Deriving the Adjoint / Tangent Linear Model for Nonlinear PDE

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The discussion focuses on deriving the adjoint or tangent linear model for a specific nonlinear partial differential equation. The user seeks clarification on the integration by parts technique used in the derivation, particularly how to transition from the continuous adjoint equation to a discretized form involving Lagrange multipliers. There is confusion regarding the origin of the Lagrange multipliers and the steps needed to express the adjoint equation in a summation format. Participants emphasize the importance of understanding the underlying mathematical principles to apply this technique effectively in future problems. The conversation highlights a collaborative effort to clarify complex mathematical concepts.
finite_diffidence
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Homework Statement
Deriving the Adjoint / Tangent Linear Model for a non-linear PDE
Relevant Equations
please see below as latex is not rendering
I am trying to derive the adjoint / tangent linear model matrix for this partial differential equation, but cannot follow the book's steps as I do not know the math. This technique will be used to solve another homework question. Rather than posting the homework question, I would like to understand the technique generally so I can forever use it. Here is the equation:

$$\frac{\partial u}{\partial t} = \frac{\partial u}{\partial y} \frac{\partial u}{\partial x}$$

Now in the book the derivation continues as follows:

1. First substitute in $$ u \rightarrow u + \delta u $$.

Then get rid of all the terms which have $$ \delta u \delta u $$

We are left with:

$$ \frac{\partial \delta u}{\partial t} = \frac{\partial \delta u}{\partial y} \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} \frac{\partial \delta u}{\partial x}$$

2. Now we can integrate by parts to move the derivatives over from the perturbation to the Lagrange multiplier. We ignore the surface term picked up as it will vanish at the boundaries:

$$ \int \lambda \left(\frac{\partial \delta u}{\partial t} - \frac{\partial \delta u}{\partial y} \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} \frac{\partial \delta u}{\partial x} \right)d\Omega dt
$$

to:

$$ \int \delta u \left( -\frac{\partial \lambda}{\partial t} + \frac{\partial}{\partial y} (\lambda \frac{\partial u}{\partial x}) + \frac{\partial u}{\partial x} (\lambda \frac{\partial u}{\partial y}) \right)d\Omega dt
$$

> How did we do this trick? I was told it is integration by parts, but could someone do it explicitly step by step?3. This gives us the continuous adjoint equation:

$$
\frac{\partial \lambda}{\partial t} = \frac{\partial}{\partial y} (\lambda \frac{\partial u}{\partial x}) + \frac{\partial u}{\partial x} (\lambda \frac{\partial u}{\partial y})
$$

4. From that equation we can discretize in time and write out the longhand summation of all the lagrange multiplies. We pick up a load of terms:

$$
\lambda_0 - \lambda_1 \left(u_1 - u_0 - \Delta t \frac{\partial u_0}{\partial y} \frac{\partial u_0}{\partial x}\right) + \lambda_1 \left(u_2 - u_1 - \Delta t \frac{\partial u_1}{\partial y} \frac{\partial u_1}{\partial x}\right)
$$

Or playing the same trick :

$$ \lambda_0 - \lambda_1 + \Delta t\left(\frac{\partial}{\partial y}\lambda_1\frac{\partial u_1}{\partial x} + \frac{\partial}{\partial x}\lambda_1\frac{\partial u_1}{\partial y} \right)
$$

> How do we get these two lines? I have no idea. If someone could do it very explicitly that would be helpful in me understanding what I am missing.

If someone with infinitely better math skills than mine could illuminate the way, that would be much appreciated.
 
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Hello @finite_diffidence , :welcome: !

finite_diffidence said:
Rather than posting the homework question
Humor us and do it anyway. The context is not as irrelevant as you seem to think .

##u(x,y,t) = x+y+t## is a solution to the DE.

Where do the Lagrange multipliers come from ? Out of the blue ?
finite_diffidence said:
Now in the book
finite_diffidence said:
I was told it is integration by parts
A talking book ?
 
BvU said:
Hello @finite_diffidence , :welcome: !

Humor us and do it anyway. The context is not as irrelevant as you seem to think .

##u(x,y,t) = x+y+t## is a solution to the DE.

Where do the Lagrange multipliers come from ? Out of the blue ?
A talking book ?

Hello BvU,

The other question is basically continuing this same equation, but I must derive the adjoint matrix.

My attempt of the problem starting at step 2. Looking at each term individually as I multiply through by ##\lambda##:

Integrating with respect to ##t##:

$$
\int \lambda \frac{\partial \delta u}{\partial t} = \lambda \delta u - \int \frac{\partial \lambda}{\partial t}\delta u
$$

Now looking at the second term, integrating with respect to space (##y##):

$$
\int -\lambda \frac{\partial \delta u}{\partial y} \frac{\partial u}{\partial x} = -\lambda \frac{\partial u}{\partial x} \delta u + \int \frac{\lambda}{\partial y} \frac{\partial u}{\partial x}
$$

Now looking at the third term, integrating with respect to space (##x##):

$$
\int -\lambda \frac{\partial \delta u}{\partial x} \frac{\partial u}{\partial y} = -\lambda \frac{\partial u}{\partial y} \delta u + \int \frac{\lambda}{\partial x} \frac{\partial u}{\partial y}
$$

The terms without the integrals disappear as we are dealing with Dirichlet boundary conditions, so we are left with the continuous adjoint equation i.e 3..

However, now I do not know how to get to 4. What are the steps to go from my continuous adjoint to the multiplied out version. I am not sure where the Lagrange multipliers come from. The book stated integration by parts which I figured out after much staring.
 
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