Possible to find a gradient system for this?

[Jamin2112]

Homework Statement

Does

x' = xe^x2tanh(x+y) + (1/2)e^x2sech²(x+y)
y' = (1/2)e^x2sech²(x+y)

contain a limit cycle? Possibly-relevant theorem below.

Homework Equations

Theorem. Closed orbits are impossible in gradient systems.

Definition. If x' = -ΔV for some cont. diff'ble, single-valued scalar func. V(x), then such a system is called a gradient system with potential function V.

The Attempt at a Solution

So the problem looks like a setup to apply this theorem ... but I can't figure out how. I've started at it for like 1 hr.

Thanks in advance.

 

[haruspex]

Definition. If x' = -ΔV for some cont. diff'ble, single-valued scalar func. V(x),

You mean -∇V, right?

 

[haruspex]

y' = (1/2)e^x2sech²(x+y)

Since you want that to look like -∂V/∂y, what does it suggest for V?

 

[Jamin2112]

Since you want that to look like -∂V/∂y, what does it suggest for V?

That V(x,y) should contain ∫-∂V/∂y dy somewhere

 

[Jamin2112]

That V(x,y) should contain ∫-∂V/∂y dy somewhere

But I don't know how to integrate that. I can't even find it on an integral chart.

 

[haruspex]

(1/2)e^x2sech²(x+y)
But I don't know how to integrate that. I can't even find it on an integral chart.

Remember, you're only trying to integrate that wrt y. You should find sech² on the chart. If not, what's the integral of sec²?

 

[Jamin2112]

Remember, you're only trying to integrate that wrt y. You should find sech² on the chart. If not, what's the integral of sec²?

Ah, okay. So I know, for one thing, that V(x,y) somehow involves (1/2)xe^x2tanh(x+y). And ... I see it now. If V(x,y) = (1/2)xe^x2tanh(x+y), then ∂V/∂x = xe^x2tanh(x+y) + (1/2)xe^x2sech²(x+y). But, I want <-∂V/∂x, -∂V/∂y> = < xe^x2tanh(x+y) + (1/2)e^x2sech²(x+y), (1/2)e^x2sech²(x+y)>.

 

[Jamin2112]

Ah, wait. I could do

V(x,y) = (1/2)e^x2tanh(-(x+y)). Then

-∂V/∂x = -[xe^x2 * tanh(-(x+y)) + (1/2)e^x2 * -sech²(-(x+y))]
= xe^x2 * tanh(x+y) + (1/2)e^x2 * sech²(x+y)

since sech is even and tanh is odd.

 

[haruspex]

That's it.