System of Linear Differential Equations

线性微分方程组 | System of Linear Differential Equations (LODEs)

This chapter we focus on

\[ \begin{equation} \frac{d \symbfit{X}(t)}{dt} = \symbfit{A}(t)\symbfit{X}(t)+\symbfit{B}(t) \label{eq-LODEs} \end{equation} \]

with initial condition

\[ \begin{equation} \symbfit{X}(t_0)=\symbfit{X}_0 \label{eq-initial-LODEs} \end{equation} \]

where \(t_0\in I=(a,b)\), \(\symbfit{X}_0 = (x_1^0,x_n^0,\cdots, x_n^0)^T\) is a given constant vector.

线性微分方程组解的存在唯一性 | Existence and Uniqueness of LODEs

This is quite similar to Picard Theorem in chapter Existence and Uniqueness Theorem, but it is still useful to give a special form of Picard Sequence for LODEs, which is also an approximation to solving it.

线性微分方程组解的存在唯一性定理 | Theorem of Existence and Uniqueness of LODEs

LODEs \(\ref{eq-LODEs}\) with initial condition \(\ref{eq-initial-LODEs}\) has only one solution on interval \(I\).

Prove it.

Hints

We have to measure the distance in matrix. Now we need to give a definition of norm of vectors and matrixes(to see more details in Norm of vectors and matrixes) in Numerical Analysis.

\[ \Vert\mathbfit{X} \Vert = \sum_{i=1}^{n}|x_i|, \quad \Vert\mathbfit{A} \Vert = \sum_{i=1}^{n}\sum_{j=1}^{n}|{a_{ij}}| \]

It is easy to see that ...

• convert LODEs into its equivalent integral equations.

\[ \mathbfit{X}(t) = \mathbfit{X}_0 + \int_{t_0}^{t}\left[ \mathbfit{A}(s)\mathbfit{X}(s)+\mathbfit{B}(s)\right] ds \]

• formulate Picard Sequence.

Define:

\[ \begin{align} \mathbfit{X}_0(t) &= \mathbfit{X}_0 \nonumber\\ \mathbfit{X}_1(t) &= \mathbfit{X}_0 + \int_{t_0}^{t}\left[ \mathbfit{A}(s)\mathbfit{X}_0(s)+\mathbfit{B}(s)\right] ds \nonumber\\ \mathbfit{X}_2(t) &= \mathbfit{X}_0 + \int_{t_0}^{t}\left[ \mathbfit{A}(s)\mathbfit{X}_1(s)+\mathbfit{B}(s)\right] ds \nonumber\\ &\vdots \nonumber\\ \mathbfit{X}_n(t) &= \mathbfit{X}_0 + \int_{t_0}^{t}\left[ \mathbfit{A}(s)\mathbfit{X}_{n-1}(s)+\mathbfit{B}(s)\right] ds \label{eq-LODEs-integral}\\ \end{align} \]

Consider similarly and we can say the above sequence is well-defined.

• Prove Picard Sequence convergent.

denote

\[ C = \sup_{s\in J}\Vert\mathbfit{A}(s)\Vert, \quad D = C\Vert\mathbfit{X}(s)\Vert + \sup_{s\in J}\Vert\mathbfit{B}(s)\Vert \]

we can get

\[ \begin{align*} \Vert \mathbfit{X}_1(t) - \mathbfit{X}_0(t) \Vert &\leq D |t-t_0|\\ \Vert\mathbfit{X}_2(t) - \mathbfit{X}_1(t) \Vert &\leq \int_{t_0}^{t} \Vert \mathbfit{A}(s) \Vert \Vert \mathbfit{X}_{1}(s)- \mathbfit{X}_{0}(s) \Vert ds \\ &\leq \int_{t_0}^{t} C D |s-t_0| ds = \frac{D}{C} \frac{(C|t-t_0|)^2}{2}\\ &\vdots\\ \Vert\mathbfit{X}_n(t) - \mathbfit{X}_{n-1}(t) \Vert &\leq \frac{D}{C} \frac{(C|t-t_0|)^{n}}{(n)!} \end{align*} \]

which shows the Picard Sequence converges.

• Prove the convergent function is solution of LODEs \(\ref{eq-LODEs}\).

If we denote \(\mathbfit{X}(t) = \lim_{n\rightarrow \infty}\mathbfit{X}_n(t)\) and let \(n\rightarrow \infty\) on both sides of integral equation \(\ref{eq-LODEs-integral}\), we get

\[ \mathbfit{X}(t) = \mathbfit{X}_0 + \int_{t_0}^{t}\left[ \mathbfit{A}(s)\mathbfit{X}(s)+\mathbfit{B}(s)\right] ds \]

which is a solution.

• prove uniqueness.

Similar to proof in Picard Theorem.

二阶方程边值问题 | Boundary Problem of Second-Order LODE

This pattern we focus on LODE

\[ \begin{equation} y''+p(x)'+q(x)y = f(x) \label{eq: BP-SecondOrder} \end{equation} \]

with Boundary Condition \(y(a)=\alpha, y(b)=\beta\), where \(p(x), q(x) \in C^1[a, b]\)

• H-LODE

共轭点 | Conjugate Point

If homogeneous LODE(H-LODE)

\[ \begin{equation} y''+p(x)'+q(x)y = 0 \label{eq: BP-SO-H} \end{equation} \]

with boundary condition \(y(a)=0, y(b)=0\), has non-zero solution, then \(\{a, b\}\) is called the Conjugate Point of the H-LODE.

We usually take use of the following method to check if the boundary point will induce indefinite solutions or no solutions.

共轭点的充要条件 | Necessary and Sufficient Condition for Conjugate Point

\(\{a, b\}\) is the Conjugate Point of H-LODE, if and only if \(\forall y_1, y_2\) of the solution of H-LODE, which are linear irrelevant, satisfies

\[ \left| \begin{array}{cc} y_1(a)& y_2(a)\\ y_1(b)& y_2(b) \end{array} \right| =0 \]

Hints

substitute the boundary condition and we get two linear equation system for parameters \(c_1, c_2\). And the above is the determinant of the system.

齐次方程存在唯一解的充要条件 | Necessary and Sufficient Condition for existing only one solution for H-LODE

H-LODE \(\ref{eq: BP-SO-H}\) with boundary point \(y(a)=\alpha, y(b)=\beta\) has only one solution, if and only if \(\{a, b\}\) is not the conjugate point of the H-LODE.

Hints

Focus on the determinant of the linear irelevantly solutions of \(y_1, y_2\).

• Non-H-LODE

We partition the solution of LODE \(\ref{eq: BP-SecondOrder}\) into three parts.

线性非齐次方程存在唯一解的充分条件 | Sufficient Condition for existing only one solution for Non-H-LODE

If \(\{a, b\}\) is not the conjugate point of H-LODE \(\ref{eq: BP-SO-H}\), then Non-H-LODE \(\ref{eq: BP-SecondOrder}\) has only one solution.

There are two ways to prove it.

Version 1Version 2

Assume \(y_1(x)\) is a solution of H-LODE

\[ \begin{equation} y'' + p(x)y' +q(x)y = 0, \quad y(a) = 0, \quad y(b) = 1 \label{eq: BP-1} \end{equation} \]

and \(y_2(x)\) is a solution of H-LODE

\[ \begin{equation} y'' + p(x)y' +q(x)y = 0, \quad y(a) = 1, \quad y(b) = 0 \label{eq: BP-2} \end{equation} \]

and \(y_3(x)\) is a solution of Non-H-LODE

\[ \begin{equation} y'' + p(x)y' +q(x)y = f(x), \quad y(a) = 0, \quad y(b) = 0 \label{eq: BP-3} \end{equation} \]

and the solution of LODE \(\ref{eq: BP-SecondOrder}\) can be represented as

\[ y = \alpha y_1 + \beta y_2 + y_3 \]

where \(y_1, y_2\) are linearly irrelevant because of existence and uniqueness theorem.

We can get \(y_3\) through Variation of Constant. That is, let \(y_3 = u_1(x) y_1(x)+u_2(x)y_2(x)\), then

\[ u_1 = \int \frac{-y_2 f}{W}dt, \quad u_2 = \int \frac{y_1 f}{W}dt \]

Substitute the boundary condition \(\ref{eq: BP-1}, \ref{eq: BP-2}, \ref{eq: BP-3}\) we get

\[ u_2(a)=0, \quad u_1(b)=0 \]

Based on this, we can transform the \(u_1(x), u_2(x)\) to definite integral whose upper limit of integral is variable

\[ u_1 = \int_{x}^{b} \frac{y_2 f}{W}dt, \quad u_2 = \int_{x}^{a} \frac{y_1 f}{W}dt \]

So we can write particular solution

\[ y_3 = y_1(x) \int_{x}^{b} \frac{y_2(t) f(t)}{W(t)}dt + y_2(x)\int_{x}^{a} \frac{y_1(x) f(x)}{W(x)}dt \]

If we define Green Function as

\[ G(x,t)= \begin{cases} \displaystyle \frac{y_2(x)y_1(t)}{W(t)}, \quad a\leq t\leq x \\ \displaystyle \frac{y_1(x)y_2(t)}{W(t)}, \quad x\leq t\leq b \end{cases} \]

then

\[ y_3(x) = \int_{a}^{b}G(x, t)f(t)dt \]

According to the textbook.

S-L 边值问题 | Sturm-Liouville Boundary Problem

Solve for PDE

\[ \begin{cases} \displaystyle u_{tt} = a^2 u_{xx} , \quad 0\leq x\leq L, t\geq 0\\ \displaystyle u|_{x=0} = u|_{t=0} = 0 \end{cases} \]

Assume we can seperate the variables \(x, t\). That is, let \(u = X(x)T(t)\) and substitute, we get

\[ \begin{align*} X(x)T''(t)&=a^2X''(x)T(t)\\ \Rightarrow \frac{T''(t)}{a^2T(t)} = \frac{X''(x)}{X(x)} &\overset{\Delta}{=}constant =-\lambda \end{align*} \]

Thus, we have to find \(T(t), X(x)\) such that

\[ T''(t)+a^2\lambda T(t)= 0 \]

\[ \begin{cases} X''(x) + \lambda X(x)= 0 \\ X(0)=X(L)=0 \end{cases} \]