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# third order differential equation

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on 16 January 2013

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#### Transcript of third order differential equation

Consider that parameters a e b have the following value: Introduction di on i u eq tial ffe at The system is now a system of first order differential equations. Now we calculate the eigenvalues in the point If we consider a=b=1 the equation but what he sees from the inside, ON THE STRUCTURE OF THE SOLUTIONS OF A TWO-PARAMETER FAMILY OF THREE-DIMENSIONAL ORDINARY DIFFERENTIAL EQUATIONS Simona Cadoni
Valentina M. Granata
Simona Olianas ren Three dimensional ordinary A DIFFERENTIAL EQUATION is an equation that contains the derivative or differentials of one or more dependent variables with respect to one or more independent variables. If the equation contains only ordinary derivatives (of one or more dependent variables) with respect to a single independent variable, the equation is called an ordinary differential equation. . A system of ordinary differential equations is a simultaneous set of equations that involves two or more dependent variables that depend on one independent variable.
A solution of the system is a set of functions that satisfies each equation on some interval Our study will focus in the analysis of the system in the neighbourhood of the equilibrium point, where it is convenient to bring stability analysis to the study of a simpler problem obtained by approximating the original system in the neighbourhood of the equilibrium points. ... 2 costant parameters y+ay+by+y=y A system is stable if its dynamics, from certain initial conditions in the neighbourhood of a critical point for the system, reaches such a point and in this case we can say that does not grow further. Steady states and the study of stability To verifying the presence of stationary point, we impose the derivatives equal to zero y'1=0 from which y2=0

y'2=0 from which y3=0

y'3=0 from which -y1+y1^2-by2-ay3=0 remember that : y2=0 y3=0 substituting this value in the equation, it becomes -y1+y1^2=0

y1(y1-1)=0

y1=0 and y1=1 so there are two steady states: P1(0,0,0) P2(1,0,0) .. . This is a third order equation. To study it, it is necessary to impose simplifications and reduce it to a first order equation system . y=y1 by analogy y^2=y1^2

y'1= y2
this is the first derivative of y1 that I will call y2

y'2=y3
the second derivative now is the first derivative of y2 that I will call y3

y'3=-y1+y1^2-by2-ay3 ....Calculating eigenvalues..... Let's start
with a clarification of a formal nature: Y'1=f(Y1,Y2,Y3)

Y'2=g(Y1,Y2,Y3)

Y'3=h(Y1,Y2,Y3) Now
it is possible to write the
Jacobian Matrix for the equation: P1 (0,0,0) To calculate the Eigenvalue let’s consider a=b=0 Now the equation is: We can obtain the same outocome using the software "Mathematica": Nsolve[y^3+1==0] {{y-> -1.},{y-> 0.5 -0.866025*i },{y-> 0.5 +0.866025*i }} Now we calculate the eigenvalues in the point: P2(1,0,0) The solution has a negative real part SADDLE FOCUS STABLE We can get the same outcome using the software "Mathematica": we obtain three roots NSolve[y^3+y^2+y+y+1==0] {{y-> -1.},{y-> -1 *i },{y-> +1*i }} A solution that brings our system to a Hopf Bifurcation. becomes: As done before, let's the parameters a e b assume the values: a=b=0 we obtain three roots: we obtain the same output with "Mathematica": NSolve[y^3 - 1 == 0] {{y -> -0.5 - 0.866025*i}, {y -> -0.5 + 0.866025*i}, {y -> 1.}} If we consider Now the equation is: becomes: from which, using Mathematica
we obtain three roots: NSolve[y^3 + y^2+y-1 == 0] {{y -> -0.771845-1.11514*i}
{y -> -0.771845+1.11514*i}
{y -> 0.5436989}} Conclusion we have supposed a and b to be real parameters when parameter values are near a=b=0 it will turn out that there is a region in which the global solution structure of the equation admits a double scroll structure when parameters values are near a=b=+-1 there is a region in which the global solution structure appears to be determinate by the periodic solution Hopfs bifurcation ...and... a=b=1 the equation The solution has one positive and two negative real part SADDLE FOCUS UNSTABLE the end The solution has one positive and two negative real part SADDLE FOCUS UNSTABLE y+ay+by+y=y ... .. . 2 Remember that: f(y1,y2,y3)= y2
g(y1,y2,y3)=y3
h(y1,y2,y3)=-y1+y1^2-by2-ay3
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