Say I have a integral equation $$\phi(x,t) = \lambda\int dx'\ dt'\int\frac{d\omega\ dk}{(2\pi)^2}\frac{e^{-ik(x-x')+i\omega(t-t')}}{\omega^2-k^2-m^2+i\epsilon}\phi^3(x',t')$$ How do I convert it into a differential equation? It's a multiple choice question with the following choices,

  1. $$\bigg(\frac{\partial^2}{\partial t^2}+\frac{\partial^2}{\partial x^2}-m^2+i\epsilon \bigg)\phi(x,t) = -\frac{1}{6}\lambda\ \phi^3(x,t)$$
  2. $$\bigg(\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}+m^2-i\epsilon \bigg)\phi(x,t) = \lambda\ \phi^2(x,t)$$
  3. $$\bigg(\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}+m^2-i\epsilon \bigg)\phi(x,t) = -3\lambda\ \phi^2(x,t)$$
  4. $$\bigg(\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}+m^2-i\epsilon \bigg)\phi(x,t) = -\lambda\ \phi^3(x,t)$$
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    $\begingroup$ As this is multiple choice, the easiest thing is to check each of the statements (just insert the expression for $\phi(x,t)$). After a few lines of calculation you will see some pattern... $\endgroup$ – Fabian Jan 24 '17 at 18:38
  • $\begingroup$ since derivatives don't act on $x',t'$ the answer will be 4. since $\phi^3$ isn't changed $\endgroup$ – tired Jan 24 '17 at 18:54
  • $\begingroup$ @tired $\phi^3$ haven't changed in (1) also. Also, the derivative is taken twice, $\partial_t^2$ but integrated once, which will decrease the power to $\phi^2$. $\endgroup$ – Ayatana Jan 24 '17 at 19:03
  • $\begingroup$ what is with the $1/6$? $\endgroup$ – tired Jan 24 '17 at 19:04
  • $\begingroup$ I don't know about -1/6, I couldn't solve the equation, how will I know that the -1/6 doesn't turn up when solving? $\endgroup$ – Ayatana Jan 24 '17 at 19:09

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