Lecture 22 Adjunct Methods

1. Lecture 22Adjunct Methods

2. Part 1 Motivation

3. Motivating scenario We want to predict tomorrow’s weather, u(t) … We have a atmospheric model chugging away to predict temperature, pressure, etc. This model depends on a forcing f(t), for example, sea surface temperature, which we known only imperfectly

4. prediction u(t) tyesterday t Yesterday’s model run ttoday f(t) ty tt But now we have new data for today prediction u(t) new data t tyesterday ttoday f(t) tt ty

5. new prediction Today’s model run should include yesterday’s data to help constrain the poorly known forcing old prediction u(t) new data t tyesterday ttoday f(t) new old tt ty How do we adjust the forcing (which was imperfectly known, anyway) to better predict yesterday’s weather?

6. Part 2 The mathematics of continuous functions inner products linear operators and their adjoints

7. Discretevectors uk and vk Continuousfunctions f(t) and g(t)

8. functions f(t) and g(t)Discrete approximation as a vectorsuk=f(kDt) and vk=g(kDt)

9. Discretedot product: c = Sk ukvk = uv scalar Continuousinner product c =  f(t) g(t) dt = (f,g)

10. discrete approximation as dot product(f,g) =  f(t) g(t) dt = Dt Sk ukvk = Dtuvwith uk=f(kDt) and vk=g(kDt) inner product (f,g)

11. Discretematrix: vj = Sk Mjkuk or v=Mu ContinuousLinear operator f = Lg

12. What is a linear operator?Linear differential operatorinvolving derivatives and known functionsLg = [ p(t) d/dt q(t) d/dt ] g(t) known

13. and/orLinear integral operatorinvolving intergral and known functionsLg =  p(t,t’) g(t’) dt’ known

14. ifL1g=f and L2f=gthenL1=L2-1 and L2=L1-1one linear operator is the inverse of the other

15. discrete approximations Sample differential operator plus b.c. 1 0 0 0 … 0 -1 1 0 0 … 0 0 -1 1 0 … 0 …… … 0 0 -1 1 Mu=v, M = Dt-1 Lg=f with L = d/dtplus b.c. g(0)=known Sample integral operator plus b.c. 1 0 0 0 … 0 1 1 0 0 … 0 1 1 1 0 … 0 …… … 1 1 1 1 Lg=f with Lg = 0tg(t’)dt’plus b.b. g(0)=known Mu=v, M = Dt

16. Question concerning a dot product … given two matrices A and B when is (Au)v = u(Bv) ? Answer: when B=AT, since (Au)v = (Au)Tv = uTATv = uT(ATv) = u (ATv)

17. Question concerning an inner product … given two linear operators L1 and L2 when is ( L1f , g ) = ( f, L2g ) ? Answer: never mind, but let’s give it a name (L1f, g) = (f, L2g) when L1 is the adjoint of L2 let’s denote the adjoint relationship L2=L1* means “adjoint”

18. Transformif A=BT then B=ATATT=AAT-1=A-1T(A+B)T= AT+BTif AT=A then A is symmetric Adjointif L1=L2* then L2=L1*L**=LL*-1=L-1*(L1+L2)*= L1*+L2* if L*=L then Lis self-adjoint

19. Calculating adjoints by integration by parts Let L = d/dt with b.c. zero at ± (Lf, g) = -+ df/dt g dt = f g |-+ --+f dg/dt dt = --+f dg/dt dt = (f, L*g) So L* = -d/dt with b.c. zero at ±

20. Three simple adjoints L* c(x) -d/dt b.c.: function 0 at ± d2/dt2 b.c.: function and its first derivative 0 at ± L c(x) d/dt b.c.: function 0 at ± d2/dt2 b.c.: function and its first derivative 0 at ±

21. Part 3 Functional derivatives How to represent the idea that a perturbation in forcing, f(t) cause a perturbation in response, u(t)

22. Here’s the differential equation L u(t) = f(t) forcing Data di linearly depends on u(t)through an inner productdi = (hi, u)

23. Differential Equation Lu=f A perturbation in f(t) causes a perturbation in u(t) f0(t)  f0(t)+df(t) u0(t)  u0(t)+du(t) Suppose df(t) was localized at time t0: df(t)=ed(t-t0) Then du(t) is a function of e and t0: du(t,e,t0) Then the function (or Fréchet)derivative is: du(t)/df(t0) = lime0 [ u(t,e,t0) – u(t,e=0,t0) ] / e

24. An impulsive perturbation in forcing ed(t-t0) Causes a perturbation in response du(t,e,t0) Then the general perturbation df in forcing causes the response du =  (du/df) df dt0 = ( du/df, df )

25. An impulsive perturbation in forcing ed(t-t0) t t0 Causes this response du t0 defines du/da A more complicated perturbation in forcing df t t0 du = (du/da, da) Causes this response du t0

26. Then the general perturbation df in forcing causes the response du =  (du/df) df dt0 = ( du/df, df ) In a discrete world: du1 du2 du3 = Dt … duN df1 df2 df3 … dfN du(t1)/df(t1) du(t1)/df(t2) du(t1)/df(t3) … du(t2)/df(t1) du(t2)/df(t2) du(t2)/df(t3) … du(t3)/df(t1) du(t3)/df(t2) du(t3)/df(t3) … … du(tN)/df(t1) du(tN)/df(t2) du(tN)/df(t3) … Might solve with least-squares …

27. Part 4 Calculating the data kernel The functional derivative of data with respect to forcing

28. The Goal to find the data kernel, gi(t)which relates a perturbation in the data, ddi, to a perturbation in the forcing df(t) through an inner product ddi = ( gi(t), df(t) )

29. Note that since the data kernel satisfies it is a functional derivative gi(t) = ddi /df(t) ddi = ( gi(t), df(t) )

30. Step 1:assume that a function u(t) solves a linear differential equation with forcing f(t) L u(t) = f(t)

31. Step 2:assume the differential equation has green function F(t,t’) so the solution can be written:note that L-1 is the inverse of L, sincef=Lu and u=L-1f u(t) =  F(t,t’) f(t’) dt = (F(t,t’), f(t) )  L-1 f(t)

32. Step 3:assume that the data, di, are related to the solution u(t) through an inner product di = ( hi(t), u(t) )

33. Step 4:do some substitutions and manipulations di = ( hi(t), u(t) ) = ( hi(t), L-1f(t) ) = (L-1*hi(t), f(t) ) = (L*-1hi(t), f(t) )

34. Step 4:since the problem is linear, this rule applies to perturbations of functions as well as to the functions themselves di = (L*-1hi(t), f(t) ) So ddi = (L*-1hi(t), df(t) )

35. Step 5:by comparing the definition of the data kernelddi = ( gi(t), df(t) )to the resultddi = (L*-1hi(t), df(t) )recognize that the data kernel is gi(t) = L*-1hi(t)

36. Step 6:since the data kernel satisfiesgi(t) = L*-1hi(t)then it must satisfy the differential equation L*gi(t) = hi(t)

37. This is the desired resultsa way of calculating the data kernel, gi(t)by solving the differential equation L*gi(t) = hi(t)

38. Part 5 An example Note: In this example I use very simple differential equations that can be solved analytically. In a reality, you would be using much more complicated differential equations that but be solved numerically ..

39. Example: Newtonian cooling equation du/dt + cu = f(t) L = d/dt + c u(t) is temperature f(t) is heating c is a constant

40. Green’s Function du/dt + cu = d(t-t’) F(t,t’) = H(t-t’) exp{ -c(t-t’) } unit step function

41. Adjoint differential equation L = d/dt + c The adjoint of d/dt is –d/dt and the adjoint of c is c So L* = -d/dt + c And so du/dt + cu = f(t) has corresponding adjoint equation -dgi/dt + cgi = hi

42. Greens Function of the Adjoint differential equation -dgi/dt + cgi = d(t-t’) has solution G(t,t’) = {1-H(t-t’)} exp{ c(t-t’) }

43. interpretation Suppose hi = d(t) so that the data di is just u(t=0), temperature at time 0 Then G(t,t’=0) is the data kernel gi(t) Now suppose that we make an impulsive perturbation of heating at time t0: df(t)=d(t-t0) Then ddi = u(t=0) = ( gi(t), df(t) ) = (G(t,t’=0) , d(t-t0) ) = G(t0,t’=0)

44. Interpretation, continued So for an impulsive perturbation of heating at time t0 du(t=0) = G(t0,t’=0) We would expect: no effect on temperature if heat applied after time t=0 large effect if applied just prior to t=0 minimal effect if it is applied way before t=1 Small effect Large effect No effect

45. example H f0 u0 f uobs duobs t

46. Forming data from u(t)Here I use an example of the data being averages of neighboring u’s d1 = u(t1) so h1 = [1, 0, 0, 0, 0, 0 … 0]T dj = ½ { u(tj-1) + u(tj) } for j>1 so hj = ½ [0, 0, 0, … 1, 1, … 0, 0, 0]T

47. d0 dobs ddobs t

48. The problem Reconstruct df from dd

49. Setup for Least Squares ddi = ( gi(t), df(t) ) dd1 dd2 dd3 … ddN g1 df1 df2 df3 … dfN g2 = g3 … gN time varies along columns …

50. results dftrue dfpre error t