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#1 malina

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Napisano 02.12.2011 - 20:22

Pewnie robię gdzieś prosty błąd, ale od godziny nie mogę go znaleźć :o Zadanko jest takie:
Funkcja F spełnia warunek \frac{\partial^2F}{\partial x^2}+\frac{\partial^2 F}{\partial y^2}=0. Pokaż, że funkcja
 f(u,v)=F(u^2-v^2,2uv) spełnia warunek:
\frac{\partial^2f}{\partial u^2}+\frac{\partial^2 f}{\partial v^2}=0

Pochodne wyszły mi takie:
\frac{\partial F}{\partial u}= \frac{\partial f}{\partial x}\cdot 2u+\frac{\partial f}{\partial y}\cdot 2v
\frac{\partial F}{\partial v}=\frac{\partial f}{\partial x}\cdot(-2v)+\frac{\partial f}{\partial y}\cdot 2u
\frac{\partial ^2F}{\partial u^2}=\frac{\partial ^2f}{\partial x^2}\cdot 2u+2\frac{\partial f}{\partial x}+\frac{\partial^2f}{\partial y^2}\cdot 2v+2\frac{\partial f}{\partial y}
\frac{\partial ^2F}{\partial v^2}=\frac{\partial^2 f}{\partial x^2}\cdot (-2v)+\frac{\partial f}{\partial x}\cdot(-2)+\frac{\partial ^2f}{\partial y^2}\cdot 2u+2\frac{\partial f}{\partial y}

Po dodaniu tego do siebie wcale nie wychodzi 0 :(
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Napisano 25.09.2011 - 17:55

#2 octahedron

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Napisano 03.12.2011 - 14:52

<br />\\x=u^2-v^2<br />\\y=2uv<br />\\ <br />\\\frac{\partial f}{\partial u}=\frac{\partial F}{\partial x}\cdot\frac{\partial x}{\partial u}+\frac{\partial F}{\partial y}\cdot\frac{\partial y}{\partial u}=\frac{\partial F}{\partial x}\cdot 2u+\frac{\partial F}{\partial y}\cdot 2v<br />\\\frac{\partial f}{\partial v}=\frac{\partial F}{\partial x}\cdot\frac{\partial x}{\partial v}+\frac{\partial F}{\partial y}\cdot\frac{\partial y}{\partial v}=-\frac{\partial F}{\partial x}\cdot 2v+\frac{\partial F}{\partial y}\cdot 2u<br />\\ <br />\\\frac{\partial^2 f}{\partial u^2}=\frac{\partial }{\partial u}\(\frac{\partial F}{\partial x}\cdot 2u+\frac{\partial F}{\partial y}\cdot 2v\)=\(\frac{\partial^2 F}{\partial x^2}\cdot\frac{\partial x}{\partial u}+\frac{\partial^2 F}{\partial x\partial y}\cdot\frac{\partial y}{\partial u}\)\cdot 2u+2\cdot\frac{\partial F}{\partial x}+\(\frac{\partial^2 F}{\partial y\partial x}\cdot\frac{\partial x}{\partial u}+\frac{\partial^2 F}{\partial y^2}\cdot\frac{\partial y}{\partial u}\)\cdot 2v=<br />\\=\(\frac{\partial^2 F}{\partial x^2}\cdot 2u+\frac{\partial^2 F}{\partial x\partial y}\cdot 2v\)\cdot 2u+2\cdot\frac{\partial F}{\partial x}+\(\frac{\partial^2 F}{\partial y\partial x}\cdot 2u+\frac{\partial^2 F}{\partial y^2}\cdot 2v\)\cdot 2v=<br />\\=4u^2\cdot\frac{\partial^2 F}{\partial x^2}+4uv\(\frac{\partial^2 F}{\partial x\partial y}+\frac{\partial^2 F}{\partial y\partial x}\)+2\cdot\frac{\partial F}{\partial x}+4v^2\cdot\frac{\partial^2 F}{\partial y^2}<br />\\ <br />\\ <br />\\\frac{\partial^2 f}{\partial v^2}=\frac{\partial }{\partial v}\(-\frac{\partial F}{\partial x}\cdot 2v+\frac{\partial F}{\partial y}\cdot 2u\)=-\(\frac{\partial^2 F}{\partial x^2}\cdot\frac{\partial x}{\partial v}+\frac{\partial^2 F}{\partial x\partial y}\cdot\frac{\partial y}{\partial v}\)\cdot 2v-2\cdot\frac{\partial F}{\partial x}+\(\frac{\partial^2 F}{\partial y\partial x}\cdot\frac{\partial x}{\partial v}+\frac{\partial^2 F}{\partial y^2}\cdot\frac{\partial y}{\partial v}\)\cdot 2u=<br />\\=-\(-\frac{\partial^2 F}{\partial x^2}\cdot 2v+\frac{\partial^2 F}{\partial x\partial y}\cdot 2u\)\cdot 2v-2\cdot\frac{\partial F}{\partial x}+\(-\frac{\partial^2 F}{\partial y\partial x}\cdot 2v+\frac{\partial^2 F}{\partial y^2}\cdot 2u\)\cdot 2u=<br />\\=4v^2\cdot\frac{\partial^2 F}{\partial x^2}-4uv\(\frac{\partial^2 F}{\partial x\partial y}+\frac{\partial^2 F}{\partial y\partial x}\)-2\cdot\frac{\partial F}{\partial x}+4u^2\cdot\frac{\partial^2 F}{\partial y^2}<br />\\ <br />\\\frac{\partial^2 f}{\partial u^2}+\frac{\partial^2 f}{\partial v^2}=4(u^2+v^2)\(\frac{\partial^2 F}{\partial x^2}+\frac{\partial^2 F}{\partial y^2}\)=0<br />\\
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