Difference between revisions of "Template:Energy contour and preliminaries"

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Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
 
Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
 
<center><math>
 
<center><math>
{  \int\int_\mathcal{U}\left(\phi^*\nabla^2\phi - \phi\nabla^2\phi^* \right)dxdz
+
{  \int\int_\mathcal{U}\left(\phi^*\nabla^2\phi - \phi\nabla^2\phi^* \right)\mathrm{d}x\mathrm{d}z
= \int_\mathcal{S}\left(\phi^*\frac{\partial\phi}{\partial n} - \phi\frac{\partial\phi^*}{\partial n} \right)dl },
+
= \int_\mathcal{S}\left(\phi^*\frac{\partial\phi}{\partial n} - \phi\frac{\partial\phi^*}{\partial n} \right)\mathrm{d}l },
 
</math></center>
 
</math></center>
 
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
 
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
 
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to  
 
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to  
 
<center><math>
 
<center><math>
  \Im\int_\mathcal{S}\phi^*\frac{\partial\phi}{\partial n} dl =  0,
+
  \Im\int_\mathcal{S}\phi^*\frac{\partial\phi}{\partial n} \mathrm{d}l =  0,
 
</math></center>
 
</math></center>

Revision as of 07:29, 24 February 2009

Based on the method used in Evans and Davies 1968, a check can be made to ensure the solutions energy balance. The energy balance equation is derived by applying Green's theorem to [math]\displaystyle{ \phi }[/math] and its conjugate. The domain of integration is shown in the figure on the right. We assume that the angle is sufficiently small that we do not get total reflection.

A diagram depicting the area [math]\displaystyle{ \mathcal{U} }[/math] which is bounded by the rectangle [math]\displaystyle{ \mathcal{S} }[/math].] {A diagram depicting the area [math]\displaystyle{ \mathcal{U} }[/math] which is bounded by the rectangle [math]\displaystyle{ \mathcal{S} }[/math]. The rectangle [math]\displaystyle{ \mathcal{S} }[/math] is bounded by [math]\displaystyle{ -h\leq z \leq0 }[/math] and [math]\displaystyle{ -\infty\leq x \leq \infty }[/math]

Applying Green's theorem to [math]\displaystyle{ \phi }[/math] and its conjugate [math]\displaystyle{ \phi^* }[/math] gives

[math]\displaystyle{ { \int\int_\mathcal{U}\left(\phi^*\nabla^2\phi - \phi\nabla^2\phi^* \right)\mathrm{d}x\mathrm{d}z = \int_\mathcal{S}\left(\phi^*\frac{\partial\phi}{\partial n} - \phi\frac{\partial\phi^*}{\partial n} \right)\mathrm{d}l }, }[/math]

where [math]\displaystyle{ n }[/math] denotes the outward plane normal to the boundary and [math]\displaystyle{ l }[/math] denotes the plane parallel to the boundary. As [math]\displaystyle{ \phi }[/math] and [math]\displaystyle{ \phi^* }[/math] satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to

[math]\displaystyle{ \Im\int_\mathcal{S}\phi^*\frac{\partial\phi}{\partial n} \mathrm{d}l = 0, }[/math]