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partial derivative

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

∂∂x(fh)+∂(gh)∂y[br]\displaystyle \frac{\partial}{\partial x} (fh) + \frac{\partial (gh)}{\partial y}[br]

is one-signed, e.g it is not equal to zero for any x and y. Given
f=x(2−x−y)f=x(2 -x -y)
g=y(4x−x2−3)[br]x≥0,y≥0g=y(4x -x^2 -3)[br]x\geq 0, y\geq 0.

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

∂∂x(fh)+∂(gh)∂y[br]\displaystyle \frac{\partial}{\partial x} (fh) + \frac{\partial (gh)}{\partial y}[br]

is one-signed, e.g it is not equal to zero for any x and y. Given
f=x(2−x−y)f=x(2 -x -y)
g=y(4x−x2−3)[br]x≥0,y≥0g=y(4x -x^2 -3)[br]x\geq 0, y\geq 0.

I have tried ex,1,xa+ybe^x, 1, x^a+y^b. I really don't know what to do now. Please help!:redface:

To clarify, we're insisting on strict positivity rather than just positivity? In the latter case, h(x,y)=0h(x,y) = 0 will do.

Expanding using the product rule, then evaluating at (1,0)(1,0) and (0,0)(0,0) gives the restrictions that ∂h∂x(1,0)>0,h(0,0)<0\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?
Reply 2
Avatar for TeeEm
TeeEm
19
Original post by Jundong
I am trying to find a function h(x,y) such that

∂∂x(fh)+∂(gh)∂y[br]\displaystyle \frac{\partial}{\partial x} (fh) + \frac{\partial (gh)}{\partial y}[br]

is one-signed, e.g it is not equal to zero for any x and y. Given
f=x(2−x−y)f=x(2 -x -y)
g=y(4x−x2−3)[br]x≥0,y≥0g=y(4x -x^2 -3)[br]x\geq 0, y\geq 0.

I have tried ex,1,xa+ybe^x, 1, x^a+y^b. I really don't know what to do now. Please help!:redface:


carry out the product rule and then group in twos

something useful may appear


EDIT IGNORE/sorry I misread the derivatives
(edited 9 years ago)
Reply 3
Avatar for Jundong
Jundong
OP
Original post by Smaug123
To clarify, we're insisting on strict positivity rather than just positivity? In the latter case, h(x,y)=0h(x,y) = 0 will do.

Expanding using the product rule, then evaluating at (1,0)(1,0) and (0,0)(0,0) gives the restrictions that ∂h∂x(1,0)>0,h(0,0)<0\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.
Reply 4
Avatar for Jundong
Jundong
OP
Original post by Smaug123
To clarify, we're insisting on strict positivity rather than just positivity? In the latter case, h(x,y)=0h(x,y) = 0 will do.

Expanding using the product rule, then evaluating at (1,0)(1,0) and (0,0)(0,0) gives the restrictions that ∂h∂x(1,0)>0,h(0,0)<0\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?


:frown:Still couldn't figure out h. It looks so easy.
Reply 5
Avatar for TeeEm
TeeEm
19
Original post by Jundong
:frown: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?
Original post by Jundong
:frown: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".
(edited 9 years ago)
Reply 7
Avatar for Jundong
Jundong
OP
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=x‾˙(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.
Reply 8
Avatar for Jundong
Jundong
OP
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−x2)h+x(2−x−y)∂∂xh+y(1−x)(3−x)∂h∂y(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 x2x^2 are negative. I have tried (a,b,c constant.)
eax[br]eaxy[br]xayb[br]xa+c[br]1[br]e^{ax}[br]e^{axy}[br]x^a y^b[br]x^a+c[br]1[br]
(edited 9 years ago)
Original post by Jundong
The expression is
(2x−1−y−x2)h+x(2−x−y)∂∂xh+y(1−x)(3−x)∂h∂y(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 x2x^2 are negative. I have tried (a,b,c constant.)
eax[br]eaxy[br]xayb[br]xa+c[br]1[br]e^{ax}[br]e^{axy}[br]x^a y^b[br]x^a+c[br]1[br]

Your expression is, I think, correct; notice that (2x−1−y−x2)=−(y+(x−1)2)(2x-1-y-x^2) = -(y+(x-1)^2), which is only ever non-negative when y=0, x=1.
Reply 10
Avatar for Jundong
Jundong
OP
Original post by Smaug123
Your expression is, I think, correct; notice that (2x−1−y−x2)=−(y+(x−1)2)(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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