ポテンシャル関数ポテンシャル関数の局所的存在ベクトル場、3次元、3変数、偏導関数、累乗、指数関数、三角関数、正弦と余弦

学習環境

続解析入門 (原書第2版) (S.ラング(著)、松坂和夫(翻訳)、片山孝次(翻訳)、岩波書店)の第5章(ポテンシャル関数)、3(ポテンシャル関数の局所的存在)の練習問題13の解答を求めてみる。

\begin{array}{l} D_{1} f_{2} (x, y, z) = 0 \\ D_{1} f_{3} (x, y, z) = 0 \\ D_{2} f_{1} (x, y, z) = 0 \\ D_{2} f_{3} (x, y, z) = 0 \\ D_{3} f_{1} (x, y, z) = 0 \\ D_{3} f_{2} (x, y, z) = 0 \end{array}

よってポテンシャル関数をもつ。

\begin{array}{l} φ (x, y, z) = \int 2 x dx + ψ (y, z) \\ = x^{2} + ψ (y, z) \end{array}

\begin{array}{l} \frac{\partial}{\partial y} φ (x, y, z) = \frac{\partial}{\partial y} ψ (y, z) = 3 y \\ ψ (y, z) = \frac{3}{2} y^{2} + u (z) \end{array}

\begin{array}{l} \frac{\partial}{\partial z} φ (x, y, z) = \frac{\partial}{\partial z} ψ (y, z) = \frac{\partial}{\partial z} u (z) = z \\ u (z) = 2 z^{2} \end{array}

よって、求める問題のベクトル場のポテンシャル関数は、

φ (x, y, z) = x^{2} + \frac{3}{2} y^{2} + 2 z^{2}

\begin{array}{l} D_{1} f_{2} (x, y, z) = 1 \\ D_{1} f_{3} (x, y, z) = 1 \\ D_{2} f_{1} (x, y, z) = 1 \\ D_{2} f_{3} (x, y, z) = 1 \\ D_{3} f_{1} (x, y, z) = 1 \\ D_{3} f_{1} (x, y, z) = 1 \\ D_{3} f_{2} (x, y, z) = 1 \end{array}

\begin{array}{l} φ (x, y, z) = \int (y + z) dx + ψ (y, z) \\ = (y + z) x + ψ (y, z) \\ = x y + x z + ψ (y, z) \end{array}

\begin{array}{l} \frac{\partial}{\partial y} φ (x, y, z) = x + \frac{\partial}{\partial y} ψ (y, z) = x + z \\ ψ (y, z) = z y + u (z) \end{array}

\begin{array}{l} \frac{\partial}{\partial z} φ (x, y, z) = x + y + \frac{\partial}{\partial z} u (z) = x + y \\ u (z) = 0 \end{array}

φ (x, y, z) = x y + x z + z y

\begin{array}{l} φ (x, y, z) = \int e^{y + 2 z} dx + ψ (y, z) \\ = x e^{y + 2 z} + ψ (y, z) \end{array}

\begin{array}{l} \frac{\partial}{\partial y} = x e^{y + 2 z} + \frac{\partial}{\partial y} ψ (y, z) = x e^{y + 2 z} \\ ψ (y, z) = 0 + u (z) \end{array}

\begin{array}{l} \frac{\partial}{\partial z} φ (x, y, z) = 2 x e^{y + 2 z} + \frac{\partial}{\partial z} u (z) = 2 x e^{y + 2 z} \\ u (z) = 0 \end{array}

φ (x, y, z) = x e^{y + 2 z}

φ (x, y, z) = \int y \sin z dx + ψ (y, z) = x y \sin z + ψ (y, z)

\begin{array}{l} \frac{\partial}{\partial y} φ (x, y, z) = x \sin z + \frac{\partial}{\partial y} ψ (y, z) = x \sin z \\ ψ (y, z) = 0 + u (z) \end{array}

\begin{array}{l} \frac{\partial}{\partial z} φ (x, y, z) = x y \cos z + \frac{\partial}{\partial z} u (z) = x y \cos z \\ u (z) = 0 \end{array}

φ (x, y, z) = x y \sin z

φ (x, y, z) = \int y z dx + ψ (y, z) = x y z + ψ (y, z)

\begin{array}{l} \frac{\partial}{\partial y} φ (x, y, z) = x z + \frac{\partial}{\partial y} ψ (y, z) = x z + z^{3} \\ ψ (y, z) = y z^{3} + u (z) \end{array}

\begin{array}{l} \frac{\partial}{\partial z} φ (x, y, z) = x y + 3 y z^{2} + \frac{\partial}{\partial z} u (z) = x y + 3 y z^{2} \\ u (z) = 0 \end{array}

φ (x, y, z) = x y z + y z^{3}

\begin{array}{l} φ (x, y, z) = \int e^{y z} dx + ψ (y, z) \\ = x e^{y z} + ψ (y, z) \end{array}

\begin{array}{l} \frac{\partial}{\partial y} φ (x, y, z) = x z e^{y z} + \frac{\partial}{\partial y} ψ (y, z) = x z e^{y z} \\ ψ (y, z) = 0 + u (z) \end{array}

\begin{array}{l} \frac{\partial}{\partial z} φ (x, y, z) = x y e^{y z} + \frac{\partial}{\partial z} u (z) = x y e^{y z} \\ u (z) = 0 \end{array}

φ (x, y, z) = x e^{y z}

\begin{array}{l} φ (x, y, z) = \int z^{2} dx + ψ (y, z) \\ = x z^{2} + ψ (y, z) \end{array}

\begin{array}{l} \frac{\partial}{\partial y} φ (x, y, z) = \frac{\partial}{\partial y} ψ (y, z) = 2 y \\ ψ (y, z) = y^{2} + u (z) \end{array}

\begin{array}{l} \frac{\partial}{\partial z} φ (x, y, z) = 2 x z + \frac{\partial}{\partial z} u (z) = 2 x z \\ u (z) = 0 \end{array}

φ (x, y, z) = x z^{2} + y^{2}

\begin{array}{l} φ (x, y, z) = \int y z \cos (x y) dx + ψ (y, z) \\ = z \sin (x y) + ψ (y, z) \end{array}

\begin{array}{l} \frac{\partial}{\partial y} φ (x, y, z) = x z \cos (x y) + \frac{\partial}{\partial y} ψ (y, z) = x z \cos (x y) \\ ψ (y, z) = 0 + u (z) \end{array}

\begin{array}{l} \frac{\partial}{\partial z} = \sin (x y) + \frac{\partial}{\partial z} u (z) = \sin (x y) \\ u (z) = 0 \end{array}

φ (x, y, z) = z \sin (x y)

\begin{array}{l} φ (x, y, z) = \int (y^{3} z + y) dx + ψ (y, z) \\ = x y^{3} z + x y + ψ (y, z) \end{array}

\begin{array}{l} \frac{\partial}{\partial y} φ (x, y, z) = 3 x y^{2} z + x + \frac{\partial}{\partial y} ψ (y, z) = 3 x y^{2} z + x + z \\ ψ (y, z) = y z + u (z) \end{array}

\begin{array}{l} \frac{\partial}{\partial z} φ (x, y, z) = x y^{3} + y + \frac{\partial}{\partial z} u (z) = x y^{3} + y \\ u (z) = 0 \end{array}

φ (x, y, z) = x y^{3} z + x y + y z

コード、入出力結果(Wolfram Language, Jupyter)

VectorPlot3D[{2x, 3y, 4z}, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}]

VectorPlot3D[{y + z, x + z, x + y}, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}]

VectorPlot3D[{Exp[y + 2z], x Exp[y+2z], 2 x Exp[y+2z]}, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}]

VectorPlot3D[{y Sin[z], x Sin[z], x y Cos[z]}, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}]

VectorPlot3D[{y z, x z + z ^3, x y + 3 y z^2}, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}]

VectorPlot3D[{Exp[y z], x z Exp[y z], x y Exp[y z]}, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}]

VectorPlot3D[{z^2, 2y, 2 x z}, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}]

VectorPlot3D[{y z Cos[x y], x z Cos[x y], Sin[x y]}, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}]

VectorPlot3D[{y^3 z + y, 3 x y^2 z + x + z, x y^3 + y}, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}]