臺灣大學數學系

九十一學年度第一學期碩博士班資格考試題

偏微分方程(Differential)

Sept 11, 2002

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Total score: $ 100$ points

$ 1$.
($ 20$ points) Solve the following Cauchy problem for $ u(x,y)$:
(a)
($ 10$ points) Equation $ (x+2) u_{x} + 2y u_{y} = 2u$ with initial condition $ u(-1,y) = \sqrt{y}$.
(b)
($ 10$ points) Equation $ x^{2} u_{x} - y^{2} u_{y} = 0$ with initial condition $ u(1,y) = f(y)$.
$ 2$.
($ 20$ points)(a) ($ 10$ points) Show that for $ n=3$ the general solution of the wave equation:

$\displaystyle u_{tt} = c^{2} \sum_{i=1}^{n} u_{x_{i} x_{i}},
$

with spherical symmetry about the origin has the form:

$\displaystyle u = \frac{f(r+ct) + g(r-ct)}{r}, \qquad r^{2} = \sum_{i=1}^{n}x_{i}^{2},
$

with suitable $ f$ and $ g$. Here $ c$ is a positive constant. (b) ($ 10$ points) Show that the solution of the above problem with initial data of the form:

$\displaystyle u = 0, \quad u_{t} = \phi(r)
$

is given by

$\displaystyle u = \frac{1}{2 c r} \int_{r-ct}^{r+ct} \xi \phi(\xi) \; d\xi.
$

Here $ \phi$ is a even function of $ r$.
$ 3$.
(20 points) Let $ u$ be harmonic in a domain D. Show that $ u$ has partial derivatives of all orders in $ D$.
$ 4$.
($ 20$ points) Consider the following one-dimensional diffusion equation in the semi-infinite interval $ 0 \leq x < \infty $:
    $\displaystyle u_{t} - u_{xx} = 0, \qquad x \geq 0 , \quad t \geq 0$  
    $\displaystyle u(x,0) = 0, \qquad
u(0,t) = a t^{n}$   if$\displaystyle \quad t > 0,$  

where $ a$ is a positive constant and $ n$ is a non-negative constant.
(a)
(10 points) Assume the solution of the problem take the form:

$\displaystyle u(x,t) = a t^{n} f(\xi),$   where$\displaystyle \quad \xi = \frac{x}{2 \sqrt{t}}.
$

Show that $ f$ satisfies the conditions:

$\displaystyle f^{''}(\xi) + 2 \xi f^{'}(\xi) - 4 n f = 0,
\\
f(0) = 1, \qquad f(\infty) = 0.
$

(b)
(10 points) Find the solution of the above ordinary differential equation, and hence the solution of the original diffusion problem.
$ 5$.
($ 20$ points) Solve the following one-dimensional diffusion equation in the unit interval $ 0 \leq x \leq 1$:
    $\displaystyle u_{t} - u_{xx} = x \sin{t}, \qquad 0 \leq x \leq 1, \quad t \geq 0$  
    $\displaystyle u(x,0) = x (1-x), \qquad
u(0,t) = u_{x}(1,t) = 0$   if$\displaystyle \quad t > 0.$  


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