Math Notes on sections 7.8,9.1, and 9.3. Derivation of a solution in the repeated roots case: 3 4 A = 1 1. x =e t : + e t w 2.

Transcription

1 Math 7 Notes on sections 7.8,9., and 9.3. Derivation of a solution in the repeated roots case We consider the eample = A where 3 4 A = The onl eigenvalue is = ; and there is onl one linearl independent eigenvector, which can be taken as One solution to the di erential equation is =e t For a second solution we tr =te t + e t w w From this we obtain Setting = A we get te t + e t = te t + e t + e t w w = e t w w and considering the two components separatel we get te t te t + e t ( + w ) = te t (6 4) + e t (3w 4w ) te t + e t ( + w ) = te t ( ) + e t (w w ) + e t w w In each equation the coe cients of te t are the same on each side and so these terms cancel out. Then dividing b e t we get + w = 3w 4w + w = w w

2 Writing this as a matri sstem we get 4 w w = This clearl has a solution because the right side is in the column space of the matri on the left. The obvious solution is w = ; but other solutions are ne also. The geeral solution is the one given in the back of the book for this problem. Section 9.. for In this section we summarize the di erent phase planes possible = A when n = Cases where det A = ; so that one eigenvalue is zero, are not discussed. Re ning the classi cation in the tet, we have the following cases. (). < < In this case all solutions tend to zero. An eample is = = with general solution c e t + c e t We see that on a solution, (t) = c (t) for some constant c; unless (t) is identicall zero. This gives the phase plane

3 The arrows should all be towards (; ) Note that there are solutions on the positive and negative -aes, pointing up and down, as well as on the positive and negative -aes, pointing left and right. This is called a stable node. (b) < <. Reverse the arrows on the diagram This is called an unstable node. Change the signs on the right side of the eample above to get an eample of this case. (). > ; <. Saddle. Eample = = Arrows point towards the right in the right half plane and towards the left in the left half plane, because of the equation = (3)(a) = < Onl one linearl independent eigenvector. This is also called a stable node. The eample discussed in the rst section of these notes is in this categor. Here are some phase plots 3

4 The arrows all head in a direction along the curve to get to (; ). This is another stable node. (4. = i (a) = The phase plane is circles or ellipses. (b) < Inward going spirals. (; ) is a stable spiral point. (c) > ; Outward going spirals. (; ) is an unstable spiral point. = i. A = 5e+7 e+8.5e+8 e+8 -.5e+8 -.5e e+8 4

5 This spiral is tightl wound, and so when plotted accuratel, ou can t even see one turn clearl. - We now move to new material. In this section we consider nonlinear equations. The rst eample is + + = If we set = get and write the equivalent rst order sstem in terms of and we = = + Both the single equation and the equivalent sstem are nonlinear because of the term It is ver important to realize that for this sstem, no formulas for solutions are known. For this reason, we start with the direction eld. steps. This is best done in several Step. Find those points (; ) where the tangent vector ( ; ) is horizontal. This means that = ; and from the equations we see that this occurs along the parabola = This parabola is called the horizontal nullcline. At points (; ) with > ; we have < ; so a tangent vector points generall downward (but possibl to the right or the left.) At points (; ) with < ; the tangent vector points generall up. Step. Find those points where the tangent vector is vertical. These are the points where = ; and from the equations, we see that this is the -ais, where = Thus, the -ais is the vertical nullcline. Plotting the horizontal and vertical nullclines, we see that the divide the plane into ve regions. 5

6 Step 3. Pick a point in each of the regions created b the nullclines, and calculate the tangent vector ( ; ) at that point, for eample b making a table of ; ; ;. Here is an eample Here I won t be able to choose eactl those points Step 4. Identif the points where and are both zero. These are called equilibrium points. This means we need = and + = 6

7 Setting = ; + = ( ) = ; we get two points (; ) and (; ). These points represent constant solutions of the sstem. For eample, (t) = ; (t) = solves the sstem, as does (t) = (t) = Step 5. Linearize the sstem around each equilibrium point. This is the newest and perhaps most complicated step, although more concrete than the previous steps. I will describe it in terms of a general sstem and then appl it to the eample. A general sstem of two rst order equations can be written as = f (; ) = g (; ) For such a sstem, an equilibrium point is a point ( ; ) such that f ( ; ) = and g ( ; ) = Suppose that ( ; ) is an We form the Jacobian matri in terms of and ; and then let A be the speci c A ; ) ; ( ; ) ; ) We then consider the so-called linearized sstem, which I will write in terms of new variables, u and v The linearized sstem is u u = A () v v We nd the eigenvalues and eigenvectors of A; and this tells us the phase plane for the linearized sstem in the u; v plane. This then tells us a lot about the phase plane of the original nonlinear sstem in a neighborhood of the critical point ( ; ). Unfortunatel it doesn t tell us everthing about the local phase plane of the non-linear sstem. A theorem about this appears in section 9.3 of the tet (Theorem 7

8 9.3.), and this is followed b a table showing the relation between the linearized phase plane and that of the nonlinear sstem (which the tet calls almost linear ). I will now appl this to the eample. In our eample we have f (; ) = g (; ) = + From this we nd that the Jacobian matri is + We have two equilibrium points, so we must consider each, separatel. At (; ) we get A = This turns out to have eigenvalues p 3i The are comple, with negative real part, so the u; v phase plane is a stable spiral. Here I will draw a spiral which is not quantitativel right but which shows the spiral nature more clearl than the eample above. This is what ou would usuall do in a hand drawing. The arrow should point inward. v u For a spiral solution we usuall do not care about the eigenvectors. Now consider the equilibrium point (; ). Substituting = ; = into the Jacobian matri gives a di erent A We get A = p eigenvectors 5 + p $ p 5 ; 5 + $ p 5. (These were calculated b m computer.) We note the following features 8

9 (i) < ; > so we have a saddle point. (ii) The eigenvector associated with associated with has positive slope. has negative slope, while the eigenvector This gives us the following phase plane v - - u - - Notice that in the last two phase planes I labeled the aes as u; v. In transferring these to the original sstem, we must understand that the describe the phase plane near the equilibrium point ( ; ). In the case of (; ) this does not make an di erence between (u; v) and (; ). In the case of (; ) ; the (u; v) picture must be copied to a neighborhood of the point (; ) We now have the following information about the phase plane The nal step is to tr to connect the two local pictures. At this stage the onl tool we have is the direction eld. M software does not allow me to put the direction eld on the same graph as the phase curves. So I will show a computer generated direction eld below. See if ou can pick out the local pictures above and how the are connected. 9

10 With a little e ort ou can, but notice that it is not eas to detect a spiral as opposed to a center.