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Given is the following oscillator $$\ddot{x} + \lambda\dot{x}=x-x^3$$ I've already rewritten this as a system of first order equations $$\begin{cases} \dot{x} = y \\ \dot{y}=x-x^3-\lambda y \end{cases}$$ Now the question is how the Hamiltonfunction $H(x,y)=\frac{1}{2}y^2+U(x)$ changes in time under our given oscillator with $\lambda \in \mathbb{R}$. We know that for $\lambda=0$, $\dot{y}=-U'(x)$. If I calculate $\dot{H}$ I find $$\dot{H}=\frac{\partial H}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial H}{\partial y}\frac{\partial y}{\partial t}=U'(x)\cdot y + y(x-x^3-\lambda y)$$ The problem I face is that I don't know if $U(x)$ depends on $\lambda$. Does anyone have an idea to get me going?

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Instead of the Hamiltonian, I can give you the Lagrangian.

In the existence of friction, it is well known to physicists that the following Lagrangian gives the correct equation of motion.

$$L[x,\dot{x},t] = e^{\lambda t/m} \bigg(\frac{1}{2}m\dot{x}^2 - U(x)\bigg)$$

$$\Rightarrow m\ddot{x} +\lambda \dot{x} +U'(x)=0 $$

In your case, $m=1$ and $U(x)=-\frac{1}{2}x^2+\frac{1}{4}x^4$.

You can easily figure out the Hamiltonian.

Otherwise, I would start from defining the first order differential equations in the following way,

\begin{align} e^{\lambda t}\dot{x} &= y \\ e^{-\lambda t} \dot{y} &= x-x^3 \end{align}

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IMO it works best to start with the original equation, multiply with $\dot x$, and integrate, identifying the complete derivative expressions. This gets $$ \left[\frac12\dot x^2+\frac14(x^2-1)^2\right]_{t_0}^t=-λ\int_{t_0}^t\dot x(s)^2\,ds. $$ This should tell you what the Hamiltonian is in the case without friction and how the friction term siphons off energy.

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