In the biochemical process of glycolysis, living cells obtain energy (ATP) by breaking down sugar (glucose)....

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In the biochemical process of glycolysis, living cells obtain energy (ATP) by breaking down sugar (glucose). Many intermediate reactions and compounds are involved. A model for the dynamics of two of these compounds is glucose → reactions involving X, Y, Z, ... ATP products dx dt = -x + ay + x²y, dy = dt It-0 =20, b-ay - x²y, ylt=o = yo, Here t is time, x is the concentration of compound X (adenosine diphosphate), y is the concentration of compound Y (fructose-6-phosphate), and a and b are positive constants that describe the reaction kinetics. Here we show that the system for compounds X and Y always has an equilibrium, and remarkably also has a periodic orbit under certain conditions on a and b. The periodic orbit is observed to be asymptotically stable, and attracts solutions from a large neighborhood into a never-ending cycle. All quantities are dimensionless. t≥ 0. (a) Sketch the nullclines and show that the system has only one equilibrium solution (x, y) for any a> 0 and b > 0. Explicitly find the equilibrium. (b) For simplicity, assume b = 1/2 is fixed. Determine the stability of the equilibrium in terms of a> 0. Show that the equilibrium is unstable for 0< a < a#, and asymptotically stable for a> a #, for an appropriate number a#. (c) Consider a shaded region R as shown, where the hole is centered at (x*, y.). Show that the straight edges of R can be chosen such that the direction field along these edges either points inward or along the edge. Using the result in (b), and the Poincaré-Bendixson Theorem, deduce that the system must contain a closed orbit in R for any 0 < a <a#. Give explicit locations for the vertices of R. [Hint: consider the nullclines; also, note that y + ≤ 0 implies -1 provided > 0.] Would same conclusion hold for a > a#? y In the biochemical process of glycolysis, living cells obtain energy (ATP) by breaking down sugar (glucose). Many intermediate reactions and compounds are involved. A model for the dynamics of two of these compounds is glucose → reactions involving X, Y, Z, ... ATP products dx dt = -x + ay + x²y, dy = dt It-0 =20, b-ay - x²y, ylt=o = yo, Here t is time, x is the concentration of compound X (adenosine diphosphate), y is the concentration of compound Y (fructose-6-phosphate), and a and b are positive constants that describe the reaction kinetics. Here we show that the system for compounds X and Y always has an equilibrium, and remarkably also has a periodic orbit under certain conditions on a and b. The periodic orbit is observed to be asymptotically stable, and attracts solutions from a large neighborhood into a never-ending cycle. All quantities are dimensionless. t≥ 0. (a) Sketch the nullclines and show that the system has only one equilibrium solution (x, y) for any a> 0 and b > 0. Explicitly find the equilibrium. (b) For simplicity, assume b = 1/2 is fixed. Determine the stability of the equilibrium in terms of a> 0. Show that the equilibrium is unstable for 0< a < a#, and asymptotically stable for a> a #, for an appropriate number a#. (c) Consider a shaded region R as shown, where the hole is centered at (x*, y.). Show that the straight edges of R can be chosen such that the direction field along these edges either points inward or along the edge. Using the result in (b), and the Poincaré-Bendixson Theorem, deduce that the system must contain a closed orbit in R for any 0 < a <a#. Give explicit locations for the vertices of R. [Hint: consider the nullclines; also, note that y + ≤ 0 implies -1 provided > 0.] Would same conclusion hold for a > a#? y

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a To sketch the nullclines we set each derivative equal to zero and solve for z and y Setting dadt 0 we have x ay x2y 0 Setting dydt 0 we have b ay x2... View the full answer