Return Boundary Value Problems
u t − α 2 ∇ 2 u = F {\displaystyle \scriptstyle u_{t}-\alpha ^{2}\nabla ^{2}u=F}
∇ 2 u = F {\displaystyle \scriptstyle \nabla ^{2}u=F}
u t t − c 2 ∇ 2 u = F {\displaystyle \scriptstyle u_{tt}-c^{2}\nabla ^{2}u=F}
d i v ( E ) = 4 π ρ {\displaystyle div(E)=4\pi \rho }
c u r l ( E ) + 1 c ∂ B ∂ t = 0 {\displaystyle \scriptstyle curl(E)+{\frac {1}{c}}{\frac {\partial B}{\partial t}}=0}
c u r l ( B ) − 1 c ∂ E ∂ t = 4 π c J {\displaystyle \scriptstyle curl(B)-{\frac {1}{c}}{\frac {\partial E}{\partial t}}={\frac {4\pi }{c}}J}
d i v ( B ) = 0 {\displaystyle \scriptstyle div(B)=0}
