Dalam Gambar 2., bentuk level set d

Dalam Gambar 2., bentuk level set dari integral (13) yang diplot diperoleh dengan memilih nilai parameter (a 0 , a 1 , b 0 , b 1 ) = (4, 0.7, 0.5, 1.0) (Gambar 2. atas-kiri) dan (a 0 , a 1 , b 0 , b 1 ) = (4, −0.875, 0.5, 1.0) (Gambar 2. atas-kanan). Sedangkan untuk diagram bifurkasi diperoleh disekitar titik 2-periodik (x, y) = (−1, 1) untuk pilihan nilai parameter a 0 = 4, (b 0 , b 1 ) = (0.5, 1.0), dan a 1 ∈ (−3, 0) (Gambar 2. bawah-kiri) dan disekitar titik tetap (x, y) = (1, 1) untuk
pilihan nilai parameter a 0 = 4, (b 0 , b 1 ) = (0.5, 1.0), dan a 1 ∈ (−1, 2) (Gambar 2. bawah-kanan). Keadaan dinamik, sebagai contoh untuk pilihan nilai parameter a 0 = 4 dan a 1 = 0.7, titik 2-periodik (x, y) = (−1, 1) dari sistem (13) bertipe centre dan kurva solusinya tertutup sebagaimana terlihat dalam Gambar 2 (atas-kiri). Oleh karena itu titik 2-periodik stabil. Sebaliknya, untuk pilihan nilai parameter yang sama, titik tetap (x, y) = (1, 1) bertipe sadle. Oleh karena itu titik tetap (x, y) = (1, 1) tidak stabil. Sedangkan untuk keadaan dinamik dengan pilihan nilai parameter a 0 = 4 dan a 1 = −0.875, titik tetap (x, y) = (1, 1) dari sistem (13) bertipe centre dan kurva solusinya tertutup sebagaimana terlihat dalam Gambar 2 (atas-kanan). Oleh karena itu titik tetap stabil. Sebaliknya, untuk pilihan nilai parameter yang sama, titik 2-periodik (x, y) = (−1, 1) bertipe sadle. Oleh karena itu titik 2-periodik (x, y) = (−1, 1) tidak stabil.

Hasil (Inggris) 1:

In Figure 2, the shape of the level sets of integral (13) who plotted retrieved by selecting parameter values (a 0, a is 1, b 0, b 1) = (4, 0.7, 0.5, 1.0) (Figure 2-a top-left) and (a 0, a is 1, b 0, b 1) = (4 − 0.875, 0.5, 1.0) (Figure 2-a top-right). As for the vicinity of the point of bifurcation diagram obtained 2-periodic (x, y) = (−1, 1) to the option value of parameters a 0 = 4, (b 0, b 1) = (0.5, 1.0), and a 1 ∈ (−3, 0) (Figure 2-a bottom-left) and the vicinity of a fixed point (x, y) = (1, 1) forthe option value of parameters a 0 = 4, (b 0, b 1) = (0.5, 1.0), and a 1 ∈ (−1, 1) (fig 2. below-right). Dynamic circumstances, for example for the selection of the value of the parameters a 0 = 4 and a = 0.7, 1 two-point with (x, y) = (−1, 1) from the system (13) type of solution curves and closed as seen in Figure 2 (top-left). Therefore point 2-periodic stable. Instead, for the same parameter value options, a fixed point (x, y) = (1, 1) type sadle. Therefore a fixed point (x, y) = (1, 1) is not stable. As for the dynamic State with a choice of parameter values with 0 = 4 and a = − 1 0.875, a fixed point (x, y) = (1, 1) from the system (13) type of solution curves and closed as seen in Figure 2 (top-right). Therefore the point remains stable. Instead, for the same parameter value options, point 2-periodic (x, y) = (−1, 1) type sadle. Therefore 2-periodic point (x, y) = (−1, 1) is not stable.

Hasil (Inggris) 2:

In Figure 2, the form level set of integral (13) is plotted obtained by selecting parameter values ​​(a 0, a 1, b 0, b 1) = (4, 0.7, 0.5, 1.0) (Figure 2. The upper-left ) and (a 0, a 1, b 0, b 1) = (4, -0875, 0.5, 1.0) (Figure 2. The upper-right). As for the bifurcation diagram obtained around the 2-periodic point (x, y) = (-1, 1) for selection of parameter values ​​a 0 = 4, (b 0, b 1) = (0.5, 1.0), and a 1 ∈ ( -3, 0) (Fig 2. The bottom-left) and around the fixed point (x, y) = (1, 1) for
selection of parameter values ​​a 0 = 4, (b 0, b 1) = (0.5, 1.0) , and a 1 ∈ (-1, 2) (Figure 2 bottom-right). Dynamic state, as an example for the selection of parameter values ​​a 0 = 4 and a 1 = 0.7, 2-periodic point (x, y) = (-1, 1) of the system (13) of type center and a closed solution curve as shown in Figure 2 (top-left). Therefore, point 2-periodic stable. Conversely, for the same selection parameter values, fixed point (x, y) = (1, 1) type sadle. Therefore the fixed point (x, y) = (1, 1) unstable. As for the dynamic state with a choice of parameter values ​​0 = 4 and a 1 = -0875, fixed point (x, y) = (1, 1) of the system (13) of type center and a closed solution curve as shown in Figure 2 (top -Right). Therefore, the point remains stable. Conversely, for the same selection parameter value, 2-periodic point (x, y) = (-1, 1) type sadle. Therefore the 2-periodic point (x, y) = (-1, 1) unstable.