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A Obtain parameter derivatives solving PDE

  1. Mar 4, 2016 #1
    I have a PDE which is the following:

    $$\frac {\partial n}{\partial t} = -G\cdot\frac {\partial n}{\partial L}$$

    with boundary condition: $$n(t,0,p) = \frac {B}{G}$$
    , where G is a constant, L is length and t is time.

    G and B depend on a set of parameters, something like $$B = k_1\cdot C^a$$ and $$G = k_2\cdot C^b$$, where C is a given variable.

    I am using finite differences method to integrate this PDE, but I also wanted to obtain the derivatives of n with respect to its parameters (k1, k2, a and b in this case) along the integration. How can this be done?

    I thought about something like integrating these derivatives like in the previous PDE. Something like:

    $$ \frac{\partial}{\partial t}\left(\frac {\partial n}{\partial p}\right) = -[\frac{\partial G}{\partial p}\frac{\partial n}{\partial L} + G\cdot\frac{\partial}{\partial L}\left(\frac {\partial n}{\partial p}\right)]$$

    and apply the respective boundary conditions: $$\frac {\partial n}{\partial p}(t,0) = \frac {\partial}{\partial p}(\frac {B}{G})$$

    I have already calculated this using Matlab, and the results look similar although not exactly equal to the ones obtained with finite differences (which I also don't trust much). Do you agree with this formulation?
     
  2. jcsd
  3. Mar 5, 2016 #2
    The wording of the question is very bad.
    In the first PE it appears that $n$ is a function of at least two variables t and L.
    In the boundary condition n(t,0,p) they are now three variables. The vatiable p is new. We don't know what variable is equal to 0.
    Then, a new "given variable" appears C with no connection to the PDE or with the preceeding variables.
    I will no longer spent time on a question raised with so flippancy.
     
  4. Mar 5, 2016 #3
    Sorry if the formulation was not the most correct one. So, just to be clear:

    $$\frac {\partial n(t,L,p)}{\partial t} = -G(t,p)\cdot\frac {\partial n(t,L,p)}{\partial L}$$

    where $$p = \{k1,k2,a,b\}$$

    One can assume that:

    $$G(t,p) = k1\cdot a \cdot t$$

    Boundary condition is:

    $$n(t,0,p) = \frac {B(t,p)}{G(t,p)}$$

    with

    $$B(t,p) = k2\cdot b \cdot t$$

    The C variable doesn't matter much for this question. What I would like to have your opinion on was concerning the validity of this equation:

    $$ \frac{\partial}{\partial t}\left(\frac {\partial n(t,L,p)}{\partial p}\right) = - [\frac{\partial G(t,p)}{\partial p}\cdot \frac{\partial n(t,L,p)}{\partial L} + G(t,p)\cdot\frac{\partial}{\partial L}\left(\frac {\partial n(t,L,p)}{\partial p}\right)]$$

    with boundary condition: $$\frac {\partial n(t,0,p)}{\partial p} = \frac {\partial}{\partial p}(\frac {B(t,p)}{G(t,p)})$$
     
    Last edited: Mar 5, 2016
  5. Mar 5, 2016 #4
    ## \frac {\partial n(t,L,p)}{\partial t} = -G(p)\cdot\frac {\partial n(t,L,p)}{\partial L} ##
    General solution : ## \quad n(t,L,p)=F\left(\left(L-G(p)\:t\;\right)\:,\:p\right)##
    where ## F ## is any differentiable function of two variables.
    The goal now is to find which function(s) ##F## complies the boundary condition ## n(t,0,p)=\frac{B(p)}{G(p)} ##
    If such a function exists, then :
    ## F\left(\left(-G(p)\:t\:\right)\:,\:p\right)= \frac{B(p)}{G(p)}##
    The term on the left side is function of ## t ## , while the term of right side is not. This is a contradiction.
    So, at first sight, there is no solution of the PDE which agrees with the boundary condition.
    This draw to think that something is wrong or missing in the first part of the wording of the question.
     
  6. Mar 5, 2016 #5
    Yes, you're right I forgot it. Both B(p) and G(p) should be B(p,t) and G(p,t). I changed it already on the previous post.

    I already solved the first PDE I showed using finite differences. However, I was interested in obtaining the derivatives of n(t,L,p) with respect to the parameter vector p. I was just wondering, if I could obtain those by solving the last PDE that I showed on the previous post.
     
    Last edited: Mar 5, 2016
  7. Mar 5, 2016 #6
    Frist you wrote that ##G## is constant :

    upload_2016-3-5_15-59-59.png
    Now, If ##G## is fonction of ##t## , the general solution of the PDE, before any further conditions is :
    ## n(t,L,p)=F \left( \left( -\int G(t)dt \right)\:,\: p\: \right) ##
    where ##F## is any differentiable function of two variables.
    But it is not possible to continue if the wording of the question always changes.
    Best regards.
     

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  8. Mar 5, 2016 #7
    First you wrote that ##G## is constant :
    upload_2016-3-5_16-13-34.png

    Now, if ##G## is function of ##t## the general solution of the PDE , before any further condition, becomes:
    ## n(t,L,p)=F\left( \left( L - \int G(p,t)dt \right)\:,\: p\: \right) ##
    where ##F## is any differentiable function of two variables.
    Well, it is not acceptable to continue with successive changes in the wording of the problem.
    Best regards.
     
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