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im just checking to see if im doing this right?

$$\frac{du}{d\tau}=u(1-u)-h $$

show this equations has 2 steady states and check their linear stability.

this is what i have done:

$u=0$ and $u=1$

$$\frac{d^2u}{d\tau^2} = -2u+1 $$

at $u=0$, $\frac{d^2u}{d\tau^2}$=1 which is unstable

at $u=1$, $\frac{d^2u}{d\tau^2}$=-1 which is stable

is this right? thanks in advance

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2 Answers 2

up vote 3 down vote accepted

The steady states occur when the RHS is zero, so

$$u-u^2-h = 0$$

which gives you steady states at

$$u^\pm = \tfrac{1}{2}(1\pm\sqrt{1-4h})$$

You can then plug these values into the second derivative, which you already computed, to determine their stability (though make sure you have the signs the right way around - positive is unstable, negative is stable!)

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thank you very much- yes i edited it straight after i realised i go tthe signs for stability the wrong way around. many thanks –  user54511 Apr 18 '13 at 12:38
    
And to avoid computing any derivative, just plot the function $u\mapsto u(1-u)-h$ and the result should jump at your face. –  Did Apr 6 at 22:33

A steady state occurs when all derivatives with respect to time are zero. That is, in this case,

$$ \frac{du}{d\tau} = u(1-u)-h = 0 $$ We can rewrite this as

$$ u^2-u+h=0 $$ It is this equation that must be solved for $u$ to get the steady states.

The approach to stability, I believe to be correct (although sign may be backwards - you need to check that), but it must be applied at the correct steady states.

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thank you very much for your help- much appreciated –  user54511 Apr 18 '13 at 12:49

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