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2

Hint Since you have $A\to B\to C$, I suppose that the kinetic model is $$\frac{dA}{dt}=-\alpha A$$ $$\frac{dB}{dt}=\alpha A-\beta B$$ $$\frac{dC}{dt}=\beta B$$ (remember that $B$ appears and disappears). Integrating the first equation is simple. Using the initial condition, you obtained $$A=M e^{-\alpha t}$$ which is perfectly correct. So, the second ...


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One possible way to model car traffic is to use min-plus or max-plus semirings. See for instance these slides: ´┐╝Fundamental Traffic Diagrams : A Maxplus Point of View, where you can find more references at the end. See also Bart De Schutter's Traffic and transportation networks publications, with 253 entries.


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I can remove the derivative for you, but my expression still involves $ \mathcal L(y(t)^2) $. Maybe you can work it as a convolution? Anyways, check my work, but using integration by parts, with $ u = e^{-st} $ and $ v = (y'(t))^2 $, and the fact that $\int (y(t))^2 dt = y'(t)y(t) + \int y'(t)y(t)dt $, and $\int y'(t)y(t)dt = y(t)^2/2 $ $$\int _0^\infty ...


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The $x_{t+1}^2$ is not a problem because if $x \in \{0,1\}$, $x^2 = x$. The $x_t x_{t+1}$ can be turned into a new binary variable $y_t$ such that $y_t \ge x_t + x_{t+1} - 1$ and $y_t \le (x_t + x_{t+1})/2$. You may need only one of these...


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Separation of variables is fine. Note $$\begin{align} \dfrac{dy}{dx} = \dfrac{y( \gamma x-\delta)}{x(\alpha - cy)} &\implies \displaystyle\int^y\left(\dfrac{\alpha}{y'} - c \right)dy' = \displaystyle\int^x \left(\gamma -\dfrac{\delta}{x'}\right)dx' \\ &\implies \alpha\log y-cy=\gamma x-\delta \log x+\log K \\ &\implies \exp[\log x^\delta ...


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I guess you are looking for autoregressive model such as $$X(t)=a_1 X(t-1)+a_2 X(t-2)+a_3 X(t-3)$$ What you can do is to use ordinary least squares method to solve $$Y = \begin{bmatrix} X_4 \\ X_5 \\ ... \\ X_{60} \\ \end{bmatrix} = \begin{bmatrix} X_1 & X_2 & X_3 \\ X_2 & X_3 & X_4 \\ . & . & . \\ X_{57} ...


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If you want a normal-type distribution, you could consider a folded normal distribution. Essentially, if $X$ is a normally distributed random variable, then the random variable $Y=|X|$ follows a folded normal distribution. In particular, if $X$ has mean $0$, then $Y=|X|$ follows a half normal distribution, which decreases monotonically. Thus $1-Y$ increases ...


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You can define a bilinear surface from 4 points as $S(u,v)=$$\begin{pmatrix} 1-u & u\end{pmatrix}$ $\begin{bmatrix} P_{00} & P_{01} \\ P_{10} & P_{11} \end{bmatrix}$ $\begin{Bmatrix} 1-v \\ v\end{Bmatrix}$ where $0 \le u \le 1, 0 \le v \le 1$, $P_{00}=(x_1,y_1)$, $P_{10}=(x_2,y_2)$,$P_{01}=(x_3,y_3)$ and $P_{11}=(x_4,y_4)$.



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