Questions about ODE definition

[WayneX-Wong]

Hi all,

I am now thinking of a lump parameter problem (or scalar variable problem), of which ODE can be simplified into the following form:
$$A \cdot \frac{d(X \cdot Y)}{dt}=B \cdot F(X)\cdot G(Y)\quad when \quad X>a$$
$$A ^ { \prime } \cdot \frac { d ( X \cdot Y ) } { d t } = B ^ { \prime } \cdot F ^ { \prime } ( X ) \cdot G ^ { \prime } ( Y ) \quad when \quad a \lt X \lt b$$
There are two variables $X$ and $Y$, and the remaining parameters $A$ and $B$ or the functions $F$ and $G$ are known, and $t$ is time. In order to close this ODE, an equation is added as
$$\xi(Y)=\chi(X)$$

The definition of general time-varying equations in MOOSE should be relatively convenient, directly through the ScalarKernels system. And if the ODE changes with time $t$, it should be realized through the Controls system. However, in the problem I studied, the ODE changes with the variable $X$ to be solved. Does MOOSE have any built-in systems that can meet such requirements?

Thanks!

[hugary1995]

Sounds like you are writing your custom scalar kernels anyways, so this should be a if-else condition. Regularize the function domain as you need.

[WayneX-Wong]

Thanks! I'll try

[WayneX-Wong]

Hi all,

Another question about the ODEs refered above,

$$A \cdot \frac { d ( X \cdot Y ) } { d t } = B \cdot F ( X , Y ) \quad X \gt a$$

Could rearrange the LHS term (Time deriviate term) into,

$$A \cdot X \cdot \frac{d(Y)}{dt}+A \cdot Y \cdot \frac{d(X)}{dt}$$

So I can define a ODETimekernel about $u$ coupled with $v$ , and use it in the input deck twice wrt $X$ and $Y$ respectively.

While what about RHS? There's no way to separate two variables. Should I define it a $X$ coupled $Y$ or conversely, or both two?

[WayneX-Wong]

@hugary1995 Thank you again for your patience Gary!
Certain it is. In order to close this ODE, another ODE funtion is added in form like:

$$ξ ( Y ) = χ ( X , Y )$$

Which form is similar to the RHS of the ODE above, so I didn‘t mention above, but still have question like this

Hi all,

Another question about the ODEs refered above,

A⋅d(X⋅Y)dt=B⋅F(X,Y)X>a

Could rearrange the LHS term (Time deriviate term) into,

A⋅X⋅d(Y)dt+A⋅Y⋅d(X)dt

So I can define a ODETimekernel about u coupled with v , and use it in the input deck twice wrt X and Y respectively.

While what about RHS? There's no way to separate two variables. Should I define it a X coupled Y or conversely, or both two?

To be more specific, I wanna refer to the example Example 3 : Multiphysics coupling
Which has PDEs written as
$$- \nabla \cdot \nabla u + \nabla v \cdot \nabla u = 0$$
$$- \nabla \cdot \nabla v=0$$
About second term of first PDE, a convection kernel was defined on $u$ with coupled variable $v$. For me it's understandable, cause $v$ can be solved independently by the second PDE. So when it comes to first PDE, $u$ seems to be more "dominant" than $v$, so $u$ coupled $v$ rather than $v$ coupled $u$ is defined.

But when it comes to my ODEs, variable $X$ and $Y$ have equal status, each variable cannot be determined independently by a single ODE. So I don't know what to do with ODEkernel definitions for terms like $χ ( X , Y )$. Should it be $X$ coupled $Y$ or conversely, or both two?

I’d like to apologize again for the statement that caused me confusion! I am not sure whether my statement is clear this time, and if so, any suggestions would be appreciated very much!

[hugary1995]

Okay, no problem. I think I understand now. I guess your confusion is coming from the definition of "variable" in the MOOSE input file. A lot of the times you will see something like this in the input file

[ScalarKernels]
  [foo]
    type = A
    variable = x
    coupled_variable = y
  []
  [bar]
    type = B
    variable = x
  []
  [baz]
    type = C
    variable = y
    coupled_variable = x
  []
[]

There isn't any difference between variable and coupled_variable for the purpose of retrieving values -- their variable values will be made available to you when you compute the residual and Jacobian. The value of the variable is retrieved on line 21 in "ScalarKernel.C" as _u. The coupled variable value can be retrieved (coupled) on demand using the method coupledScalarValue.

There is one important difference, however, for the purpose of assembly. One natural question to ask is: when there are multiple equations and multiple scalar variables, where do their corresponding residuals (and Jacobians) get assembled into the system of equations? The MOOSE convention is that the residual of the scalar kernels are associated with the variable of the kernel. In your case, you have two scalar equations in the following form:

Equation 1: f(X, Y) = 0
Equation 2: g(X, Y) = 0

with two unknowns X and Y. You need to tell MOOSE how to assemble the residuals. One possibility is to associate the first equation with X, and the second equation with Y, i.e.

[ScalarKernels]
  [equation1]
    type = Eq1
    variable = X
    coupled_variable = Y
  []
  [equation2]
    type = Eq2
    variable = Y
    coupled_variable = X
  []
[]

Another possibility is apparently to switch the variable and coupled_variable in each kernel. But that's a choice you have to make when you write the source code.

[WayneX-Wong]

Okay, I'll try again. In fact, I have already defined according custom scalerkernel.

To be specific

1. $$A \cdot X \cdot \frac { d ( Y ) } { d t } + A \cdot Y \cdot \frac { d ( X ) } { d t }$$
   Defined according kernel, time derivative term as variable and another as coupled_variable, using twice in the input respectively

2. $$x ( X , Y )$$ $$F ( X , Y )$$
   Defined according kernel, using once in the input with $X$ as variable and $Y$ as coupled_variable

But the calculated resudial doesn't deccrease, looks like the equation is not closed, so I have above question.

I will try more, this may take some time, I will update if there is any result.

[hugary1995]

It's a system of 2 dofs, so I'd just print out the Jacobian (a 2-by-2 matrix) to verify.

[WayneX-Wong]

@hugary1995 My custom kernels runs successfully and get correct result according your guidance

with two unknowns X and Y. You need to tell MOOSE how to assemble the residuals. One possibility is to associate the first equation with X, and the second equation with Y, i.e.

There were other non-convergence problems among the process, which turned out to be due to zero jacobian. I solved the problem by defining zero Jacobian as a small value as Issue #23594 suggests.

Thank you again for your professional guidance and patience Gary!