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For $\beta \in \Bbb R$ we're given the system $$\begin{cases}x'=-\frac{1}{2}x^3+xy^2 \\ y'=\beta x^2y -\frac{1}{2}y^3\end{cases}$$

find a lyapunov-function of the form $V(x,y)=\alpha x^2+y^2$

We know that $$\begin{align}(1) \dot V(x,y) &\le0 \\ (2) V(x,y) &\gt0 , (x,y)\neq(0,0) \\(3) V(x,y)&=0 , (x,y)=(0,0)\end{align}$$ What I did:

Define $f(x,y)= (-\frac{1}{2}x^3+xy^2,\beta x^2y -\frac{1}{2}y^3)$than

$\begin{align}\dot V &= <\nabla V,f> \\&= <\begin{pmatrix}2\alpha x\\2y\end{pmatrix},\begin{pmatrix}-\frac{1}{2}x^3+xy^2 \\\beta x^2y -\frac{1}{2}y^3\end{pmatrix}>\\ &=-x^4\alpha+\alpha 2x^2y^2+2\beta x^2y^2-y^4\\ & = -x^4\alpha+ 2x^2y^2(\alpha+\beta )-y^4\\& = -(x^4\alpha+y^4)+ 2x^2y^2(\alpha+\beta )\end{align}$

For $\alpha=-\beta$ weg get:$$V(x,y)=-\beta x^2+y^2$$

For $\beta=\frac{1}{8}$and $(1,\frac{1}{4})$ isn't $$V(1,\frac{1}{4})=-\frac{1}{8} \cdot 1^2+(\frac{1}{4})^2=-\frac{1}{8}+\frac{1}{16}=-\frac{1}{16}<0$$

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    Let's fix this error first: $\nabla V = (2 \alpha x, 2 y)$, not $(\alpha x^2, y^2)$. And $\dot{V} = \langle \nabla V, f \rangle$, not $\langle V, f \rangle$. – 2017-02-05
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    @Evgeny: OMG ! Give me a sec. :D – 2017-02-05
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    It looks much better now :) And the answer is pretty straightforward when you have a linear combinations of terms with even degrees of $x$ and $y$. – 2017-02-05

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