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根据习题(1.2.3)已知下列非齐次两点边值问题(1.2.28)
\[\begin{cases} \mathbf{L}u=-\frac{d}{dx}(p\frac{du}{dx})+qu=f, x\in(a,b)\\ u(a)=\alpha, u'(b)=\beta \end{cases}\]与下列变微分方程等价:求$u\in H^1, u(a)=\alpha$ ,使
\[J(u_*)=\min_{u\in H^1,u(a)=\alpha}J(u)\]其中
\[J(u)=\frac{1}{2}a(u,u)-(f,u)-p(b)\beta u(b)\\ a(u,v)=\int_a^{b}(p\frac{du}{dx}\frac{dv}{dx})d\\ (g,u)=\int_a^b gudx\]设$[a,b]=[-1,1],p(x)\equiv-(\pi^2-1)^{-1},q(x)\equiv1,\alpha=0,\beta=-e\pi$,以及
\[f(x)=\frac{2\pi}{\pi^2-1}\cos(x\pi)e^x\]
- 分别取$h\in\lbrace 0.20,0.10,0.05,0.02\rbrace$,将求解域等分为长度为$h$的单元或子空间
- 根据上述剖分,就边值问题(1.2.28)和基函数(2.1.16)中$\varphi_i$而设计$\varphi_0(x)$,编程构建相应的 Ritz-Galekin 方程(即有限元方程)
- 分别使用高斯消元法和雅克比迭代法(迭代 30 次),求解上述有限元方程
- 计算得到有限元解 $u_h$ 并绘制其函数图像
- 已知$u(x)=\sin(x\pi)e^x$是上述边值问题的解析解,针对不同的步长$h$和线性方程组解法得到的数值解$u_h$,绘制误差函数$(u_h-u)$的函数图像,且进行观察分析。
首先构建第 2 问的有限元方程。仿照(2.1.16)设计$\varphi_0(x)$如下:
\[\phi_i(x)=\begin{cases} 1+\frac{1-x_i}{h}, x\in[x_{i-1},x_i]\\ 1-\frac{1-x_i}{h}, x\in[x_i,x_{i+1}]\\ 0, else \end{cases}\\ \phi_n(x)=\begin{cases} 1+\frac{x-x_n}{h}, x\in[x_{i-1},x_i]\\ 0, else \end{cases}\\\]试探函数
\[u_h(x)=\sum_{i=0}^n\sigma_i\varphi_i(x), \, \sigma_i=u_h(x_i)\]因为$\alpha=0,\beta\ne 0$,故右边值条件非齐次,根据(2.1.4)可得有限元方程为:
\[\sum_{i=1}^nu_i\int_a^b [p\varphi_i'\varphi_j'+q\varphi_i\varphi_j]\, dx=(f,\varphi_j)+p(b)\beta\varphi_j(b),\, j\in \lbrace 1,2,\dots,n\rbrace\]由于$\mid i-j\mid >1$时,$a(\varphi(i)-\varphi(j))=0,且$$p(b)\beta\varphi_j(b)=0$ 对于 $j\ne n$ 恒成立,故方程组形式如下:
\[\begin{bmatrix} a(\varphi_1,\varphi_1) & a(\varphi_2,\varphi_1) & 0 & 0 & \dots & 0 \\ a(\varphi_1,\varphi_2) & a(\varphi_2,\varphi_2) & a(\varphi_3,\varphi_2) & 0 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ 0 & 0 & \dots & a(\varphi_{n-2},\varphi_{n-1}) & a(\varphi_{n-1},\varphi_{n-1}) & a(\varphi_{n},\varphi_{n-1}) \\ 0 & 0 & \dots & 0 & a(\varphi_{n-1},\varphi_{n}) & a(\varphi_{n},\varphi_{n}) \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ \dots\\ v_{n-1}\\ v_n \end{bmatrix}\\=\begin{bmatrix} (f,\varphi_{1})\\ (f,\varphi_{1})\\ (f,\varphi_{1})\\ (f,\varphi_{n})+p(b)\beta\varphi_n(b) \end{bmatrix}\]带入$a(u,v)=\int_a^b(p\frac{du}{dx}+quv)dx$有
\[a(\varphi_{i-1},\varphi_i) = \int_{x_{i-1}}^{x_i}[p\varphi_i'\varphi_j'+q\varphi_i\varphi_j]\, dx \\ = \int_{x_{i-1}}^{x_i} [-\frac{p(x)}{h^2}+q(x)\varphi_i(x)\varphi_{i-1}(x)]\, dx\]通过仿射变换到$[0,1]$上的标准山形函数
\[\varphi_i(x)=\begin{cases} \frac{x-x_{i-1}}{h}, \, x\in[x_{i-1},x_i]\\ 1-\frac{x-x_{i-1}}{h}, \, x\in[x_,x_{i+1}]\\ 0,\, else \end{cases} \\ \varphi_n(x)=\begin{cases} \frac{x-x_{i-1}}{h}, \, x\in[x_{i-1},x_i]\\ 0,\, else \end{cases} \\\]有
\[a(\varphi_{i-1},\varphi_i) = \int_0^1[-\frac{p(x_{i-1}+h\theta)}{h}+hq(x_{i-1}+h\theta)(1-\theta)\theta]\,d\theta\]带入$p(x)\equiv-(\pi^2-1)^{-1},q(x)\equiv1$有
\[a(\varphi_{i-1},\varphi_i)=\int_0^1\frac{1}{h(\pi^2-1)}+h(1-\theta)\, d\theta \\ =\frac{1}{h(\pi^2-1)}+\frac{h}{6}\]类似可得$a(\varphi_i,\varphi_i),a(\varphi_i,\varphi_{i+1})$,于是对$x\in[2,n-1]$总结有
\[a(\varphi_i,\varphi_j)= \begin{cases} \frac{h}{6}-\frac{p}{h} & i=j-1 \\ 2(\frac{p}{h}+\frac{h}{3}) & i=j \\ \frac{h}{6}-\frac{p}{h} & i=j+1 \\ 0 & \mathit{else} \end{cases}\]上式对$i=1$时仍满足条件,仍然成立;$i=n$时山形函数仅满足一半部分$x\in[x_{n-1},x_n]$,不含右半部分,于是其值应为上式的二分之一。
以下编写 python 程序求解各问题。
import altair as alt
import numpy as np
import scipy
import sys
def p(x):
return -1 / (np.pi * np.pi - 1)
def q(x):
return 0
def f(x):
return ((2 * np.pi) / (np.pi * np.pi - 1)) * np.cos(np.pi * x) * np.exp(x)
def phi(a, b, n, i, x):
h = (b - a) / n
xi = a + i * h
if xi - h <= x and x <= xi:
return 1 + (x - xi) / h
if xi <= x and x <= xi + h:
return 1 - (x - xi) / h
return 0
def dphi(a, b, n, i, x):
h = (b - a) / n
xi = a + i * h
if xi - h <= x and x <= xi:
return 1 / h
if xi <= x and x <= xi + h:
return -1 / h
return 0
def a_phi(a, b, n, i, j):
h = (b - a) / n
if i == j - 1:
return h / 6 - p(0) / h
if i == j:
return 2 * (p(0) / h + h / 3)
if i == j + 1:
return h / 6 - p(0) / h
return 0
def vec_b(a, b, n, j, beta):
h = (b - a) / n
xi = a + j * h
tmp = scipy.integrate.quad(
lambda x: f(x) * phi(a, b, n, j, x), max(a, xi - h), min(b, xi + h)
)[0]
return p(b) * beta * phi(a, b, n, j, b) + tmp
def get_A(a, b, n):
A = np.ndarray((n, n), dtype=np.float64)
for i in range(n):
for j in range(n):
if i == n - 1 and j == n - 1:
A[i][j] = a_phi(a, b, n, i + 1, j + 1) * 0.5
else:
A[i][j] = a_phi(a, b, n, i + 1, j + 1)
return A
def get_B(a, b, n, beta):
ret = np.ndarray((n,), dtype=np.float64)
for j in range(n):
ret[j] = vec_b(a, b, n, j + 1, beta)
return ret
def gaussian(A, b):
n = len(A)
for i in range(0, n - 1):
for j in range(i + 1, n):
tmp = A[j, i] / A[i, i]
A[j, i:n] = A[j, i:n] - tmp * A[i, i:n]
b[j] -= tmp * b[i]
for i in range(n - 1, -1, -1):
b[i] = (b[i] - np.dot(A[i, i + 1 : n], b[i + 1 : n])) / A[i, i]
return b
def jacobi(A, b, iter):
n = len(A)
L = np.array(np.tril(A, -1))
U = np.array(np.triu(A, 1))
D_inv = np.diag(1 / np.diag(A))
x = np.zeros(n)
for _it in range(iter):
x = D_inv.dot(b - L.dot(x) - U.dot(x))
return x.flatten()
def problem1():
print("Problem 1:")
a, b = (-1, 1)
for h in [0.20, 0.10, 0.05, 0.02]:
range = np.arange(a, b + h, h)
print(h)
print(np.around(range, 2))
def problem2():
print("Problem 2:")
a, b, alpha, beta = (-1, 1, 0, -np.pi * np.e)
for h in [0.20, 0.10, 0.05, 0.02]:
print(h)
n = int((b - a) / h)
A = get_A(a, b, n)
B = get_B(a, b, n, beta)
print(np.around(A, 2))
print(np.around(B, 2))
def problem3():
print("Problem 3:")
a, b, alpha, beta = (-1, 1, 0, -np.pi * np.e)
for h in [0.20, 0.10, 0.05, 0.02]:
print(h)
n = int((b - a) / h)
A = get_A(a, b, n)
B = get_B(a, b, n, beta)
x0 = gaussian(A.copy(), B.copy())
x1 = jacobi(A.copy(), B.copy(), 30)
print(np.around(x0, 2))
print(np.around(x1, 2))
def problem4():
print("Problem 4:")
chart = []
a, b, alpha, beta = (-1, 1, 0, -np.pi * np.e)
for h in [0.20, 0.10, 0.05, 0.02]:
print(h)
n = int((b - a) / h)
A = get_A(a, b, n)
B = get_B(a, b, n, beta)
x0 = gaussian(A.copy(), B.copy())
x1 = jacobi(A.copy(), B.copy(), 30)
for idx in range(n):
x = a + idx * h
chart.append({"h": h, "x": x, "y": x0[idx], "method": "gaussian"})
chart.append({"h": h, "x": x, "y": x1[idx], "method": "jacobi"})
idx = idx + 1
chart = (
alt.Chart(data=alt.InlineData(chart))
.encode(
x=alt.X("x:Q"),
y=alt.Y("y:Q", scale=alt.Scale(domain=(-5, 2))),
column=alt.Column("method:N"),
color=alt.Color("h:N"),
)
.mark_line(clip=True)
)
chart.save("problem4.html")
def problem5():
print("Problem 5:")
chart = []
a, b, alpha, beta = (-1, 1, 0, -np.pi * np.e)
for h in [0.20, 0.10, 0.05, 0.02]:
print(h)
n = int((b - a) / h)
A = get_A(a, b, n)
B = get_B(a, b, n, beta)
x0 = gaussian(A.copy(), B.copy())
x1 = jacobi(A.copy(), B.copy(), 30)
for idx in range(n):
x = a + idx * h
fx = np.sin(np.pi * x) * np.e**x
chart.append({"h": h, "x": x, "diff": x0[idx] - fx, "method": "gaussian"})
chart.append({"h": h, "x": x, "diff": x1[idx] - fx, "method": "jacobi"})
idx = idx + 1
chart = (
alt.Chart(data=alt.InlineData(chart))
.encode(
x=alt.X("x:Q"),
y=alt.Y("diff:Q", scale=alt.Scale(domain=(-5, 2))),
column=alt.Column("method:N"),
color=alt.Color("h:N"),
)
.mark_line(clip=True)
)
chart.save("problem5.html")
def main(*argv):
problem1()
problem2()
problem3()
problem4()
problem5()
return 0
if __name__ == "__main__":
sys.exit(main(*sys.argv))
以下是该程序的屏幕输出,对应问题 1 ~ 3 的解答。注意为保持输出格式整齐,输出时仅保留了两位小数,实际计算结果和计算精度均使用六十四位浮点类型,可认为保持了高精度。
Problem 1:
0.2
[-1. -0.8 -0.6 -0.4 -0.2 -0. 0.2 0.4 0.6 0.8 1. ]
0.1
[-1. -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 -0. 0.1 0.2 0.3
0.4 0.5 0.6 0.7 0.8 0.9 1. ]
0.05
[-1. -0.95 -0.9 -0.85 -0.8 -0.75 -0.7 -0.65 -0.6 -0.55 -0.5 -0.45
-0.4 -0.35 -0.3 -0.25 -0.2 -0.15 -0.1 -0.05 0. 0.05 0.1 0.15
0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75
0.8 0.85 0.9 0.95 1. ]
0.02
[-1. -0.98 -0.96 -0.94 -0.92 -0.9 -0.88 -0.86 -0.84 -0.82 -0.8 -0.78
-0.76 -0.74 -0.72 -0.7 -0.68 -0.66 -0.64 -0.62 -0.6 -0.58 -0.56 -0.54
-0.52 -0.5 -0.48 -0.46 -0.44 -0.42 -0.4 -0.38 -0.36 -0.34 -0.32 -0.3
-0.28 -0.26 -0.24 -0.22 -0.2 -0.18 -0.16 -0.14 -0.12 -0.1 -0.08 -0.06
-0.04 -0.02 0. 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18
0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42
0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6 0.62 0.64 0.66
0.68 0.7 0.72 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9
0.92 0.94 0.96 0.98 1. ]
Problem 2:
0.2
[[-0.99 0.6 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0.6 -0.99 0.6 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0.6 -0.99 0.6 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0.6 -0.99 0.6 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0.6 -0.99 0.6 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0.6 -0.99 0.6 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0.6 -0.99 0.6 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0.6 -0.99 0.6 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0.6 -0.99 0.6 ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0.6 -0.5 ]]
[-0.05 -0.02 0.03 0.09 0.14 0.13 0.06 -0.08 -0.25 0.79]
0.1
[[-2.19 1.14 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
0. 0. 0. 0. 0. 0. 0. 0. ]
[ 1.14 -2.19 1.14 0. 0. 0. 0. 0. 0. 0. 0. 0.
0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 1.14 -2.19 1.14 0. 0. 0. 0. 0. 0. 0. 0.
0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 1.14 -2.19 1.14 0. 0. 0. 0. 0. 0. 0.
0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 1.14 -2.19 1.14 0. 0. 0. 0. 0. 0.
0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 1.14 -2.19 1.14 0. 0. 0. 0. 0.
0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 1.14 -2.19 1.14 0. 0. 0. 0.
0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 1.14 -2.19 1.14 0. 0. 0.
0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 1.14 -2.19 1.14 0. 0.
0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 1.14 -2.19 1.14 0.
0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 1.14 -2.19 1.14
0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1.14 -2.19
1.14 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1.14
-2.19 1.14 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
1.14 -2.19 1.14 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
0. 1.14 -2.19 1.14 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
0. 0. 1.14 -2.19 1.14 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
0. 0. 0. 1.14 -2.19 1.14 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
0. 0. 0. 0. 1.14 -2.19 1.14 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
0. 0. 0. 0. 0. 1.14 -2.19 1.14]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
0. 0. 0. 0. 0. 0. 1.14 -1.09]]
[-0.03 -0.03 -0.02 -0.01 0. 0.01 0.03 0.05 0.06 0.07 0.07 0.07
0.06 0.03 -0. -0.04 -0.08 -0.13 -0.16 0.87]
0.05
[[-4.48 2.26 0. ... 0. 0. 0. ]
[ 2.26 -4.48 2.26 ... 0. 0. 0. ]
[ 0. 2.26 -4.48 ... 0. 0. 0. ]
...
[ 0. 0. 0. ... -4.48 2.26 0. ]
[ 0. 0. 0. ... 2.26 -4.48 2.26]
[ 0. 0. 0. ... 0. 2.26 -2.24]]
[-0.01 -0.01 -0.01 -0.01 -0.01 -0.01 -0.01 -0.01 -0. 0. 0. 0.01
0.01 0.02 0.02 0.02 0.03 0.03 0.03 0.04 0.04 0.04 0.04 0.03
0.03 0.03 0.02 0.02 0.01 -0. -0.01 -0.02 -0.03 -0.04 -0.05 -0.06
-0.07 -0.08 -0.09 0.92]
0.02
[[-11.26 5.64 0. ... 0. 0. 0. ]
[ 5.64 -11.26 5.64 ... 0. 0. 0. ]
[ 0. 5.64 -11.26 ... 0. 0. 0. ]
...
[ 0. 0. 0. ... -11.26 5.64 0. ]
[ 0. 0. 0. ... 5.64 -11.26 5.64]
[ 0. 0. 0. ... 0. 5.64 -5.63]]
[-0.01 -0.01 -0.01 -0.01 -0.01 -0.01 -0.01 -0.01 -0.01 -0.01 -0.01 -0.
-0. -0. -0. -0. -0. -0. -0. -0. -0. -0. -0. -0.
0. 0. 0. 0. 0. 0. 0. 0. 0. 0.01 0.01 0.01
0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01
0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01
0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.
0. 0. -0. -0. -0. -0. -0.01 -0.01 -0.01 -0.01 -0.01 -0.01
-0.02 -0.02 -0.02 -0.02 -0.02 -0.03 -0.03 -0.03 -0.03 -0.03 -0.03 -0.03
-0.04 -0.04 -0.04 0.94]
Problem 3:
0.2
[-0.32 -0.62 -0.75 -0.57 -0.05 0.72 1.47 1.83 1.44 0.14]
[ -21.27 -34.61 -64. -68.79 -104.79 -99.25 -138.33 -121.45 -158.9
-131.79]
0.1
[-0.13 -0.28 -0.42 -0.55 -0.63 -0.66 -0.62 -0.5 -0.3 -0.01 0.34 0.72
1.1 1.43 1.67 1.76 1.66 1.34 0.79 0.04]
[-0. -0.04 -0.12 -0.23 -0.39 -0.56 -0.75 -0.91 -1.09 -1.17 -1.32 -1.3
-1.45 -1.4 -1.66 -1.7 -2.23 -2.51 -3.38 -3.94]
0.05
[-0.06 -0.13 -0.2 -0.27 -0.34 -0.41 -0.47 -0.53 -0.58 -0.61 -0.64 -0.64
-0.63 -0.61 -0.56 -0.49 -0.4 -0.28 -0.15 -0. 0.16 0.34 0.53 0.72
0.91 1.09 1.27 1.42 1.55 1.65 1.72 1.74 1.71 1.64 1.5 1.32
1.07 0.77 0.41 0.01]
[ 0.02 0.04 0.05 0.05 0.05 0.05 0.04 0.02 0. -0.02 -0.04 -0.07
-0.09 -0.12 -0.14 -0.17 -0.19 -0.21 -0.22 -0.23 -0.24 -0.24 -0.23 -0.21
-0.19 -0.16 -0.13 -0.1 -0.07 -0.03 -0.02 -0. -0.04 -0.07 -0.2 -0.31
-0.55 -0.78 -1.16 -1.51]
0.02
[-0.02 -0.05 -0.07 -0.1 -0.13 -0.15 -0.18 -0.21 -0.24 -0.26 -0.29 -0.32
-0.35 -0.38 -0.4 -0.43 -0.45 -0.48 -0.5 -0.52 -0.54 -0.56 -0.58 -0.59
-0.61 -0.62 -0.63 -0.63 -0.64 -0.64 -0.64 -0.63 -0.62 -0.61 -0.6 -0.58
-0.56 -0.54 -0.51 -0.48 -0.45 -0.41 -0.37 -0.33 -0.28 -0.23 -0.18 -0.12
-0.06 -0. 0.06 0.13 0.2 0.27 0.34 0.41 0.49 0.57 0.64 0.72
0.79 0.87 0.95 1.02 1.09 1.16 1.23 1.3 1.36 1.42 1.47 1.53
1.57 1.61 1.65 1.68 1.7 1.72 1.73 1.73 1.73 1.72 1.7 1.67
1.63 1.58 1.53 1.46 1.39 1.31 1.22 1.12 1.01 0.89 0.76 0.63
0.48 0.33 0.17 0. ]
[ 0. 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01
0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0. 0. 0. 0.
-0. -0. -0. -0. -0.01 -0.01 -0.01 -0.01 -0.01 -0.02 -0.02 -0.02
-0.02 -0.02 -0.02 -0.03 -0.03 -0.03 -0.03 -0.03 -0.03 -0.03 -0.04 -0.04
-0.04 -0.04 -0.04 -0.04 -0.04 -0.04 -0.04 -0.04 -0.04 -0.04 -0.04 -0.04
-0.04 -0.03 -0.03 -0.03 -0.03 -0.03 -0.02 -0.02 -0.02 -0.02 -0.01 -0.01
-0.01 -0. 0. 0.01 0.01 0.01 0.02 0.02 0.03 0.03 0.04 0.04
0.04 0.05 0.05 0.05 0.05 0.05 0.03 0.02 -0.01 -0.04 -0.1 -0.17
-0.27 -0.36 -0.51 -0.66]
Problem 4:
0.2
0.1
0.05
0.02
Problem 5:
0.2
0.1
0.05
0.02
以下是问题 4 的输出图像。注意到 jacobi 方法在 $h=0.20$ 时极值远远超过了坐标轴范围的十倍,因此并没有在图中显示出来。
{
"config": {"view": {"continuousWidth": 400, "continuousHeight": 300}},
"data": {"name": "data-8179d865bf4dea9647cb306c153f1a05"},
"mark": {"type": "line", "clip": true},
"encoding": {
"color": {"field": "h", "type": "nominal"},
"column": {"field": "method", "type": "nominal"},
"x": {"field": "x", "type": "quantitative"},
"y": {"field": "y", "scale": {"domain": [-5, 2]}, "type": "quantitative"}
},
"$schema": "https://vega.github.io/schema/vega-lite/v4.17.0.json",
"datasets": {
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]
}
}
以下是问题 5 的输出图像。与问题 4 相同地,jacobi 方法在 $h=0.20$ 时没有在图中显示出来。由图像可以看出,gaussian 法随着 h 的减小逐渐收敛,但 jacobi 法波动不稳定,总体趋势略有减少。
{
"config": {"view": {"continuousWidth": 400, "continuousHeight": 300}},
"data": {"name": "data-cb6cd9b7596cdf937fb83fc34af5ad01"},
"mark": {"type": "line", "clip": true},
"encoding": {
"color": {"field": "h", "type": "nominal"},
"column": {"field": "method", "type": "nominal"},
"x": {"field": "x", "type": "quantitative"},
"y": {"field": "diff", "scale": {"domain": [-5, 2]}, "type": "quantitative"}
},
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