Fitting Parameters in a Differential Equation With a Non-Analytical Solution

Transcription

1 NDSolve u.nb 1 Fiing Parameers in a Dierenial Equaion Wih a Non-Analyical Soluion Mahew Smih 9-Jan-005 In[]:= Remove "Global` " O General::"spell" O General::"spell1" Graphics` Oen dierenial equaions do no have analyical soluions, however such equaions can be solved using numerical echniques. As was shown in he "Dierenial Equaions in Mahemaica" noes he NDSolve command in Mahemaica can be used o solve dierenial equaions numerically. For example consider he equaion: i: < 3 d = 10 -, i: 3 d = - where = 5 a = 0. This can be solved in Mahemaica using he ollowing command: In[37]:= sol NDSolve ' I 3, 10,, 0 5,,, 0, 5 1 Ou[37]= InerpolaingFuncion 0., 5., The resul is an inerpolaing uncion which has been ied o he numerical daa. The uncion is only valid in he speciied range o he independan variable i.e or 0 5. We can deine a uncion n() ha is equal o he polynomial using he ollowing command: In[3]:= n _. sol Ou[3]= InerpolaingFuncion 0., 5., And hen plo he resul using:

2 NDSolve u.nb In[39]:= p1 Plo n,, 0, 5 ; Imagine we have conduced an experimen where we have measured as a uncion o wih he ollowing resuls: In[]:= exp 0.5, 1.1, 1.5,.1,.3, 3.1 ; exp.,.1,.,.3,.,.9 ; In[1]:= p LisPlo Transpose exp, exp, PloSyle AbsoluePoinSize, Frame True, Axes False, FrameLabel "", "", TexSyle FonSize 0, FormaType Bold, RoaeLabel False, ImageSize 00, 300 ; We can compare our numerical soluion or as ollows :

3 NDSolve u.nb 3 In[3]:= Show p, p1 ; The soluion clearly isn' such a grea i. The goodness o he i can be characerised by calculaing he sum o he square dierences beween he soluion and he daa : R = exp To do his irs we evaluae our numerical uncion a he experimenal values o : In[]:= n1 n. exp Ou[]=.9735,.335,.35, 9.377, 9.971,.313 Then we calculae ind he dierence beween he experimenal values and values rom he model and square hem: In[5]:= exp n1 Ou[5]= 0.51, , , , 7.301, Finaly we sum he numbers using: In[]:= R Apply Plus, Ou[]=.011 These comands can be nealy bundled ogeher o give: In[7]:= R Apply Plus, exp. sol. exp Ou[7]=.011

4 NDSolve u.nb Imagine ha we decide o modiy our iniial dierenial equaion such ha we have: i: < p i: p d = 10 -, d = - Where p has o be deduced experimenally. Also we realise ha in our hase o do he experimen we orgo o record he iniial value or a = 0, les call his q, so we don' know ha consan eiher. How would we go abou inding values or hese? One way o do his would be o guess a iniial values or p and q, hen use NDSolve o ind he soluion or () and hen calculae he sum o he square dierences. One could hen proceed o calculae he leas squares dierence ieraively over a large range o values o p and q unil he minimum in he sum o he square dierence is ound. Forunaley or us we can wrie a shor piece o code in Mahemaica ha will do his or us. We deine he uncion sse(p,q) as ollows: In[71]:= sse p_? NumberQ, q_? NumberQ : Block sol,, sol NDSolve ' I p, 10,, 0 q,,, 0, 5 1 ; Apply Plus, exp. sol. exp The exac synax is perhaps a lile conusing bu wha he uncion is doing is acually quie simple. The _?NumberQ ollowing he leers p and q signiies ha he uncion needs numerical inpus or boh p and q. The ucion is hen deined as a block, his allows us o combine Mahemaica commands which will all be evaluaed ogeher. The declaraion {sol,} ells Mahemaica ha he block conains wo seperae uncions; sol and. The uncion hen evaluaes he NDSolve command o ind () wih he supplied values o q and p beore calculaing he leas squares dierence beween () and he experimenal values o. The semi-colon aer he NDSolve comand supresses is oupu so ha he uncion jus reurns he leas squares dierence. For example i we evalue he uncion or p =3 and q=5 we ge: In[7]:= sse 3, 5 Ou[7]=.011 Which is he same as we calculaed above. The FindMinimum in Mahemaica command uses numerical mehods o ind he minimum o a uncion. We can use i o ind he minimum value o sse(p,q) in he range p 3 and q 5 as ollows: In[73]:= b FindMinimum sse p, q, p,, 3, q,, 5 Ou[73]= , p 1.99, q The resuls ell us ha he minimum occurs a p = 1.99 and q = We can now ind () a his value and plo i agains our daa: In[7]:= p b, 1, Ou[7]= 1.99 In[75]:= q b,, Ou[75]=

5 NDSolve u.nb 5 In[77]:= solb NDSolve ' I p, 10,, 0 q,,, 0, 5 1 Ou[77]= InerpolaingFuncion 0., 5., In[79]:= p3 Plo x. solb, x, 0, 5 ; In[3]:= Show p, p3 ; The resul seems o i he daa very well.