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Thread: Gradient system

  1. #1
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    Gradient system

    Given that a system is a gradient system, here's how to find it's potential function V. Suppose that $\displaystyle x'=f(x,y)$ and $\displaystyle y'=g(x,y).$ Then $\displaystyle x'=-\nabla V$ implies $\displaystyle f(x,y)=-\frac{\partial V}{\partial x}$ and $\displaystyle g(x,y)=-\frac{\partial V}{\partial y}$.

    Use this procedure to find V for the following gradient system:
    $\displaystyle x'=y^2+y cos x$
    $\displaystyle y'=2xy+sin x$
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  2. #2
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    Quote Originally Posted by splash View Post
    Given that a system is a gradient system, here's how to find it's potential function V. Suppose that $\displaystyle x'=f(x,y)$ and $\displaystyle y'=g(x,y).$ Then $\displaystyle x'=-\nabla V$ implies $\displaystyle f(x,y)=-\frac{\partial V}{\partial x}$ and $\displaystyle g(x,y)=-\frac{\partial V}{\partial y}$.

    Use this procedure to find V for the following gradient system:
    $\displaystyle x'=y^2+y cos x$
    $\displaystyle y'=2xy+sin x$
    Can you state your question differently.
    You want to find a vector field $\displaystyle F(x,y)$ such as what?
    $\displaystyle \nabla F=<y^2+y\cos x,2xy+\sin x>$
    In that case,
    $\displaystyle F_x=y^2+y\cos x$
    $\displaystyle F_y=2xy+\sin x$
    Thus, (in first equation)
    $\displaystyle F=xy^2+y\sin x+f(y)$
    Where $\displaystyle f(y)$ is to be determined.
    But then,
    $\displaystyle F_y=2xy+\sin x+f'(y)$
    So, $\displaystyle f'(y)=0$
    So, $\displaystyle f(y)=C$
    Thus, it seems the function,
    $\displaystyle F(x,y)=xy^2+y\sin x+C$
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