Euler-Lagrange Equation Derivation

The problem asks for the Euler-Lagrange equation associated with the functional \(I\left( u \right) = \int_B {{{\left( {1 + {{\left| {\triangledown u} \right|}^2}} \right)}^{\frac{1}{2}}}dx}\). This functional is related to finding surfaces with minimal area, where the integrand represents the element of surface area.

Lagrangian Identification

The functional is given in the form \(I(u) = \int_\Omega L(x, u, \nabla u) dx\). In this case, the domain \(\Omega\) is the unit ball \(B \subset \mathbb{R}^3\), and the integrand (Lagrangian) is:

\(L(x, u, \nabla u) = {\left( {1 + {{\left| {\triangledown u} \right|}^2}} \right)}^{\frac{1}{2}}\)

Notice that the Lagrangian \(L\) does not explicitly depend on the spatial variable \(x\) or the function \(u\), only on its gradient \(\nabla u\).

General Euler-Lagrange Equation

For a functional of the form \(I(u) = \int_\Omega L(x, u, \nabla u) dx\), the Euler-Lagrange equation is given by:

\(\frac{\partial L}{\partial u} - \sum_{i=1}^n \frac{\partial}{\partial x_i} \left( \frac{\partial L}{\partial u_{x_i}} \right) = 0\)

Here, \(n=3\) as the domain is in \(\mathbb{R}^3\), and \(u_{x_i} = \frac{\partial u}{\partial x_i}\).

Simplifying the Equation

Since \(L\) does not depend on \(u\), \(\frac{\partial L}{\partial u} = 0\). The Euler-Lagrange equation simplifies to:

\(\sum_{i=1}^3 \frac{\partial}{\partial x_i} \left( \frac{\partial L}{\partial u_{x_i}} \right) = 0\)

This can be written in vector notation as \(div(\nabla_{\nabla u} L) = 0\), where \(\nabla_{\nabla u} L\) is the gradient of \(L\) with respect to the components of \(\nabla u\).

Calculating Partial Derivatives of L

Let's calculate the partial derivatives of \(L\) with respect to \(u_{x_i}\). Let \(p_i = u_{x_i}\). Then \(L = (1 + p_1^2 + p_2^2 + p_3^2)^{1/2}\). The square of the gradient magnitude is \(|\nabla u|^2 = \sum_{j=1}^3 u_{x_j}^2\).

\(\frac{\partial L}{\partial u_{x_i}} = \frac{\partial}{\partial p_i} {\left( {1 + \sum_{j=1}^3 p_j^2} \right)}^{\frac{1}{2}}\)

Using the chain rule:

\(\frac{\partial L}{\partial u_{x_i}} = \frac{1}{2} {\left( {1 + \sum_{j=1}^3 p_j^2} \right)}^{-\frac{1}{2}} \cdot \frac{\partial}{\partial p_i} \left( {1 + \sum_{j=1}^3 p_j^2} \right)\)

\(\frac{\partial L}{\partial u_{x_i}} = \frac{1}{2} {\left( {1 + |\nabla u|^2} \right)}^{-\frac{1}{2}} \cdot (2 p_i)\)

\(\frac{\partial L}{\partial u_{x_i}} = \frac{p_i}{{\left( {1 + |\nabla u|^2} \right)}^{\frac{1}{2}}} = \frac{u_{x_i}}{{\left( {1 + |\nabla u|^2} \right)}^{\frac{1}{2}}}\)

Form 1 of the Euler-Lagrange Equation

Substituting this back into the simplified Euler-Lagrange equation \(\sum_{i=1}^3 \frac{\partial}{\partial x_i} \left( \frac{\partial L}{\partial u_{x_i}} \right) = 0\), we get:

\(\sum_{i=1}^3 \frac{\partial}{\partial x_i} \left( \frac{u_{x_i}}{{\left( {1 + |\nabla u|^2} \right)}^{\frac{1}{2}}} \right) = 0\)

This is exactly the divergence form:

\(div\left( {\frac{{\nabla u}}{{{{\left( {1 + {{\left| {\nabla u} \right|}^2}} \right)}^{\frac{1}{2}}}}}} \right) = 0\)

This matches option 1.

Form 2 of the Euler-Lagrange Equation

We can expand the divergence form using the product rule. Let \(w = {\left( {1 + |\nabla u|^2} \right)}^{-\frac{1}{2}}\). The equation is \(\sum_{i=1}^3 \frac{\partial}{\partial x_i} (w u_{x_i}) = 0\).

\(\sum_{i=1}^3 \left( \frac{\partial w}{\partial x_i} u_{x_i} + w \frac{\partial u_{x_i}}{\partial x_i} \right) = 0\)

\(\sum_{i=1}^3 \left( \frac{\partial w}{\partial x_i} u_{x_i} \right) + w \sum_{i=1}^3 u_{x_i x_i} = 0\)

The second term is \(w \Delta u\). Now let's calculate \(\frac{\partial w}{\partial x_i}\). \(w = (1 + |\nabla u|^2)^{-1/2}\)

\(\frac{\partial w}{\partial x_i} = -\frac{1}{2} (1 + |\nabla u|^2)^{-3/2} \frac{\partial}{\partial x_i} (1 + |\nabla u|^2)\)

\(\frac{\partial}{\partial x_i} (1 + |\nabla u|^2) = \frac{\partial}{\partial x_i} \left( \sum_{j=1}^3 u_{x_j}^2 \right) = \sum_{j=1}^3 2 u_{x_j} \frac{\partial u_{x_j}}{\partial x_i} = 2 \sum_{j=1}^3 u_{x_j} u_{x_j x_i}\)

So, \(\frac{\partial w}{\partial x_i} = -\frac{1}{2} (1 + |\nabla u|^2)^{-3/2} \left( 2 \sum_{j=1}^3 u_{x_j} u_{x_j x_i} \right) = - (1 + |\nabla u|^2)^{-3/2} \sum_{j=1}^3 u_{x_j} u_{x_j x_i}\).

Substitute this back into the expanded equation:

\(\sum_{i=1}^3 \left( - (1 + |\nabla u|^2)^{-3/2} \sum_{j=1}^3 u_{x_j} u_{x_j x_i} \right) u_{x_i} + (1 + |\nabla u|^2)^{-1/2} \Delta u = 0\)

Multiply by \((1 + |\nabla u|^2)^{3/2}\):

\(- \sum_{i=1}^3 \sum_{j=1}^3 u_{x_i} u_{x_j} u_{x_j x_i} + (1 + |\nabla u|^2) \Delta u = 0\)

Rearranging the terms and swapping dummy indices \(i\) and \(j\) in the sum \(\sum_{i=1}^3 \sum_{j=1}^3 u_{x_i} u_{x_j} u_{x_j x_i}\) to match the form in option 4 (\(\sum_{i=1}^3 \sum_{j=1}^3 u_{x_j} u_{x_i} u_{x_i x_j}\)), which is the same sum with swapped indices, we get:

\((1 + |\nabla u|^2) \Delta u = \sum_{i=1}^3 \sum_{j=1}^3 u_{x_i} u_{x_j} u_{x_j x_i}\)

This can also be written as:

\(\left( {1 + {{\left| {\nabla u} \right|}^2}} \right)\Delta u = \sum\nolimits_{i,j = 1}^3 {{u_{xi}}{u_{xj}}{u_{xi,xj}}}\)

This matches option 4.

Conclusion

Both option 1 and option 4 represent the Euler-Lagrange equation for the given functional. The first form is the standard divergence form, and the second form is an expansion of the first one, often referred to as the minimal surface equation in non-parametric form.