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Partial differentiation of integral

  1. Jun 14, 2015 #1

    blue_leaf77

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    If I have a function

    ##f(u,u^*) = \int u^* \hat{O} u d^3\mathbf{r}##

    both ##u## and ##u^*## are functions of ##\mathbf{r}## where ##\mathbf{r}## position vector, ##\hat{O}## some operation which involves ##\mathbf{r}## (e.g. differentiation), and the star sign denotes complex conjugate. Now I want to find the differential expression for ##f##, namely ##df = \frac{\partial f}{\partial u}du + \frac{\partial f}{\partial u^*}du^*##. I think ##u## and ##u^*## are independent, right? When I have to calculate the partial derivative of ##f## w.r.t. ##u^*##, it seems that I simply need to bring the differentiation inside the integral and differentiate only ##u^*## which gives me

    ##\int \hat{O} u d^3\mathbf{r}##,

    is this also justified? If yes how should I calculate ##\frac{\partial f}{\partial u}##?
    Any help is appreciated.
     
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  3. Jun 14, 2015 #2

    fzero

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    You are computing what is known as the functional derivative. There is a formal definition given in the wiki, but the subsequent development leads to the chain rule and the practical rule that

    $$ \frac{\delta u(\mathbf{r})}{\delta u(\mathbf{r}')} = \delta^3(\mathbf{r}-\mathbf{r}').$$

    The derivative of a function by an independent function is zero, e.g., ##\delta u^*(\mathbf{r})/\delta u(\mathbf{r}')=0##.

    Applying these to your functional above gives

    $$ \frac{\delta f( u,u^*) }{\delta u^*(\mathbf{r})} = \int \delta^3(\mathbf{r}'-\mathbf{r}) \hat{O} u(\mathbf{r}')d^3\mathbf{r}' = \hat{O} u(\mathbf{r}).$$

    However

    $$ \frac{\delta f( u,u^*) }{\delta u(\mathbf{r})} = \int u^*(\mathbf{r}')\hat{O} \delta^3(\mathbf{r}'-\mathbf{r}) d^3\mathbf{r}' ,$$

    where if ##\hat{O}## is a differential operator it will generally act on the delta function, so we can't automatically simplify this. In practice, the function ##u^*## will satisfy some integrability condition so that we can integrate this by parts, deriving an expression where an operator acts on ##u^*## and we can perform the integral using the delta function (without any operator acting on it).

    You might want to try this out for some examples where ##\hat{O}## is a gradient or Laplacian to see how it works.
     
  4. Jun 14, 2015 #3

    blue_leaf77

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    Thanks that very helpful, but the expression
    doesn't seem to have correctly matched dimensions on both sides, the LHS is dimensionless, while the RHS has a dimension of inverse volume.
     
  5. Jun 14, 2015 #4

    fzero

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    This is a consequence of the definition of the operation. The operator ##\delta/\delta f## carries units which are the product of the inverse units of ##f## with the units of the inverse volume. You should be able to see this by looking at the rigorous definition that uses the test function ##\phi##.
     
  6. Jun 14, 2015 #5

    blue_leaf77

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    Alright, I got it.
     
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