# Partial differentiation of integral

1. Jun 14, 2015

### blue_leaf77

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.

2. Jun 14, 2015

### fzero

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.

3. Jun 14, 2015

### blue_leaf77

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.

4. Jun 14, 2015

### fzero

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 $\phi$.

5. Jun 14, 2015

### blue_leaf77

Alright, I got it.