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Poisson's equation in electrostatic does not satisfy the linear superposition principle. Can I say that since Laplacian operator is a non-linear operator so it does not follow the linear superposition principle or something more else?

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    $\begingroup$ Laplacian operator is a non-linear operator?? $\endgroup$ – Qmechanic Sep 6 '18 at 15:25
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    $\begingroup$ Surely $\nabla^2(f+g)=\nabla^2 f+\nabla^2 g$... $\endgroup$ – ZeroTheHero Sep 6 '18 at 15:27
  • $\begingroup$ sorry I was wrong $\endgroup$ – no one Sep 6 '18 at 17:24
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Poisson's equation is

$$ \nabla^2V=\rho $$

It is true that if you put in $V_1$ and $V_2$ as solutions to Poisson's equations, then $V_1+V_2$ would not satisfy Poisson's equation, because

$$ \nabla^2(V_1+V_2)=\nabla^2 V_1+\nabla^2 V_2=\rho+\rho=2\rho\neq\rho $$

Note that this is not because $\nabla^2$ is not a linear operator. $\nabla^2$ is just a bunch of derivatives, so emphatically is a linear operator. $\nabla^2(V_1+V_2)=\nabla^2 V_1+\nabla^2 V_2$, so it's linear. The problem is that the other side of the equation is not linear in $V$. The other side of the equation gives you back $\rho$ whether you put in $V_1$, $V_2$, or $V_1+V_2$ into the equation.

If you want to think of this in terms of operators, the left side of the equation has the $\nabla^2$ operator. When you specify a $V$, it gives you back $\nabla^2V$. This is a linear operator. The right side of the equation has the $\rho$ "operator". When you specify a $V$, it gives you back $\rho$. No matter what $V$ you plug in, it always gives you back the same $\rho$. This is not a linear operator. Thus, since your equation doesn't look like $(\text{linear operator})V=(\text{other linear operator})V$, it is not a linear equation.

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  • $\begingroup$ Should the second equation equal $\rho_1 +\rho_1 $ rather than $2\rho$? $\endgroup$ – user45664 Sep 6 '18 at 17:22
  • $\begingroup$ @user45664 we're assuming $V_1$ and $V_2$ are solutions to the SAME equation. So they have identical $\rho$. $\endgroup$ – Jahan Claes Sep 6 '18 at 17:29
  • $\begingroup$ But then your second equation shows that $\nabla^2$ is linear? Am I confused with the wording? $\endgroup$ – user45664 Sep 6 '18 at 17:33
  • $\begingroup$ @user45664 $\nabla^2$ IS linear, that's the point. The point is that Poisson's equation as a whole is not linear, because if $V_1$ and $V_2$ are solutions, then $V_1+V_2$ is NOT a solution. That's precisely because you have $2\rho$ on the RHS when you wanted just $\rho$. $\endgroup$ – Jahan Claes Sep 6 '18 at 17:34
  • $\begingroup$ @user45664 Perhaps the confusion is coming from what exactly I'm defining to be Poisson's equation. I'm saying someone hands you a fixed $\rho$, the charge density of your system, and you want to solve $\nabla^2 V=\rho$. That's Poisson's equation. Then $V_1+V_2$ is a solution to a DIFFERENT equation, one where the charge density in your system was doubled. It DOESN'T satisfy your equation, which was given for a certain, fixed $\rho$. $\endgroup$ – Jahan Claes Sep 6 '18 at 17:37

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