# Solving the given equation

How to find general solution of matrix differential equation $\textbf{X}'(t)=\textbf{A}\textbf{X}(t)+\textbf{B}(t)$ by Green's Function/Matrix Method?

In here, $0<t<t_{max}$ \begin{aligned} &\textbf{X}(t)=\left( \begin{array}{cccc} x_{1}(t) \\ x_{2}(t) \\ x_{3}(t)\\ x_{4}(t) \end{array} \right), \textbf{A} = \left( \begin{array}{cccc} a_{11} & a_{12} &a_{13} & a_{14}\\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32}& a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{array} \right),\\ &\textbf{B}(t)=\left( \begin{array}{cccc} b_{1}(t)\\ b_{2}(t)\\ b_{3}(t)\\ b_{4}(t) \end{array} \right). \end{aligned} initial conditions: $$x_1(0)=m_1$$ $$x_2(0)=m_2$$ $$x_3(0)=m_3$$ $$x_4(0)=m_4.$$

$\lvert A \rvert >0$ and $A$ has four distinct complex eigenvalues $\lambda_1=c i$, $\lambda_2=-c i$, $\lambda_3=d i$, $\lambda_4=-d i$ where $a_{ij},c,d\in \mathbb{R}$ for $i,j=1,2,3,4.$ Respectively, the eigenvectors are $v_k$ for $k=1,2,3,4.$

• – LutzL Jun 5 '17 at 7:49

It is rather simple, however, there will be a Green's matrix (tensor) not a green's function. The solution to the equation $$X'-AX=B$$ May be written as $$X(t)=X_{0}(t)+\int\mathcal{G}(t, t')B(t')dt'$$ Where $$X'_{0}(t)-AX_{0}(t)=0$$ and $\mathcal{G}(t, t')$ is a $4\times4$ matrix satisfying $$\frac{d}{dt}\mathcal{G}(t, t')-A\mathcal{G}(t, t')=I\delta(t-t')$$ Where $I$ is the identity. One way to find $\mathcal{G}(t, t')$ is via the Fourier transform $$\mathcal{G}(t, t')=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat{\mathcal{G}}(\omega, t')e^{i\omega{t}}d\omega$$ and $$\delta(t-t')=\frac{1}{2\pi}\int_{\mathbb{R}}e^{i\omega(t-t')}d\omega$$ Which gives $$[i\omega{I}-A]\hat{\mathcal{G}}(\omega, t')=I\frac{1}{\sqrt{2\pi}}e^{-i\omega{t'}}$$ Hence $$\hat{\mathcal{G}}(\omega, t')=\frac{1}{\sqrt{2\pi}}[i\omega{I}-A]^{-1}e^{-i\omega{t'}}$$ and $$\mathcal{G}(t, t')=\frac{1}{2\pi}\int_{\mathbb{R}}[i\omega{I}-A]^{-1}e^{i\omega(t-t')}d\omega$$ Where you also need to invert the matrix $[i\omega{I}-A]$. Another approach would be to diagonalize the matrix $A$, $A=S\Lambda{S}^{-1}$ (which you've already done), then you do the transformation $\mathcal{G}\rightarrow\mathcal{G}'=S^{-1}\mathcal{G}S$ and you obtain $$\frac{d}{dt}\mathcal{G}'(t, t')-\Lambda{\mathcal{G}'(t, t')}=\delta(t-t')$$ Or component wise $$\mathcal{G}'_{lm}(t, t')=X_{0}(t), \ l\neq{m}$$ and $$\frac{d}{dt}\mathcal{G}'_{ll}(t, t')-\Lambda_{ll}\mathcal{G}'_{ll}(t, t')=\delta(t-t')$$ Which is easy to solve...
• Is it not much simpler to look at the solution formula $X(t)=U(t)X(0)+U(t)\int_0^tU(t')^{-1}B(t')dt'$ to find $G(t,t')=U(t)U(t')^{-1}$? – LutzL Jun 5 '17 at 12:13