Exact Differential Equation and Green's theorem

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SUMMARY

This discussion focuses on the relationship between Exact Differential Equations and Green's theorem, particularly in the context of equations involving multiple variables. An Exact Differential Equation of the form dF=P(x,y,z,w)dx+Q(x,y,z,w)dy+R(x,y,z,w)dz+S(x,y,z,w)dw satisfies specific conditions, including the equality of mixed partial derivatives. The discussion also highlights the relevance of Stokes' theorem and the curl operator in understanding these equations in three or more variables.

PREREQUISITES
  • Understanding of Exact Differential Equations
  • Familiarity with Green's theorem
  • Knowledge of Stokes' theorem
  • Concept of mixed partial derivatives
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  • Study the application of Stokes' theorem in higher dimensions
  • Explore the properties of mixed partial derivatives in multivariable calculus
  • Learn about the curl operator and its implications in vector calculus
  • Investigate the conditions for Exact Differential Equations in four or more variables
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Mathematicians, physics students, and engineers who are working with multivariable calculus and seeking to deepen their understanding of differential equations and vector calculus.

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We can use Green's theorem to understand why the Exact Differential Equation satisfy the conditions it should have ...

How about a DE for more than two variables ?

Eg.dF=P(x,y,z,w)dx+Q(x,y,z,w)dy+R(x,y,z,w)dz+S(x,y,z,w)dw

IF the equation above is an Exact Differential Equation , what condition it would satisfy ?
 
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An equation of the form f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz= 0 is "exact" if and only if there exist some F(x,y,z) such that
dF= \frac{\partial F}{\p<br /> artial x}dx+ \frac{\partial F}{\partial y}dy+ \frac{\partial F}{\partial z}= f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz

That means that we must have
\frac{\partial F}{\partial x}= f(x,y,z)
\frac{\partial F}{\partial y}= g(x,y,z)
and
\frac{\partial F}{\partial z}= h(x,y,z)

So, as long as those functions are continuous, we must have the "mixed partials" equal:
\frac{\partial^2 F}{\partial x\partial y}= \frac{\partial f}{\partial y}= \frac{\partial g}{\partial x}= \frac{\partial^2 F}{\partial y\partial x}
etc.

Since you mention "Green's theorem" (for two variables) you might want to look at \nabla\times f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k} and Stoke's theorem.
 
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What u mentioned contains three variables ~and I've seen such a case in the book I read...

Suppose \vec{F}=P(x,y,z)\hat{i}+Q(x,y,z)\hat{j}+R(x,y,z)\hat{k}

dT=P(x,y,z)dx+Q(x,y,z)dy+R(x,y,z)dz is exact if only if \vec{F}\bullet(\nabla\times\vec{F})=0

I'm now requesting for the case which contains more than three variables~


HallsofIvy said:
An equation of the form f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz= 0 is "exact" if and only if there exist some F(x,y,z) such that
dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy+ \frac{\partial F}{\partial z}= f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz

That means that we must have
\frac{\partial F}{\partial x}= f(x,y,z)
\frac{\partial F}{\partial y}= g(x,y,z)
and
\frac{\partial F}{\partial z}= h(x,y,z)

So, as long as those functions are continuous, we must have the "mixed partials" equal:
\frac{\partial^2 F}{\partial x\partial y}= \frac{\partial f}{\partial y}= \frac{\partial g}{\partial x}= \frac{\partial^2 F}{\partial y\partial x}
etc.

Since you mention "Green's theorem" (for two variables) you might want to look at \nabla\times f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k} and Stoke's theorem.
 

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