## what are the meanings of adjoint model's input and output

Discussion about tangent linear and adjoint models, variational data assimilation, and other related issues.

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LiZhou
Posts: 7
Joined: Tue Dec 26, 2017 5:39 pm
Location: Institute of Oceanography,CAS

### what are the meanings of adjoint model's input and output

Dear all,
I have a modeling test about sensitivty analysis of Southern Ocean's MKE to initial conditions. Therefor, I had to use ROMS ADM to preduct of the gradient of cost function J to initial perturbations(x0'). I'm confusing that what shoule be inputted at ADM modeling step. Is it the gradident dJ/dxt' at starting time, or only the perturbation xt' at starting time? And I have mentioned the unit of output is 1/x'. I have some troulbe to understand thest.
Have somebody had done similar work? Expecting kind discussion.Thanks for your suggestions!
Sincerely.
Li Zhou

ezaron
Posts: 12
Joined: Mon Oct 26, 2009 3:06 am
Location: Portland State University

### Re: what are the meanings of adjoint model's input and outpu

Hi Li,

Suppose you are interested in a linear function of the model's trajectory. Let the value of this function be denoted, y. Then there exists another function, call it K(x,t), which permits you to compute y using an inner product:
y = \int_D \int_0^T K(x,t) u(x,t) dx dt.
The model's trajectory is denoted u(x,t), where one could image u to be a vector containing components of velocity, temperature, salinity, etc., the usual prognostic variables computed by ROMS. Of course, (x,t) denote (space,time) coordinates.

You indicated "MKE" which I will guess refers to "mean kinetic energy at time t=T". This is not a linear function, but presumably the sensitivity of this quantity interests you for a perturbation around some basic state trajectory,
y = \int_D \int_0^T \bar{u}(x,t) \delta(t-T) u(x,t) dx dt,
so the function K is given by \bar{u}(x,t) \delta(t-T), i.e., it equals the basic state (the velocity vector field) at the time of interest, times a delta-function at that time of interest.

The adjoint model can be used to compute the quantity A=dy/du(x,t=0). For a given perturbation u'(x) given at t=0, the adjoint computes the quantity A(x) such that
y' = \int_D A(x) u'(x) dx. (Note that I am being a little vague about a factor related to the area of the domain, but you can figure this out by verifying this equation for a very small test perturbation field, u'(x), once you have computed A(x) using the adjoint model.)

Thus, you would take your velocity field, \bar{u}(x,t), at time t=T and input this to the adjoint (and zero out all the other components (temp and salinity) of the input fields). Then integrate the adjoint model backwards in time to t=0. The adjoint fields at the initial time are A(x). You should convince yourself they approximately satisfy the equation above. (Since y' and u' are perturbations, you will compute these from the differences of two model runs.)

Good luck!

LiZhou
Posts: 7
Joined: Tue Dec 26, 2017 5:39 pm
Location: Institute of Oceanography,CAS

### Re: what are the meanings of adjoint model's input and outpu

hi ezaron,

LiZhou
Posts: 7
Joined: Tue Dec 26, 2017 5:39 pm
Location: Institute of Oceanography,CAS

### Re: what are the meanings of adjoint model's input and outpu

hi ezaron,
ezaron wrote:Hi Li,

Suppose you are interested in a linear function of the model's trajectory. Let the value of this function be denoted, y. Then there exists another function, call it K(x,t), which permits you to compute y using an inner product:
y = \int_D \int_0^T K(x,t) u(x,t) dx dt.
The model's trajectory is denoted u(x,t), where one could image u to be a vector containing components of velocity, temperature, salinity, etc., the usual prognostic variables computed by ROMS. Of course, (x,t) denote (space,time) coordinates.

You indicated "MKE" which I will guess refers to "mean kinetic energy at time t=T". This is not a linear function, but presumably the sensitivity of this quantity interests you for a perturbation around some basic state trajectory,
y = \int_D \int_0^T \bar{u}(x,t) \delta(t-T) u(x,t) dx dt,
so the function K is given by \bar{u}(x,t) \delta(t-T), i.e., it equals the basic state (the velocity vector field) at the time of interest, times a delta-function at that time of interest.

The adjoint model can be used to compute the quantity A=dy/du(x,t=0). For a given perturbation u'(x) given at t=0, the adjoint computes the quantity A(x) such that
y' = \int_D A(x) u'(x) dx. (Note that I am being a little vague about a factor related to the area of the domain, but you can figure this out by verifying this equation for a very small test perturbation field, u'(x), once you have computed A(x) using the adjoint model.)

Thus, you would take your velocity field, \bar{u}(x,t), at time t=T and input this to the adjoint (and zero out all the other components (temp and salinity) of the input fields). Then integrate the adjoint model backwards in time to t=0. The adjoint fields at the initial time are A(x). You should convince yourself they approximately satisfy the equation above. (Since y' and u' are perturbations, you will compute these from the differences of two model runs.)

Good luck!