HA10-P4

[Victor Ivrii]

Problem 4 http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter10/S10.P.html#problem-10.P.4

[Xi Yue Wang]

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...

[Victor Ivrii]

Ouch! What are? $L_{u_x}$ and $L_{u_y}$?

[Xi Yue Wang]

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}

[Victor Ivrii]

(\ref{K}) follows from (\ref{L}).

[Xi Yue Wang]

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 $$-f - \frac{\partial}{\partial x}(\frac{ku_x}{\sqrt{1+u_x^2+u_y^2}}) - \frac{\partial}{\partial y}(\frac{ku_y}{\sqrt{1+u_x^2+u_y^2}}) = 0$$