# Greens Function Partial Derivative Symmetry

I am trying to understand if, in general, there is some sort of symmetry in partial derivatives of a greens function w.r.t. the first and second (unprimed and primed) variable.

For a self-adjoint linear differential operator $$L$$,

$$Lg(x,x') = \delta(x,x'),$$
$$L=L^\dagger, g = g^\dagger$$

we have in general (Hilbert inner-product)

$$g(x,x') = g(x',x)^*$$.

If the coefficients of the underlying DE are real, than

$$g(x,x') = g(x',x)$$,

so $$g(x,x')$$ itself is symmetric.

Now I am wondering if we can say something similar when it comes to derivatives. Here it is shown that

$$\frac{\partial g(x,x')}{\partial x} = \frac{\partial g(x',x)}{\partial x'}$$,

but no statement is made about the relationship between $$\frac{\partial g(x,x')}{\partial x}$$ and $$\frac{\partial g(x,x')}{\partial x'}$$. However, looking at the specific greens function that solves the wave equation with point-source excitation

$$(\nabla^2+k^2)g(r,r') = -\delta(r,r')$$,
$$g(r,r')=\frac{e^{jkR}}{4\pi R}$$,
$$R=\sqrt{(x_1-x_1')^2+(x_2-x_2')^2+(x_3-x_3')^2}$$,

it is true that

$$\frac{\partial g(x_1,x_2,x_3;x_1',x_2',x_3')}{\partial x_j} = - \frac{\partial g(x_1,x_2,x_3;x_1',x_2',x_3')}{\partial x'_j}$$,
e.g.,
$$\nabla g(x_1,x_2,x_3;x_1',x_2',x_3') = -\nabla' g(x_1,x_2,x_3;x_1',x_2',x_3')$$.

Is this true for any greens function in case we have an self-adjoint operator with real coefficients?

No. For example, consider the Green's function for the Laplacian $$\nabla^2 u = 0$$ on $$[0,1]$$, subject to the Dirichlet boundary conditions $$u(0) = u(1) = 0$$. We have $$G(x,x') = \begin{cases} x'(1 - x) & x > x' \\ x(1 - x') & x \leq x' \end{cases}$$ and so $$\frac{\partial G(x, x')}{\partial x} = \begin{cases} -x' & x > x' \\ 1 - x' & x \leq x' \end{cases}$$ but $$\frac{\partial G (x, x')}{\partial x'} = \begin{cases} 1-x & x > x' \\ x & x \leq x' \end{cases}.$$
On an intuitive level, the property $$\partial G/\partial x = - \partial G/\partial x'$$ in your example is a consequence of the property $$G(x,x') = f(x - x')$$, i.e., the Green's function only depends on the separation of $$x$$ and $$x'$$. If a PDE problem has translational symmetry, then there will (usually) exist a Green's function with this property. But in the case I've outlined above, the boundary conditions break the translational symmetry, and so the conjecture fails.
• @3ak: Not necessarily. Periodic boundary conditions on a compact domain (like $u(0) = u(1)$, $u'(0) = u'(1)$ on $[0,1]$) could also yield the "translational symmetry" needed for your property to hold. Commented Aug 4 at 13:48