(a) Considerando a continuidade de todas as derivadas parciais, mostre que
\boldsymbol{\nabla} \times(\boldsymbol{\nabla} \times \boldsymbol{F})=-\boldsymbol{\nabla}^{2} \boldsymbol{F}+\operatorname{grad}(\operatorname{div} \mathbf{F})
onde
\nabla^{2} \boldsymbol{F}=\nabla^{2}(P \boldsymbol{i}+Q \boldsymbol{j}+R \boldsymbol{k})=\nabla^{2} P \boldsymbol{i}+\nabla^{2} Q \boldsymbol{j}+\nabla^{2} R \boldsymbol{k}
(b) utilize a identidade do item (a) para obter o resultado no problema 23.
25. Se \mathbf{E}=\mathbf{E} (x, y, z, t) e \mathbf{H}=\mathbf{H} (x, y, z, t) representam os campos elétrico e magnético no espaço livre, então as equações de Maxwell são
\begin{aligned} & \operatorname{div} \boldsymbol{E}=0, \nabla \times \boldsymbol{E}=-\frac{1}{c} \frac{\partial \boldsymbol{H}}{\partial t} \\\ & \operatorname{div} \boldsymbol{H}=0, \nabla \times \boldsymbol{H}=-\frac{1}{c} \frac{\partial \boldsymbol{E}}{\partial t} \end{aligned}
onde c é a velocidade da luz. Use a identidade do problema 24 para mostrar que \boldsymbol{E} e \boldsymbol{H} satisfazem
(a) \quad \nabla^{2} \boldsymbol{E}=-\frac{1}{c^{2}} \frac{\partial^{2} \boldsymbol{E}}{\partial t^{2}}
(b) \quad \nabla^{2} \boldsymbol{H}=-\frac{1}{c^{2}} \frac{\partial^{2} \boldsymbol{H}}{\partial t^{2}}

Passo 1

O enunciado nos pede:
(a) provar a identidade vetorial
\nabla\times(\nabla\times \boldsymbol{F}) =-\nabla^{2}\boldsymbol{F}+\nabla(\nabla\cdot\boldsymbol{F})
para \boldsymbol{F}=P\,\boldsymbol{i}+Q\,\boldsymbol{j}+R\,\boldsymbol{k} com derivadas parciais contínuas;
(b) usar essa identidade nas equações de Maxwell no vácuo
\begin{aligned} \nabla\cdot \boldsymbol{E}&=0, &\nabla\times \boldsymbol{E}&=-\tfrac{1}{c}\,\frac{\partial \boldsymbol{H}}{\partial t},\\\ \nabla\cdot \boldsymbol{H}&=0, &\nabla\times \boldsymbol{H}&=-\tfrac{1}{c}\,\frac{\partial \boldsymbol{E}}{\partial t} \end{aligned}
para deduzir as equações de onda
\nabla^{2}\boldsymbol{E}=-\tfrac{1}{c^{2}}\frac{\partial^{2}\boldsymbol{E}}{\partial t^{2}}, \quad \nabla^{2}\boldsymbol{H}=-\tfrac{1}{c^{2}}\frac{\partial^{2}\boldsymbol{H}}{\partial t^{2}}.
O conteúdo é cálculo vetorial componente a componente e aplicação a PDEs.

Passo 2

Definimos
\boldsymbol{F}=P\,\boldsymbol{i}+Q\,\boldsymbol{j}+R\,\boldsymbol{k},
\nabla\cdot\boldsymbol{F} =\frac{\partial P}{\partial x} +\frac{\partial Q}{\partial y} +\frac{\partial R}{\partial z},
\nabla\times\boldsymbol{F} =\begin{vmatrix} \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\\ \partial/\partial x&\partial/\partial y&\partial/\partial z\\\ P&Q&R \end{vmatrix},
\nabla^{2}\boldsymbol{F} =(\nabla^{2}P)\,\boldsymbol{i}+(\nabla^{2}Q)\,\boldsymbol{j}+(\nabla^{2}R)\,\boldsymbol{k},
com \nabla^{2}=\partial^{2}/\partial x^{2}+\partial^{2}/\partial y^{2}+\partial^{2}/\partial z^{2}.

Passo 3

Vamos calcular a componente x de \nabla\times(\nabla\times\boldsymbol{F}):
(\nabla\times(\nabla\times \boldsymbol{F}))_{x} =\frac{\partial}{\partial y}\bigl((\nabla\times\boldsymbol{F})_{z}\bigr) -\frac{\partial}{\partial z}\bigl((\nabla\times\boldsymbol{F})_{y}\bigr)
=\frac{\partial}{\partial y}\Bigl(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Bigr) -\frac{\partial}{\partial z}\Bigl(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\Bigr)
=\frac{\partial^{2}Q}{\partial x\partial y} -\frac{\partial^{2}P}{\partial y^{2}} -\frac{\partial^{2}P}{\partial z^{2}} +\frac{\partial^{2}R}{\partial z\partial x}.
Por outro lado,
\frac{\partial}{\partial x}(\nabla\cdot\boldsymbol{F}) =\frac{\partial^{2}P}{\partial x^{2}} +\frac{\partial^{2}Q}{\partial y\partial x} +\frac{\partial^{2}R}{\partial z\partial x}.
Logo
(\nabla\times(\nabla\times \boldsymbol{F}))_{x} =\frac{\partial}{\partial x}(\nabla\cdot\boldsymbol{F}) -\Bigl(\frac{\partial^{2}P}{\partial x^{2}} +\frac{\partial^{2}P}{\partial y^{2}} +\frac{\partial^{2}P}{\partial z^{2}}\Bigr)
=(\nabla(\nabla\cdot\boldsymbol{F}))_{x} -(\nabla^{2}\boldsymbol{F})_{x}.
Análogas contas para as componentes y e z dão o mesmo padrão, estabelecendo
\nabla\times(\nabla\times \boldsymbol{F}) =-\nabla^{2}\boldsymbol{F}+\nabla(\nabla\cdot\boldsymbol{F}).

Passo 4

Aplicamos agora a \boldsymbol{E}. Da identidade temos
\nabla\times(\nabla\times \boldsymbol{E}) =-\nabla^{2}\boldsymbol{E}+\nabla(\nabla\cdot\boldsymbol{E}).
Mas \nabla\cdot\boldsymbol{E}=0 pelas equações de Maxwell, logo
\nabla\times(\nabla\times \boldsymbol{E}) =-\nabla^{2}\boldsymbol{E}.
Por outro lado \nabla\times\boldsymbol{E}=-\tfrac{1}{c}\,\partial\boldsymbol{H}/\partial t, então
\nabla\times(\nabla\times \boldsymbol{E}) =-\frac{1}{c}\,\frac{\partial}{\partial t}(\nabla\times\boldsymbol{H}) =-\frac{1}{c}\,\frac{\partial}{\partial t}\Bigl(\tfrac{1}{c}\,\frac{\partial\boldsymbol{E}}{\partial t}\Bigr) =-\frac{1}{c^{2}}\frac{\partial^{2}\boldsymbol{E}}{\partial t^{2}}.
Comparando com -\nabla^{2}\boldsymbol{E} obtemos
\nabla^{2}\boldsymbol{E} =-\frac{1}{c^{2}}\frac{\partial^{2}\boldsymbol{E}}{\partial t^{2}}.

Passo 5

Para \boldsymbol{H} o raciocínio é idêntico: \nabla\cdot\boldsymbol{H}=0 e \nabla\times\boldsymbol{H}=-\tfrac{1}{c}\,\partial\boldsymbol{E}/\partial t
implicam
\nabla^{2}\boldsymbol{H} =-\frac{1}{c^{2}}\frac{\partial^{2}\boldsymbol{H}}{\partial t^{2}}.

Resposta

\nabla\times(\nabla\times \boldsymbol{F}) =-\nabla^{2}\boldsymbol{F} +\nabla(\nabla\cdot\boldsymbol{F}),
\nabla^{2}\boldsymbol{E} =-\frac{1}{c^{2}}\frac{\partial^{2}\boldsymbol{E}}{\partial t^{2}}, \quad \nabla^{2}\boldsymbol{H} =-\frac{1}{c^{2}}\frac{\partial^{2}\boldsymbol{H}}{\partial t^{2}}.

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