Using Mathematica to Teach Linear Differential Operators and the Method of Undetermined Coefficients

Transcription

1 Using Mathematica to Teach Linear Differential Operators and the Method of Undetermined oefficients Itai Seggev Knox ollege / Wolfram Research Joint Mathematics Meetings January 9, 2011 New Orleans, Louisiana

2 2 JMM2011.nb Problem Statement Given a linear differential operator PD and a polynomial-exponential forcing function f, we wish to find the general solution to the homogeneous equation PDy 0 one solution to the driven equation PDy f by means of an ansatz We can then solve any IVP for PD, but we need to enter PD into Mathematica

3 JMM2011.nb Ingredients In[11]:= Solve[] oefficientlist[] omplexexpand[], Re[], Im[] (if dealing with complex and or trigonometric equations) Some magic code: leardpluss DPlusS : PowerDPlusSt_, s_:0, n_ : Functionf,NestD, t s &, f, n DPlusSt_, s_:0^n_f ^: DPlusSt, s n f DPlusSt_,s_:0f_: Df,ts f UnprotectPlus, Times; Plus : Plus a_:1 o1_function, cruft f_ : a o1f Pluscruftf Plus : Plus a_:1dplusst_, s, cruft f_ : a DPlusSt, sf Pluscruftf Times : Timesa_, DPlusSt_,s f_ : a DPlusSt,sf Times : Timesa_, o1_function f_ : a o1f ProtectPlus, Times;

4 4 JMM2011.nb Playing with DPlusSt, s D s The basic operator In[21]:= Out[21]= DPlusSt, syt s yt y t Monomials Positive powers: In[22]:= DPlusSt, 0^2yt DPlusSt^2yt Out[22]= Out[23]= y t y t Positive powers with shift: In[24]:= Out[24]= DPlusSt, s^2yt Expand s 2 yt 2 s y t y t The identity operator In[25]:= Out[25]= DPlusSt, s^0yt yt A polynomial In[26]:= Out[28]= In[29]:= learpofd PofDt_, s_ : 0 : L DPlusSt, s^2 R DPlusSt, s PofDt, 0yt yt R y t L y t PofDt, Ωyt Simplify DPlusSt, s0 Out[29]= 1 Ω R L Ω yt R 2 L Ω y t L y t

5 JMM2011.nb Step 1: Finding the basic solution set Simple roots Assuming a constant coefficient operator, we know the solution is given by the characteristic roots In[30]:= PofDt_, s_ : 0 : DPlusSt, s 2 5 DPlusSt, s 6 DPlusSt, s 0 The characteristic polynomial is obtain with D 0, i.e., computing PD1 In[31]:= PofDt, s1 Solve 0, s Exps t. Out[31]= 6 5 s s 2 Out[32]= Out[33]= s 3, s 2 3 t, 2 t So, we have our basic solution set. Repeated roots In[34]:= PofDt_, s_ : 0 : DPlusSt, s 2 4 DPlusSt, s 4 DPlusSt, s 0 PofDt, s1 Solve 0, s Exps t. Out[35]= 4 4 s s 2 Out[36]= Out[37]= s 2, s 2 2 t, 2 t Our set is linearly dependent, we need to replace one vector with t e 2 t

6 6 JMM2011.nb Step 2: Find the particular solution (real case) In[38]:= learpofd PofDt_, s_ : 0 : DPlusSt, s 2 5 DPlusSt, s 6 DPlusSt, s 0 A couple of forcing functions In[40]:= f1t_ : t 2 3 f2t_ : Exp7 t t 2 f 1 We know that the particular solution has the same form as the driving polynomial in the case of a simple polynomial driving term. In[42]:= y1t_ : a2 t 2 a1 t a0 We must substitute in and solve for the coefficients. oefficientlist[] to the rescue! In[43]:= Out[43]= oefficientlistf1t, t oefficientlistpofdty1t, t Solve, a0, a1, a2 y1t. First 3, 0, 1 Out[44]= 6 a0 5 a1 2 a2, 6 a1 10 a2, 6 a2 Out[45]= a , a1 5 18, a2 1 6 Out[46]= t t f 2 When there is an exponential term, it is conventional to separate the solution into its polynomial and exponential parts. In[47]:= leary2, h2 h2t_ : a1 t a0 y2t_ : h2t Exp7 t We then equate the polynomial part of the driving terms with the shifted operator acting on the polynomial ansatz:

7 JMM2011.nb In[50]:= Out[50]= oefficientlistf2t Exp7 t, t oefficientlistpofdt, 7h2t, t Solve, a0, a1 y2t. First 2, 1 Out[51]= 90 a0 19 a1, 90 a1 Out[52]= a0 161 Out[53]= 7 t t 8100, a Alternate solution While it is conventional to do the separation, it is not strictly required. oefficientlist[] will consider the exponential part of the "coefficient", allowing its cancellation. In[54]:= oefficientlistf2t, t oefficientlistpofdty2t, t Solve, a0, a1 y2t. First Out[54]= Out[55]= 2 7 t, 7 t 90 a0 7 t 19 a1 7 t, 90 a1 7 t Out[56]= a0 161 Out[57]= 7 t t 8100, a

8 8 JMM2011.nb Step 2: Find the particular solution (trig/complex cases) Sinusoidal driving term In[58]:= learpofd, f PofDt_, s_ : 0 : DPlusSt, s 2 5 DPlusSt, s 6 DPlusSt, s 0 ft_ : Sin7 t t 2 Since it we have a sinusoid, we need to complexify the problem, solving with a driving term f c t t 2 7 t. In[61]:= learyc, hc hct_ : a1 t a0 fct_ : Exp7 t t 2 yct_ : hct Exp7 t Finding the complex solution is as before. In[65]:= Out[65]= oefficientlistfct Exp7 t, t oefficientlistpofdt, 7 hct, t Solve, a0, a1 yct_ yct. First 2, 1 Out[66]= a a1, a Out[67]= a Out[68]= 7 t , a t We can then find the solution by taking the imaginary part of our complex solution. In[69]:= Out[69]= yt_ omplexexpandimyct Simplify t os7 t t Sin7 t Notice that the omplexexpand[] is essential, since the value above is not valid if t is complex. In[70]:= Out[70]= Imyct Simplify Im t t For free, we also get the solution for a cosine driving term. In[71]:= Out[71]= omplexexpandreyct Simplify t os7 t t Sin7 t Alternate solution As before, we are not obligated to separate out the polynomial if we don't want to.

9 JMM2011.nb In[72]:= yct_ : hct Exp7 t oefficientlistfct, t oefficientlistpofdtyct, t Solve, a0, a1 Out[73]= Out[74]= 2 7 t, 7 t a0 7 t 5 14 a1 7 t, a1 7 t Out[75]= a , a

10 10 JMM2011.nb Bonus slide: additional cases Degeneracy A degeneracy occurs when the characteristic polynomial has a repeated root, and/or the shift equals one of the roots. In[76]:= In[79]:= learpofd, f PofDt_, s_ : 0 : DPlusSt, s 2 4 DPlusSt, s 4 DPlusSt, s 0 ft_ : t^2 2 t 3 Exp2 t leary, h ht_ : a2 t^2 a1 t a0 yt_ : ht Exp2 t Now, what happens if we proceed with our normal ansatz without paying attention to the degeneracy? In[82]:= Out[82]= oefficientlistft Exp2 t, t oefficientlistpofdt, 2ht, t Solve, a0, a1, a2 3, 2, 1 Out[83]= Out[84]= 2 a2 We get no solution: something has gone wrong. But, there is more information. The two lowest terms have dropped out entirely: we have a double degeneracy (repeated root equal to shift) Hence, we use a polynomial of degree two higher (but possessing the same number of terms). In[85]:= Out[86]= ht_ : a4 t^4 a3 t^3 a2 t^2 oefficientlistft Exp2 t, t oefficientlistpofdt, 2ht, t Solve, a2, a3, a4 yt. First 3, 2, 1 Out[87]= 2 a2, 6 a3, 12 a4 Out[88]= a2 3 2, a3 1 1, a Out[89]= 2 t 3 t 2 2 t3 3 t4 12 Higher order operators In[90]:= learpofd, f PofDt_, s_ : 0 : DPlusSt, s 4 4 DPlusSt, s 3 3 DPlusSt, s 2 4 DPlusSt, s 4 DPlusSt, s 0 ft_ : t^2 2 t 3 Expt haracteristic roots are -2, -2, 1, and -1.

11 JMM2011.nb In[93]:= PofDt, s1 Simplify Out[93]= 2 s 2 1 s 2 We have single degenerac: shift equal to a simple root. So we need a polynomial of one degree higher. Rest of process is unchanged. In[94]:= Out[97]= leary, h ht_ : a3 t^3 a2 t^2 a1 t yt_ : ht Expt oefficientlistft Expt, t oefficientlistpofdt, 1ht, t Solve, a1, a2, a3 yt. First 3, 2, 1 Out[98]= 2 a1 6 a2, 4 a2 18 a3, 6 a3 Out[99]= a1 9 4, a2 1 4, a3 1 6 Out[100]= t 9 t 4 t2 t3 4 6 RL circuit (complete example from lecture notes) An RL circuit obeys L q.. R q q t (1) If we wanted to express the relationship in terms of current, we would need to differentiate the previous equation to get: L İ. R I I t. (2) Getting equation (2) is helpful because expressing charge in terms of current involves a definite integral: qt qt 0 t0 t Is s. If we substitute this equation into equation (1), we'd get a so-called integro-differential equation. Differentiating both sides with respect to t gives us a plain old differential equation. For convenience, we put a bunch of assumptions into the global variable $Assumptions which will shorten many Simplify[] commands. In[101]:= Out[101]= $Assumptions a Reals, R 0, Ω 0, L 0, 0, t Reals a Reals, R 0, Ω 0, L 0, 0, t Reals We define our differential operator In[102]:= learpofd PofDt_, s_ : 0 : L DPlusSt, s^2 R DPlusSt, s DPlusSt, s 0 We want to solve equation (2) assuming t a cosωt. First, we must find the characteristic roots:

12 12 JMM2011.nb In[104]:= SolvePofDt, s1 0, s Simplify Out[104]= s R 4 L 2 L R2, s R 4 L 2 L R2 As long as R is non-zero (i.e., there actually is a resistor in the circuit), the characteristic roots have non-zero real part and hence cannot equal s Ω. This eliminates degenerate cases of undetermined coefficients. We use Mathematica to find the solution, but as in example 3 this is unnecessary. Since the polynomial multiplying the cosine is a constant, the polynomial ht in the driven solution is a ht P Ω We ask Mathematica to verify: In[105]:= Out[107]= learh, y, z ht_ : a0 SolveoefficientListPofDt, Ωht, t a, a0 zt_ ht Ω t. 1 a a0 1 R Ω L Ω2 Out[108]= 1 a t Ω R Ω L Ω2 Since we're focused on a cosine driving term, we need to take the real part of z. The following would be the steps corresponding to finding the real part by hand. We make the denominator real by multiplying by its complex conjugate: In[109]:= Out[109]= 1 R Ω L 1 Ω2 R Ω L Ω2 1 2 L Ω2 2 R 2 Ω 2 L 2 Ω 4 Simplify We do the same thing to the numerator and use Euler's identity to convert to sines and cosines. 1 In[110]:= ExpToTriga t Ω R Ω L Ω2 Expand Out[110]= a ost Ω a Sint Ω a R Ω ost Ω a L Ω 2 ost Ω a R Ω Sint Ω a L Ω 2 Sint Ω We then take combine the numerator and denominator. Since the denominator is now real, this simply means take the real part of the numerator and keep the denominator unchanged. In[111]:= Out[111]= yprovisional FullSimplify Re a 1 L Ω 2 ost Ω R Ω Sint Ω 1 Ω 2 R 2 L 2 L Ω 2 Notice that we could have gotten to y provisional in one step using omplexexpand[]:

13 JMM2011.nb In[112]:= yprovisional omplexexpandrezt Out[112]= a ost Ω a L Ω2 ost Ω a R Ω Sint Ω R 2 Ω 2 1 L Ω2 2 R 2 Ω 2 1 L Ω2 2 R 2 Ω 2 1 L Ω2 2 Also, here is the imaginary part. It looks similar, but the roles of sine and cosine have been reversed: In[113]:= Out[113]= omplexexpandimzt a R Ω ost Ω R 2 Ω 2 1 L Ω2 2 a Sint Ω a L Ω2 Sint Ω R 2 Ω 2 1 L Ω2 2 R 2 Ω 2 1 L Ω2 2 Let Θ arctan 1 L Ω2, R Ω. In[114]:= φ ArcTan 1 L Ω2, R Ω Out[114]= ArcTan 1 L Ω2, R Ω The two-variable arctangent gives an angle between Π and Π, taking into account the quadrant of the point. Then the expression in the Evaluate[] in the cell below is equivalent to y provisional. Notice that they have the same denominator, and the equality of the numerators follows from the angle addition formulas and then basic facts cosarctanx, y x, sinarctanx, y y. Just to be sure, we ask Mathematica to verify: In[115]:= Out[115]= a osω t φ FullSimplifyEvaluate TrigExpand yprovisional, 1 L Ω2 2 R Ω 2 TransformationFunctions Automatic,. osarctand_, n_ d, SinArcTand_, n_ n & True So, the general solution is one of the following The only remaining issue is what is the size of 4 L versus R 2, which affects the homogeneous solution. When 4 L is smaller, we have two decaying exponentials. When 4 L is larger, we have two oscillating, decaying exponentials. When they are equal, we have an exponential and a polynomial exponential. This is the electrical equivalent of a critically damped harmonic oscillator, in which the system returns to equilibrium as quickly as possible. learly it is a very interesting case from an engineering standpoint. So here are the three possible solutions: In[116]:= yoverdampedt_ : a osω t φ 1 L Ω2 2 R Ω 2 c1 R 4 L R2 2 L t c2 R 4 L R2 2 L t yriticallydampedt_ : yunderdampedt_ : a osω t φ 1 L Ω2 2 R Ω 2 c1 R t R 2 L c2 t t 2 L a osω t φ 1 L Ω2 2 R Ω 2 c1 R t 2 L os 4 L 2 L R2 t c2 R 2 L t Sin 4 L 2 L R2 t

14 14 JMM2011.nb All in three cases, however, the homogeneous solution decays to zero, leaving just the driven solution asymptotically. This piece has its largest amplitude when the denominator is minimized, which for fixed R occurs at 1 L Ω2, or Ω 1 L. Recall that 1 L is what we found for the natural frequency of an L circuit, so the amplitude is maximized at resonance. Indeed, if there is no resistance, the amplitude will grow without bound, as we found on problem Being finished with our assumptions, we reset them by assigning True to $Assumptions In[119]:= Out[119]= $Assumptions True True

15 JMM2011.nb Is that a parameter which I see varying before me? Who said we need constant coefficients? In[120]:= learpofd, y PofDt_ : DPlusSt DPlusSt t t^2 DPlusSt0 Basic solution set We can construct a basic solution set in the standard manner In[122]:= DSolvePofDtyt 0, y1 1, y'1 0, yt, t DSolvePofDtyt 0, y1 0, y'1 1, yt, t Wronskianyt. Join,, t Out[122]= Out[123]= yt 2 t t 2 yt t t 2 Out[124]= t 2 By adding and subtracting the first to/from the second, we would get t, t 2 is a basic solution set In[125]:= Out[128]= 0 Out[129]= 0 leary1, y2 y1t_ : t y2t_ : t^2 PofDty1t PofDty2t Wronskiany1t, y2t, t Out[130]= t 2 Variation of Parameters onsider a logarithmic forcing function In[131]:= ft_ : Logt 2 We can construct the

16 16 JMM2011.nb In[132]:= y2s fs ygent_ SimplifyIntegrate, s, 1, t y1t Wronskiany1s, y2s, s y1s fs Integrate, s, 1, t y2t Wronskiany1s, y2s, s c1 y1t c2 y2t, Assumptions t 0 Expand Out[132]= 2 t c1 t 2 t 2 c2 t 2 2 t 2 Logt t 2 Logt t2 Logt 3 In[133]:= PofDtyGent Simplify ygen1 ygen'1 Out[133]= Logt 2 Out[134]= Out[135]= In[136]:= c1 c2 c1 2 c2 PlotEvaluateyGent. c1 0, c2 0, t, 0, Out[136]=

17 JMM2011.nb Use and Reception ode given as black box to students advanced S students can analyze/appreciate it Students assigned these problems, including higher order equations Students had little difficulty mastering the use these operators and showed they could self correct if started with wrong polynomial Emphasize operator/linear algebra view There is a single theory of linear equations!