θ \\ = - \frac{i}{4 a} \int_{- π}^{π} e^{- i a \sin θ + i 2 θ} + e^{- i a \sin θ - i 2 θ} d θ \\ = - \frac{i π}{a} J_{2} (a) \end{matrix}

所以

\frac{1}{2 i} \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} θ \sin θ (e^{i a \sin θ} - e^{- i a \sin θ}) d θ = \frac{π}{a} J_{2} (a)

2.利用生成函数证明下列等式 (a)

\cos x = J_{0} (x) - 2 J_{2} (x) + 2 J_{4} (x) - 2 J_{6} (x) + \dots

\sin x = 2 J_{1} (x) - 2 J_{3} (x) + 2 J_{5} (x) - 2 J_{7} (x) + \dots

证明：

e^{i x} = \sum_{k = - \infty}^{\infty} J_{k} (x) (i)^{k}

e^{- i x} = \sum_{k = - \infty}^{\infty} J_{k} (x) (- i)^{k}

所以

\cos x = \frac{1}{2} (e^{i x} + e^{- i x}) = J_{0} (x) - 2 J_{2} (x) + 2 J_{4} (x) - 2 J_{6} (x) + \dots

\sin x = \frac{1}{2 i} (e^{i x} - e^{- i x}) = 2 J_{1} (x) - 2 J_{3} (x) + 2 J_{5} (x) - 2 J_{7} (x) + \dots

(b)

1 = J_{0}^{2} (x) + 2 [J_{1}^{2} (x) + J_{2}^{2} (x) + \dots]

1 = e^{i x} e^{- i x} = (\sum_{k = - \infty}^{\infty} J_{k} (x) (i)^{k}) (\sum_{k = - \infty}^{\infty} J_{k} (x) (- i)^{k}) = \sum_{k = - \infty}^{\infty} J_{k}^{2} (x) = J_{0}^{2} (x) + 2 [J_{1}^{2} (x) + J_{2}^{2} (x) + \dots]

3.计算如下积分 (a)

\int_{0}^{x} x^{- 1} J_{4} (x) d x, \int_{0}^{x} x^{3} J_{0} (x) d x

\int_{0}^{x} x^{- 1} J_{4} (x) d x = \int_{0}^{x} \frac{1}{8} (J_{3} (x) + J_{5} (x)) d x = \frac{1}{8} (2 J_{4} (x) + 2 \int_{0}^{x} J_{5} (x) d x)

利用递推关系

J_{5} = J_{3} - 2 J_{4}^{'} = J_{1} - 2 J_{2}^{'} - 2 J_{4}^{'}

所以

\int_{0}^{x} J_{5} (x) d x = \int_{0}^{x} J_{1} (x) d x - 2 J_{2} (x) - 2 J_{4} (x)

\int_{0}^{x} J_{1} (x) d x = - \int_{0}^{x} J_{0}^{'} (x) d x = - (J_{0} (x) - 1)

所以

\int_{0}^{x} x^{- 1} J_{4} (x) d x = \frac{1}{8} (2 (1 - J_{0} (x)) - 4 J_{2} (x) - 2 J_{4} (x)) = \frac{1}{4} (1 - J_{0} (x) - 2 J_{2} (x) - J_{4} (x))

\begin{matrix} \int_{0}^{x} x^{3} J_{0} (x) d x = \int_{0}^{x} 2 x^{2} J_{1} (x) - x^{3} J_{2} (x) d x \\ = \int_{0}^{x} 2 \frac{d}{d x} (x^{2} J_{2} (x)) - \frac{d}{d x} (x^{3} J_{3} (x)) d x = 2 x^{2} J_{2} (x) - x^{3} J_{3} (x) \end{matrix}

(b)

\int_{0}^{x} J_{0} (x) \cos x d x, \int_{0}^{x} x^{n} J_{n} (x) \cos x d x

令

F (t) = t J_{0} (t) \cos t + t J_{1} (t) \sin t

则

\begin{matrix} F^{'} (t) = J_{0} (t) \cos t + t J_{0}^{'} (t) \cos t - t J_{0} (t) \sin t + J_{1} (t) \sin t + t J_{1}^{'} (t) \sin t + t J_{1} (t) \cos t \\ = J_{0} (t) \cos t + t (- J_{1} (t)) \cos t - t J_{0} (t) \sin t + J_{1} (t) \sin t + t (J_{0} (t) - \frac{J_{1} (t)}{t}) \sin t + t J_{1} (t) \cos t \\ = J_{0} (t) \cos t - t J_{1} (t) \cos t - t J_{0} (t) \sin t + J_{1} (t) \sin t + t J_{0} (t) \sin t - J_{1} (t) \sin t + t J_{1} (t) \cos t \\ = J_{0} (t) \cos t \end{matrix}

因此

\int J_{0} (t) \cos t d t = t J_{0} (t) \cos t + t J_{1} (t) \sin t + C

所以

\int_{0}^{x} J_{0} (t) \cos t d t = {t J_{0} (t) \cos t + t J_{1} (t) \sin t |}_{0}^{x} = x J_{0} (x) \cos x + x J_{1} (x) \sin x .

对于第二个积分，考虑 $\frac{d}{d x} (x^{n} J_{n} (x)) = x^{n} J_{n - 1} (x)$ ，所以

F (t) = t^{n + 1} J_{n} (t) \cos t + t^{n + 1} J_{n + 1} (t) \sin t

\begin{matrix} F^{'} (t) = (n + 1) t^{n} J_{n} (t) \cos t + t^{n + 1} J_{n}^{'} (t) \cos t - t^{n + 1} J_{n} (t) \sin t + t^{n + 1} J_{n} (t) \sin t + t^{n + 1} J_{n + 1} (t) \cos t \\ = (n + 1) t^{n} J_{n} (t) \cos t + t^{n + 1} (J_{n}^{'} (t) + J_{n + 1} (t)) \cos t = (n + 1) t^{n} J_{n} (t) \cos t + \frac{1}{2} t^{n + 1} (J_{n - 1} (t) + J_{n + 1} (t)) \cos t \\ = (n + 1) t^{n} J_{n} (t) \cos t + t^{n} J_{n} (t) \cos t = (2 n + 1) t^{n} J_{n} (t) \cos t \end{matrix}

所以

\int_{0}^{x} x^{n} J_{n} (x) \cos x d x = \frac{1}{2 n + 1} x^{n} J_{n} (x) \cos x

7.求解下列定解问题

{\begin{cases} \frac{\partial u}{\partial t} - κ [\frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial u}{\partial r}) + \frac{1}{r^{2}} \frac{\partial^{2} u}{\partial φ^{2}}] = 0 \\ {u |}_{r = a} = 0, {u |}_{t = 0} = u_{0} \sin 2 φ \end{cases}

应用分离变量法

u (r, φ, t) = R (r) Φ (φ) T (t)

满足

\frac{d}{d t} T = - ω^{2} T

\frac{d [}{d 2}] φ Φ = - μ Φ

R^{″} + \frac{R^{'}}{ρ} + (1 - \frac{μ}{ρ^{2}}) R = 0

其中

ρ = \sqrt{\frac{ω^{2}}{κ}} r

显然当 $ω = 0$ 时解不满足初始条件

所以解为

Φ (φ) = \sin 2 φ

得到 $μ = 4$ ，再利用原点处的有界性，得到

R (r) = B_{2, n} J_{2} (ρ)

带入 $u |_{r = a} = 0$

ω_{i} = \frac{μ_{i}^{(2)}}{a} \sqrt{κ}

所以本征值问题的解为

u (r, φ, t) = \sin 2 φ \sum_{n = 1}^{\infty} B_{2, n} J_{2} (\frac{μ_{n}^{(2)}}{a} r) \exp (- κ {(\frac{μ_{n}^{(2)}}{a})}^{2} t)

其系数 $B_{2, n}$ 由初始条件确定

B_{2, n} = \frac{u_{0} \int_{0}^{a} r J_{2} (\frac{μ_{n}^{(2)}}{a} r) d r}{\int_{0}^{a} r {[J_{2} (\frac{μ_{n}^{(2)}}{a} r)]}^{2} d r}

计算积分得系数为

B_{2, n} = \frac{2 u_{0} [2 - 2 J_{0} (μ_{n}^{(2)}) - μ_{n}^{(2)} J_{1} (μ_{n}^{(2)})]}{{μ_{n}^{(2)}}^{2} {[J_{3} (μ_{n}^{(2)})]}^{2}}

所以最终解为

u (r, φ, t) = 2 u_{0} \sin 2 φ \sum_{n = 1}^{\infty} \frac{2 - 2 J_{0} (μ_{n}^{(2)}) - μ_{n}^{(2)} J_{1} (μ_{n}^{(2)})}{{μ_{n}^{(2)}}^{2} {[J_{3} (μ_{n}^{(2)})]}^{2}} J_{2} (\frac{μ_{n}^{(2)}}{a} r) \exp (- κ {(\frac{μ_{n}^{(2)}}{a})}^{2} t)