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# Mathematical Biol I MATH 6770

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This 14 page Class Notes was uploaded by Miss Noel Mertz on Monday October 26, 2015. The Class Notes belongs to MATH 6770 at University of Utah taught by James Keener in Fall. Since its upload, it has received 30 views. For similar materials see /class/229925/math-6770-university-of-utah in Mathematics (M) at University of Utah.

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Date Created: 10/26/15

Stochastic birth death processes September 8 2006 Here is the problem Suppose we have a nite population of for example radioactive particles with decay rate A When will the population disappear go extinct 1 Poisson process as a birth process To illustrate the ideas in a simple problem consider a waiting time problem Poisson process How long does it take for 71 events to occur if all events are independent and if the probability of an event in time dt is Adt The equations dPO 7AP dt 07 and dt APn71 7 APT 1 2 for n gt 1 with initial data P0O 1 Pn0 O for n gt 1 The conjecture is that the solution is of the form 1 Pnt jknt eXp7t 71 This is certainly correct for P0t We check it inductively dP 1 1 7 n7AWH exp7gtt 7 A7W exp7At APH 7 AP dt 71 n as desired It is also easy to nd the generating function Set 92 t 220 P1921 so that 89 X X E 2 APsz 7 AZszk k1 k0 X X zAZszk 7 AZszk k0 k0 M2 i 19 3 4 with initial data 92 0 1 This is actually an ordinary differential equation with solution Wit eXpZ 1W 8 and Taylor series 0 1 9Zyt eXPPM Z jWVC 9 k0 so that Pnt t exp7t as stated above 2 stochastic birth process Start with the stochastic birth problem d bn71pn17bnpm 7113 10 and 13110 1 Then the solution is the negative binomial distribution mltniWMmmekmwwmmahu an I The proof uses induction Mean and variance are u aexpbt and 72 aexpbtexpbt 7 1 We can also use a generating function 0 9 Zmzk 12 k1 and then observe that i ibacil zhibk 2k 13 at 131971 Pk k1 k1 gt0 0 222 20 7131971219 2 7 b2 2 Iqasz 14 k2 k1 bzzgip 219717sz ip 2k 15 32 kil 32 k k2 k1 8 bwi am 17 with initial data g0z z The solution can be found using the method of characteristics Set f 517 1 and then dg 7 39 dt agdz EquotEE m 7 39 39 2 7 7 Ebz7z70 19 with 117 bz 7 22 This we solve easily 9 23 along 20 1mg 11120 71m 20 or along 7 zexp7bt T 17 217 exp7bt39 From this we easily nd the solution using the negative binomial expansion 172a lt f11gt2k 22 190 20 The problem is that with the usual continuous population model du 7 7 23 w u ltgt the population never goes extinct because the solution ummmam w cannot become zero Of course the problem is that we are using a continuous model when only a discrete model can work So we write the master equation for P the probability that at time t the population has 71 members dPn dt with initial conditions PNO 1 and Pn0 O for n 7 N Mn 1Pn1 7 MP 25 It takes some work but one can show by induction that the solution to this problem is N N7 Pjt j exp7Ntexpt71 3 26 The calculation is left to the reader This is exactly the binomial distribution with pt exp7t Try a generating function as well gzt 220 P192 and observe that X X 7 k k at 7 AklPk1z 71Aksz X X A 20 1Pk1zk 7 A2 2 kszk l k0 k0 A3 i Pk 12k 7 A23 i PM 32 1670 32 1670 7 a 0 a 0 A7 P k7gt7 P k 321 k2 282 k2 59 A 1 7 7 27 lt z 2 lt gt The initial data for this are gz0 z 28 and the solution is found using the method of characteristics The solution is constant along characteristics satisfying dz a A2 71 29 It follows easily that lnz 71 lnzo 71t 30 or 20 1z 71exp7t 31 Thus the solution is 927t 1 eXP t ZexpgttN7 32 which can easily be expanded using the binomial expansion The mean and variance are easily calculated using facts about the binomial distribution namely ut Nexp7t vart Nexp7t 7 exp72t 33 which is the deterministic answer for the population size Now the extinction probability is Pom 7 lt17 explt7AtgtgtN7 lt34gt The expected time of going extinct is given by but I can t calculate this explicitly More generally if A is a function of time lnz71ln20710 Asds 36 20 1z71exp7O Asds 37 Thus the solution is t we 7 lt1lt2 71gtexplt7 AltsgtdsgtgtN 38gt which can easily be expanded using the binomial expansion More generally 3 The full birthdeath problem Suppose we have a birth death process dPn dt Past experience tells us that there is a generating function g2t 2160 P1921 7 and that an 1Pn1 7 047113 71Pn1 7 lan 39 89 gt0 0 gt0 0 7 19 19 19 19 E 7 a 19 1Pk12 7 Zakpkz Zk 71Pk12 7 8 19Pk2 190 190 191 190 043 i031c 21ch1 7 0423 ipl k 1223 ipk 216 1 7 P1 21c 32 1 32 32 71 32 7 190 190 191 190 59 17204782a 40 Of course7 the solution is found using the method of characteristics7 g 26 say7 where 7417aw27m an which we can solve to nd 82 7 04 7 820 7 04 ln 2 71 7 1H 1a Wt 421 from which we nd that 2 7 04 i 20 7 04 271gtM7 1771 ea We solve for 20 to nd wz7annma7 ww7az7u 32 7 aeXpa 7 W 7 M2 7139 It is possible to nd the Taylor series expansion of g however7 the original problem was to nd P0t7 which we can easily do In fact7 N aamm7 m7agt P0 t lt 45 awmm7 m7 1 It is interesting to calculate the probability of extinction If 04 gt then 44 lim P0t 17 46 t7gtltgto whereas if 8 gt a then lim P0t 7 31 47 t71ltgto Some questions to answer maybe 0 What is the expected time of extinction in a birth death process as a function of initial population size7 N at least I can calculate this numeically7 if not analytically o What is the variance in population size in a birth death process can the moment equations be solved How about the moment generating function 4 Ion Channels A similar model describes the behavior of an ion channel with k independent subunits all of which must be open in order for the ion channel to conduct ions Let pj be the probability that 9 subunits are open Then dp 67 10 7 9 1p71 lt9 1p71 y 10 Wm 48 with appropriate restriction on the indices There are two interesting observations about this model First it is relatively easy to show that there is an invariant manifold given by p lt gt730 en 49gt with dn 7 a 7 I don t know how to show that this is a stable invariant manifold but I believe it is I can show it in a few easy cases like k 23 a1 n n 50 Second we can nd the equation for the generating function 16 9t72 223727 51 70 to be 89 k1 I k1 I k I a age j1Pj1Z j1j1p7123 20987 M may27 k1 I k1 I 161 I k I k I ak 213712 1 20702712 9 013N123 3 1211372 M 213723 71 71 7 1 70 70 k k k k k ak 213327 04 ijjsz ijjzjil 7 a ijjzj ak 213327 70 70 70 70 70 1M2 1 21972 1 2W W 291372371 70 70 akzilgtgliz azgt 3239 52 It is not hard to check that one solution of this pde is go 72 1 7V2 53gt provided 50 holds And of course 49 follows from the binomial expansion Notice also that if 710 0 then gz0 1 which implies all subunits are initially closed The general solution in the case that 04 and 8 are constant can be found using the method of charac teristics We suppose that gz0 1 all subunits are initially closed and then the characteristic equations are g aweug jijalwx m 54gt We nd that the characteristic curves are a lt Z 712gtexpltelta gttgt W 2071 gt 55 It is also possible to integrate the 77phase plane77 equation for g dg akg i 7 56 dz 8 042397 gt to nd k 8 0420 57 g 8 w lt gt To nd 92 t we now solve 55 for 20 and substitute into 57 20 a2 expltelta gttgt m2 71gt 58 azexpa gtt 7 W 1 Something seems wrong here since the answer is supposed to be a polynomial of degree k Oh well this is not the interesting case anyway On Kinetic Rate Theory Math 6770 Fall 2005 It is common practice when building kinetic models of chemical reactions to invoke the famous Arrhenius formula AG koff KO eXPm7 1 where AG is the free energy of the chemical bond k3 is Boltzmann s constant and T is absolute temperature When a bond is externally forced the off rate is increased lf AG is thought of as the height of the potential well out of which the binding molecule must escape then an applied force can be viewed as tilting the potential well With tilting the height of the barrier is decreased by the amount FL where L is an appropriate scale factor Thus according to this reasoning the off rate from the modi ed potential well should be AGiFL koff 0 exp7 kBT gt koexp keexplt 0gt lt2 where F is the external force This is referred to as Bell s law The purpose of these notes is to examine the off rate of a forced potential well as a function of the external force F It is shown using rigorous estimates that for Aggy gtgt 1 k m 2 alt17 gtexplt7alt17 if lt3 I 7 AG 7 FL whereaimdfim 1 Mean First Exit Times A detailed understanding of off rates is found by examining the Fokker Planck equation We suppose that a molecule is subject to three forces namely the force from the potential well of a binding site the applied force F and Brownian forcing from the environment The Fokker Planck equation describes the evolution of the probability pw t that the molecule is at position x at time t and is given by up are 7 mp mm 4 where 1 is the molecular viscosity kBT is the thermal energy We have ignored molecular inertia For the problem at hand we suppose that the binding site is located in the interval 7L lt w lt L and that the boundary at w 7L is re ecting The off rate is de ned as the inverse of the mean rst exit time from the binding region at w 0 with w starting at the minimal point of the potential According to Gardner the function pwt is more precisely given as the conditional probability pw th t and the problem at hand is to determine how long the particle remains in the domain We de ne L Gm Lpltwtyogtdw lt5 1 as the probability that the particle is in the domain at time t and observe that if Ty is the random variable for the time at which the particle rst leaves the domain having started at y then PTy gt t amt 7 0 Gtytdt 6 PTy lt t 7 0th3tgtdt 7 Thus Gtyt is the pdf for the random variable Furthermore the expected value of the random variable Ty is EltTltygtgt 7 w tamw w await 8 We also observe that for a time autonomous process v 0mm Q ptltwtyogtdy Q ptltwoyrtgtdy 9 Now pw 0y7 satis es the backward Fokker Planck equation 71237 U W Fm kBpryy 10 so that lGt 7U y 7 FGy kBTny 11 Finally integrate this equation with respect to t to get the ordinary differential equation kBTTyy W31 Fm 7V 12 where 731 ETy subject to boundary conditions TLgt 0 and 7L O 11 A quadratic potential well We suppose that the potential well is exactly the quadratic polynomial 3 W Ami 13gt i i i i 2 i i We introduce nondimensional variables x 739 1ng in terms of which 12 becomes a 72aw7fail 14 with a 71 0 01 0 where a 84 and f 577 20 i l i i l i l 5 m 15 20 an 35 AD 45 5 A T Figure 1 Plot of koff solid curve and 0 exp7kABif dashed curve on a logarithmic scale plotted as functions of 13 12 The numerical solution It is easy to nd the numerical solution of this boundary value problem Let 00w be the solution of the differential equation 14 subject to initial data 0071 0671 0 Then 0w 0090 7 701 This is easily found using a standard numerical integrator The classic result 1 was derived for a quadratic potential Uw AG2 7LltwltL 15 and is an approximate result that applies only if 172 is suf ciently large We can readily verify this by solving 12 numerically In Fig 1 are shown the results of this computation In Fig 1 is shown the ratio of the computed koff to the asymptotic off rate so exp7kABf plotted on a logarithmic scale as functions of 13 It is clear from these plots that the asymptotic formula is not valid for small arguments In fact with 17 less than 5 the Arrhenius formula is too large by an order of magnitude Given that the Arrhenius formula is not correct for small 13 it is not surprising that the effect of a constant load is not as stated by 7 In Fig 2 is shown koff plotted as a function of the unloading force 15 for several values of 133 For small F each of these curves is nearly linear on a log scale at least for suf ciently large indicating that for small enough F and large enough the effect of F is approximately exponential as hypothesized However the range of validity of this approximation is quite limited Off Rate from a forced quadratic potential 1 i i i 1 I IV I I I II I I in I I I I I39 l I I I I I I ll 5 I m 7 II I I 39 I I I g 39 39 x I I I m 7 I I I I I I I m is m 20 an AD so 70 an an 1 n SD Force Figure 2 Plot of k0 on a logarithmic scale as a function of shown for 1 11 21 31 B B 41 and 51 top to bottom 121 Large 198 Asymptotics We begin by examining the unforced problem f 0 We use an integrating factor so that 14 becomes expiaw2a 7 exp7aw2 16 we iexpltaw2gt 3 expan2dn7 17gt so that 1 z 0w expaw21 exp7an2dndw 18 and 1 x 170 expaw2 exp7an2dndw 19 D for This is the integral we wish to approximate There are several approximations we make Observe rst that 1 1 1 1 lt700 expaw271exp7an2dndwiA z expaw27n2dndw 20 The second of the integrals in 20 is small This can be established by introducing the change of variables 2 w 7 n g w n It follows that 01zlexpax2n2dn S jjgexpwzadzd 21 2 17 exp7a 2 7d 0 1 2 a i 01 1 7st ln4a 7 EU E14agt7 S where E1w is the exponential integral 006 s 67 67 E1w 8 d5 0 2 I Thus 1 1 0 lt expltaltw2 7722dn In olt1gt 0 3 4a a for large a A useful formula will be that mew 2gtd 1 exlt 82 1olt1gt 7048 S g 7 704 i 5 p 2a p 1 Now we make two approximations First 1 2 00 2 00 2 1expan dn exp7an dn721 exp7an dn 7 00 7 exp7a 1 0 Second we use the change of variables 32 1 7 z and Watson s Lemma to show that 01expaw2dw expa 01 wdz xl 7 z expa 1 1 1 7 7 7 O 7 2 a2a2 13 Putting all of these together we nd that lna 1 expa 1 7 O 7 0 7 7 1 O 7 0 4 agtlt0gt 2 132 a lt 7 so that the leading order approximation is 00 a ea 1olt1gtgt 2 132 a well known result 22 23 24 25 26 30 31 Now we turn to the forced problem f 7 0 and use the same method That is after multiplying by an integrating factor expeawz may 7 7 explt7aw2 m 32 I a wgt 7 7expawz 7 m exp7an2 mom 33gt 71 so that 1 z aogt 7 expa32 7 22gt 7 m 7 7dndwr 34gt 0 71 This is the integral we wish to approximate First we break this expression into two 1 1 1 1 aogt 7 expa32 7 22gt 7 m 7 WWW 7 expa32 7 22gt 7 m 7 ngtdndwgt 35gt 0 71 0 x We can show that the second of these is small using the Cauchy Schwarz inequality That is 01 A1 expaac2 7 712 fw 7 71dndw s 01 I exp2aw2 7 n2dndw 01 I exp72fx 7 ndndwgt12 S ltng 12 36 l I Now we approximate two integrals 1 f2 1 f 2 7 2 Lexpean an 7 expEgt1exp7an7Egt gt37 22 1 I ew 7 W exp7a fgtgt 7 m exp7a f provided 2f7a lt 1 To approximate the second we assume that lt 1 to write 1 f2a 1 expaw2 7 fwdw expaw2 7 fwdw expaw2 7 fwdw 37 0 0 f2a 6 Now we introduce the Change of variables 2 i2 7 w 7 if in the rst of these and z 2a 2a 17 2 7 w 7 2 in the second7 so that 1 fZa expa4c2 7 fwdw expa4c2 7 fwdw expa4c2 7 fwdw D D a 1 f2 1gt2 eXp7az i 0 Hiv7z 1 kid expafO 2 Ma Mo7ev7z dz A useful observation is that It follows that 1 expa4c2 7 fwdw m 7 0 f Finally7 we obtain the leading order approximation N1 7r eXpa17 1a dm g7757 77 which holds provided a gtgt 1 and f lt 2a 43

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