# Functional differential equation (from Quantum Field Theory).

I have a certain differential equation that includes functional derivatives. I know the solution, but I'm having a hard time to show that the equation is indeed solved by the solution. The background for this question is quantum field theory (in particular, scalar $\phi^4$).

The equation, called the Schwinger-Dyson equation, reads $$\nabla^2 \delta_xF[f]-\frac{g}{3!}\delta_x^3 F[f]-f(x)F[f]=0\tag{1}$$ where $\delta_x$ is a functional derivative: $$\delta_x\equiv\frac{\delta}{\delta f(x)}\tag{2}$$ and $F[f]$ is a functional of $f(x)$, a scalar function of $x\in\mathbb R^n$. Also, $\nabla^2$ is the Laplacian on a certain manifold.

The standard method to solve this equation is through a "functional Fourier transform", where we formally integrate over the space of functions (also known as functional integration). The solution supposedly reads (cf. [1]): $$F[f]\propto \exp\left[\frac{g}{4!}\int\mathrm dx\ \delta_x^4\right]\exp\left[\frac{1}{2}\int\mathrm dy\,\mathrm dz\ f(y)G(y-z)f(z)\right]\tag{3}$$ where $G$ is a Green function of $\nabla^2$, $$\nabla^2G(x)\equiv\delta(x)\tag{4}$$

Question

I know how to use the Fourier transform to solve $(1)$, and the solution is indeed $(3)$. But then I tried to plug $(3)$ back into $(1)$ and show that the equation is solved without using functional integration, and I failed. The expansion of the differential operator $\exp\left[\frac{g}{4!}\int\mathrm dx\ \delta_x^4\right]$ in power series in $g$ leads to a somewhat complicated combinatoric problem that I don't seem to be able to solve.

My attempt was as follows: take $(1)$ and let us suppose that $F$ can be expanded in power series in $g$: $$F[f]=\sum_{k=0}^\infty g^kF^{(k)}[f]\tag{5}$$

Plugging this into $(1)$ and insisting that the equation is solved order by order in $g$, we get the system of equations \begin{aligned} \nabla^2 \delta_xF^{(0)}[f]-f(x)F^{(0)}[f]&=0\\ \nabla^2 \delta_xF^{(k)}[f]-\frac{1}{3!}\delta_x^3 F^{(k-1)}[f]-f(x)F^{(k)}[f]&=0 \end{aligned}\tag{6}

Now, the solution for $F^{(0)}$ is straightforward: $$F^{(0)}[f]\propto \exp\left[\frac{1}{2}\int\mathrm dy\,\mathrm dz\ f(y)G(y-z)f(z)\right]\tag{7}$$

If we trust the solution $(3)$, then the solution for $F^{(k)}$ should be \begin{aligned} F^{(k)}[f]&\overset?= \left(\frac{1}{4!}\int\mathrm dx\ \delta_x^4\right)^kF^{(0)}[f]\\ &\overset?=\frac{1}{4!}\int\mathrm dx\ \delta_x^4 F^{(k-1)}[f] \end{aligned}\tag{8} but here I fail: there is no way I can show (say, with induction) that this indeed satisfies $(6)$. Thus, here I am. Can anyone help me to show that $(8)$ satisfies $(6)$?

[1]: http://www.scl.rs/papers/QFT2notes.pdf, equation (2.23). Here the author uses functional integrals to solve the (generalised) Schwinger-Dyson equation, equation (2.12). In this post we use the notation $F[f]$ instead of $Z[J]$, and also the mass $m$ is zero to keep the analysis simple. Also, I have dropped some factors of $i$ to clean up the notation.

I think in this case, the proof that equation $(8)$ solves equation $(6)$ is indeed by induction. However, note first that there is a factor $1/k!$ (from the Taylor expansion of the exponential) missing in the first line of equation $(8)$, and a factor $1/k$ in the second line.

The inductive step starts by multiplying the second line of equation $(6)$ by$$\left({1\over{k + 1}}\right)\left({1\over{4!}} \int dz\,\delta_x^4\right)$$from the left. The first term immediately takes the desired form. The second term almost takes the desired form, but with an extra prefactor $k/(k + 1)$. The third term gives two contributions, by virtue of the product rule of (functional) differentiation. One contribution has the desired form $f(x) F^{(k + 1)}[f]$, the other one has a form similar to the second term, but with a prefactor $1/(k + 1)$ instead of $k/(k + 1)$. The two prefactors combine to $1$, bringing also the second term to the desired form.