ODE and PDE, Systems of ODEs
April 11, 2021
1. Systems of first-order linear equations
Equations of higher order can always be transformed into systems of first order equations. To transform an arbitray \(n^{th}\) order equation:
\[y^{(n)}=F\left(t, y, y^{\prime}, \ldots, y^{(n-1)}\right)\tag{1}\]into a system of \(n\) first-order differential equations by introducing the variables \(x_{1}, x_{2}, \ldots, x_{n}\) defined by:
\[x_{1}=y, x_{2}=y^{\prime}, x_{3}=y^{\prime \prime}, \ldots, x_{n}=y^{(n-1)} \tag{2}\]It follows immediately that:
\[\begin{array}{c} x_{1}^{\prime}=x_{2} \\ x_{2}^{\prime}=x_{3} \\ \vdots\\ x_{n-1}^{\prime}=x_{n} \\ x_{n}^{\prime}=F\left(t, x_{1}, x_{2}, \ldots, x_{n}\right) \end{array} \tag{3}\]A more general case of such n first-order linear differential equation has the form:
\[\begin{aligned} x_{1}^{\prime} &=p_{11}(t) x_{1}+\cdots+p_{1 n}(t) x_{n}+g_{1}(t), \\ x_{2}^{\prime} &=p_{21}(t) x_{1}+\cdots+p_{2 n}(t) x_{n}+g_{2}(t), \\ &\vdots \\ x_{n}^{\prime} &=p_{n 1}(t) x_{1}+\cdots+p_{n n}(t) x_{n}+g_{n}(t) . \end{aligned} \tag{4}\]If each of the functions \(g_{1}(t), \ldots, g_{n}(t)\) is zero for all \(t\) in the interval \(I\), then the system (4) is said to be homogeneous; otherwise, it is nonhomogeneous.
A homogeneous first-order linear system can be further expressed in a matrice form:
\[\mathbf{x}^{\prime}=\mathbf{A x} \tag{5}\]2. Homogeneous Linear System with Constant Coefficients
The general solution to \(\mathbf{x}^{\prime}=\mathbf{A x}\) is \(\mathbf{x} = \mathbf{\xi}e^{rt}\), where \(r\) is an eigenvalue and \(\boldsymbol{\xi}\) an associated eigenvector of the coefficient matrix \(\mathbf{A}\).
- Real and different eigenvalues
If the \(n\) eigenvalues of matrix \(\mathbf{A}\) are all real and different. Thus, associated with each eigenvalue \(r_{i}\) is a real eigenvector \(\boldsymbol{\xi}^{(i)}\), and the \(n\) eigenvectors \(\boldsymbol{\xi}^{(1)}, \ldots, \boldsymbol{\xi}^{(n)}\) are linearly independent. The corresponding general solution of the differential system equation (5) is:
\[\mathbf{x}=c_{1} \boldsymbol{\xi}^{(1)} e^{r_{1} t}+\cdots+c_{n} \boldsymbol{\xi}^{(n)} e^{r_{n} t} \tag{6}\]- Complex-valued eigenvalues
Suppose that there is one pair of complex conjugate eigenvalues of \(\mathbf{A}\), \(r_{1}=\lambda+i \mu\) and \(r_{2}=\lambda-i \mu .\) Then the corresponding eigenvectors \(\boldsymbol{\xi}^{(1)}\) and \(\boldsymbol{\xi}^{(2)}\) are also complex conjugates. Let us write \(\boldsymbol{\xi}^{(1)}=\mathbf{a}+i \mathbf{b}\), where \(\mathbf{a}\) and \(\mathbf{b}\) are real; then we can write: \(\mathbf{x}^{(1)}(t)=\mathbf{u}(t)+i \mathbf{v}(t)\), where:
\[\begin{array}{l} \mathbf{u}(t)=e^{\lambda t}(\mathbf{a} \cos (\mu t)-\mathbf{b} \sin (\mu t)) \\ \mathbf{v}(t)=e^{\lambda t}(\mathbf{a} \sin (\mu t)+\mathbf{b} \cos (\mu t)) \end{array}\]Assume all other eigenvalues \(r_3,\dots,r_n\) are real and distinct, then the general solution is:
\[\mathbf{x}=c_{1} \mathbf{u}(t)+c_{2} \mathbf{v}(t)+c_{3} \boldsymbol{\xi}^{(3)} e^{r_{3} t}+\cdots+c_{n} \boldsymbol{\xi}^{(n)} e^{r_{n} t} \tag{7}\]- Repeated eigenvalues
When the matrix \(\mathbf{A}\) has a repeated eigenvalue, say that \(r=\rho\) is an \(m\)-fold eigenvalue. In this event, there are two possibilities: either there are \(m\) linearly independent eigenvectors corresponding to the eigenvalue \(\rho\), or else, there are fewer than \(m\) linearly independent eigenvectors.
In the first case, it makes no difference that the eigenvalue \(r = \rho\) is repeated as the distinct case, still has the general solution like equation (6).
In the second case, if the matrix \(\mathbf{A}\) is not Hermitianfor example \(\mathbf{A}= \left(\begin{array}{rr} 1 & -1 \\ 1 & 3 \end{array}\right)\) , then there may be fewer than \(m\) idependent eigenvevtors corresponding to the eigenvalue \(\rho\). We have dealt with similar situations for linear equation here.
Suppose that \(r=\rho\) is a double eigenvalue of \(\mathbf{A}\), but that there is only one corresponding eigenvector \(\boldsymbol{\xi} .\) Then one solution is:
\[\mathbf{x}^{(1)}(t)=\boldsymbol{\xi} e^{\rho t}\]We need to find a second solution:
\[\mathbf{x}^{(2)}(t)=\boldsymbol{\xi} t e^{\rho t}+\boldsymbol{\eta} e^{\rho t}\]where \(\eta\)The vector \(\eta\) is called a generalized eigenvector of the matrix \(\mathbf{A}\) corresponding to the eigenvalue \(\rho\). is determined from:
\[(\mathbf{A}-\rho \mathbf{I}) \boldsymbol{\eta}=\boldsymbol{\xi}\]