non-homogeneous linear ode

给定线性变换 $\mathcal{L}:{\mathbb{R}}^n\to {\mathbb{R}}^n$$\mathcal{L}:\mathbb{R}^n\to\mathbb{R}^n$，以及 $f(t):\mathbb{R}\to {\mathbb{R}}^n$$f(t):\mathbb{R}\to\mathbb{R}^n$，待求 $y(t):\mathbb{R}\to {\mathbb{R}}^n$$y(t):\mathbb{R}\to\mathbb{R}^n$ 且 $y(0)=y_0$$y(0)=y_0$，
解出满足条件的 $y$$y$: $y^{\prime }(t)=\mathcal{L}(y(t))+f(t)$$\,y'(t)=\mathcal{L}(y(t))+f(t)$

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记 $\mathcal{L}$$\mathcal{L}$ 所对应的矩阵为 $A$$A$，令 $z(t):=e^{-tA}y(t)$$z(t):=e^{-tA}y(t)$

因为

$$\frac{d}{dt}(e^{-tA}y(t))=\frac{d}{dt}{e}^{-tA}y(t)+e^{-tA}{y}^{\prime }(t),\frac{d}{dt}{e}^{tA}=A{e}^{tA}=e^{tA}A,$$

所以

$$z^{\prime }(t)=(-A{e}^{-tA})y(t)+e^{-tA}(Ay(t)+f(t))=e^{-tA}f(t).$$

$0\to t$$0\to t$ 积分：

$$e^{-tA}y(t)-y_0={\int }_0^t{e}^{-sA}f(s)ds.$$

变形，两边同乘 $e^{tA}$$e^{tA}$：

$$\boxed{{\displaystyle y(t)=e^{tA}{y}_0+{\int }_0^t{e}^{(t-s)A}f(s)ds}}.$$

现在问题转化为计算 $e^{tA}$$e^{tA}$ 和 ${\displaystyle {\int }_0^t{e}^{(t-s)A}f(s)ds}$$\displaystyle \int_{0}^{t}e^{(t-s)A}f(s)\,ds$：

对 $A$$A$ 做 Jordan 分解，$A=PJ{P}^{-1}$$A=PJP^{-1}$，$J=\mathrm{diag}(J_{{\lambda }_1},J_{{\lambda }_2},\dots ,J_{{\lambda }_L})$$J=\mathrm{diag}\big(J_{\lambda_1},J_{\lambda_2},\ldots,J_{\lambda_L}\big)$

$$\begin{array}{rl}{x}_{{\lambda }_{\ell }}^{\prime }(t)& =J_{{\lambda }_{\ell }}{x}_{{\lambda }_{\ell }}(t)+{\tilde{f}}_{{\lambda }_{\ell }}(t),\ell =1,\dots ,L,\\ e^{tJ}& =\mathrm{diag}(e^{t{J}_{{\lambda }_1}},e^{t{J}_{{\lambda }_2}},\dots ,e^{t{J}_{{\lambda }_L}}),\\ e^{t{J}_{{\lambda }_{\ell }}}& =\underset{k=1}{\overset{m_{\ell }}{⨁}}{e}^{t{J}_{{\lambda }_{\ell }}^{(k)}},J_{{\lambda }_{\ell }}^{(k)}={\lambda }_{\ell }{I}_{r_{\ell ,k}}+N_{\ell ,k},(N_{\ell ,k}{)}^{r_{\ell ,k}}=0,\\ e^{t{J}_{{\lambda }_{\ell }}^{(k)}}& =e^{{\lambda }_{\ell }t}(I_{r_{\ell ,k}}+t{N}_{\ell ,k}+\frac{t^2}{2!}{N}_{\ell ,k}^2+\cdots +\frac{t^{r_{\ell ,k}-1}}{(r_{\ell ,k}-1)!}{N}_{\ell ,k}^{r_{\ell ,k}-1}),\\ f(t)& =e^{\alpha t}q,\tilde{q}:=P^{-1}q=({\tilde{q}}_{{\lambda }_1}|{\tilde{q}}_{{\lambda }_2}|\cdots |{\tilde{q}}_{{\lambda }_L}{)}^{\mathrm{\top }},\\ {\tilde{q}}_{{\lambda }_{\ell }}& =({\tilde{q}}_{{\lambda }_{\ell }}^{(1)};{\tilde{q}}_{{\lambda }_{\ell }}^{(2)};\dots ;{\tilde{q}}_{{\lambda }_{\ell }}^{(m_{\ell })}),{\tilde{q}}_{{\lambda }_{\ell }}^{(k)}\in {\mathbb{R}}^{r_{\ell ,k}},\\ {\int }_0^t{e}^{(t-s)J}{e}^{\alpha s}\tilde{q}ds& =\underset{\ell =1}{\overset{L}{⨁}}{\int }_0^t{e}^{(t-s)J_{{\lambda }_{\ell }}}{e}^{\alpha s}{\tilde{q}}_{{\lambda }_{\ell }}ds,\\ {\int }_0^t{e}^{(t-s)J}{e}^{\alpha s}\tilde{q}ds& =\underset{\ell =1}{\overset{L}{⨁}}\sum _{k=1}^{m_{\ell }}{e}^{\alpha t}\sum _{r=0}^{r_{\ell ,k}-1}\frac{N_{\ell ,k}^r}{r!}{I}_r(t;{\lambda }_{\ell }-\alpha ){\tilde{q}}_{{\lambda }_{\ell }}^{(k)},\\ I_0(t;c)& =\frac{e^{ct}-1}{c}(c\ne 0),I_r(t;c)=\frac{e^{ct}{t}^r}{c}-\frac{r}{c}{I}_{r-1}(t;c)(r\ge 1,c\ne 0),\\ I_r(t;0)& =\frac{t^{r+1}}{r+1}.\end{array}$$

注 : 利用Cayley–Hamilton定理可以绕过$A$$A$的Jordan分解 :

$${\chi }_A(\lambda )={\lambda }^n+c_{n-1}{\lambda }^{n-1}+\cdots +c_0,{\chi }_A(A)=0.$$

$$e^{tA}=\sum _{k=0}^{n-1}{\varphi }_k(t)A^k,A^n=-c_{n-1}{A}^{n-1}-\cdots -c_{0}I.$$

$$\{\begin{array}{l}{\varphi }_0^{\prime }(t)=-c_0{\varphi }_{n-1}(t),\\ {\varphi }_1^{\prime }(t)={\varphi }_0(t)-c_1{\varphi }_{n-1}(t),\\ ⋮\\ {\varphi }_{n-2}^{\prime }(t)={\varphi }_{n-3}(t)-c_{n-2}{\varphi }_{n-1}(t),\\ {\varphi }_{n-1}^{\prime }(t)={\varphi }_{n-2}(t)-c_{n-1}{\varphi }_{n-1}(t),\end{array}{\varphi }_0(0)=1,{\varphi }_1(0)=\cdots ={\varphi }_{n-1}(0)=0.$$

$$e^{tA}={\varphi }_0(t)I+{\varphi }_1(t)A,{\varphi }_0^{\prime }=-\delta {\varphi }_1,{\varphi }_1^{\prime }={\varphi }_0-\tau {\varphi }_1,{\varphi }_0(0)=1,{\varphi }_1(0)=0,$$

$$e^{tA}=e^{\frac{\tau t}{2}}[\cosh \left(\frac{\mathrm{\Delta }t}{2}\right)I+\frac{2}{\mathrm{\Delta }}\sinh \left(\frac{\mathrm{\Delta }t}{2}\right)(A-\frac{\tau }{2}I)],\mathrm{\Delta }=\sqrt{{\tau }^2-4\delta },\tau =\mathrm{tr}A,\delta =detA.$$