Finding a Liapunov function. Is my analysis correct? Consider the following system of differential equations $x'=f(x)$:
\begin{equation}
x_1' = -x_1 -x_2^3 \\
x_2' = x_1-x_2
\end{equation}
Here is the Liapunov function I wish to use: $V(x_1, x_2) = \alpha x_1^2 + \beta x_2^4$. So I need to find $\alpha$ and $\beta$ that satisfy the following 3 Liapunov conditions:
The conditions to be a Liapunov funtion w.r.t the origin $x_0$:
1) $V(0, 0) = 0$ 
Condition 1) is clearly true.
2) $V(x) > 0$ for all $x\neq x_0$ This is clearly true for $\alpha \geq 0$ and $\beta \geq 0$ but not both equal to $0$.
3) $\nabla V\cdot f \leq 0$ for all $x$:
\begin{align}
\nabla V\cdot f &= 2\alpha x_1(-x_1-x_2^3) + 4\beta x_2 (x_1-x_2)\\
                &= -2\alpha x_1^2 -2\alpha x_1x_2^3 + 4\beta x_1x_2 - 4\beta x_2^2\\
                &= -4\beta x_2^2 -2\alpha x_1^2 -2\alpha x_1x_2^3 +4\beta x_1x_2\\
                &= -4\beta x_2^2 -2\alpha + x_1x_2(-2\alpha x_2^2+4\beta)
\end{align}
In order for condition 3) to hold, that is $\nabla V(x)\cdot f \leq 0$ for all $x$, then the third term must equal $0$. That is we need:
\begin{equation}
-2\alpha x_2^2+4\beta =0
\end{equation}
If this is the case, all solutions depend on $x_2$ unless both $\alpha$ and $\beta$ equal $0$ which cannot happen as condition 2) would not be satisfied.
This is where I am stuck. Please help! 

We are given
$$x_1' = -x_1 -x_2^3 \\
x_2' = x_1-x_2$$
with
$$V(x_1, x_2) = \alpha x_1^2 + \beta x_2^4$$
We have:
\begin{align}
 V' &= 2 \alpha x_1 x'_1 + 4 \beta x_2^3 x'_2 \\
 &=2 \alpha x_1(-x_1 -x_2^3) + 4 \beta x^3_2(x_1-x_2)\\
 &=-2 \alpha x_1^2 -2 \alpha x_1 x^3_2 + 4 \beta x_1 x^3_2 - 4 \beta x^4_2 \\
 &=-2 \alpha x^2_1 + x_1 x^3_2(4 \beta -2 \alpha) -4 \beta x^4_2
\end{align}
Now, what if we let $4 \beta - 2 \alpha = 0 \rightarrow 4 \beta = 2 \alpha$ to get rid of the $x_1 x^3_2$ term, we have:
$$V' = -2 \alpha x^2_1 - 4 \beta x^4_2 = -2 \alpha x^2_1 - 2 \alpha x^4_2 = -2 \alpha (x^2_1 + x^4_2)$$
Clearly this is negative $\forall (x_1,x_2)$, so we have (since $4 \beta = 2 \alpha \rightarrow \beta = \dfrac{\alpha}{2}$):
$$V(x_1, x_2) = \alpha x^2_1 + \beta x^4_2 = \alpha x^2_1 + \dfrac{\alpha}{2} x^4_2$$
Lets verify this:
\begin{align}
V' &= 2 \alpha x_1 x'_1 + 2 \alpha x^3_2 x'_2\\ 
&= 2 \alpha x_1(-x_1-x^3_2)+ 2 \alpha x^3_2(x_1-x_2) \\
&= -2 \alpha x^2_1 -2 \alpha x_1 x^3_2 + 2 \alpha x_1 x^3_2 - 2 \alpha x^4_2\\
&= -2 \alpha (x^2_1 + x^4_2)
\end{align}
That checks out and meets all the criteria for a Liapunov function.
Note: do you now see where you made a slight error with the derivative?
This is telling us there are no closed orbits and we can see this in a phase portrait plot as follows (we could have also found the Jacobian and critical points to determine this).

