Analiza matematyczna, zadanie nr 3761

Autor	Zadanie / Rozwiązanie
magda2219 postów: 19	2015-11-03 18:38:33 Wykazać, że równanie Laplace'a w R^2 postaci: (delta^2u)/(deltax^2)+(delta^2u)/(deltay^2)=0 we współrzednych biegunowych x=rcosfi y=rsinfi wyraza sie wzorem: (delta^2u)/(deltar^2)+(1/r^2)(delta^2u)/(deltay^2)+(1/r)(deltau)/(deltar)=0

janusz78 postów: 820	2015-11-04 22:00:20 $\frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial y^2}=0$ $ x =r\cos(\phi), y = r\sin(\phi)$ $ r^2 = x^2 +y^2, \ \ \tan(\phi) = \frac{y}{x}$ $ r = \sqrt{x^2+y^2},\ \ \phi = \arctan(\frac{y}{x}),$ $\frac{\partial r}{\partial x}= \frac{x}{\sqrt{x^2+y^2}}= \frac{r\cos(\phi)}{r}= \cos(\phi).$ $\frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial y^2}=0$ $ x =r\cos(\phi), y = r\sin(\phi)$ $ r^2 = x^2 +y^2, \ \ \tan(\phi) = \frac{y}{x}$ $ r = \sqrt{x^2+y^2},\ \ \phi = \arctan(\frac{y}{x}),$ $\frac{\partial r}{\partial x}= \frac{x}{\sqrt{x^2+y^2}}= \frac{r\cos(\phi)}{r}= \cos(\phi).$ $\frac{\partial r}{\partial y}= \frac{y}{\sqrt{x^2+y^2}}= \frac{r\cos(\phi)}{r}= \cos(\phi).$ $ \frac{\partial \phi}{\partial x}= \frac{1}{1+(\frac{y}{x})^2}(-\frac{y}{x^2})=\frac{-y}{x^2+y^2}= -\frac{\sin(\phi)}{r}.$ $ \frac{\partial \phi}{\partial y}= \frac{1}{1+(\frac{y}{x})^2}(\frac{1}{x})=\frac{x}{x^2+y^2}= \frac{\cos(\phi)}{r}.$ Ze wzorów na pochodne cząstkowe funkcji złożonej $\frac{\partial u}{\partial x}= \frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial u}{\partial \phi}\frac{\partial \phi}{\partial x},$ $\frac{\partial u}{\partial x}= \cos(\phi)\frac{\partial u}{\partial r} - \frac{\sin(\phi)}{r}\frac{\partial u}{\partial \phi}$ (1) $\frac{\partial^2 u}{\partial x^2}= \frac{\partial}{\partial x}(\frac{\partial u}{\partial x})= $ $\cos(\phi)\frac{\partial}{\partial r}(\frac{\partial u}{\partial x})- \frac{\sin(\phi)}{r}\frac{\partial}{\partial \phi}(\frac{\partial u}{\partial x}).$ Stosując ponownie wzór (1) $\frac{\partial^2 u}{\partial x^2}=\frac{\partial}{\partial r}\left(\cos(\phi)\frac{\partial u}{\partial r} - \frac{\sin(\phi)}{r}\frac{\partial u}{\partial \phi}\right)- \frac{\sin(\phi)}{r}\frac{\partial}{\partial \phi}\left(\cos(\phi)\frac{\partial u}{\partial r} - \frac{\sin(\phi)}{r}\frac{\partial u}{\partial \phi}\right)= cos^2(\phi) \frac{\partial^2 u}{\partial r^2}-cos(\phi)\left(-\frac{\sin(\phi)}{r^2}\frac{\partial u}{\partial \phi} +\frac{\sin(\phi)}{r}\frac{\partial^2 u}{\partial r \partial \phi}\right)+$ $-\frac{\sin(\phi)}{r} \left(-\sin(\phi)\frac{\partial u}{\partial r}+\cos(\phi)\frac{\partial^2 u}{\partial \phi \partial r}\right)+ \frac{\sin(\phi)}{r}\left(\frac{\cos(\phi)}{r}\frac{\partial u}{\partial \phi} +\frac{\sin(\phi)}{r}\frac{\partial^2 u}{\partial \phi^2}\right).$ Uwzględniając równość pochodnych cząstkowych mieszanych otrzymujemy $\frac{\partial^2 u}{\partial x^2}= \cos^2(\phi)\frac{\partial^2 u}{\partial r^2}-\frac{2\sin(\phi)\cos(\phi)}{r}\frac{\partial^2 u}{\partial r \partial \phi}+ \frac{\sin^2(\phi)}{r^2}\frac{\partial u^2}{\partial \phi^2}+\frac{\sin^2(\phi)}{r}\frac{\partial u}{\partial r}+ \frac{2\sin(\phi)\cos(\phi)}{r^2}\frac{\partial u}{\partial \phi}$ (2) Podobnie obliczamy pochodną cząstkową $\frac{\partial^2 u}{\partial y^2}.$ $\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial u}{\partial \phi}\frac{\partial \phi}{\partial y}.$ $\frac{\partial u}{\partial y}=\sin(\phi)\frac{\partial u}{\partial r}+ \frac{\cos(\phi)}{r}\frac{\partial u}{\partial \phi}.$ Postepujemy tak jak wyżej tzn. podstawiamy w miejsce $u, \ \ \frac{\partial u}{\partial y}$ i po uporządkowaniu składników sumy otrzymujemy ostatecznie $\frac{\partial^2 u}{\partial y^2}= \sin^2(\phi)\frac{\partial^2 u}{\partial r^2}+\frac{2\sin(\phi)\cos(\phi)}{r}\frac{\partial^2 u}{\partial r \partial \phi}+ \frac{\cos^2(\phi)}{r^2}\frac{\partial u^2}{\partial \phi^2}+\frac{\cos^2(\phi)}{r}\frac{\partial u}{\partial r}- \frac{2\sin(\phi)\cos(\phi)}{r^2}\frac{\partial u}{\partial \phi}$ (3) Po dodaniu stronami (2) i (3) $\frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial y^2}= (\cos^2(\phi)+\sin^2(\phi))\frac{\partial^2 u}{\partial r^2}+ \frac{\sin^2(\phi)+\cos^2(\phi)}{r^2}\frac{\partial^2 u}{\partial \phi^2}+ \frac{\sin^2(\phi)+\cos^2(\phi)}{r}\frac{\partial u}{\partial r}= \frac{\partial^2 u}{\partial r^2} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \phi^2} + \frac{1}{r}\frac{\partial u}{\partial r} =0.$

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