Numerical Methods in Cancer Models

Transcription

1 Numerical Methods in Cancer Models Doron Levy Department of Mathematics and Center for Scientific Computation and Mathematical Modeling (CSCAMM) University of Maryland, College Park

2 Plan Motivation 1 What is cancer? 2 Delayed di erential equations 3 Agent-based models 4 PDEs

3 Delayed Di erential Equations in Cancer Models Analysis & Numerical Methods Doron Levy Department of Mathematics and Center for Scientific Computation and Mathematical Modeling (CSCAMM) University of Maryland, College Park

4 Outline Motivation 1 Motivation Cancer Immunology Stem Cell Transplantation 2 Zero Crossings Time Scales 3 Numerical Methods 4

5 Outline Motivation The immune response to leukemia Stem cell transplantation 1 Motivation Cancer Immunology Stem Cell Transplantation 2 Zero Crossings Time Scales 3 Numerical Methods 4

6 Motivation The immune response to leukemia Stem cell transplantation What is leukemia? Normal cells: stem cells turn into mature cells Leukemia: A malignant transformation of a stem cell or a progenitor cell Myeloid or Lymphocytic Acute or Chronic Doron Levy Montre al, May 2013

7 CML Motivation The immune response to leukemia Stem cell transplantation 3phases Chronic: uncontrolled proliferation Accelerated Acute: Uncontrolled proliferations. Cells do not mature Philadelphia chromosome Translocation (9;22) Oncogenic BCR-ABL gene fusion The ABL gene expresses a tyrosine kinase. Growth mechanisms Easy to diagnose Drug targeting this genetic defect (a tyrosine kinase inhibitor)

8 Treating leukemia Motivation The immune response to leukemia Stem cell transplantation Chemotherapy Bone Marrow or Stem Cell transplant Chemo + radiotherapy + transplantation Imatinib (Gleevec) Molecular targeted therapy - suppresses the corrupted control system $32K-$98K/year

9 Problems with existing therapies The immune response to leukemia Stem cell transplantation Remission vs. Cure: Can CML be cured? Yes! but only with a bone marrow (or stem cell) transplant Requires a (matching) donor Ariskyprocedure(+unpredictablesidee ects) Imatinib? Does not cure the disease: stopping it causes a relapse New Medical Data: There is an anti-leukemia immune response (Lee lab) The strength and dynamics of the specific anti-leukemia immune response can be measured Number of cells Activity (count signaling molecules)

10 The observed immune response The immune response to leukemia Stem cell transplantation A di erent immune response for each patient. However: Question: At the beginning of the treatment: no immune response Peak: around 6-12 months (after starting the drug treatment) Later: waning immune response What is the relation between the dynamics of the cancer, the drug, and the immune response?

11 Amathematicalmodel The immune response to leukemia Stem cell transplantation r y divide x2 n nτ Leukemic stem cell (y 0 ) a y (a y ) Progenitor b y (b y ) Differentiated c y (c y ) cell (y 1 ) cell (y 2 ) Terminal cell (y 3 ) s T T cells (T) p 0 e -cnc kc d 0 + q C p(c,t) d 1 + q C p(c,t) d 2 + q C p(c,t) d 3 + q C p(c,t) d T Ingredients: Leukemia cells: stem cells,, fully functional cells Mutations, Drug (Imatinib), Anti leukemia immune response Kim, Lee, Levy: PLoS Computational Biology, 08 Michor et al. (Nature 05). Cronkite and Vincent (69), Rubinow (69), Rubinow & Lebowitz (75), Fokas, Keller, and Clarkson (91), Mackey et al (99,...), Neiman (00), Moore & Li (04), Michor et al (05), Komarova & Woodarz (05).

12 Michor s model + immune response The immune response to leukemia Stem cell transplantation Cells without mutations: Anti-cancer T cells: ẏ 0 = [r y (1 u) d 0 ]y 0 q c p(c, T )y 0, ẏ 1 = a y y 0 d 1 y 1 q c p(c, T )y 1, ẏ 2 = b y y 1 d 2 y 2 q c p(c, T )y 2, ẏ 3 = c y y 2 d 3 y 3 q c p(c, T )y 3. T = s t d t T p(c, T )C +2 n q T p(c n, T n )C n, p(c, T ) = p 0 e cnc kt, C = X (y i + z i ), C n = C(t n ).

13 The immune response to leukemia Stem cell transplantation Accounting for the immune response Cell Concentration (k/µl) Leukemia No immune response 0.01 T cells Time (months)

14 Stopping imatinib (simulation) The immune response to leukemia Stem cell transplantation Cell Concentration (k/µl) x T cell concentration Imatinib removed at month 12 Leukemia Time (months)

15 The immune response to leukemia Stem cell transplantation Cancer vaccines: a mathematical design Cell Concentration (k/µl) Leukemia cells Inactivated leukemia cells (vaccines) Anti-leukemia T cells Time (months) log 10 [concentration] log 10 [concentration] SC DC Time (months) PC TC Time (months) Inactivated leukemia cells Anti-cancer T cells V = d V V q c p(c, T )V + s v (t) T = s t d t T p(c, T )(C + V )+2 n p(c n T n )(q T C n + V n )

16 Model populations Motivation The immune response to leukemia Stem cell transplantation Host Cells Cancer Anti-donor T cells General blood cells Donor cells Anti-cancer T cells (cancer-specific) Anti-host T cells General blood cells

17 Everything takes time Motivation The immune response to leukemia Stem cell transplantation

18 Anti-Cancer T Cells Motivation The immune response to leukemia Stem cell transplantation σ σ TC/C Interaction Ignore React Proliferate nτ q 1 TC/C TC/C p 2 TC/C p 1 ρ Reload q 2 TC/C q 3 TC/C kct C x2 n υ T C d TC Death Die or Become anergic kt D T C p 2 TD/TC p 1 TD/TC Survive Perish TD/TC Interaction

19 Anti-Host T Cells Motivation The immune response to leukemia Stem cell transplantation TH /C Interaction Ignore React q 3 TH/C Die or become anergic TH/C p 2 pth/c 1 TH /H Interaction σ kct H Proliferate qth/c 1 nτ ρ Reload x2 n υ q 2 TH/C σ kht H nτ Ignore React pth/h 2 pth/h 1 ρ q 1 TH/H x2 n υ q 2 TH/H T H d TH Death kt D T H σ Proliferate nτ x2 n Reload p 2 TD/TH p 1 TD/TH υ Survive Perish q 2 TH/TD q 1 TH/TD p 2 TH/TD p 1 TH/TD ρ No flow qth/td 3 =0 Killed by T D Ignore React T H /T D Interaction

20 Anti-Donor T Cells Motivation The immune response to leukemia Stem cell transplantation TD/TC Interaction Ignore React No flow qtd/tc 3 = 0 T D /D Interaction ptd/tc 2 ptd/tc 1 q 2 TD/TC σ kt C T D Proliferate qtd/tc 1 nτ ρ Reload x2 n υ σ kdt D nτ Ignore React ptd/d 2 ptd/d 1 ρ q 1 TD/D x2 n υ q 2 TD/D T D d TD Death p 2 TH/TD σ Survive kt H T Perish D pth/td 1 Proliferate nτ x2 n υ Reload q 2 TD/TH ptd/th 2 ptd/th 1 q 1 TD/TH ρ No flow qtd/th 3 =0 Killed by T H Ignore React T D /T H Interaction

21 General Donor and Host Blood Cells The immune response to leukemia Stem cell transplantation Stem Cells Stem Cells S D S H D H d D p TD/D 1 kdt D Perish ρ TD/D Interaction d H p 1 TH/H kht H Perish ρ TH/H Interaction Death Death

22 Cancer Cells Motivation The immune response to leukemia Stem cell transplantation C/TH Interaction Perish pc/th 1 kt H C logistic growth r C C p 1 C/TC kt C C Perish C/TC Interaction Death rate is included in net logistic growth term ρ Death ρ

23 Equations... Motivation The immune response to leukemia Stem cell transplantation dt H dt = d TH T H kct H kt D T H kht H + p T H /C 2 kc(t )T H (t )+p T D /T H 2 p T H /T D 2 kt D (t )T H (t ) + p T H /H 2 kh(t )T H (t ) +2 n p T H /C 1 q T H /C 1 kc(t n )T H (t n ) +2 n p T H /H 1 q T H /H 1 kh(t n )T H (t n ) +2 n p T D /T H 2 p T H /T D 1 q T H /T D 1 kt D (t n )T H (t n ) + p T H /C 1 q T H /C 2 kc(t )T H (t ) + p T H /H 1 q T H /H 2 kh(t )T H (t ) + p T D /T H 2 p T H /T D 1 q T H /T D 2 kt D (t )T H (t ).

24 Time Delays Motivation The immune response to leukemia Stem cell transplantation Time for reactive T cell-antigen interaction = 5min Time for unreactive interactions = 1min Time for cell division = day T cell recovery time after killing another cell = 1 day

25 Relapse Motivation The immune response to leukemia Stem cell transplantation Cell Concentration in 10 3 cells/µl General host cells H Anti host T cells T H Cell Concentration in 10 3 cells/µl 6 x Cancer cells C eventually overwhelms T H Time in Days never goes to Time in Days

26 Remission Motivation The immune response to leukemia Stem cell transplantation Cell Concentration in 10 3 cells/µl General host cells H Anti host cells T H Cell Concentration in 10 3 cells/µl 6 x Cancer cells C C = 0 at time Time in Days Time in Days

27 Oscillations Motivation The immune response to leukemia Stem cell transplantation Cell Concentration in 10 3 cells/µl A: Stable oscillation Anti host T cells Cancer cells General host cells Cell Concentration in 10 3 cells/µl B: Unstable Oscillation Time in Days Time in Days

28 Extinction instead of stability The immune response to leukemia Stem cell transplantation Cell Concentration in 10 3 cells/µl Without state constraint Anti host T cells T H Cancer cells C The value of C crosses 0 at time Time in Days Cell Concentration in 10 3 cells/µl With state constraint Cancer cells C Anti host T cells T H The value of C crosses 0 at time , and does not recover Time in Days

29 The immune response to leukemia Stem cell transplantation Initial anti-host cells vs. initial host cells Initial anti host T cell concentration T H,0 (cells/µl) Results up to time 2000 Successful Unsuccessful Unresolved Higher initial host blood cell concentrations improve the chances of a successful cure Greater initial anti-host T cell concentrations slightly favor the chances of cure Initial host cell concentration H 0 (10 3 cells/µl)

30 The immune response to leukemia Stem cell transplantation Number of cell divisions vs. cancer growth rate Average # of T cell divisions n Results up to time 1000 Successful Unsuccessful Unresolved A higher average number of T cell divisions favor complete remission Higher cancer growth rate make complete remission slightly more likely Cancer growth rate r C (1/day)

31 Outline Motivation Zero Crossing Time Scales 1 Motivation Cancer Immunology Stem Cell Transplantation 2 Zero Crossings Time Scales 3 Numerical Methods 4

32 Zero crossing: Example Zero Crossing Time Scales AsimpleDDE: dx dt = rx(t 1), x(t) =1, t < 0. Solve: t 2 [0, 1). Then dx dt = rx(t 1) = r x = rt + c =1 rt. If r > 0thenx(t) =0forT 2 [0, 1)

33 Zero crossing: example Zero Crossing Time Scales AsimpleDDE: dx dt = rx(t 1), x(t) =1, t < 0. Proceed: t 2 [1, 2) Then dx dt = rx(t 1) = r + r 2 t r 2 x =1 rt + r 2 2 (t 1)2.

34 Zero crossing: example Zero Crossing Time Scales AsimpleDDE: dx dt = rx(t 1), x(t) =1, t < 0. The general solution: t 2 [n, n + 1) Xn+1 k (t k + 1)k x(t) = ( r) k! k=0 Question: For what r does that exist a T 2 [n, n + 1) such that Xn+1 k (T k + 1)k ( r) k! k=0 = 0?

35 Zero crossing: example Zero Crossing Time Scales AsimpleDDE: dx dt = rx(t 1), x(t) =1, t < 0. The general solution: t 2 [n, n + 1) Xn+1 k (t k + 1)k x(t) = ( r) k! k=0 Question: For what r does that exist a T 2 [n, n + 1) such that Xn+1 k (T k + 1)k ( r) k! k=0 = 0?

36 Time Scales Motivation Zero Crossing Time Scales A toy problem x n+1 = (1 y n )x n y n+1 = x 2 n + k + y n The map iterated twice x n+2 = (1+x 2 n k y n )(1 y n )x n y n+2 = ((1 y n ) 2 + 1)x 2 n +2k + y n

37 Time Scales Motivation Zero Crossing Time Scales A toy problem x n+1 = (1 y n )x n y n+1 = x 2 n + k + y n The map iterated twice x n+2 = (1+x 2 n k y n )(1 y n )x n y n+2 = ((1 y n ) 2 + 1)x 2 n +2k + y n

38 This corresponds to... Motivation Zero Crossing Time Scales

39 Outline Motivation 1 Motivation Cancer Immunology Stem Cell Transplantation 2 Zero Crossings Time Scales 3 Numerical Methods 4

40 Approach #1 Motivation A mesh in time that is based on the delay. Numerical methods for ODEs that use the mesh points only (multistep methods) Example: y 0 (t) =f (t, y(t), y(t (t))), t 0 apple t apple t f, y(t) = (t), t apple t 0. Asetofmeshpoints: ={t 0, t 1...,t N = t f }, such that t n (t n) 2. Forward Euler: y n+1 = y n + h n+1f (t n, y n, y q), q < n. Same idea with Adams-like methods, Heun, etc.

41 Approach #2: Feldstein Free the mesh selection from the delay Use extranodal points for the approximation of the delayed term y(t (t)). Example: (t 0 apple (t) apple t) y 0 (t) =f (t, y(t), y( (t))), t 0 apple t apple t f, Assume h =(t f y(t 0 )=y 0, t 0)/m. Foranyt n = t 0 + nh, define (tn) t 0 q(n) =floor. h Anumericalmethod: yn+1 = y n + hf (t n, y n, z n), y 0 = y(t 0), where z n = y q(n), i.e., a piecewise-constant approximation of y( (t)). Alternatively, a piecewise-linear approximation: z n =(1 r(n))y q(n) + r(n)y q(n)+1.

42 Approach #3: Bellman s method of steps Assume a constant delay. In the first interval [t 0, t 0 + ] theddehasthe form: y 0 (t) =f (y(t), (t )), y(t 0 )= (t 0 ). In the second interval [t 0 +,t 0 +2 ], define y 1 = y(t y 2 (t) =y(t). Then: 8 y1 >< 0(t) =f (t,y 1(t), (t 2 )), y2 0(t) =f (t, y 2(t), y 1 (t)), y 1 (t 0 + ) = (t 0 ), >: y 2 (t 0 + ) =y(t 0 + ) ) and And so on... For every timestep, a larger system. However, this system can be solved using standard methods for ODEs.

43 Approach #4: Methods based on continuous extensions Continuous Extension: very low cost method to get an accurate approximation of the solution at every point in the interval (t n + h n+1 ) = n,1( )y n n,in+1( )y n in 0 +h n+1 (y n,...,y n in ;, g, n), 0 apple apple 1. where g (f, y) =f (t, y, (t (t, y))). This is how every Matlab routine provides solutions at the sampled points

44 Approach #4: Methods based on continuous extensions Consider the DDE y 0 (t) =f (t, y(t), y(t (t, y(t)))), t 0 apple t apple t f, y(t) = (t), t apple t 0. Using continuous extensions, solving the DDE amounts to solving the ODE: w 0 n+1 (t) =f (t, w n+1 (t), x(t (t, w n+1 (t)))), t n apple t apple t n+1 where w n+1 (t n )=y n, 8 < (s), s apple t 0, x(s) = (s), t 0 apple s apple t n, : w n+1 (s), t n apple s apple t n+1, and is the continuous extension interpolant.

45 Additional Reading Bellen and Zennaro, Numerical Methods for Delay Di erential Equations, Oxford Shampine and Thompson, Numerical Solution of Delay Di erential Equations

46 Outline Motivation 1 Motivation Cancer Immunology Stem Cell Transplantation 2 Zero Crossings Time Scales 3 Numerical Methods 4

47 SQRT = ^ ; = D X DT DX T T DT DX DT DX T T DT X T! T T # d 2x dx dx(t 1 ) dx(t 2 )! x = 0 2 dt" dt dt dt #ROSSING SET 7 ;SQRT SQRT SQRT DX T T DT DX DT X DX T T DT # 5NMARKED AREAS IN THE FIGURE ABOVE REGIONS STABLE REPRESENT! DENOTE THE NUMBER OF 4HE NUMBERS T 5NMARKED AREAS IN THE FIGURE ABOVE REPRESENT STABLE REGIONS 4HE NUMBERS DENOTE THE NUMBER OF EIGENVALUES IN THE RIGHT HALF PLANE.OTICE THAT # REPRESENTS AN ISOLATED STABLE REGION " " # # X CORRESPONDING "LUE LINESDENOTE BRANCHES TO INDICES OF TYPE U V 'REEN LINES DENOTETBRANCHES CORRESPONDING T TO INDICES OF TYPE U V "!! T T T T T T T T 5NMARKED IN THE FIGURE ABOVE 5NMARKED IN THE FIGURE ABOVE " AREASAREAS REPRESENT STABLE REGIONS REPRESENT STABLE REGIONS 4HE4HE NUMBERS DENOTE THE THE NUMBER OF OF NUMBERS DENOTE NUMBER EIGENVALUES IN THE RIGHTRIGHT HALFHALF PLANE EIGENVALUES IN THE PLANE X ISOLATED.OTICE THAT # REPRESENTS AN STABLE REGION EIGENVALUES IN THE RIGHT HALF PLANE!! SQRT = ^ ; = #ROSSING SET 7 ;SQRT SQRT SQRT SQRT = ^ = ; = #ROSSING SET 7 SQRT SQRT SQRT ^ ; = T ;SQRT T T D X DT DX T T DT DX T T X X D X DTDX DT DX DT DX T T DT DX T T DT DT T D X DT T Motivation #ROSSING SET 7 ;SQRT SQRT SQRT numerical methods X X T TT T T T " "#.OTICE THATTHAT # REPRESENTS AN ISOLATED.OTICE # REPRESENTS AN ISOLATED STABLE REGION STABLE REGION "LUE LINES DENOTE BRANCHES CORRESPONDING TO INDICES OF TYPE U V 'REEN LINES DENOTE BRANCHES CORRESPONDING TO INDICES OF TYPE U V "LUE LINES BRANCHES CORRESPONDING "LUEDENOTE LINES DENOTE BRANCHES CORRESPONDING TO INDICES OF TYPE TO INDICES OF U TYPEV U V 'REEN'REEN LINES DENOTE BRANCHES CORRESPONDING LINES DENOTE BRANCHES CORRESPONDING TO INDICES OF TYPE V U V TO INDICES OF U TYPE T T # Doron Levy T T T T Montre al, May 2013

48 Stability crossing curves Ref: Gu, Niculescu, Chen, On stability crossing curves for general systems with two delays, J. Mathematical Analysis & Applications, 311 (2005), pp Stability crossing curves: The set of delays for which the characteristic equation has at least one imaginary zero (or pair of imaginary zeros). Associated with change of stability

49 The two delays case A DDE with two constant delays x(t)+c 1 x(t 1 )+c2x(t 2 )+c 3 x 0 (t)+c 4 x 0 (t 1 )+c5x 0 (t 2 )=0. The characteristic equation: h(s) =h 0 (s)+h 1 (s)e 1s + h 2 (s)e 2s. Let a k (s) =h k (s)/h 0 (s). Then a(s, 1, 2 )=1+a 1 (s)e 1s + a 2 (s)e 2s =0. Stability: a question of the number of the roots of the characteristic equation with a real part on the right hand side of the plane.

50 The two delays case For an imaginary s = i! to satisfy a(s, 1, 2 ) = 0, the vector corresponding to the three terms must form a triangle: a(s, 1, 2 )=1+a 1 (s)e 1s + a 2 (s)e 2s =0. Hence, their magnitudes must satisfy the triangle inequalities: a 1 (i!) + a 2 (i!) 1, 1 apple a 1 (i!) a 2 (i!) apple1.

51 The two delays case The triangle inequalities determine which i! may be zeros of a(s) The set of all such! are the crossing set. Any given! defines a collection of pairs ( 1, 2 ). 1 = \a 1(i!)+(2u 1) ± 1! 2 = \a 2(i!)+(2v 1) 2! where from the law of cosine: 0, u = u ± 0, u± 0 +1,... 0, u = v ± 0, v ± 0 +1,... 1,2 = cos 1 1+ a1,2 (i!) 2 a 2,1 (i!) 2 2 a 1,2 (i!), and u ± 0, v ± 0 are the smallest possible integers such that the corresponding 1,2 are nonnegative.

52 The two delays case The crossing set always consists of a finite number of intervals of finite length Any interval of! s defines a collection of curves in R 2 The general case is a union of the following sets:

53 Example Motivation DDE: dx dt = 2x(t 1)+x(t 2 ).