1.（上册）Page 2-2-10：2.3.3 三维射线方程（即从Fermat 原理出 发导出三维射线追踪方程，用变分方法）
射线方程为： x = x ( τ ) x=x(\tau) x=x(τ)，则 d s = ∣ x ˙ ( τ ) ∣ d τ ds=|\dot x(\tau)|d\tau ds=∣x˙(τ)∣dτ
三维射线模型：
J ( x ) = ∫ μ ( x ) d s = ∫ τ 0 τ 1 μ ( x ) ∣ x ˙ ( τ ) ∣ d τ = min \begin{aligned} J(x)&=\int \mu(x)ds \\ &=\int_{\tau_0}^{\tau_1}\mu(x)|\dot x(\tau)|d\tau=\min \end{aligned} J(x)=∫μ(x)ds=∫τ0τ1μ(x)∣x˙(τ)∣dτ=min
Euler 方程为： ∂ L ∂ x i − d d τ ∂ L ∂ x ˙ i = 0 \displaystyle \frac{\partial{L}}{\partial x_i}-\frac d{d\tau}\frac{\partial L}{\partial \dot x_i}=0 ∂xi∂L−dτd∂x˙i∂L=0
L 函数为： L = μ ( x ) ∣ x ˙ ( τ ) ∣ L=\mu(x)|\dot x(\tau)| L=μ(x)∣x˙(τ)∣，则 ∂ L ∂ x i = ∂ μ ∂ x i ∣ x ˙ ( τ ) ∣ \displaystyle \frac{\partial L}{\partial x_i}=\frac{\partial \mu}{\partial x_i}|\dot x(\tau)| ∂xi∂L=∂xi∂μ∣x˙(τ)∣， ∂ L ∂ x ˙ i = μ ( x ) x ˙ i ∣ x ˙ ∣ \displaystyle \frac{\partial L}{\partial \dot x_i}=\mu(x)\frac{ \dot x_i}{ |\dot x|} ∂x˙i∂L=μ(x)∣x˙∣x˙i
代入 Euler 方程得到： ∂ μ ∂ x i ∣ x ˙ ( τ ) ∣ − d d τ [ μ ( x ) x ˙ i ∣ x ˙ ∣ ] = 0 \displaystyle \frac{\partial \mu}{\partial x_i}|\dot x(\tau)|-\frac d{d\tau}\left[\mu(x)\frac{ \dot x_i}{|\dot x|}\right]=0 ∂xi∂μ∣x˙(τ)∣−dτd[μ(x)∣x˙∣x˙i]=0
d d τ [ μ ( x ( τ ) ) t ^ ] = ∣ x ˙ ( τ ) ∣ ∇ μ ( t ^ = x ˙ ∣ x ˙ ∣ 为单位切向量 ) \displaystyle \frac{d}{d\tau}\left[ \mu(x(\tau))\hat t \right] = |\dot x(\tau)|\nabla\mu\quad\quad(\hat t=\frac{\dot x}{|\dot x|} 为单位切向量) dτd[μ(x(τ))t^]=∣x˙(τ)∣∇μ(t^=∣x˙∣x˙为单位切向量)
把 τ \tau τ 当作弧长 s s s，则 ∣ x ˙ ( s ) ∣ = 1 |\dot x(s)|=1 ∣x˙(s)∣=1
d s = ∣ x ˙ ( τ ) ∣ d τ ds=|\dot x(\tau)|d\tau ds=∣x˙(τ)∣dτ 把 τ \tau τ 替换为 s s s 可得： d s = ∣ x ˙ ( τ ) ∣ d s ds=|\dot x(\tau)|ds ds=∣x˙(τ)∣ds，于是有 ∣ x ˙ ( s ) ∣ = 1 |\dot x(s)|=1 ∣x˙(s)∣=1
则有
d d s [ μ ( x ( s ) ⋅ t ^ ] = ∇ μ \frac{d}{ds}\left[ \mu(x(s)\cdot\hat t \right] = \nabla \mu dsd[μ(x(s)⋅t^]=∇μ
把 τ \tau τ 替换为 s s s ，并代入 ∣ x ˙ ( s ) ∣ = 1 |\dot x(s)|=1 ∣x˙(s)∣=1
2. （上册）Page 4-2-6 至 4-2-7： 例3 (Hamilton力学)
在极坐标系下，质点在势场 V ( x , y ) = K cos θ r 2 \displaystyle V(x,y)=\frac{K\cos\theta}{r^2} V(x,y)=r2Kcosθ 下，已知 m = 1 , r ∣ t = 0 = a , r ˙ ∣ t = 0 = 0 m=1,\ r|{t=0}=a,\ \dot r|{t=0}=0 m=1, r∣t=0=a, r˙∣t=0=0，求 r ( t ) r(t) r(t)
① 平面直角坐标系转换极坐标系：
x = r cos θ , y = r sin θ , x ˙ = r ˙ cos θ − r θ ˙ sin θ , y ˙ = r ˙ sin θ + r θ ˙ cos θ x=r\cos\theta,\quad y=r\sin\theta,\quad \dot x=\dot r\cos\theta-r\dot\theta\sin\theta,\quad \dot y=\dot r\sin\theta+r\dot\theta\cos\theta x=rcosθ,y=rsinθ,x˙=r˙cosθ−rθ˙sinθ,y˙=r˙sinθ+rθ˙cosθ
x ˙ 2 + y ˙ 2 = r ˙ 2 cos 2 θ − 2 r r ˙ θ ˙ cos θ sin θ + r 2 θ ˙ 2 sin 2 θ + r 2 sin 2 θ + 2 r r ˙ θ ˙ cos θ sin θ + r 2 θ ˙ 2 cos 2 θ = r ˙ 2 + r 2 θ ˙ 2 \begin{aligned} \dot x^2+\dot y^2=\ &\dot r^2\cos^2\theta-2r\dot r\dot\theta\cos\theta\sin\theta + r^2\dot\theta^2\sin^2\theta \\ +\ & r^2\sin^2\theta + 2r\dot r\dot\theta\cos\theta\sin\theta + r^2\dot\theta^2\cos^2\theta \\ =\ &\dot r^2 + r^2\dot\theta^2 \end{aligned} x˙2+y˙2= + = r˙2cos2θ−2rr˙θ˙cosθsinθ+r2θ˙2sin2θr2sin2θ+2rr˙θ˙cosθsinθ+r2θ˙2cos2θr˙2+r2θ˙2
动能： T = 1 2 m ( x ˙ 2 + y ˙ 2 ) = 1 2 ( r ˙ 2 + r 2 θ ˙ 2 ) \displaystyle \Tau = \frac12m(\dot x^2+\dot y^2)=\frac12(\dot r^2+ r^2\dot\theta^2) T=21m(x˙2+y˙2)=21(r˙2+r2θ˙2)
势能： Π = V ( x , y ) = K cos θ r 2 \displaystyle \Pi = V(x,y) = \frac{K\cos\theta}{r^2} Π=V(x,y)=r2Kcosθ
Hamilton 函数为： H = T + Π \Eta =\Tau+\Pi H=T+Π ，L 函数为： L = T − Π L=\Tau-\Pi L=T−Π
代入动能和势能有： H = T + Π = 1 2 ( r ˙ 2 + r 2 θ ˙ 2 ) + K cos θ r 2 \displaystyle \Eta = \Tau+\Pi=\frac12(\dot r^2 + r^2\dot\theta^2) + \frac{K\cos\theta}{r^2} H=T+Π=21(r˙2+r2θ˙2)+r2Kcosθ
利用 p r = ∂ L ∂ r ˙ = r ˙ , p θ = ∂ L ∂ θ ˙ = r 2 θ ˙ \displaystyle p_r = \frac{\partial L}{\partial \dot r} = \dot r, \quad p_\theta = \frac{\partial L}{\partial\dot\theta} = r^2\dot\theta pr=∂r˙∂L=r˙,pθ=∂θ˙∂L=r2θ˙ 于是有 r ˙ = p r \dot r = p_r r˙=pr， θ ˙ = p θ r 2 \displaystyle \dot\theta = \frac{p_\theta}{r^2} θ˙=r2pθ
代入 Hamilton 函数：
H = 1 2 ( p r 2 + p θ 2 r 2 ) + K cos θ r 2 \displaystyle \Eta =\frac12 \left(p_r^2 + \frac{p_\theta^2}{r^2}\right) + \frac{K\cos\theta}{r^2} H=21(pr2+r2pθ2)+r2Kcosθ
② 写出 Hamilton 正则方程
r ˙ = ∂ H ∂ p r = p r , θ ˙ = ∂ H ∂ p θ = p θ r 2 , p ˙ r = − ∂ H ∂ r = 1 r 3 p θ 2 + 2 K cos θ r 3 , p ˙ θ = − ∂ H ∂ θ = K sin θ r 2 \begin{aligned} \displaystyle &\dot r =\frac{\partial \Eta}{\partial p_r} = p_r,\quad \dot\theta=\frac{\partial \Eta}{\partial p_\theta}=\frac{p_\theta}{r^2}, \\ \displaystyle &\dot p_r = -\frac{\partial \Eta}{\partial r} = \frac1{r^3}p_\theta^2 + \frac{2K\cos\theta}{r^3},\quad \dot p_\theta = -\frac{\partial \Eta}{\partial \theta} = \frac{K\sin\theta}{r^2} \end{aligned} r˙=∂pr∂H=pr,θ˙=∂pθ∂H=r2pθ,p˙r=−∂r∂H=r31pθ2+r32Kcosθ,p˙θ=−∂θ∂H=r2Ksinθ
从而运动方程为：
{ r ¨ = p ˙ r = 1 r 3 p θ 2 + 2 K cos θ r 3 = r θ ˙ 2 + 2 K cos θ r 3 θ ¨ = − 2 r ˙ p θ r 3 + K sin θ r 4 = − 2 1 r r ˙ θ ˙ + K sin θ r 4 (4-2-18) \begin{cases} \displaystyle \ddot r = \dot p_r=\frac1{r^3}p_\theta^2 + \frac{2K\cos\theta}{r^3}=r\dot\theta^2+ \frac{2K\cos\theta}{r^3} \\\\ \displaystyle\ddot\theta = -\frac{2\dot rp_\theta}{r^3} + \frac{K\sin\theta}{r^4} = -2\frac1r\dot r\dot\theta+\frac{K\sin\theta}{r^4} \end{cases} \tag{4-2-18} ⎩ ⎨ ⎧r¨=p˙r=r31pθ2+r32Kcosθ=rθ˙2+r32Kcosθθ¨=−r32r˙pθ+r4Ksinθ=−2r1r˙θ˙+r4Ksinθ(4-2-18)
由（4-2-18-2）可知 r 2 θ ¨ + 2 r r ˙ θ ˙ = K sin θ r 2 \displaystyle r^2\ddot\theta + 2r\dot r\dot\theta=\frac{K\sin\theta}{r^2} r2θ¨+2rr˙θ˙=r2Ksinθ，从而运动方程
{ r ¨ = r ˙ θ 2 + 2 K cos θ r 3 r 2 θ ¨ + 2 r r ˙ θ ˙ = K sin θ r 2 r ∣ t = 0 = a , r ˙ ∣ t = 0 = 0 (4-2-19) \begin{cases} &\ddot r = \dot r\theta^2+\frac{2K\cos\theta}{r^3} \\ &r^2\ddot\theta+2r\dot r\dot\theta = \frac{K\sin\theta}{r^2} \\ &r|{t=0}=a,\quad\dot r|{t=0}=0 \\ \end{cases} \tag{4-2-19} ⎩ ⎨ ⎧r¨=r˙θ2+r32Kcosθr2θ¨+2rr˙θ˙=r2Ksinθr∣t=0=a,r˙∣t=0=0(4-2-19)
③： H \Eta H 函数解方程
H ( r , θ , r ˙ , θ ˙ ) = 1 2 ( r ˙ 2 + r 2 θ ˙ 2 ) + K cos θ r 2 = E ( 总能量 ) \Eta(r,\theta,\dot r,\dot \theta) = \frac12(\dot r^2+r^2\dot\theta^2)+\frac{K\cos\theta}{r^2} = E\ (总能量) H(r,θ,r˙,θ˙)=21(r˙2+r2θ˙2)+r2Kcosθ=E (总能量)
解出 θ ˙ 2 \dot\theta^2 θ˙2
θ ˙ 2 = 2 E r 2 − 2 K cos θ r 4 − r ˙ 2 r 2 \dot\theta^2 = \frac{2E}{r^2} - \frac{2K\cos\theta}{r^4} - \frac{\dot r^2}{r^2} θ˙2=r22E−r42Kcosθ−r2r˙2
代入（4-2-19-1）式
r ¨ = 2 E r − 2 K cos θ r 3 − r ˙ 2 r + 2 K cos θ r 3 = 2 E r − r ˙ 2 r \ddot r=\frac{2E}r-\frac{2K\cos\theta}{r^3}-\frac{\dot r^2}{r}+\frac{2K\cos\theta}{r^3} = \frac{2E}r - \frac{\dot r^2}r r¨=r2E−r32Kcosθ−rr˙2+r32Kcosθ=r2E−rr˙2
利用 r ¨ = d d t r ˙ = d r ˙ d r ⋅ d r d t = 1 2 d d r r ˙ 2 \displaystyle \ddot r =\frac d{dt}\dot r = \frac{d\dot r}{dr}\cdot\frac{dr}{dt}=\frac12\frac d{dr}\dot r^2 r¨=dtdr˙=drdr˙⋅dtdr=21drdr˙2 于是
1 2 d d t r ˙ 2 = 2 E r − r ˙ 2 r \frac12\frac d{dt}\dot r^2=\frac{2E}{r} - \frac{\dot r^2}{r} 21dtdr˙2=r2E−rr˙2
两边乘以 2 r 2 2r^2 2r2
r 2 d ( r ˙ 2 ) + 2 r r ˙ 2 d r = 4 E r d r r^2d(\dot r^2) + 2r\dot r^2dr=4Erdr r2d(r˙2)+2rr˙2dr=4Erdr
即
d ( r 2 r ˙ 2 ) = 2 E d r 2 d(r^2\dot r^2)=2Edr^2 d(r2r˙2)=2Edr2
积分
r 2 r ˙ 2 = 2 E r 2 − A ( 利用 r ∣ t = 0 = a , r ˙ ∣ t = 0 = 0 ) r^2\dot r^2=2Er^2-A\quad(利用\ r|{t=0}=a,\ \dot r|{t=0}=0) r2r˙2=2Er2−A(利用 r∣t=0=a, r˙∣t=0=0)
A = 2 E a 2 A=2Ea^2 A=2Ea2，则
r ˙ = 2 E ( r 2 − a 2 ) r \dot r=\frac{\sqrt{2E(r^2-a^2)}}{r} r˙=r2E(r2−a2)
即
d t = r d r 2 E ( r 2 − a 2 ) dt = \frac{rdr}{\sqrt{2E(r^2-a^2)}} dt=2E(r2−a2) rdr
从而
t = r 2 − a 2 2 E 或 r 2 = 2 E t 2 + a 2 t=\sqrt{\frac{r^2-a^2}{2E}}\quad 或 \quad r^2=2Et^2+a^2 t=2Er2−a2 或r2=2Et2+a2
3.（下册）Page 9-1-1至9-1-3： 离散Kalman滤波公式及推导
Kalman 滤波的递推式
Step1：利用 k − 1 k-1 k−1 时刻得到最优估计 x ^ k − 1 \hat x_{k-1} x^k−1 作出状态一步预测估计 x ^ k , k − 1 \hat x_{k,k-1} x^k,k−1
x ^ k , k − 1 = ϕ k , k − 1 x ^ k − 1 \hat x_{k,k-1} = \phi_{k,k-1} \hat x_{k-1} x^k,k−1=ϕk,k−1x^k−1
Step2：计算一步预测误差协方差：
P k , k − 1 = ϕ k , k − 1 P k − 1 ϕ k , k − 1 T + Γ k Q k − 1 Γ k − 1 T \displaystyle \begin{aligned} P_{k,k-1}=\phi_{k,k-1}P_{k-1}\phi_{k,k-1}^T+\Gamma_kQ_{k-1}\Gamma_{k-1}^T \end{aligned} Pk,k−1=ϕk,k−1Pk−1ϕk,k−1T+ΓkQk−1Γk−1T
Step3：计算增益矩阵
K k = P k , k − 1 H k T ( R k + H k P k , k − 1 H k T ) − 1 \Kappa_k = P_{k,k-1}H_k^T(R_k+H_kP_{k,k-1}H_k^T)^{-1} Kk=Pk,k−1HkT(Rk+HkPk,k−1HkT)−1
Step4：状态滤波 x ^ k \hat x_k x^k
x ^ k = x ^ k , k − 1 + K k ( y k − H k x ^ k , k − 1 ) \hat x_k=\hat x_{k,k-1} + \Kappa_k(y_k - H_k\hat x_{k,k-1}) x^k=x^k,k−1+Kk(yk−Hkx^k,k−1)
Step5：计算滤波误差
P k = ( I − K k H k ) P k , k − 1 ( I − K k H k ) T + K k R k K k T P_k = (I-\Kappa_kH_k)P_{k,k-1}(I-\Kappa_kH_k)^T + \Kappa_kR_k\Kappa_k^T Pk=(I−KkHk)Pk,k−1(I−KkHk)T+KkRkKkT
4.（下册）Page 10-1-3： 例3，Abel变换及逆变换
正问题： Γ \Gamma Γ 给定，求 T A → B T_{A\to B} TA→B 的时间
反问题： T A → B T_{A\to B} TA→B 的时间固定，求 Γ \Gamma Γ 的形状，设 x = ψ ( y ) x=\psi(y) x=ψ(y)，则
1 2 m v 2 + m g y = m g h ( v = d s d t ) \frac12mv^2+mgy=mgh\quad(v=\frac{ds}{dt}) 21mv2+mgy=mgh(v=dtds)
于是
v = 2 g ( h − y ) v=\sqrt{2g(h-y)} v=2g(h−y)
T A → B = ∫ A B 1 v d s = ∫ 0 h 1 + ψ 2 ( y ) 2 g ( h − y ) d y = ∫ 0 h Φ ( y ) h − y d y ( 其中 Φ ( y ) = 1 + ψ 2 ( y ) 2 g ) (10-1-3) \begin{aligned} T_{A\to B} &= \int_A^B \frac1vds=\int_0^h \frac{\sqrt{1+\psi^2(y)}}{\sqrt{2g(h-y)}}dy \\ &= \int_0^h \frac{\Phi(y)}{\sqrt{h-y}}dy\quad(其中 \Phi(y)=\frac{\sqrt{1+\psi^2(y)}}{\sqrt{2g}}) \end{aligned} \tag{10-1-3} TA→B=∫ABv1ds=∫0h2g(h−y) 1+ψ2(y) dy=∫0hh−y Φ(y)dy(其中Φ(y)=2g 1+ψ2(y) )(10-1-3)
从而已知 T A → B = T ( h ) T_{A\to B}=T(h) TA→B=T(h)，求 Φ ( y ) \Phi(y) Φ(y) 或 ψ ( y ) \psi(y) ψ(y)
（10-1-3）为 Abel 方程，一般形式的 Abel 方程为：
∫ 0 x Φ ( x ) ( x − y ) α d y = f ( x ) ( 0 < α < 1 , f ( 0 ) = 0 ) (10-1-4) \int_0^x \frac{\Phi(x)}{(x-y)^\alpha}dy=f(x) \quad (0<\alpha<1,f(0)=0)\tag{10-1-4} ∫0x(x−y)αΦ(x)dy=f(x)(0<α<1,f(0)=0)(10-1-4)
（10-1-4）式两边乘以 1 ( z − x ) 1 − α \displaystyle \frac1{(z-x)^{1-\alpha}} (z−x)1−α1 再 x x x 从 0 0 0 到 z z z 积分可以得到 Abel 逆变换
Φ ( x ) = sin α π π d d x ∫ 0 x f ( y ) ( x − y ) 1 − α d y (10-1-5) \Phi(x) = \frac{\sin\alpha\pi}{\pi}\frac d{dx}\int_0^x \frac{f(y)}{(x-y)^{1-\alpha}}dy \tag{10-1-5} Φ(x)=πsinαπdxd∫0x(x−y)1−αf(y)dy(10-1-5)
对于 Abel 积分 f ( x ) f(x) f(x) 有微小变分，引起 Φ \Phi Φ 很大变化，故求 Abel 变换逆变换是不稳定的
5. 2024. 3. 5 讲稿，Page2，Paga3：例2，变分的逆问题
求证 J [ u ( x ) ] = ∫ ( ∣ ∇ u ∣ 2 2 − f u ) d x = min \displaystyle J[u(x)]=\int\left( \frac{|\nabla u|^2}{2}-fu \right)dx=\min J[u(x)]=∫(2∣∇u∣2−fu)dx=min 。
推导：
∫ Ω ( − Δ u − f ) δ u d x = 0 \int_\Omega(-\Delta u-f)\delta u dx=0 ∫Ω(−Δu−f)δudx=0
将括号拆分得到：
− ∫ Ω Δ u δ u d x − ∫ Ω f δ u d x = 0 -\int_\Omega\Delta u\delta udx-\int_\Omega f\delta udx=0 −∫ΩΔuδudx−∫Ωfδudx=0
其中
∫ Ω Δ u δ u d x = − ∫ Ω ∇ u ⋅ ∇ δ u d x = − ∫ Ω δ ∇ u 2 2 d x \begin{aligned} \int_\Omega\Delta u\delta udx=-\int_\Omega\nabla u \cdot\nabla\delta udx \\\\ =-\int_\Omega\delta\frac{\nabla u^2}{2}dx \\ \end{aligned} ∫ΩΔuδudx=−∫Ω∇u⋅∇δudx=−∫Ωδ2∇u2dx
∴ ∫ Ω ( − Δ u − f ) δ u d x = ∫ Ω δ ∇ u 2 2 d x − ∫ Ω f δ u d x = δ ∫ Ω ( ∣ ∇ u ∣ 2 2 − f u ) d x = 0 \begin{aligned} \displaystyle \therefore \int_\Omega(-\Delta u-f)\delta udx&=\int_\Omega\delta\frac{\nabla u^2}{2}dx-\int_\Omega f\delta udx \\\\ &=\delta\int_\Omega \left( \frac{|\nabla u|^2}{2}-fu \right)dx=0 \end{aligned} ∴∫Ω(−Δu−f)δudx=∫Ωδ2∇u2dx−∫Ωfδudx=δ∫Ω(2∣∇u∣2−fu)dx=0
由此得证
6. 2024.3.26 讲稿，Page 1：Legendre变换的概念
u = u ( x , y ) u=u(x,y) u=u(x,y)
u x = ξ , u y = η , x u x + y u y − u = ω u_x=\xi,\quad u_y=\eta,\quad xu_x+yu_y-u=\omega ux=ξ,uy=η,xux+yuy−u=ω 即为 Legendre 变换
ω ξ = x , ω e t a = y , ξ ω ξ + η ω η − ω = u \omega_\xi=x,\quad \omega_eta=y,\quad \xi\omega_\xi+\eta\omega_\eta-\omega=u ωξ=x,ωeta=y,ξωξ+ηωη−ω=u 即为 Legendre 逆变换
7. 2024. 4.18 讲稿，Page 7，Page 8：带正则化的最小二乘方法
由于 H x = y Hx=y Hx=y 或 H x + ϵ = y Hx+\epsilon=y Hx+ϵ=y ⟹ \Longrightarrow ⟹ J ( x ) = ∥ H x − y ∥ 2 = min J(x)=\|Hx-y\|^2=\min J(x)=∥Hx−y∥2=min 的 最小二乘解不唯一，故需在 J ( x ) J(x) J(x) 引入约束项：
J ( x ) = ∥ H x − y ∥ 2 + α ∥ L ( x − x b ) ∥ 2 = min J(x)=\|Hx-y\|^2+\alpha\|L(x-x_b)\|^2=\min J(x)=∥Hx−y∥2+α∥L(x−xb)∥2=min
其中 L L L 是算子， x b x_b xb 是背景场， α \alpha α 是权重因子
**求解过程：**取 x b = 0 x_b=0 xb=0，则
J ( x ) = ∥ H x − y ∥ 2 + α ∥ L x ∥ 2 = ( y − H x ) T ( y − H x ) + α ( L x ) T L x = ( y T − x T H T ) ( y − H x ) + α x T L T L x = min \begin{aligned} J(x)&=\|Hx-y\|^2+\alpha\|Lx\|^2=(y-Hx)^T(y-Hx)+\alpha(Lx)^TLx \\\\ &=(y^T-x^TH^T)(y-Hx)+\alpha x^TL^TLx=\min \end{aligned} J(x)=∥Hx−y∥2+α∥Lx∥2=(y−Hx)T(y−Hx)+α(Lx)TLx=(yT−xTHT)(y−Hx)+αxTLTLx=min
则 ( δ 指对 x 求导 ) (\delta\ 指对 x 求导) (δ 指对x求导)
δ J = − ( δ x ) T H T ( y − H x ) − ( y T − x T H T ) H δ x + α ( δ x ) T L T L x + α x T L T L δ x = − 2 ( y T − x T H T ) H δ x + 2 α x T L T L δ x = 2 [ ( x T H T − y T ) H + α x T L T L ] δ x = 0 \begin{aligned} \delta J&=-(\delta x)^TH^T(y-Hx)-(y^T-x^TH^T)H\delta x+\alpha(\delta x)^TL^TLx + \alpha x^TL^TL\delta x \\\\ &= -2(y^T-x^TH^T)H\delta x + 2\alpha x^TL^TL\delta x \\\\ &=2\left[ (x^TH^T-y^T)H + \alpha x^TL^TL \right]\delta x=0 \end{aligned} δJ=−(δx)THT(y−Hx)−(yT−xTHT)Hδx+α(δx)TLTLx+αxTLTLδx=−2(yT−xTHT)Hδx+2αxTLTLδx=2[(xTHT−yT)H+αxTLTL]δx=0
第 2 步只留了 δ x \delta x δx 项并翻了个2倍
则 ( x T H T − y T ) H + α x T L T L = 0 (x^TH^T-y^T)H+\alpha x^TL^TL=0 (xTHT−yT)H+αxTLTL=0
转置得
H T ( H x − y ) + α L T L x = 0 ( H T H + α L T L ) x = H T y \begin{aligned} H^T(Hx-y)+\alpha L^TLx&=0 \\ (H^TH+\alpha L^TL)x&=H^Ty \\ \end{aligned} HT(Hx−y)+αLTLx(HTH+αLTL)x=0=HTy
于是 x x x 的稳定近似解为：
x ^ = ( H T H + α L T L ) − 1 H T y \hat x=(H^TH+\alpha L^TL)^{-1}H^Ty x^=(HTH+αLTL)−1HTy