# 数学代写|常微分方程代写ordinary differential equation代考|SEPARATED BOUNDARY CONDITIONS

#### Doug I. Jones

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## 数学代写|常微分方程代写ordinary differential equation代考|SEPARATED BOUNDARY CONDITIONS

In this section we study the existence and behavior of eigenvalues for problem (P). Our first task is to generalize the oscillation results of Section 3.6.
Given the equation
$$\left(k(t) y^{\prime}\right)^{\prime}+g(t) y=0$$
and letting $x=k(t) y^{\prime}$, we obtain the first order system
\begin{aligned} & y^{\prime}=x / k(t), \ & x^{\prime}=-g(t) y . \end{aligned}
(Reference should be made to the preceding section for the assumptions on the functions $k$ and $g$.) We can transform (2.2) using polar coordinates to obtain
\begin{aligned} & x^{\prime}=r^{\prime} \cos \theta-(r \sin \theta) \theta^{\prime}=-g(t) r \sin \theta, \ & y^{\prime}=r^{\prime} \sin \theta+(r \cos \theta) \theta^{\prime}=r \cos \theta / k(t), \end{aligned}
or
\begin{aligned} & r^{\prime}=[1 / k(t)-g(t)] r \cos \theta \sin \theta, \ & \theta^{\prime}=g(t) \sin ^2 \theta+\cos ^2 \theta / k(t) . \end{aligned}
If $y \neq 0$, then $y$ and $y^{\prime}$ cannot simultaneously vanish and $r^2=x^2+y^2>0$. Thus we can take $r(t)$ as always positive [or else as $r(t) \equiv 0$ ]. Therefore, Eq. (2.1) is equivalent to Eq. (2.2) or to Eq. (2.3).
We now state and prove our first result.
Theorem 2.1. Let $k_i \in C^{\prime}[a, b]$ and $g_i \in C[a, b]$ for $i=1,2$ with $0<k_2 \leq k_1$ and $g_1 \leq g_2$. Let $\phi_i$ be a solution of $\left(k_i y^{\prime}\right)^{\prime}+g_i y=0$ and let $r_i$ and $\theta_i$ satisfy the corresponding problem in polar coordinates, i.e.,
\begin{aligned} r_i^{\prime} & =\left[1 / k_i(t)-g_i(t)\right] r_i \cos \theta_i \sin \theta_i, \ \theta_i^{\prime} & =g_i(t) \sin ^2 \theta_i+\cos ^2 \theta_i / k_i(t) . \end{aligned}

## 数学代写|常微分方程代写ordinary differential equation代考|ASYMPTOTIC BEHAVIOR OF EIGENVALUES

In the present section we shall require the notation $O(\cdot)$ and $o(\cdot)$ encountered in the calculus. Recall that for a function $g: R \rightarrow R$, and for $\beta \geq 0$, the notation $g(x)=O\left(|x|^\beta\right)$ as $|x| \rightarrow \infty$ means that
$$\lim {|x| \rightarrow \infty} \sup \frac{|g(x)|}{|x|^\beta}<\infty .$$ Also, recall that $g(x)=o\left(|x|^\beta\right)$ as $|x| \rightarrow \infty$ means that $$\lim {|x| \rightarrow \infty} \frac{|g(x)|}{|x|^\beta}=0$$
If in the above, the continuous variable $x$ is replaced by an integer valued variable $m \geq 0$, then $g(m)=O\left(m^\beta\right)$ as $m \rightarrow \infty$ and $g(m)=o\left(m^\beta\right)$ as $m \rightarrow \infty$ are defined in the obvious way.
In this section we study in detail the behavior as $m \rightarrow \infty$ of the eigenvalues $\lambda_m$ and eigenfunctions $\phi_m$ of problem (P). We assume here that $k, \rho \in C^3[a, b], g \in C^1[a, b]$ and that the constants $\beta$ and $\delta$ in (BC) are not zero.
Let $K$ be the constant defined by
$$K=\pi^{-1} \int_a^b[\rho(v) / k(v)]^{1 / 2} d v .$$
Then under the Liouville transformation
$$s=K^{-1} \int_a^1[\rho(v) / k(v)]^{1 / 2} d v, \quad Y=K^{1 / 2}(\rho k)^{1 / 4} y, \quad \mu^2=K^2 \lambda,$$
Eq. (1.1) assumes the form
$$\left(d^2 Y / d s^2\right)+\left[\mu^2-q(s)\right] Y=0 .$$

# 常微分方程代写

## 数学代写|常微分方程代写ordinary differential equation代考|SEPARATED BOUNDARY CONDITIONS

$$\left(k(t) y^{\prime}\right)^{\prime}+g(t) y=0$$

\begin{aligned} & y^{\prime}=x / k(t), \ & x^{\prime}=-g(t) y . \end{aligned}
(有关功能的假设，请参阅上一节 $k$ 和 $g$)我们可以用极坐标变换(2.2)得到
\begin{aligned} & x^{\prime}=r^{\prime} \cos \theta-(r \sin \theta) \theta^{\prime}=-g(t) r \sin \theta, \ & y^{\prime}=r^{\prime} \sin \theta+(r \cos \theta) \theta^{\prime}=r \cos \theta / k(t), \end{aligned}

\begin{aligned} & r^{\prime}=[1 / k(t)-g(t)] r \cos \theta \sin \theta, \ & \theta^{\prime}=g(t) \sin ^2 \theta+\cos ^2 \theta / k(t) . \end{aligned}

\begin{aligned} r_i^{\prime} & =\left[1 / k_i(t)-g_i(t)\right] r_i \cos \theta_i \sin \theta_i, \ \theta_i^{\prime} & =g_i(t) \sin ^2 \theta_i+\cos ^2 \theta_i / k_i(t) . \end{aligned}

## 数学代写|常微分方程代写ordinary differential equation代考|ASYMPTOTIC BEHAVIOR OF EIGENVALUES

$$\lim {|x| \rightarrow \infty} \sup \frac{|g(x)|}{|x|^\beta}<\infty .$$ 同样，回想一下 $g(x)=o\left(|x|^\beta\right)$ as $|x| \rightarrow \infty$ 意思是 $$\lim {|x| \rightarrow \infty} \frac{|g(x)|}{|x|^\beta}=0$$

$$K=\pi^{-1} \int_a^b[\rho(v) / k(v)]^{1 / 2} d v .$$

$$s=K^{-1} \int_a^1[\rho(v) / k(v)]^{1 / 2} d v, \quad Y=K^{1 / 2}(\rho k)^{1 / 4} y, \quad \mu^2=K^2 \lambda,$$

$$\left(d^2 Y / d s^2\right)+\left[\mu^2-q(s)\right] Y=0 .$$

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