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Chapter 6 / Re: Laplace Equation Section 6.4
« on: March 19, 2020, 07:45:52 PM »
I understand now, there is another constraint $$\Theta'(0)=\Theta'(2\pi)$$, but what does this condition mean physically?
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Chapter 6 / Laplace Equation Section 6.4
« on: March 19, 2020, 07:22:02 PM »
In section 6.4, the topic about separation of variables.
I don't exactly understand why the constraint $$\Theta(0)=\Theta(2\pi)$$ gives the Eigen functions $\Theta_{s,n}(\theta)=sin(n\pi \theta)$and $\Theta_{c,n}(\theta)=cos(n\pi \theta)$. Shouldn't the condition produce the Eigen functions $\Theta_{s,n}(\theta)=sin(\frac{n \pi \theta}{2})$and $\Theta_{c,n}(\theta)=cos(\frac{n\pi \theta}{2})$ instead?
Since previously a similar problem $$\Theta'' + \lambda \Theta = 0 $$ $$\Theta(0)=\Theta(\alpha)$$ give the Eigen functions $sin(\frac{n\pi \theta}{\alpha} )$ $cos(\frac{n\pi \theta}{\alpha} )$ and the Eigenvalue$\lambda_n = (\frac{n \pi}{\alpha})$
P.S. I couldn't get the Quercus discussion to work
I don't exactly understand why the constraint $$\Theta(0)=\Theta(2\pi)$$ gives the Eigen functions $\Theta_{s,n}(\theta)=sin(n\pi \theta)$and $\Theta_{c,n}(\theta)=cos(n\pi \theta)$. Shouldn't the condition produce the Eigen functions $\Theta_{s,n}(\theta)=sin(\frac{n \pi \theta}{2})$and $\Theta_{c,n}(\theta)=cos(\frac{n\pi \theta}{2})$ instead?
Since previously a similar problem $$\Theta'' + \lambda \Theta = 0 $$ $$\Theta(0)=\Theta(\alpha)$$ give the Eigen functions $sin(\frac{n\pi \theta}{\alpha} )$ $cos(\frac{n\pi \theta}{\alpha} )$ and the Eigenvalue$\lambda_n = (\frac{n \pi}{\alpha})$
P.S. I couldn't get the Quercus discussion to work
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