Primitiva de una forma exacta

Question

Tengo el siguiente diferencial $3$ -forma, $$ H = \cos \zeta \sin \zeta\, \mathrm{d}\zeta \wedge \mathrm{d}\varphi_1 \wedge \mathrm{d}\varphi_2 $$ que sé que es exacta por las propiedades de mi problema. Ahora me gustaría encontrar una forma primitiva $B$ tal que $H = \mathrm{d}B$ .

He averiguado que dependerá del coseno de la diferencia $\varphi_1 -\varphi_2$ (de manera que la derivada con respecto a los dos ángulos desaparecerá al final. Pero no consigo hacer el cálculo formal completo. ¿Alguna idea?

Answer 1

Usted puede fijar: $$B = f(\zeta ,{\varphi _1},{\varphi _2}){\kern 1pt} {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2}$$ Entonces la derivada exterior da: $$\begin{gathered} dB = df(\zeta ,{\varphi _1},{\varphi _2}){\kern 1pt} \wedge {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} \hfill \\ = (\frac{{\partial f}}{{\partial \zeta }}{\text{d}}\zeta + \frac{{\partial f}}{{\partial {\varphi _1}}}{\text{d}}{\varphi _1} + \frac{{\partial f}}{{\partial {\varphi _2}}}{\text{d}}{\varphi _2}) \wedge {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} \hfill \\ = \frac{{\partial f}}{{\partial \zeta }}{\text{d}}\zeta \wedge {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} \hfill \\ \end{gathered} $$ Es decir: $$\frac{{\partial f}}{{\partial \zeta }}(\zeta ,{\varphi _1},{\varphi _2}) = \cos \zeta \sin \zeta $$ Así que..: $$f(\zeta ,{\varphi _1},{\varphi _2}) = - \frac{1}{2}{\cos ^2}(\zeta )$$ y $$B = - \frac{1}{2}{\cos ^2}(\zeta ){\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2}$$ con: $$H = dB = \cos \zeta \sin \zeta {\kern 1pt} {\text{d}}\zeta \wedge {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2}$$ Más general: Si establecemos: $$A = f \cdot {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} + g \cdot {\kern 1pt} {\text{d}}\zeta \wedge {\text{d}}{\varphi _2} + h{\kern 1pt} \cdot {\text{d}}\zeta \wedge {\text{d}}{\varphi _1}$$ entonces: $$\begin{gathered} dA = \frac{{\partial f}}{{\partial \zeta }}{\text{d}}\zeta \wedge {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} + \frac{{\partial g}}{{\partial {\varphi _1}}}{\text{d}}{\varphi _1} \wedge {\text{d}}\zeta \wedge {\text{d}}{\varphi _2} + \frac{{\partial h}}{{\partial {\varphi _2}}}{\text{d}}{\varphi _2} \wedge {\text{d}}\zeta \wedge {\text{d}}{\varphi _1} \hfill \\ dA = (\frac{{\partial f}}{{\partial \zeta }} + \frac{{\partial h}}{{\partial {\varphi _2}}} - \frac{{\partial g}}{{\partial {\varphi _1}}}){\text{d}}\zeta \wedge {\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} \hfill \\ \end{gathered}$$ Comparar da: $$\frac{{\partial f}}{{\partial \zeta }} + \frac{{\partial h}}{{\partial {\varphi _2}}} - \frac{{\partial g}}{{\partial {\varphi _1}}} = \cos \zeta \sin \zeta $$ Así que concluimos: $$\begin{gathered} \frac{{\partial f}}{{\partial \zeta }} = \cos \zeta \sin \zeta \hfill \\ \frac{{\partial h}}{{\partial {\varphi _2}}} = \frac{{\partial g}}{{\partial {\varphi _1}}} \hfill \\ \end{gathered}$$ Ahora fijando: $$\begin{gathered} h = \frac{{\partial P}}{{\partial {\varphi _1}}} \hfill \\ g = \frac{{\partial P}}{{\partial {\varphi _2}}} \hfill \\ f(\zeta ,{\varphi _1},{\varphi _2}) = - \frac{1}{2}{\cos ^2}(\zeta ) \hfill \\ \end{gathered}$$ para una función $P$ que tenemos: $$\begin{gathered} A = - \frac{1}{2}{\cos ^2}(\zeta ){\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} - \left( {\frac{{\partial P}}{{\partial {\varphi _1}}}{\text{d}}{\varphi _1} + \frac{{\partial P}}{{\partial {\varphi _2}}}{\text{d}}{\varphi _2}} \right) \wedge {\text{d}}\zeta \hfill \\ A = - \frac{1}{2}{\cos ^2}(\zeta ){\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} - \left( {\frac{{\partial P}}{{\partial \zeta }}{\text{d}}\zeta + \frac{{\partial P}}{{\partial {\varphi _1}}}{\text{d}}{\varphi _1} + \frac{{\partial P}}{{\partial {\varphi _2}}}{\text{d}}{\varphi _2}} \right) \wedge {\text{d}}\zeta \hfill \\ A = - \frac{1}{2}{\cos ^2}(\zeta ){\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} - {\text{d}}P \wedge {\text{d}}\zeta \hfill \\ A = - \frac{1}{2}{\cos ^2}(\zeta ){\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} + {\text{d}}\zeta \wedge {\text{d}}P \hfill \\ \end{gathered} $$ Eso es: $$B = - \frac{1}{2}{\cos ^2}(\zeta ){\text{d}}{\varphi _1} \wedge {\text{d}}{\varphi _2} = A - {\text{d}}\zeta \wedge {\text{d}}P$$ $B$ difiere de $A$ mediante una forma cerrada ${\text{d}}\zeta \wedge {\text{d}}P$

Answer 2

¿Por qué no intentar escribir algo como \begin{equation} B=B_{1} (\zeta,\varphi_1,\varphi_2)\mathrm{d}\varphi_1 \wedge \mathrm{d}\varphi_2+B_{2}(\zeta,\varphi_1,\varphi_2)\mathrm{d}\zeta \wedge \mathrm{d}\varphi_1+B_{3}(\zeta,\varphi_1,\varphi_2)\mathrm{d}\zeta \wedge \mathrm{d}\varphi_2 \end{equation}

Si $H=dB$ es exacta, puede encontrar restricciones en cada $B_{i}$ .

Answer 3

En general, asuma $f$ es una función continua de una variable y $F$ es una antiderivada de $f$ (es decir, $F' = f$ ). Si $$ H = f(x)\, dx \wedge dy,\qquad B = F(x)\, dy, $$ entonces $dB = H$ .

Observaciones similares son válidas para las formas de grado superior, y la $3$ -forma $$ H = \cos\zeta \sin\zeta\, d\zeta \wedge d\varphi_{1} \wedge d\varphi_{2} $$ es de este tipo. (En particular, $H$ es exacto a priori por inspección. :)