Question

Se z = f(x,y) tem derivadas parciais de segunda ordem continuas e satisfaz a equação: ∂2f/( ∂x2)+∂2f/∂y2 = 0 então f é dita harmônica . A função z = In (x2 + y2 ) é harmônica

Answer 1

A função

$z=\mathrm{\ell n}(x^2+y^2)$

será harmônica somente se

$\dfrac{\partial^2 z}{\partial x^2}+\dfrac{\partial^2 z}{\partial y^2}=0$

( equação de Laplace )

_________

Calculando as derivadas de 1ª ordem em relação a $x$ e a $y:$

$\bullet\;\;\dfrac{\partial z}{\partial x}=\dfrac{\partial}{\partial x}\big[\mathrm{\ell n}(x^2+y^2)\big]\\\\\\ =\dfrac{1}{x^2+y^2}\cdot \dfrac{\partial}{\partial x}(x^2+y^2)\\\\\\ =\dfrac{1}{x^2+y^2}\cdot 2x\\\\\\ =\dfrac{2x}{x^2+y^2}$


$\bullet\;\;\dfrac{\partial z}{\partial y}=\dfrac{\partial}{\partial y}\big[\mathrm{\ell n}(x^2+y^2)\big]\\\\\\ =\dfrac{1}{x^2+y^2}\cdot \dfrac{\partial}{\partial y}(x^2+y^2)\\\\\\ =\dfrac{1}{x^2+y^2}\cdot 2y\\\\\\ =\dfrac{2y}{x^2+y^2}$

____________

Calculando as derivadas de 2ª ordem para a equação de Laplace:

$\bullet\;\;\dfrac{\partial^2 z}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial z}{\partial x} \right )\\\\\\ =\dfrac{\partial}{\partial x}\left(\dfrac{2x}{x^2+y^2} \right )\\\\\\ =\dfrac{\frac{\partial}{\partial x}(2x)\cdot (x^2+y^2)-2x\cdot \frac{\partial}{\partial x}(x^2+y^2)}{(x^2+y^2)^2}\\\\\\ =\dfrac{2\cdot (x^2+y^2)-2x\cdot (2x)}{(x^2+y^2)^2}\\\\\\ =\dfrac{2x^2+2y^2-4x^2}{(x^2+y^2)^2}\\\\\\ =\dfrac{-2x^2+2y^2}{(x^2+y^2)^2}$


$\bullet\;\;\dfrac{\partial^2 z}{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{\partial z}{\partial y} \right )\\\\\\ =\dfrac{\partial}{\partial y}\left(\dfrac{2y}{x^2+y^2} \right )\\\\\\ =\dfrac{\frac{\partial}{\partial y}(2y)\cdot (x^2+y^2)-2y\cdot \frac{\partial}{\partial y}(x^2+y^2)}{(x^2+y^2)^2}\\\\\\ =\dfrac{2\cdot (x^2+y^2)-2y\cdot (2y)}{(x^2+y^2)^2}\\\\\\ =\dfrac{2x^2+2y^2-4y^2}{(x^2+y^2)^2}\\\\\\ =\dfrac{2x^2-2y^2}{(x^2+y^2)^2}$

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Verificando se $z$ satisfaz a equação de Laplace:

$\dfrac{\partial^2 z}{\partial x^2}+\dfrac{\partial^2 z}{\partial y^2}\\\\\\ =\dfrac{-2x^2+2y^2}{(x^2+y^2)^2}+\dfrac{2x^2-2y^2}{(x^2+y^2)^2}\\\\\\ =\dfrac{-2x^2+2y^2+2x^2-2y^2}{(x^2+y^2)^2}\\\\\\ =\dfrac{0}{(x^2+y^2)^2}\\\\\\ =0~~~~~~\mathbf{(\checkmark)}$


Portanto,

$z=\mathrm{\ell n}(x^2+y^2)$

é função harmônica.


Bons estudos! :-)