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    I am trying to find a function h(x,y) such that

    \displaystyle \frac{\partial}{\partial x} (fh) + \frac{\partial (gh)}{\partial y}

    is one-signed, e.g it is not equal to zero for any x and y. Given
    f=x(2 -x -y)
    g=y(4x -x^2 -3)

x\geq 0, y\geq 0.

    I have tried e^x, 1, x^a+y^b. I really don't know what to do now. Please help!
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    (Original post by Jundong)
    I am trying to find a function h(x,y) such that

    \displaystyle \frac{\partial}{\partial x} (fh) + \frac{\partial (gh)}{\partial y}

    is one-signed, e.g it is not equal to zero for any x and y. Given
    f=x(2 -x -y)
    g=y(4x -x^2 -3)

x\geq 0, y\geq 0.

    I have tried e^x, 1, x^a+y^b. I really don't know what to do now. Please help!
    To clarify, we're insisting on strict positivity rather than just positivity? In the latter case, h(x,y) = 0 will do.

    Expanding using the product rule, then evaluating at (1,0) and (0,0) gives the restrictions that \frac{\partial h}{\partial x}(1,0) > 0, h(0,0) < 0. (Wlog we restrict to requiring positivity rather than allowing negativity.) Do those help?
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    (Original post by Jundong)
    I am trying to find a function h(x,y) such that

    \displaystyle \frac{\partial}{\partial x} (fh) + \frac{\partial (gh)}{\partial y}

    is one-signed, e.g it is not equal to zero for any x and y. Given
    f=x(2 -x -y)
    g=y(4x -x^2 -3)

x\geq 0, y\geq 0.

    I have tried e^x, 1, x^a+y^b. I really don't know what to do now. Please help!
    carry out the product rule and then group in twos

    something useful may appear


    EDIT IGNORE/sorry I misread the derivatives
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    (Original post by Smaug123)
    To clarify, we're insisting on strict positivity rather than just positivity? In the latter case, h(x,y) = 0 will do.

    Expanding using the product rule, then evaluating at (1,0) and (0,0) gives the restrictions that \frac{\partial h}{\partial x}(1,0) > 0, h(0,0) < 0. (Wlog we restrict to requiring positivity rather than allowing negativity.) Do those help?
    It does help thank you. I will keep trying.
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    (Original post by Smaug123)
    To clarify, we're insisting on strict positivity rather than just positivity? In the latter case, h(x,y) = 0 will do.

    Expanding using the product rule, then evaluating at (1,0) and (0,0) gives the restrictions that \frac{\partial h}{\partial x}(1,0) > 0, h(0,0) < 0. (Wlog we restrict to requiring positivity rather than allowing negativity.) Do those help?
    Still couldn't figure out h. It looks so easy.
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    (Original post by Jundong)
    Still couldn't figure out h. It looks so easy.
    I will try to have a look this afternoon but I cannot think of anything else but to carry out the differentiation and solve the PDE.

    Is this question from an analysis course or a methods course?
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    (Original post by Jundong)
    Still couldn't figure out h. It looks so easy.
    Mind giving us your expression expanded using the product rule? You should find that one term is always strictly negative, one term is strictly negative except for a small range of x and for y=0, and one term is strictly negative except for x=0 and a small range of x,y. There's only a small amount of the domain which you actually need to pay attention to.

    ETA: by "always strictly negative", I mean "almost always strictly negative".
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    (Original post by TeeEm)
    I will try to have a look this afternoon but I cannot think of anything else but to carry out the differentiation and solve the PDE.

    Is this question from an analysis course or a methods course?
    It is the final step of my question, derived from Bendixson–Dulac theorem.
    (f,g)^T=\dot{\underline{x}} and the question is to show there is no periodic orbit in the first quadrant. So I think one example of h should be sufficient.
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    (Original post by Smaug123)
    Mind giving us your expression expanded using the product rule? You should find that one term is always strictly negative, one term is strictly negative except for a small range of x and for y=0, and one term is strictly negative except for x=0 and a small range of x,y. There's only a small amount of the domain which you actually need to pay attention to.

    ETA: by "always strictly negative", I mean "almost always strictly negative".

    The expression is
    (2x-1-y-x^2)h+x(2-x-y)\frac{\partial}{\partial x} h +y(1-x)(3-x)\frac{\partial h}{\partial y}

    I was going for strictly negative because the terms like x^2 are negative. I have tried (a,b,c constant.)
    e^{ax}

e^{axy}

x^a y^b

x^a+c

1
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    (Original post by Jundong)
    The expression is
    (2x-1-y-x^2)h+x(2-x-y)\frac{\partial}{\partial x} h +y(1-x)(3-x)\frac{\partial h}{\partial y}

    I was going for strictly negative because the terms like x^2 are negative. I have tried (a,b,c constant.)
    e^{ax}

e^{axy}

x^a y^b

x^a+c

1
    Your expression is, I think, correct; notice that (2x-1-y-x^2) = -(y+(x-1)^2), which is only ever non-negative when y=0, x=1.
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    (Original post by Smaug123)
    Your expression is, I think, correct; notice that (2x-1-y-x^2) = -(y+(x-1)^2), which is only ever non-negative when y=0, x=1.
    I think I found the correct h(x,y). Thank you for your help.
    Actually the exponential almost works except one single point. So I just need to add one linear term.
 
 
 
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