Scharze space asked in 科學數學 · 1 decade ago

ODE(Sturm's comparison)

(a)Consider the differential equation:

(*) u''+q(t)u=0

Let q(t) be real valued ,continuous,and satisies 0<m<=q(t)<=M. If u=u(t)≠0 is a solution with a pair of zeros t=t1,t2(>t1), then

π/√m>=t2-t1>=π/√M

(b)Let q(t) be continuous for t>=0 and q(t)->1 as t->∞ . show that if u=u(t)≠0, is a real-valued solution of (*)

then the zeros of u(t) form a sequence 0<=t1<t2<.......such that t(n)-t(n-1)

->π as n->∞

(c)Consider the Bessel equation

(**) v''+(v'/t)+(1-(μ^2/t^2))v=0

where μ is a real parameter. The change of variable u=t^(1/2)v transforms(**) into

u''+(1-α/t^2)u=0 where α=μ^2-1/4

Show that the zeros of a real valued solution v(t) of (**) on t>0 form a sequence t1<t2<...... such that t(n)-t(n-1) ->π as n->∞

1 Answer

Rating
  • Eric
    Lv 6
    1 decade ago
    Favorite Answer

    圖片參考:http://img246.imageshack.us/img246/8206/latexrw1.p...

    Let $t_1$ and $t_2$ be a pair of successive zeros of $u$ with $t_1 < t_2$. Let $\epsilon \in (0,m)$, $m_\epsilon = m - \epsilon$, and $M_\epsilon = M + \epsilon$, and consider the ODEs

    \begin{align}

    v'' + m_\epsilon v &= 0,\\

    w'' + M_\epsilon w &= 0.

    \end{align}

    Since

    \begin{align}

    v(t) &= \sin ((t - t_1)\sqrt{m_\epsilon}) \\ &= \sin(t_1\sqrt{m_\epsilon}) \cos (t\sqrt{m_\epsilon}) + \cos(t_1\sqrt{m_\epsilon})\sin(t \sqrt {m_\epsilon})

    \end{align}

    is a solution of (1) that vanishes at $t_1$ and $t_1+\pi/\sqrt{m_\epsilon}$, by the comparison theorem, $u$ must vanish at $t_3 \in (t_1,t_1+\pi/\sqrt{m_\epsilon})$, and hence $t_2 - t_1 < \pi/\sqrt{m_\epsilon}$.

    Suppose $t_2 -t_1 < \pi/\sqrt{M_\epsilon}$. Then

    \begin{align}

    w(t) &= \sin ((t - t_1)\sqrt{M_\epsilon}) \\&= \sin(t_1\sqrt{M_\epsilon}) \cos (t\sqrt{M_\epsilon}) + \cos(t_1\sqrt{M_\epsilon})\sin(t \sqrt{M_\epsilon}),

    \end{align}

    which is a solution of (2) by the comparsion theorem, would vanish at some $t_4 \in (t_1,t_2)$, which is absurd. Therefore,

    \begin{equation}

    \pi/\sqrt{M_\epsilon} \leq t_2 - t_1 < \pi/\sqrt{m_\epsilon}

    \end{equation}

    for all $\epsilon \in (0,m)$, and the result follows.

    Let $\epsilon \in (0,1/2)$, and let $T \geq 0$ be such that $1 - \epsilon \leq q(t) \leq 1+\epsilon$ for all $t \geq T$. Consider the ODEs

    \begin{gather}

    v'' + q(t+T)v = 0, \quad t \geq 0,\\

    w'' + (1-\epsilon)w = 0, \quad t \geq 0.

    \end{gather}

    Then $v(t) = u(t+T)$ solves (8), $w(t) = \sin(t\sqrt{1-\epsilon})$ solves (9) and has (countably) infinitely many zeros, so by a comparison theorem, $v$, and hence $u$, must have (countably) infinitely many zeros. (Since $u$ is a nontrivial solution, $u$ cannot have uncountably many zeros.) Order the zeros of $u$ in increasing order as $t_1 < t_2 < \cdots$.

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