现代偏微分方程计算方法课程项目

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

\[f(x)=\frac{2\pi}{\pi^2-1}\cos(x\pi)e^x\]
  1. 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
  2. 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)
  3. 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
  4. 计算得到有限元解 $u_h$ 并绘制其函数图像
  5. 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

\[\varphi_i(x)=\begin{cases} \frac{x-x_{i-1}}{h}, \, x\in[x_{i-1},x_i]\\ 1-\frac{x-x_{i-1}}{h}, \, x\in[x_,x_{i+1}]\\ 0,\, else \end{cases} \\ \varphi_n(x)=\begin{cases} \frac{x-x_{i-1}}{h}, \, x\in[x_{i-1},x_i]\\ 0,\, else \end{cases} \\\]

\[a(\varphi_{i-1},\varphi_i) = \int_0^1[-\frac{p(x_{i-1}+h\theta)}{h}+hq(x_{i-1}+h\theta)(1-\theta)\theta]\,d\theta\]

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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      {"h": 0.2, "x": -0.8, "y": -0.6234543499346311, "method": "gaussian"},
      {"h": 0.2, "x": -0.8, "y": -34.611601079696825, "method": "jacobi"},
      {"h": 0.2, "x": -0.6, "y": -0.7496666984401675, "method": "gaussian"},
      {"h": 0.2, "x": -0.6, "y": -64.00110253299722, "method": "jacobi"},
      {
        "h": 0.2,
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      },
      {
        "h": 0.2,
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      {
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      {
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      {"h": 0.1, "x": -0.9, "y": -0.039577232851867254, "method": "jacobi"},
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      {"h": 0.1, "x": -0.8, "y": -0.11781675194164949, "method": "jacobi"},
      {"h": 0.1, "x": -0.7, "y": -0.5468176430082995, "method": "gaussian"},
      {"h": 0.1, "x": -0.7, "y": -0.2324682346123158, "method": "jacobi"},
      {"h": 0.1, "x": -0.6, "y": -0.6335891717352686, "method": "gaussian"},
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      {"h": 0.05, "x": -0.6, "y": -0.5763656232085925, "method": "gaussian"},
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      {
        "h": 0.05,
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      {
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      {"h": 0.05, "x": 0, "y": 0.1622742270157864, "method": "gaussian"},
      {"h": 0.05, "x": 0, "y": -0.23956147740464964, "method": "jacobi"},
      {
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      {
        "h": 0.05,
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      {
        "h": 0.05,
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      },
      {
        "h": 0.05,
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      {
        "h": 0.05,
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      {
        "h": 0.05,
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      {
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      {
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      {
        "h": 0.05,
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        "method": "jacobi"
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      {"h": 0.05, "x": 0.5, "y": 1.717307545354916, "method": "gaussian"},
      {"h": 0.05, "x": 0.5, "y": -0.018887642852077162, "method": "jacobi"},
      {"h": 0.05, "x": 0.55, "y": 1.738939347683239, "method": "gaussian"},
      {"h": 0.05, "x": 0.55, "y": -0.0027796607727244233, "method": "jacobi"},
      {
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      {
        "h": 0.05,
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      },
      {
        "h": 0.05,
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      {
        "h": 0.05,
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      {
        "h": 0.05,
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      {
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        "method": "jacobi"
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      {
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      {
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        "method": "jacobi"
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      {
        "h": 0.02,
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      {
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        "method": "jacobi"
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      {"h": 0.02, "x": 0.94, "y": 0.3287161117477186, "method": "gaussian"},
      {"h": 0.02, "x": 0.94, "y": -0.3641779437333099, "method": "jacobi"},
      {"h": 0.02, "x": 0.96, "y": 0.16869053482613744, "method": "gaussian"},
      {"h": 0.02, "x": 0.96, "y": -0.5111397029898572, "method": "jacobi"},
      {"h": 0.02, "x": 0.98, "y": 0.0013893655744645797, "method": "gaussian"},
      {"h": 0.02, "x": 0.98, "y": -0.6567299340209907, "method": "jacobi"}
    ]
  }
}

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"}
  },
  "$schema": "https://vega.github.io/schema/vega-lite/v4.17.0.json",
  "datasets": {
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