abla^2 f = ext{div}( ext{grad}(f)) = abla cdot ( abla f) If you work this out, you get $ abla^2 f = f_{xx} + f_{yy} + f_{zz}$. If $vec{F} = (P, Q, R)$ is a vector field, we can also define the Laplace operator on it by $ abla^2 vec{F} = ( abla^2 P, abla^2 Q, abla^2 R)$. Prove the identity $ ext{curl} ( ext{curl}(vec{F})) = ext{grad} ( ext{div}(vec{F})) - abla^2 vec{F}$
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First, we need to find the curl of F, which is given by: curl(F) = (∂R/∂y - ∂Q/∂z, ∂P/∂z - ∂R/∂x, ∂Q/∂x - ∂P/∂y) Now, we need to find the curl of curl(F): curl(curl(F)) = (∂(∂Q/∂x - ∂P/∂y)/∂y - ∂(∂P/∂z - ∂R/∂x)/∂z, ∂(∂R/∂x - ∂P/∂z)/∂z - ∂(∂Q/∂x - ∂P/∂y)/∂x, Show more…
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In Exercises $42-45,$ the Laplace operator $\Delta$ is defined by $$\Delta \varphi=\frac{\partial^{2} \varphi}{\partial x^{2}}+\frac{\partial^{2} \varphi}{\partial y^{2}}$$ For any vector field $\mathbf{F}=\left\langle F_{1}, F_{2}\right\rangle,$ define the conjugate vector field $\mathbf{F}^{*}=\left\langle- F_{2}, F_{1}\right\rangle$ Show that if $\mathbf{F}=\nabla \varphi,$ then $\operatorname{curl}_{z}\left(\mathbf{F}^{*}\right)=\Delta \varphi$
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