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# 数学代写|实分析代写Real Analysis代考|MA50400

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## 数学代写|实分析代写Real Analysis代考|Homogeneous Equations with Constant Coefficients

In this section and the next, we discuss first-order homogeneous linear systems with constant coefficients. The system is of the form $y^{\prime}=A y$ with $A$ a matrix of constants. A single homogeneous $n^{\text {th }}$-order linear equation with constant coefficients can be converted into such a first-order system and can therefore be handled by the method applicable to all first-order homogeneous linear systems with constant coefficients. But such an equation can be handled more simply in a direct fashion, and we therefore isolate in this section the case of a single $n^{\text {th }}$-order equation. This section and the next will make use of material on polynomials from Section A8 of the appendix.
The equation to be studied in this section is of the form
$$y^{(n)}+a_{n-1} y^{(n-1)}+\cdots+a_1 y^{\prime}+a_0 y=0$$
with coefficients in $\mathbb{C}$. Let us write this equation as $L(y)=0$ for a suitable linear operator $L$ defined on functions $y$ of class $C^n$ :
$$L=\left(\frac{d}{d t}\right)^n+a_{n-1}\left(\frac{d}{d t}\right)^{n-1}+\cdots+a_1\left(\frac{d}{d t}\right)+a_0 .$$
The term $a_0$ is understood to act as $a_0$ times the identity operator. Since $\frac{d}{d t} e^{r t}=$ $r e^{r t}$, we immediately obtain
$$L\left(e^{r t}\right)=\left(r^n+a_{n-1} r^{n-1}+\cdots+a_1 r+a_0\right) e^{r t} .$$
The polynomial
$$P(\lambda)=\lambda^n+a_{n-1} \lambda^{n-1}+\cdots+a_1 \lambda+a_0$$
is called the characteristic polynomial of the equation, and the formula $L\left(e^{r t}\right)=$ $P(r) e^{r t}$ shows that $y(t)=e^{r t}$ is a solution of $L(y)=0$ if and only if $r$ is a root of the characteristic polynomial. From Section A8 of the appendix, we know that the polynomial $P(\lambda)$ factors into the product of linear factors $\lambda-r$, the factors being unique apart from their order. Let us list the distinct roots, i.e., the distinct such complex numbers $r$, as $r_1, \ldots, r_k$ with $k \leq n$, and let us write $m_j$ for the number of times that $\lambda-r_j$ occurs as a factor of $P(\lambda)$, i.e., the multiplicity of $r_j$ as a root of $P$. Then we have $\sum_{j=1}^k m_j=n$ and
$$P(\lambda)=\prod_{j=1}^k\left(\lambda-r_j\right)^{m_j} .$$

## 数学代写|实分析代写Real Analysis代考|Homogeneous Systems with Constant Coefficients

Having discussed linear homogeneous equations with constant coefficients, let us pass to the more general case of first-order homogeneous linear systems with constant coefficients. We write the system as $y^{\prime}=A y$ with $A$ an $n$-by- $n$ matrix of constants. In principle we can solve the system immediately. Namely, Proposition $3.13 \mathrm{c}$ tells us that $\frac{d}{d t}\left(e^{t A}\right)=A e^{t A}$, so that each of the $n$ columns of $e^{t A}$ is a solution of $y^{\prime}=A y$. At $t=0, e^{t A}$ reduces to the identity matrix, and thus these $n$ solutions are linearly independent at $t=0$. By Theorem 4.6 these $n$ solutions form a basis of all solutions on any subinterval $(a, b)$ of $(-\infty,+\infty)$. The solution satisfying the initial condition $y\left(t_0\right)=y_0$ is $y(t)=e^{t A} e^{-t_0 A} y_0$, which is the particular linear combination $\sum_{j=1}^n c_j e^{t A} e_j$ of the columns of $e^{t A}$ in which $c_j$ is the number $c_j=\left(e^{-t_0 A} y_0\right)_j$.
In practice it is not so obvious how to compute $e^{t A}$ except in special cases in which the exponential series can be summed entry by entry. Let us write down three model cases of this kind, and ultimately we shall see that we can handle general $A$ by working suitably with these cases.
MODEL CASES.
(1) Let
$$C=\left(\begin{array}{ccccccc} 0 & 1 & 0 & 0 & \cdots & 0 & 0 \ & 0 & 1 & 0 & \cdots & 0 & 0 \ & & 0 & 1 & \cdots & 0 & 0 \ & & & \ddots & \ddots & \vdots & \vdots \ & & & & 0 & 1 & 0 \ & & & & & 0 & 1 \ & & & & & & 0 \end{array}\right)$$
be of size $m$-by- $m$ with 0 ‘s below the main diagonal. Raising $C$ to powers, we see that the $(i, j)^{\text {th }}$ entry of $A^k$ is 1 if $j=i+k$ and is 0 otherwise. Hence
$$e^{t C}=\left(\begin{array}{ccccccc} 0 & t & \frac{1}{2 !} t^2 & \frac{1}{3 !} t^3 & \cdots & \frac{1}{(m-2) !} t^{m-2} & \frac{1}{(m-1) !} t^{m-1} \ & 0 & t & \frac{1}{2 !} t^2 & \cdots & \frac{1}{(m-3) !} t^{m-3} & \frac{1}{(m-2) !} t^{m-2} \ & & 0 & t & \cdots & \vdots & \frac{1}{(m-3) !} t^{m-3} \ & & & \ddots & \ddots & & \vdots \ & & & 0 & t & \frac{1}{2 !} t^2 \ & & & & 0 & t \ & & & & & & 0 \end{array}\right)$$
with 0 ‘s below the main diagonal.
(2) Let
$$A=\left(\begin{array}{ccccccc} a & 1 & 0 & 0 & \cdots & 0 & 0 \ & a & 1 & 0 & \cdots & 0 & 0 \ & & a & 1 & \cdots & 0 & 0 \ & & & \ddots & \ddots & \vdots & \vdots \ & & & & a & 1 & 0 \ & & & & & a & 1 \ & & & & & & a \end{array}\right),$$

## 数学代写|实分析代写Real Analysis代考|Homogeneous Equations with Constant Coefficients

$$y^{(n)}+a_{n-1} y^{(n-1)}+\cdots+a_1 y^{\prime}+a_0 y=0$$

$$L=\left(\frac{d}{d t}\right)^n+a_{n-1}\left(\frac{d}{d t}\right)^{n-1}+\cdots+a_1\left(\frac{d}{d t}\right)+a_0 .$$

$$P(\lambda)=\prod_{j=1}^k\left(\lambda-r_j\right)^{m_j} .$$

## 数学代写|实分析代写Real Analysis代考|Homogeneous Systems with Constant Coefficients

(1)让
$$C=\left(\begin{array}{ccccccc} 0 & 1 & 0 & 0 & \cdots & 0 & 0 \ & 0 & 1 & 0 & \cdots & 0 & 0 \ & & 0 & 1 & \cdots & 0 & 0 \ & & & \ddots & \ddots & \vdots & \vdots \ & & & & 0 & 1 & 0 \ & & & & & 0 & 1 \ & & & & & & 0 \end{array}\right)$$

$$e^{t C}=\left(\begin{array}{ccccccc} 0 & t & \frac{1}{2 !} t^2 & \frac{1}{3 !} t^3 & \cdots & \frac{1}{(m-2) !} t^{m-2} & \frac{1}{(m-1) !} t^{m-1} \ & 0 & t & \frac{1}{2 !} t^2 & \cdots & \frac{1}{(m-3) !} t^{m-3} & \frac{1}{(m-2) !} t^{m-2} \ & & 0 & t & \cdots & \vdots & \frac{1}{(m-3) !} t^{m-3} \ & & & \ddots & \ddots & & \vdots \ & & & 0 & t & \frac{1}{2 !} t^2 \ & & & & 0 & t \ & & & & & & 0 \end{array}\right)$$

(2)让
$$A=\left(\begin{array}{ccccccc} a & 1 & 0 & 0 & \cdots & 0 & 0 \ & a & 1 & 0 & \cdots & 0 & 0 \ & & a & 1 & \cdots & 0 & 0 \ & & & \ddots & \ddots & \vdots & \vdots \ & & & & a & 1 & 0 \ & & & & & a & 1 \ & & & & & & a \end{array}\right),$$

## Matlab代写

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