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I am trying to find the stationary points of the potential $U(x,y)=x^2+y^2$ with constraint $x^2-2y^2=1$

So I set the Augmented potential $U^*=x^2+y^2+m(x^2-2y^2)$ where $m$ is the Lagrange multiplier

Then we need ${\partial U^*\over \partial x}={\partial U^*\over \partial y}={\partial U^*\over \partial z}=0$

In other words $(1+m)x=(1-2m)y=0$

together with the constraint equation $x^2-2y^2=1$ should give $m$ but all I get is $x^2+y^2+m=0$ but $m$ is supposed to be obtainable as a constant.

Could somebody tell me what has gone wrong? Thank you.

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What's $z$? If you include the entire constraint, $U^*=x^2+y^2+m(x^2-2y^2-1)$, you can recover it as $\partial U^*/\partial m=0$, but that doesn't work if you omit the $1$. Most importantly, however, you should describe how you got $x^2+y^2+m=0$, since we can't tell you where you went wrong if you don't say what you did. –  joriki May 24 '12 at 0:28
    
If $(1+m)x = 0$ then either $1+m = 0$ or $x = 0$. Consider the two cases one by one. –  Rahul May 24 '12 at 0:28
    
Thank you, @joriki –  hatch May 24 '12 at 5:34

1 Answer 1

up vote 0 down vote accepted

Your equations are fine. To satisfy the constraint $x^2-2y^2=1$, we must have $x\ne 0$. It follows that $m=-1$. (I had a less helpful answer. Thanks, joriki!)

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In fact it must be $x$ that's non-zero to satisfy $x^2-2y^2=1$. –  joriki May 24 '12 at 0:33

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