cs205a hw8

课程主页：https://graphics.stanford.edu/courses/cs205a-13-fall/schedule.html

这次回顾作业8。

Problem 1

(a)代码如下：

M = zeros(nParticles);
[m, n] = size(edges);
for i = 1: m
    x = edges(i, 1);
    y = edges(i, 2);
    M(x, x) = M(x, x) - 1;
    M(x, y) = M(x, y) + 1;
    M(y, x) = M(y, x) + 1;
    M(y, y) = M(y, y) - 1;
end

(b)令

$Y_1 =X ,Y_2 =X', Y= left[ begin{matrix} Y_1 \ Y_2 end{matrix} right]$

那么方程为

$Y' = left[ begin{matrix} X' \ X'' end{matrix} right]= left[ begin{matrix} 0 & I_n\ F&0 end{matrix} right] left[ begin{matrix} X\ X' end{matrix} right]triangleq A Y$

所以代码如下：

[m, n] = size(secondOrderMatrix);
matrix = zeros(2 * m);
matrix(1: m, m+1: 2*m) = eye(m);
matrix(m+1: 2*m, 1: m) = secondOrderMatrix;

(c)

(i)前向欧拉

$y_{k+1}= y_{k}+h Fleft(y_{k}right) =(I_n + h A)y_k$

代码如下

1	x = x + dt * firstOrder * x;

(ii)后向欧拉

$begin{aligned} y_{k}&=y_{k+1}-h Fleft(y_{k+1}right) =(I_n-hA)y_{k+1}\ y_{k+1}&=(I_n-hA)^{-1}y_k end{aligned}$

代码如下

1	x = inv(eye(2 * n) - dt * firstOrder) * x;

(iii)梯形法

$begin{aligned} y_{k+1}&=y_{k}+h frac{Fleft(y_{k+1}right)+Fleft(y_{k}right)}{2} =(I_n +frac 1 2 h A) y_k +frac 1 2 h A y_{k+1}\ y_{k+1}&=(I_n-frac 12 hA)^{-1}(I_n +frac 1 2 h A) y_k end{aligned}$

代码如下

1	x = inv(eye(2 * n) - dt * firstOrder / 2) * (eye(2 * n) + dt * firstOrder / 2) * x;

(d)leapfrog

$begin{aligned} vec{y}_{k+1} &=vec{y}_{k}+h vec{v}_{k+1 / 2} \ vec{a}_{k+1} &=Fleft[t_{k+1}, vec{y}_{k+1}right]=Avec y_{k+1} \ vec{v}_{k+3 / 2} &=vec{v}_{k+1 / 2}+h vec{a}_{k+1} =vec{v}_{k+1 / 2}+ Avec y_{k+1} end{aligned}$

代码如下

1 2	positions = positions + dt * velocities; velocities = velocities + dt * force * positions;

Problem 2

(a)求导可得

$begin{aligned} E'(t) &=theta' theta'' +theta' sin theta\ &=theta'(theta''+ sin theta)\ &=0 end{aligned}$

所以$E(t)$关于$t$是常数。

(b)

$begin{aligned} E_{k+1} -E_k &=frac 1 2 w_{k+1}^2 -cos theta_{k+1} -frac 1 2 w_{k}^2+cos theta_{k}\ &=frac 1 2 (w_k-hsin theta_{k+1})^2 -cos( theta_k +hw_k)-frac 1 2 w_{k}^2+cos theta_{k}\ &=frac 1 2 (-hsin theta_{k+1}) (2w_k -hsintheta_{k+1}) + 2sin left(frac {hw_k} 2 right)sin left(theta_k +frac {hw_k} 2 right)\ &= frac 1 {2} h^2 sin^2 theta_{k+1} -h w_ksintheta_{k+1} +2sin left(frac {hw_k} 2 right) left(sintheta_k cos left(frac {hw_k} 2 right) +costheta_k sinleft(frac {hw_k} 2 right)right)\ &=frac 1 {2} h^2 sin^2 theta_{k+1} +2sin^2left(frac {hw_k} 2 right)costheta_k + sintheta_{k} sin(hw_k)- hw_k sin(theta_k+h w_k) end{aligned}$

因为

$begin{aligned} sintheta_{k} sin(hw_k)- hw_k sin(theta_k+h w_k) &= sintheta_{k} sin(hw_k) -hw_k sin theta_k cos(hw_k) -hw_k cos theta_k sin(hw_k)\ &=-hw_k cos theta_k sin(hw_k) +sintheta_{k} cos(hw_k) left(tan(hw_k)- hw_kright)\ &=O(h^2) end{aligned}$

以及

$frac 1 {2} h^2 sin^2 theta_k +2sin^2left(frac {hw_k} 2 right)costheta_k= O(h^2)$

所以

$E_{k+1}=E_k +O(h^2)$

(c)因为

$begin{aligned} w_{k+1} &=w_k -htheta_{k+1}\ &=w_k -h(theta_k+ hw_k)\ &=-htheta_k +(1-h^2) w_k end{aligned}$

所以

$left( begin{array}{c}theta_{k+1} \ w_{k+1}end{array}right) = left( begin{matrix} 1 & h\ -h & 1-h^2 end{matrix} right) left( begin{array}{c}theta_{k} \ w_{k}end{array}right)$

(d)

$begin{aligned} E_{k+1} &= w_{k+1}^2 + hw_{k+1} theta_{k+1}+ theta_{k+1}^2\ &=left( begin{array}{c}theta_{k+1} \ w_{k+1}end{array}right)^{top} left( begin{matrix} 1 & frac 1 2h\ frac 1 2h &1 end{matrix} right)left( begin{array}{c}theta_{k+1} \ w_{k+1}end{array}right)\ &= left( begin{array}{c}theta_{k} \ w_{k}end{array}right)^{top} left( begin{matrix} 1 & -h\ h & 1-h^2 end{matrix} right) left( begin{matrix} 1 & frac 1 2h\ frac 1 2h &1 end{matrix} right)left( begin{matrix} 1 & h\ -h & 1-h^2 end{matrix} right) left( begin{array}{c}theta_{k} \ w_{k}end{array}right)\ &=left( begin{array}{c}theta_{k} \ w_{k}end{array}right)^{top} left( begin{matrix} 1-frac 12 {h^2}& -frac 1 2h\ frac 3 2 h-frac 1 2 h^3 & 1-frac 1 2 h^2 end{matrix} right) left( begin{matrix} 1 & h\ -h & 1-h^2 end{matrix} right) left( begin{array}{c}theta_{k} \ w_{k}end{array}right)\ &=left( begin{array}{c}theta_{k} \ w_{k}end{array}right)^{top} left( begin{matrix} 1& frac 1 2h\ frac 1 2h & 1 end{matrix} right) left( begin{array}{c}theta_{k} \ w_{k}end{array}right)\ &= E_k end{aligned}$

Problem 1

Problem 2

近期文章

近期评论

标签

热门

文章归档

分类目录

功能