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# 数学代写|多变量微积分和常微分方程代考Multivariate Calculus & Ordinary Differential Equations代写|MATH422 Systems with Constant Coeﬃﬃﬃcients

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## 数学代写|多变量微积分和常微分方程代考Multivariate Calculus & Ordinary Differential Equations代写|Systems with Constant Coeﬃﬃﬃcients

Our discussion in Lecture 18 has restricted usage of obtaining explicit solutions of homogeneous and, in general, of nonhomogeneous differential systems. This is so because the solution (18.4) involves an infinite series with repeated integrations and (18.14) involves its inversion. In fact, even if the matrix $A(x)$ is of second order, no general method of finding the explicit form of the fundamental matrix is available. Further, if the matrix $A$ is constant, then the computation of the elements of the fundamental matrix $e^{A x}$ from the series (18.4) may turn out to be difficult, if not impossible. However, in this case the notion of eigenvalues and eigenvectors of the matrix $A$ can be used to avoid unnecessary computation. For this, the first result we prove is the following theorem.

Theorem 19.1. Let $\lambda_1, \ldots, \lambda_n$ be the distinct eigenvalues of the matrix $A$ and $v^1, \ldots, v^n$ be the corresponding eigenvectors. Then the set
$$u^1(x)=v^1 e^{\lambda_1 x}, \quad \cdots \quad, u^n(x)=v^n e^{\lambda_n x}$$
is a fundamental set of solutions of (18.6).
Proof. Since $v^i$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda_i$, we find
$$\left(u^i(x)\right)^{\prime}=\left(v^i e^{\lambda_i x}\right)^{\prime}=\lambda_i v^i e^{\lambda_i x}=A v^i e^{\lambda_i x}=A u^i(x)$$
and hence $u^i(x)$ is a solution of (18.6). To show that (19.1) is a fundamental set, we note that $W(0)=\operatorname{det}\left[v^1, \ldots, v^n\right] \neq 0$, since $v^1, \ldots, v^n$ are linearly independent from Theorem 14.1. Thus, the result follows from Theorem 17.1.
Obviously, from Theorem $19.1$ it follows that
$$e^{A x}=\left[v^1 e^{\lambda_1 x}, \ldots, v^n e^{\lambda_n x}\right]\left[v^1, \ldots, v^n\right]^{-1}$$
and the general solution of $(18.6)$ can be written as
$$u(x)=\sum_{i=1}^n c_i v^i e^{\lambda_i x} .$$

## 数学代写|多变量微积分和常微分方程代考Multivariate Calculus & Ordinary Differential Equations代写|Periodic Linear Systems

A function $y(x)$ is called periodic of period $\omega>0$ if for all $x$ in the domain of the function
$$y(x+\omega)=y(x) .$$
Geometrically, this means that the graph of $y(x)$ repeats itself in successive intervals of length $\omega$. For example, the functions $\sin x$ and $\cos x$ are periodic of period $2 \pi$. For convenience, we shall assume that $\omega$ is the smallest positive number for which (20.1) holds. If each component $u_i(x), 1 \leq i \leq n$ of $u(x)$ and each element $a_{i j}(x), 1 \leq i, j \leq n$ of $A(x)$ are periodic of period $\omega$, then $u(x)$ and $A(x)$ are said to be periodic of period $\omega$. Periodicity of solutions of differential systems is an interesting and important aspect of qualitative study. Here we shall provide certain characterizations for the existence of such solutions of linear differential systems.

To begin with we shall provide necessary and sufficient conditions for the differential system (17.1) to have a periodic solution of period $\omega$.

Theorem 20.1. Let the matrix $A(x)$ and the function $b(x)$ be continuous and periodic of period $\omega$ in $\mathbb{R}$. Then the differential system (17.1) has a periodic solution $u(x)$ of period $\omega$ if and only if $u(0)=u(\omega)$.

Proof. Let $u(x)$ be a periodic solution of period $\omega$, then by definition it is necessary that $u(0)=u(\omega)$. To show sufficiency, let $u(x)$ be a solution of (17.1) satisfying $u(0)=u(\omega)$. If $v(x)=u(x+\omega)$, then it follows that $v^{\prime}(x)=u^{\prime}(x+\omega)=A(x+\omega) u(x+\omega)+b(x+\omega)=A(x) v(x)+b(x)$; i.e., $v(x)$ is a solution of (17.1). However, since $v(0)=u(\omega)=u(0)$, the uniqueness of the initial value problems implies that $u(x)=v(x)=u(x+\omega)$, and hence $u(x)$ is periodic of period $\omega$.

Corollary 20.2. Let the matrix $A(x)$ be continuous and periodic of period $\omega$ in $\mathbb{R}$. Further, let $\Psi(x)$ be a fundamental matrix of the differential system (17.3). Then the differential system (17.3) has a nontrivial periodic solution $u(x)$ of period $\omega$ if and only if $\operatorname{det}(\Psi(0)-\Psi(\omega))=0$.

Proof. We know that the general solution of the differential system (17.3) is $u(x)=\Psi(x) c$, where $c$ is an arbitrary constant vector. This $u(x)$ is periodic of period $\omega$ if and only if $\Psi(0) c=\Psi(\omega) c$, i.e., the system $(\Psi(0)-$ $\Psi(\omega)) c=0$ has a nontrivial solution vector $c$. But, from Theorem $13.2$ this system has a nontrivial solution if and only if $\operatorname{det}(\Psi(0)-\Psi(\omega))=0$.

# 多变量微积分和常微分方程代考

## 数学代写多变量微积分和常微分方程代考Multivariate Calculus \& Ordinary Differential Equations代写|Systems with Constant Coeffiffifficients

$$u^1(x)=v^1 e^{\lambda_1 x}, \quad \cdots \quad, u^n(x)=v^n e^{\lambda_n x}$$

$$\left(u^i(x)\right)^{\prime}=\left(v^i e^{\lambda_i x}\right)^{\prime}=\lambda_i v^i e^{\lambda_{i x}}=A v^i e^{\lambda_{i x}}=A u^i(x)$$

$$e^{A x}=\left[v^1 e^{\lambda_1 x}, \ldots, v^n e^{\lambda_n x}\right]\left[v^1, \ldots, v^n\right]^{-1}$$

$$u(x)=\sum_{i=1}^n c_i v^i e^{\lambda_i x} .$$

## 数学代写多变量凱积分和常微分方程代坒Multivariate Calculus \& Ordinary Differential Equations代胃|Periodic Linear Systems

$$y(x+\omega)=y(x) .$$

$v^{\prime}(x)=u^{\prime}(x+\omega)=A(x+\omega) u(x+\omega)+b(x+\omega)=A(x) v(x)+b(x) ; \mathbb{I}$ 。 $v(x)$ 是 (17.1) 的解。然而，由于 $v(0)=u(\omega)=u(0)$ ，初始值问题的唯一性意味着 $u(x)=v(x)=u(x+\omega)$ ，因此 $u(x)$ 是周期的周期 $\omega$.

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