我想分享关於ode的variation of parameters中 不需要预先猜测把原本线性组合的常数换成函数 也能推导出相同结果的想法: 考虑一阶ode形如: x'(t) = A(t)x(t)+b(t), 其中x, b为n维向量，A为nxn矩阵 唯一的变数为t (Case i) 如果A(t)是常数，即A(t)≡A for some nxn matrix A. 那我们可以模仿解纯量一阶ode的想法: x'(t)+a(t)x(t)=b(t) ∫a(t)dt 两边同乘积分因子: I(t)=e 原方程式变为: (I(t)x(t))'=I(t)b(t) 然後积分後除上I(t)就是答案 向量版本也一样: x'(t)-Ax(t) = b(t) -At -At -At 同乘 I(t) = e => (e x(t))' = e b(t) At -At => x(t) = e ∫e b(t) dt 这个长得跟variation of parameters的公式是一样的 那麽... A(t)若不是常数的话呢? 我们试看看找出积分因子: x'(t)-A(t)x(t) = b(t) I(t)x'(t)-I(t)A(t)x(t) = I(t)b(t) 我们希望左式变成(I(t)x(t))'= I(t)x'(t)+I'(t)x(t) 可能的选择是I'(t) = -I(t)A(t) -∫A(t)dt 这让我们也许会想说是不是I(t)=e 呢? 但这不总是可以这样，只有: (Case ii) 若A(s)A(t) = A(t)A(s)成立 -∫A(t)dt ∞ (-∫A(t)dt)^n I(t) = e = Σ --------------- n=0 n! ∞ n(-∫A(t)dt)^(n-1) ∞ (-∫A(t)dt)^n =>I'(t) = -Σ ------------------- A(t) = -Σ -------------- A(t)=-I(t)A(t) n=1 n! n=0 n! (注意:可以这样做的原因是因为A(s)A(t) = A(t)A(s)，在套n次product rule时， 这些含有积分的n-1项才会相等) -∫A(t)dt 所以可以发现这个情况下，真的就是上面说的e 也就是我们可以把原式化成: (I(t)x(t))'=I(t)b(t) -1 ∫A(t)dt -∫A(t)dt => x(t) = I (t)∫I(t)b(t)dt = e ∫e b(t)dt 这个也跟variation of parameters的公式长得很像 但若是A(s)A(t) = A(t)A(s)不成立，我们不能把它合成像上面那样，例如n=2时， t t d(-∫A(s)ds)^2 d(∫A(s)ds)(∫A(t)dt) t t --------------- = ---------------------- = A(t)∫A(s)ds+∫A(s)ds A(t) dt dt t t 这不能变成2A(t)∫A(s)ds或是2∫A(s)dsA(t) 因为A(t)A(s)≠A(s)A(t) 那这时候要怎麽办呢? 答案是把积分因子换成fundamental matrix A fundamental matrix of system X'(t) = A(t)X(t) is a solution to this system with each columns being linearly independent.(X,A都是nxn矩阵) 也可以看出fundamenal matrix的column构成x'(t)=A(t)x(t)(A是上面那个矩阵， x是n维向量)解空间的basis 对於x'(t)=A(t)x(t)+b(t)，x,b为n维向量，A为nxn矩阵.令Φ(t)是fundamental matrix => Φ'(t)=A(t)Φ(t) -1 => A(t) = Φ'(t)Φ (t) (因为根据定义Φ的column线性独立，所以可逆) 照前面所说的，我们想要找出I'(t)=-I(t)A(t)的I，把上面的关系式带进去， -1 我们想要找出I，使得I'(t)=-I(t)Φ'(t)Φ (t)成立 => I'(t)Φ(t)=-I(t)Φ'(t) => I'(t)Φ(t)+I(t)Φ'(t) = O => (I(t)Φ(t))' = O (by product rule) => I(t)Φ(t) = C for some nxn constant matrix C -1 => I(t) = CΦ (t) -1 Now set C = I, then I(t) = Φ (t) -1 -1 所以原方程式可变成(Φ (t)x(t))' = Φ (t)b(t) -1 => x(t) = Φ(t)∫Φ (t)b(t)dt 这个就是我们熟悉的variation of parameters的公式 注意到我没有先假设特解要长成Φ(t)c(t)然後把c(t)解出来 我只假设了I'(t) = -I(t)A(t)成立(矩阵基本上不能直接消AC=BC不一定A=B， 所以这个是我假设可以这样，然後做看看) 这也可以解释为什麽书本要介绍fundamental matrix的概念 fundamental matrix不只是为了解X'(t)=A(t)X(t), X,A为矩阵而要引进 也不是为了能把x'(t)=A(t)x(t), A为矩阵, x为向量表示得比较好看而引进 -∫A(t)dt 我觉得最重要的原因是因为在找积分因子的过程，无法用e 当积分因子 但我们还是希望有如纯量一阶ode的结果，所以把那个积分因子换成fundamental matrix 来解决这个问题，所以才需要引进 而这个矩阵向量版的variation of parameters也可以帮助我们推出n阶线性方程的 variation of parameters，并且不用突然设定奇怪的 (k) (k) (k) c'(t)x (t)+c'(t)x (t)+...+c'(t)x (t) = 0, k < n-2 1 1 2 2 n n 这种条件，然後解出c1,...cn 过程如下: (n) (n-1) (n-2) 考虑x (t) +a (t)x (t)+a (t)x (t) + ... a (t)x(t) = f(t) n n-1 1 已知x (t),x (t), ... ,x (t)为n linearly independent solutions of the 1 2 n corresponding homogeneos equation: (n) (n-1) (n-2) x (t) + a (t)x (t) + a x (t) + ... + a (t)x(t) = 0 n n-1 1 Then, set (n-1) y(t) = [y (t)], where y (t) = x(t), y (t) = x'(t),..., y (t) = x (t) 1 1 2 n [y (t)] 2 ... [y (t)] n A(t) = [ 0 1 0 ... 0 ] [ 0 ] [ 0 0 1 ... 0 ] [ 0 ] [ ... ], b(t) = ... [ 0 0 ...0 1 ] [ 0 ] [-a (t) -a (t) ... -a (t)] [f(t)] 1 2 n 则观察可得: (n) y'(t) = x'(t) = y (t), y'(t) = x''(t) = y (t), ... y'(t) = x (t) 1 2 2 3 n (n) (n-1) x (t) = -a (t)x(t)-a (t)x'(t)-...-a (t)x (t) + f(t) 1 2 n =-a (t)y (t)-a (t)y (t)-...-a (t)y (t) + f(t) 1 1 2 2 n n y'(t) = A(t)y(t)+b(t) 而[x (t) ] [x (t) ] [x (t) ] 1 2 n [x'(t) ] [x'(t) ] [x'(t) ] 1 2 n 为 n linearly independent solution of ... ... ... ... (n-1) (n-1) (n-1) [x (t)] [x (t)] [x (t)] 1 2 n the system y'(t) = A(t)y(t) 所以Φ(t) = [x (t) x (t) ... x (t) ] 1 2 n [x'(t) x'(t) ... x'(t) ] 为fundamental matrix of the system 1 2 n ... (n-1) (n-1) (n-1) [x (t) x (t) ... x (t)] 1 2 n Y'(t) = A(t)Y(t), Y(t)为nxn矩阵 利用variation of parameters: -1 y(t) = Φ(t)∫Φ (t)b(t)dt -1 -1 Set c(t) = ∫Φ (t)b(t)dt => c'(t) = Φ (t)b(t) => Φ(t)c'(t) = b(t) 用矩阵乘法乘开，第一个分量就是: y(t) = c (t)x (t) + c (t)x (t) + ... + c (t)x (t) 1 1 2 2 n n 而Φ(t)c'(t) = b(t)把左式用矩阵乘法乘开和右边比较系数可得: c '(t)x (t) + c'(t)x (t) + ... + c'(t)x (t) = 0 1 1 2 2 n n c'(t)x' (t) + c'(t)x'(t) + ... + c'(t)x'(t) = 0 1 1 2 2 n n ... (n-2) (n-2) (n-2) c'(t)x (t) + c'(t)x (t) + ... + c'(t)x (t) = 0 1 1 2 2 n n (n-1) (n-1) (n-1) c'(t)x (t) + c'(t)x (t) + ... + c'(t)x (t) = f(t) 1 1 2 2 n n 这个就是我们在variation of parameters所设定的n个条件 我认为这种讲法会让做variation of parameters变得更加自然 不需要做过多的猜测 希望这篇文章对各位有帮助~~ --

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