# Is the system of equations of electrostatics underdetermined or overdetermined? [duplicate]

The following equations are equations of electrostatics: $$\nabla \times \vec E=0$$ $$\nabla\cdot\vec E=\dfrac{\rho}{\epsilon_0}.$$

These are 4 independent equations, while $\vec E$ has only 3 independent components. Yet these equations do not completely specify the field, as after adding the gradient of a scalar $\nabla \lambda$ that satisfies Laplace equation ($\nabla^2 \lambda$=0) to $\vec E$ leaves the equations unchanged: $$\cases{\nabla \times \vec E=0\\\nabla\cdot\vec E=\dfrac{\rho}{\epsilon_0}}\xrightarrow[\nabla^2\lambda=0]{\vec E'=\vec E+\nabla \lambda}\cases{\nabla \times \vec {E'}=0\\\nabla\cdot\vec {E'}=\dfrac{\rho}{\epsilon_0}}$$ (note the primes on the RightHandSide $\vec E$s)

The system should be overdetermined (4 equations, 3 unknowns) but apparently it is underdetermined.

• Is the system overdetermined or underdetermined?
• How do we usually choose the arbitrary $\lambda$ in a problem with $\rho$ given and (say) Neumann boundary condition?
• Why the first equation (curl) is not enough?
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## marked as duplicate by Brandon Enright, Emilio Pisanty, John Rennie, Chris White, Kyle KanosFeb 5 '14 at 15:26

–  Qmechanic Feb 4 '14 at 20:07
@Qmechanic's link then points to physics.stackexchange.com/q/20071 –  dmckee Feb 4 '14 at 20:46
@Qmechanic The static case is completely different from that question. –  user215721 Feb 4 '14 at 21:06
The equations in question are partial differential equations, NOT algebraic equations. It is therefore not surprising that the system is underdetermined. For example, consider a function $f(x,y)$ satisfying $\partial_x f = \partial_y f = 0$. We have one function, but two equations. However, the two equations only specify $f(x,y) = c$ (constant). The system is underdetermined upto a constant shift of $f(x,y)$. A similar argument holds here. –  Prahar Feb 4 '14 at 21:16
@Prahar So the statement for the each system of (P or O)DEs the total number of equations must be equal to the unknown variables is not generally correct? –  user215721 Feb 4 '14 at 21:23

We normally choose $\lambda$ in a way that makes our calculations easiest purely because nothing says we can't do that.
The first equation is not enough to determine the system for the same reason that $\frac{\partial y(x)}{\partial x}=f(x)$ can never uniquely determine $y(x)$. We can always add a constant offset that changes $y(x)$ but not $f(x)$.