Consider the logistic model (5): $$ begin{equation*} u^{prime} (t)=alpha u(t)left( 1-frac {u(t)} {R} ight). quad u(0)=U_0tpend{equation*}...

 

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Consider the logistic model (5): $$ \begin{equation*} u^{\prime} (t)=\alpha u(t)\left( 1-\frac {u(t)} {R} \right). \quad u(0)=U_0\tp\end{equation*} SS This problem involves three input parameters: \(U_0 \) \( R), and \(\alpha \). Learning how (u) varies with (U_0). (R), and \(\alpha \) requires much experimentation where we vary all three parameters and observe the solution. A much more effective approach is to scale the problem. By this technique the solution depends only on one parameter: \(U_0/R). This exercise tells how the scaling is done. The idea of scaling is to introduce dimensionless versions of the independent and dependent variables: SS \begin{equation*} v = \frac {u} {u_c}, \quad \tau = \frac {t} {t_c}.\end{equation*} SS where \(u_c \) and \(t_c\) are characteristic sizes of \(u \) and \(t\), respectively, such that the dimensionless variables (v) and \(\tau \) are of approximately unit size. Since we know that \(u \rightarrow R \) as \( t\rightarrow\infty). (R) can be taken as the characteristic size of \(u). Insert \(u-Rv \) and \(t=t_c\tau \) in the governing ODE and choose \(t_c = 1/\alpha \). Show that the ODE for the new function (v(\tau) ) becomes $$ \begin{equation) {dv\over d\tau} = v(1-v), \quad v(0)=v_0\tp \tag {64} \end{equation) SS We see that the three parameters (U_0\). \(R), and \(\alpha \) have disappeared from the ODE problem, and only one parameter (v_0=U_0/R) is involved. Show that if \(v(\tau) ) is computed, one can recover \(u(t)) by SS \begin{equation) u(t) = Rv(\alpha t) \tp \tag {65} \end{equation} SS Geometrically, the transformation from (v) to \(u) is just a stretching of the two axis in the coordinate system. Make a program logistic_scaled.py where you compute (v(\tau) ), given \(v_0=0.05), and then you use (65) to plot (u(t)\) for \(R=100,500,1000) and \(\alpha=1 \) in one figure, and \(u(t)\) for \( \alpha=1, 5, 10 \) and (R=1000\) in another figure. Note how effectively you can generate \(u(t)) without needing to solve an ODE problem, and also note how varying \(R) and \(\alpha \) impacts the graph of \(u(t)\). Filename: logistic_scaled. Consider the logistic model (5): $$ \begin{equation*} u^{\prime} (t)=\alpha u(t)\left( 1-\frac {u(t)} {R} \right). \quad u(0)=U_0\tp\end{equation*} SS This problem involves three input parameters: \(U_0 \) \( R), and \(\alpha \). Learning how (u) varies with (U_0). (R), and \(\alpha \) requires much experimentation where we vary all three parameters and observe the solution. A much more effective approach is to scale the problem. By this technique the solution depends only on one parameter: \(U_0/R). This exercise tells how the scaling is done. The idea of scaling is to introduce dimensionless versions of the independent and dependent variables: SS \begin{equation*} v = \frac {u} {u_c}, \quad \tau = \frac {t} {t_c}.\end{equation*} SS where \(u_c \) and \(t_c\) are characteristic sizes of \(u \) and \(t\), respectively, such that the dimensionless variables (v) and \(\tau \) are of approximately unit size. Since we know that \(u \rightarrow R \) as \( t\rightarrow\infty). (R) can be taken as the characteristic size of \(u). Insert \(u-Rv \) and \(t=t_c\tau \) in the governing ODE and choose \(t_c = 1/\alpha \). Show that the ODE for the new function (v(\tau) ) becomes $$ \begin{equation) {dv\over d\tau} = v(1-v), \quad v(0)=v_0\tp \tag {64} \end{equation) SS We see that the three parameters (U_0\). \(R), and \(\alpha \) have disappeared from the ODE problem, and only one parameter (v_0=U_0/R) is involved. Show that if \(v(\tau) ) is computed, one can recover \(u(t)) by SS \begin{equation) u(t) = Rv(\alpha t) \tp \tag {65} \end{equation} SS Geometrically, the transformation from (v) to \(u) is just a stretching of the two axis in the coordinate system. Make a program logistic_scaled.py where you compute (v(\tau) ), given \(v_0=0.05), and then you use (65) to plot (u(t)\) for \(R=100,500,1000) and \(\alpha=1 \) in one figure, and \(u(t)\) for \( \alpha=1, 5, 10 \) and (R=1000\) in another figure. Note how effectively you can generate \(u(t)) without needing to solve an ODE problem, and also note how varying \(R) and \(\alpha \) impacts the graph of \(u(t)\). Filename: logistic_scaled.

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statements are both correct Its interesting to consider how optimizing different parts of a prog... View the full answer