Author Topic: HA10-P4  (Read 791 times)

Victor Ivrii

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Xi Yue Wang

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Re: HA10-P4
« Reply #1 on: December 02, 2015, 12:32:03 AM »
For Part a),
     The Euler-Lagrange PDE of
$S=\iint_D \sqrt{1+u_x^2+u_y^2} dxdy$ is
\begin{gather}
-\frac{\partial }{\partial x}(u_x \sqrt{1+u_x^2+u_y^2}) - \frac{\partial }{\partial y}(u_y \sqrt{1+u_x^2+u_y^2}) = 0\\(1+u_x^2)u_{yy}-2u_xu_yu_{xy}+(1+u_y^2)u_{xx} = 0\label{K}\end{gather}
Given the boundary condition is $u(x,y)|_\Gamma = \phi(x,y)$.
And...
« Last Edit: December 03, 2015, 07:40:00 AM by Victor Ivrii »

Victor Ivrii

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Re: HA10-P4
« Reply #2 on: December 02, 2015, 12:27:08 PM »
Ouch! What are? $L_{u_x}$ and $L_{u_y}$?

Xi Yue Wang

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Re: HA10-P4
« Reply #3 on: December 02, 2015, 09:38:19 PM »
Oh, $$L= \sqrt{1+u_x^2+u_y^2}\\L_{u_x} = \frac{u_x}{\sqrt{1+u_x^2+u_y^2}},\ L _{u_y} = \frac{u_y}{\sqrt{1+u_x^2+u_y^2}}$$
The Euler-Lagrange PDE of S is
\begin{equation}
-\frac{\partial}{\partial x}(\frac{u_x}{\sqrt{1+u_x^2+u_y^2}}) - \frac{\partial}{\partial y}(\frac{u_y}{\sqrt{1+u_x^2+u_y^2}}) = 0\label{L}
\end{equation}
« Last Edit: December 03, 2015, 04:56:14 PM by Xi Yue Wang »

Victor Ivrii

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Re: HA10-P4
« Reply #4 on: December 03, 2015, 07:41:20 AM »
(\ref{K}) follows from (\ref{L}).

Xi Yue Wang

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Re: HA10-P4
« Reply #5 on: December 03, 2015, 04:55:24 PM »
For part b), (...not sure)
     Given $$E = k\iint_D \sqrt{1+u_x^2+u_y^2} dx dy - \iint_D fu dxdy\\=\iint_D (k\sqrt{1+u_x^2+u_y^2} - fu) dx dy$$
Then we have, $$L=k\sqrt{1+u_x^2+u_y^2} - fu\\L_{u_x} = \frac{ku_x}{\sqrt{1+u_x^2+u_y^2}},\ L_{u_y} = \frac{ku_y}{\sqrt{1+u_x^2+u_y^2}},\ L_u = -f$$
Then, the Euler-Lagrange is $$-f - \frac{\partial}{\partial x}(\frac{ku_x}{\sqrt{1+u_x^2+u_y^2}}) - \frac{\partial}{\partial y}(\frac{ku_y}{\sqrt{1+u_x^2+u_y^2}}) = 0$$
« Last Edit: December 03, 2015, 05:00:30 PM by Xi Yue Wang »