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    • Laplace's Equation and Steady State

      SighHillDoveN SighHillDoveN

      Find the solution to the following system consisting of Laplace's equation with boundary conditions:

      $\nabla^2u=0, u(0)=-1,u(2)=3$

      Your solution should be quite a simple algebraic expression. Graph the solution and interpret it in the context of steady state heat flow.

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    • Exam Question

      mega_late mega_late

      Check it 1,2,
      recall that azimuthal projection satisfy the equation $T(\phi,\theta)=(r(\phi)cos(\theta),r(\phi)sin(\theta))$. Then $T_{\phi}=((r'(\phi)cos(\theta),r'(\phi)sin(\theta))$ and $T_{\theta}=((r(\phi)(-sin(\theta)),r(\phi)cos(\theta))$. Then $T_{\phi} \cdot T_{\theta} = ((r'(\phi)cos(\theta),r'(\phi)sin(\theta)) \cdot ((r(\phi)(-sin(\theta)),r(\phi)cos(\theta)) = (r'(\phi)cos(\theta))(r(\phi)(-sin(\theta))) + (r'(\phi)sin(\theta))(r(\phi)cos(\theta)) = -(r'(\phi)cos(\theta))(r(\phi)(sin(\theta))) + (r'(\phi)sin(\theta))(r(\phi)cos(\theta)) = 0$

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    • Green's Theorem #3

      Red Red

      So basically we are given $$\int_C F\cdot dn=\int \int divF=\int \int 2x+3y$$ It is a lot easier to do this in polar coordinates so we have $x=0.1cos(\theta)$ and $y=0.1sin(\theta)$ thus $divF=0.2cos(\theta)+0.3sin(\theta)$. This just becomes a matter of integration so:

      $\int_C F\cdot dn= \int_{0}^{2\pi} \int_{0}^{1} 0.2cos(\theta)+0.3sin(\theta) \space drd\theta=$

      $\int_{0}^{2\pi} 0.2cos(\theta)+0.3sin(\theta) \space d\theta=0.2 sin(\theta)-0.3cos(\theta)=0$

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