Seja z=f(x, y). Se x(t, s)=e^s cos t, e y(t, s)=e^s sin t, mostre que
partial^2 zpartial x^2+partial^2 zpartial y^2=e^-2 s[partial^2 zpartial s^2+partial^2 zpartial t^2]

Passo 1

O enunciado define
x = e^scos t, y = e^ssin t
e pede para mostrar que
partial^2zpartial x^2+partial^2zpartial y^2 = e^-2sl[partial^2zpartial s^2+partial^2zpartial t^2r]
O conteúdo é a transformação do laplaciano nabla^2z de coordenadas cartesianas (x,y) para as coordenadas (r,theta) com r=e^s, theta=t.

Passo 2

Em coordenadas polares (r,theta) sabe-se que
partial^2zpartial x^2+partial^2zpartial y^2 = partial^2zpartial r^2 + 1r,partial zpartial r + 1r^2,partial^2zpartial theta^2

Passo 3

Fazendo r=e^s e theta=t, temos
partialpartial r = partial spartial r,partialpartial s = e^-s,partialpartial s
Logo
partial^2partial r^2 = partialpartial rl(e^-spartialpartial sr) = e^-spartialpartial sl(e^-spartialpartial sr) = e^-2sl(partial^2partial s^2 - partialpartial sr)

Passo 4

Também
1r,partialpartial r = e^-s,e^-s,partialpartial s = e^-2s,partialpartial s, 1r^2,partial^2partial theta^2 = e^-2s,partial^2partial t^2

Passo 5

Substituindo na fórmula polar:
partial^2zpartial x^2+partial^2zpartial y^2 = e^-2sl(partial^2zpartial s^2 - partial zpartial sr) + e^-2spartial zpartial s + e^-2spartial^2zpartial t^2 = e^-2sl(partial^2zpartial s^2 + partial^2zpartial t^2r)

Resposta

partial^2zpartial x^2+partial^2zpartial y^2 = e^-2sl[partial^2zpartial s^2+partial^2zpartial t^2r]

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