Solutions to Dynamical Systems 2010 exam. Each question is worth 25 marks.

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1 Solutions to Dynamical Systems exam Each question is worth marks [Unseen] Consider the following st order differential equation: dy dt Xy yy 4 a Find and classify all the fixed points of Hence draw the phase diagram associated to We just need to find the solutions to Xy which are y ± In order to classify the fixed points we have to either look at the sign of the rhs of in the various regions between fixed points or use the criterium coming from linearisation which means looking at the sign of the derivative of the rhs of at each of the fixed points Following this last criterium we find that So we find that: X y y 4 yy 3y + 4 X 4 > therefore y is a repellor X 8 > therefore y is an attractor X 8 < therefore y is also an attractor Hence the phase diagram is b Solve for each of the following initial conditions: y 3 y indicating the range of values of t for which each solution is defined The equation can be solved by separating variables in the usual way: t y dy y dt t t t y yy y + y 8 + y 4 y y 8 log y 4y y 4y with y yt the initial condition This can be solved to [ mark] yt ± e 4t t e 8t t + 4 y The ± sign will be + for y > and for y < in order to satisfy the initial condition Let us now solve for each initial condition: for y 3 we have yt e 4t e 8t 9 [ marks] This solution is well defined only when e 8t 9 > This gives the condition t > 8 log Thus the range of values for which the solution is defined is t 8 log 9

2 For y the solution becomes [4 marks] e4t yt e 8t + 3 where we had to introduce a minus sign to guarantee that the initial condition is fulfilled In this case the denominator is never zero and e 8t + 3 can never become negative so the solution is valid for all values of t The interval of definition is t [4 marks] c Provide a sketch of your solutions against the variable t clearly showing the behaviour of the various functions as t ± In order to carry out the sketch it is important to study the behaviour of both solutions as t ± The solution yt e 4t y 3 e 8t 9 is clearly always positive for those values of t for which it is defined and as t it tends to the value which is the fixed point we identified before This was to be expected as the point is an attractor In this case it does not make sense to consider t as the solution is only defined for t 8 log 9 Finally since the denominator vanishes exactly at t 8 log 9 the function should diverge as it gets closer to that value Figure : The red figure corresponds to the initial condition y 3 As we can see it diverges at a small negative value of t which was computed above The blue graph is the solution with y The green lines are the fixed points y ± Concerning the other solution e4t yt y e 8t + 3 it is clearly a negative function for all values of t As t it tends to the value - which is the fixed point identified before We should have expected that since - is an attractor As t the solution tends to the value which is also compatible with our fixed point analysis as zero was identified as a repellor A plot of the two solutions is given above d Write down the linearized version of equation about a generic fixed point y a At a fixed point Xa and therefore a Taylor expansion of Xy to first order in y gives dy dt X ay a 3a + 4y a [ marks] Page

3 [ mark] e Use the result of d to find the solution of the linearized equation about the smallest of the fixed points of if the initial condition is y 3 We already found in a that X 8 therefore the linearized equation is dz dt 8z for z y + The solution to this equation is z Ae 8t where A is a constant which will be fixed by the initial condition The final answer is: yt e 8t [Seen] Consider the following system of linear differential equations: dx dt where α β γ c c are real constants ẋ αx + γx + c dx dt ẋ βx + αx + c a Write the equations in the standard form ẋ Ax + b Write down the vector a which is a fixed point of ẋ α γ x c + β α c The fixed point is the solution to the vector equation that is ẋ Ax + b x A α γ b α βγ β α x αc γc c α βγ βc c +αc α βγ [ mark] [4 marks] b Compute the eigenvalues and eigenvectors of the matrix A The eigenvalues of A are obtained by finding the zeroes of the characteristic polynomial of A: the solutions to this equation are A λi α λ βγ α βγ + λ αλ and λ α + 4α 4α βγ λ α 4α 4α βγ α + βγ α βγ The eigenvectors corresponding to these eigenvalues are obtained by solving: α γ a α ± a βγ β α b b In components this gives two equations for each eigenvector: aα + bγ α ± βγa aβ + bα α ± βγb Page 3

4 simplifying bγ ± βγa aβ ± βγb The two equations are not independent and setting a this gives b ± β eigenvectors would be: E β γ E c Define the vector z in terms of which the equation can be re-written in the form ż Az We just need to change variables to z x a where a is the fixed point obtained in section a d Indicate the conditions if any that α β and γ must satisfy so that the fixed point is: an improper node a focus a star node β γ γ Therefore the [ marks] [ mark] An improper node arises when the eigenvalues of the matrix A are equal and real In our case the eigenvalues can only be equal if βγ which corresponds to β or γ [ mark] A focus occurs when the eigenvalues are complex conjugated to each other This will happen only if β > and γ < or viceversa [ mark] Finally a star node occurs when the matrix A is proportional to the identity matrix This will only happen if β γ αc γc α βγ βc +αc α βγ [ mark] e If α β and γ classify the fixed point Find the values of c and c so that the fixed point is a We found the fixed point in section a c c a c +c f This is solved by c 3 and c For α β and γ show that the matrix A is already in its Jordan Normal form For these values of α β γ and the values of c c obtained in e find the general solution of equations For the values of α β γ given in e the eigenvalues of A are λ + i and λ i The Jordan form is therefore: which coincides with the matrix A itself J [ mark] Page 4

5 We will use the change of variables of section c so that the equation becomes of the form ż Az where and z x a a from section e The equation in z can be written in components as: ż z + z ż z + z and can be solved by using the standard change of variables z r cos θ z r sin θ This leads to the equations where A is a constant and ṙ r r Ae t θ θ t + B where B is an integration constant Therefore the general solutions are z Ae t cos t + B z Ae t sin t + B Therefore for the original variables x x the general solutions are x + Ae t cos t + B x + Ae t sin t + B or in vector form x + Ae t cos t + B + Ae t sin t + B [ marks] 3 [Seen] Consider the following system of first order nonlinear differential equations: ẋ x x x ẋ x x x 3 a Show that the fixed points of the system are either at the origin or on a line In order to find the fixed points we have to solve the simultaneous equations: x x x x x x which give solutions x x and x x which is indeed a line b Calculate the Jacobian matrix for a general point x x and deduce that the fixed point at the origin is simple whereas the fixed points on the line are not The Jacobian matrix is obtained as usual by computing the derivatives A xx X x X X x X x x x x x x x x with X x x x x x and X x x x x x Page

6 Evaluating this matrix at the origin we obtain A which is the identity matrix The determinant of this matrix is obviously and therefore nonvanishing So the origin is a simple fixed point For any point α α lying on the line x x we find that the Jacobian matrix is α α A α α + α + α which has determinant deta therefore the line of fixed points x x contains only non-simple fixed points c By writing down the linearization about the fixed point at the origin classify its nature Solve the linearized equation We have already almost answered this question above when we computed the Jacobian matrix at the origin The linearized equation about the origin is: x A x x which x and since A x is the identity matrix the fixed point at the origin is an unstable star node The equation is very easy to solve In components we have that which is solved by ẋ x ẋ x x C e t x C e t with C C arbitrary constants Since x /x C /C phase space trajectories near the fixed point will be straight lines going by the origin d By dividing the two equations in 3 obtain a new equation of the form dx dx Xx x Find general solution to this equation Explain why your solution is what one would expect given the nature of the fixed point at the origin The resulting equation is dx x x x dx x x x x x which only is well defined when x x and x The equation can be solved by separation of variables dx dx x x which gives or ln x ln x + C x e C x where C is an integration constant Therefore in general phase space trajectories are straight lines going by the origin Page 6

7 This is what one expects from the fact that the origin is an unstable star node which is precisely characterized by trajectories which are straight lines with various slopes depending on the initial condition e Use the results of c and d to sketch the phase space diagram associated to 3 Pay special attention to the behaviour of phase space trajectories near the line of fixed points found in a Does the direction of the trajectories change at this line? Why? The phase space diagram is given in the figure below Figure : The phase space diagram of equation 3 The blue line is the line of fixed points x x The phase space trajectories change direction exactly at the line x x this is because the sign of the derivatives ẋ and ẋ changes at this line Below the line that is for x < x we have that x x > therefore ẋ x x x > ẋ x x x > if both x x > This is compatible with trajectories moving away from the origin in the direction of increasing x and x However for x > x that is above the line of fixed points we have exactly the opposite behaviour that is ẋ x x x < ẋ x x x < if both x x > Therefore here the trajectories have to move towards the origin at time grows that is in the direction of decreasing values of x and x 4 [Unseen] Consider the following second order linear differential equation: where ẍ d x dt and ẋ dx dt ẍ 4ẋ x 4 a Consider now the following two first order linear differential equations ẋ ax + bx + c ẋ dx + ex + f Page 7

8 where x x above Find the values of the constants a b c d e and f so that the system is equivalent to 4 By defining x x and x ẋ ẋ we can rewrite 4 as: ẋ x ẋ 4x + x so a c f and b d e 4 b Using the result of a write equation 4 in the standard matrix form ẋ Ax + b ẋ x ẋ 4 x with b in this case [ mark] c Find the eigenvalues and eigenvectors of the matrix A Hence construct the Jordan normal form of A and the matrix P which relates A to its Jordan form J as A P JP The eigenvalues of A are obtained by finding the zeroes of the characteristic polynomial of A: A λi λ4 λ λ λ The eigenvectors corresponding to these eigenvalues are obtained by solving: a a 4 b b In components this gives two equations: b a a + 4b b which are not independent Taking a we find that the eigenvector associated to λ is E and for λ we obtain the equation 4 In components this gives two equations: a b a b b a a + 4b b which are not independent Taking a we find that the second eigenvector is E The Jordan normal form of A is just the diagonal matrix J and the matrix P is built just by setting its columns to the eigenvectors above: P Page 8

9 d Find the fixed point of the system of equations and classify its nature In this case the fixed point is just at the origin [ mark] Because the eigenvalues are real and one eigenvalue is negative and the other positive this would be a saddle e Find the general solution of the system of equations ẏ Jy Hence find the general solution for x In this case the canonical system of equations is simply: ẏ y ẏ y which is solved by y C e t and y C e t where C and C are arbitrary constants Now we need to take the solutions from part e and multiply them by the matrix P to obtain x: y y + y x P y C e t + C e t x y y + y C e t + C e t x f This can be rewritten as x C e t E + C e t E Sketch the phase diagram both in the y y plane and in the x x plane Figure 3: The y y diagram is a typical saddle As expected the x x diagram would be a rotated and slightly stretched version of the first diagram where the symmetry axes are now the directions determined by the eigenvectors E E [Unseen] Consider the following first order nonlinear differential equations [4 marks] ẋ x x x ẋ + x x x 6 which can be interpreted as modelling the populations of two competing species Page 9

10 a Find the four fixed points of the system To find the fixed points we have to solve the simultaneous equations: x x x + x x x An obvious solution is x x For x we can also have x and for x we can also have x Finally if both x and x are not zero there is also the solution x x 6 is also a fixed point Therefore we have the four points: 6 b Find the general form of the Jacobian matrix of the system and evaluate it at each fixed point Classify the fixed points according to their linearization The Jacobian matrix is obtained by computing the derivatives of the rhs of the equations 6 wrt x and x Defining X x x x / x x and X x x + x / x x we obtain A X x X x X X x x At the various fixed points it becomes: A x x x x x + x which is diagonal with real eigenvalues of different signs This means that the origin is a saddle A [ mark] In this case we need to find the eigenvalues of A in order to classify the fixed point They are the solution to the equation λ λ that is λ and λ Therefore as before this would also be a saddle A 3 The eigenvalues of A are the solutions to the equation 3 λ λ that is λ and λ 3 Therefore both eigenvalues are real and positive which corresponds to an unstable node Finally 3 A 6// 6 The eigenvalues of A 6// are the solutions to the equation 3 λ λ + λ + 4λ + 3 that is λ ± i These are complex conjugated eigenvalues with negative real part which would describe a stable focus Page

11 c For only those fixed points lying in the region x and x solve the linearized equations The only fixed points lying on the indicated region are and 6/ / 4 The first fixed point is a saddle and the linearized equation is already in the canonical form Therefore the general solution to this linearized equation is: where C C are arbitrary constants x C e t + C e t [ mark] The fixed point 6/ / 4 is a stable focus In order to solve the linearized equation about the fixed point we need to carry out the standard transformations First we change variables to z x a where a is the fixed point so that we bring the fixed point to the origin Second we rewrite the resulting linearized equation ż Az in the canonical form and solve this equation: ẏ Jy where J is the Jordan form of A and z P y where P is a nonsingular matrix that relates A and J as A P JP [ mark] In order to construct the matrix P we need the eigenvectors of A The eigenvalues λ ± i mean that the Jordan form of the matrix A is J The eigenvectors of A are obtained by solving the equations 3 6 a a b ± i b which can be written as 3a 6b ± i a a b ± i b As usual the two equations for each eigenvalue are not independent b 6 ± i 6 therefore the eigenvectors and eigenvalues are and E E 6 i i 6 λ + i Taking a we obtain λ i The matrix P that relates A to its Jacobian form is obtained by taking the real and imaginary parts of the first eigenvector and setting them as columns of J P 6 6 and we can check that indeed A P JP Page

12 [ mark] The equation ẏ Jy can be solved in the usual way for J of the form above see section g of question The solution takes the form y C e t cos t + C + C e t sin t + C d where C C are constants In order to get the solutions for the variable z we need to multiply the vector above by the matrix P and finally to get x we have to add the fixed point x a + P y 6 6 In components: + C e t cos t + C P + C e t cos t + C 6 + C e t sin t + C P + C e t sin t + C x 6 + C e t cos t + C x C t e cos 6 t + C C e t sin 6 t + C 6 In the same region of the x x plane considered in section c and using the solutions to the linearized equations sketch the phase diagram associated to the nonlinear system of equations You may employ the vector field diagram below as a guiding tool in order to produce your sketch Figure 4: The sketch here is not entirely accurate as I have ignored the presence of the other fixed points at - and Specially - would affect the shape of the trajectories on the lower right hand corner of the picture as one can see from the vector field diagram The students were not expected to notice this Page