# Are Green's functions generally symmetric?

1. Nov 30, 2015

### ShayanJ

In case of the Green's functions for the Laplace equation, we know that they're all symmetric under the exchange of primed and un-primed variables. But is it generally true for the Green's functions of all differential equations?
Thanks

2. Dec 1, 2015

### Geofleur

The Green's function will not always be symmetric.

The Green's function formalism really involves considering two problems,

$\mathbf{L}_\mathbf{x}u = f(\mathbf{x})$

and

$\mathbf{L}_\mathbf{x}^{\dagger} v = h(\mathbf{x})$,

where $\mathbf{L_x}$ and its adjoint, $\mathbf{L}^\dagger_{\mathbf{x}}$, are differential operators. These operators will satisfy the generalized Green's identity

$\int \left[ v^{*} \mathbf{L_x} u - u \left( \mathbf{L}^{\dagger}_{\mathbf{x}}v \right)^{*} \right] d^3 x =$ term that vanishes only if $\mathbf{L_x} = \mathbf{L}_\mathbf{x}^{\dagger}$.

Let's assume that $\mathbf{L_x} = \mathbf{L}_\mathbf{x}^{\dagger}$, so that the right hand side really does vanish. Corresponding to the two problems we are trying to solve, there will be two different Green's functions, $G(\mathbf{x},\mathbf{y})$ and $g(\mathbf{x},\mathbf{y}')$ that satisfy

$\mathbf{L}_\mathbf{x}G = \delta(\mathbf{x}-\mathbf{y})$

and

$\mathbf{L}_\mathbf{x}^{\dagger} g = \delta(\mathbf{x}-\mathbf{y}')$.

If we set $u = G(\mathbf{x},\mathbf{y})$, $v = g(\mathbf{x},\mathbf{y}')$, and put these into Green's identity (keeping the above two equations in mind), we get

$\int \left[ g^{*} \mathbf{L_x}G(\mathbf{x},\mathbf{y}) - G(\mathbf{x},\mathbf{y})\left( \mathbf{L}^{\dagger}_{\mathbf{x}}g(\mathbf{x},\mathbf{y}') \right)^{*} \right] d^3 x = \int \left[ g^{*}(\mathbf{x},\mathbf{y}') \delta(\mathbf{x}-\mathbf{y}) - G(\mathbf{x},\mathbf{y}) \delta(\mathbf{x} - \mathbf{y}') \right] d^3 x = 0$.

Evaluating the integral in the last equality gives

$g^{*}(\mathbf{y},\mathbf{y}')-G(\mathbf{y}',\mathbf{y}) = 0$,

so that

$g^{*}(\mathbf{y},\mathbf{y}') = G(\mathbf{y}',\mathbf{y})$.

Because we have assumed that $\mathbf{L_x} = \mathbf{L}_\mathbf{x}^{\dagger}$, we really have $g(\mathbf{y},\mathbf{y}') = G(\mathbf{y},\mathbf{y}')$, so that

$G^{*}(\mathbf{y},\mathbf{y}') = G(\mathbf{y}',\mathbf{y})$.

Finally, if the differential equations do not involve any complex coefficients, the complex conjugation will not change anything, and we will have

$G(\mathbf{y},\mathbf{y}') = G(\mathbf{y}',\mathbf{y})$.

So only if the differential operator $\mathbf{L}_{\mathbf{x}}$ is equal to its own adjoint and has no complex coefficients will the Green's function be symmetric. On the other hand, this is the case in a wide variety of circumstances.

3. Dec 1, 2015

### ShayanJ

Can you point to a book that treats Green's functions this way? I've seen Sadri-Hassani's but he considers equations with real coefficients and also doesn't consider the adjoint equation.
Thanks

4. Dec 2, 2015

### Geofleur

I really like the way Dennery and Krzywicki talk about Green's functions in their book, Mathematics for Physicists. Also the first two chapters, on complex analysis and linear algebra, are wonderful.