# 数学代写|常微分方程代写ordinary differential equation代考|MATH2410

## 数学代写|常微分方程代写ordinary differential equation代考|MATH2410

Suppose $\mathfrak{H}_0$ is a vector space. A map $\langle., .\rangle:. \mathfrak{H}_0 \times \mathfrak{H}_0 \rightarrow \mathbb{C}$ is called skew linear form if it is conjugate linear in the first and linear in the second argument, that is,
\begin{aligned} &\left\langle\lambda_1 f_1+\lambda_2 f_2, g\right\rangle=\lambda_1^\left\langle f_1, g\right\rangle+\lambda_2^\left\langle f_2, g\right\rangle \ &\left\langle f, \lambda_1 g_1+\lambda_2 g_2\right\rangle=\lambda_1\left\langle f, g_1\right\rangle+\lambda_2\left\langle f, g_2\right\rangle \end{aligned} \quad \lambda_1, \lambda_2 \in \mathbb{C} .
A skew linear form satisfying the requirements
(i) $\langle f, f\rangle>0$ for $f \neq 0$
(ii) $\langle f, g\rangle=\langle g, f\rangle^*$
is called inner product or scalar product. Associated with every scalar product is a norm
$$|f|=\sqrt{\langle f, f\rangle} .$$
(We will prove later that this is indeed a norm.) The pair $\left(\mathfrak{H}_0,\langle., .\rangle.\right)$ is called inner product space. If $\mathfrak{H}_0$ is complete with respect to the above norm, it is called a Hilbert space. It is usually no restriction to assume that $\mathfrak{H}_0$ is complete since one can easily replace it by its completion $\mathfrak{H}$. However, for our purpose this is not necessary and hence we will not do so here to avoid technical complications later on.

A vector $f \in \mathfrak{H}0$ is called normalized or unit vector if $|f|=1$. Two vectors $f, g \in \mathfrak{H}_0$ are called orthogonal or perpendicular $(f \perp g)$ if $\langle f, g\rangle=0$ and parallel if one is a multiple of the other. If $f$ and $g$ are orthogonal we have the Pythagorean theorem: $$|f+g|^2=|f|^2+|g|^2, \quad f \perp g,$$ which is one line of computation. Suppose $u$ is a unit vector. Then the projection of $f$ in the direction of $u$ is given by $$f{|}=\langle u, f\rangle u$$
and $f_{\perp}$ defined via
$$f_{\perp}=f-\langle u, f\rangle u$$
is perpendicular to $u$ since $\left\langle u, f_{\perp}\right\rangle=\langle u, f-\langle u, f\rangle u\rangle=\langle u, f\rangle-\langle u, f\rangle\langle u, u\rangle=$ 0.

## 数学代写|常微分方程代写ordinary differential equation代考|Regular Sturm-Liouville problems

Now we want to apply the theory of inner product spaces to the investigation of Sturm-Liouville problems. But first let us look at the corresponding differential equation
$$-\left(p(x) y^{\prime}\right)^{\prime}+(q(x)-z r(x)) y=0, \quad z \in \mathbb{C}, x \in I=(a, b),$$
for $y \in C^2(I, \mathbb{C})$, which is equivalent to the first-order system
\begin{aligned} y^{\prime} &=\frac{1}{p(x)} w \ w^{\prime} &=(q(x)-z r(x)) y \end{aligned}
where $w(x)=p(x) y^{\prime}(x)$. Hence we see that there is a unique solution if $p^{-1}(x), q(x)$, and $r(x)$ are continuous in $I$. In fact, as noted earlier, it even suffices to assume that $p^{-1}(x), q(x)$, and $r(x)$ are integrable over each compact subinterval of $I$. I remark that essentially all you have to do is to replace differentiable by absolutely continuous (respectively differentiable in the weak sense) in the sequel. However, we will assume that
$$r, q \in C^0([a, b], \mathbb{R}), p \in C^1([a, b], \mathbb{R}), \quad p(x), r(x)>0, x \in[a, b],$$
for the rest of this chapter and call the differential equation (5.39) regular in this case.

Denote the principal matrix solution of $(5.40)$ by
$$\Pi\left(z, x, x_0\right)=\left(\begin{array}{cc} c\left(z, x, x_0\right) & s\left(z, x, x_0\right) \ p(x) c^{\prime}\left(z, x, x_0\right) & p(x) s^{\prime}\left(z, x, x_0\right) \end{array}\right), \quad z \in \mathbb{C},$$
where $c\left(z, x, x_0\right)$ is the solution of $(5.39)$ corresponding to the initial condition $c\left(z, x_0, x_0\right)=1, p\left(x_0\right) c^{\prime}\left(z, x_0, x_0\right)=0$ and similarly for $s\left(z, x, x_0\right)$ but corresponding to the initial condition $s\left(z, x_0, x_0\right)=0, p\left(x_0\right) s^{\prime}\left(z, x_0, x_0\right)=1$.
We know that $\Pi\left(z, x, x_0\right)$ is continuous with respect to $x$ and $x_0$ by Theorem 2.9. But with respect to $z$ a much stronger result is true.

Lemma 5.6. The principal matrix solution $\Pi\left(z, x, x_0\right)$ is analytic with respect to $z \in \mathbb{C}$ for every fixed $\left(x, x_0\right) \in I \times I$.

Proof. It suffices to show that every solution is analytic with respect to $z \in \mathbb{C}$ in a neighborhood of $x_0$ if the initial conditions are constant. In this case each of the iterations (2.14) is analytic (in fact even polynomial) with respect to $z \in \mathbb{C}$. Moreover, for $z$ in a compact set, the Lipschitz constant can be chosen independent of $z$. Hence the series of iterations converges uniformly for $z$ in a compact set, implying that the limit is again analytic by the Weierstraß convergence theorem.

# 常微分方程代考

## 数学代写|常微分方程代写ordinary differential equation代考|MATH2410

$$f_{\perp}=f-\langle u, f\rangle u$$

## 数学代写|常微分方程代写ordinary differential equation代考|Regular Sturm-Liouville problems

$$-\left(p(x) y^{\prime}\right)^{\prime}+(q(x)-z r(x)) y=0, \quad z \in \mathbb{C}, x \in I=(a, b),$$

$$y^{\prime}=\frac{1}{p(x)} w w^{\prime} \quad=(q(x)-z r(x)) y$$

$$r, q \in C^0([a, b], \mathbb{R}), p \in C^1([a, b], \mathbb{R}), \quad p(x), r(x)>0, x \in[a, b],$$

$$\Pi\left(z, x, x_0\right)=\left(c\left(z, x, x_0\right) \quad s\left(z, x, x_0\right) p(x) c^{\prime}\left(z, x, x_0\right) \quad p(x) s^{\prime}\left(z, x, x_0\right)\right), \quad z \in \mathbb{C},$$

(2.14) 都是解析的 (实际上甚至多项式) 关于 $z \in \mathbb{C}$. 此外，对于 $z$ 在紧集中， Lipschitz 常数可以独立于 $z$. 因此，这一系列迭代一致地收敛于 $z$ 在一个紧集中，这意味着极限再次由 Weierstraß 收敛定理分析。

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