Let B be the unit ball in $\mathbb{R}$3 centered at origin. The Euler-Lagrange equation corresponding to the functional $I(u)=\frac{1}{2} \int_B|\nabla u|^2 d x-\frac{1}{5} \int_B|u|^5 d x$ is

Question

Prepp Team · Accepted Answer

Euler-Lagrange Equation Derivation

The problem asks for the Euler-Lagrange equation associated with the given functional $I(u)$. The functional is defined over the unit ball $B$ in $\mathbb{R}^3$ and depends on a function $u(x)$ and its gradient $\nabla u(x)$. The functional is: \[ I(u) = \int_B \left( \frac{1}{2} |\nabla u|^2 - \frac{1}{5} |u|^5 \right) dx \] This functional is in the standard form for the Calculus of Variations: $I(u) = \int_B L(x, u, \nabla u) dx$, where the Lagrangian $L$ is given by: \[ L(x, u, \nabla u) = \frac{1}{2} |\nabla u|^2 - \frac{1}{5} |u|^5 \] Here, $|\nabla u|^2 = (\frac{\partial u}{\partial x_1})^2 + (\frac{\partial u}{\partial x_2})^2 + (\frac{\partial u}{\partial x_3})^2$.

The Euler-Lagrange equation for a functional of this type is given by: \[ \frac{\partial L}{\partial u} - \sum_{i=1}^3 \frac{\partial}{\partial x_i} \left( \frac{\partial L}{\partial (\frac{\partial u}{\partial x_i})} \right) = 0 \] We need to compute the two parts of this equation: $\frac{\partial L}{\partial u}$ and $\frac{\partial L}{\partial (\frac{\partial u}{\partial x_i})}$.

Partial Derivatives Calculation

First, let's compute $\frac{\partial L}{\partial u}$. The term $\frac{1}{2} |\nabla u|^2$ does not depend on $u$, only on its derivatives. The term $-\frac{1}{5} |u|^5$ depends on $u$. \[ \frac{\partial}{\partial u} \left( -\frac{1}{5} |u|^5 \right) \] To differentiate $|u|^5$ with respect to $u$, we can write $|u|^5 = (u^2)^{5/2}$. Using the chain rule: \[ \frac{\partial}{\partial u} (u^2)^{5/2} = \frac{5}{2} (u^2)^{3/2} \cdot \frac{\partial}{\partial u}(u^2) = \frac{5}{2} |u|^3 \cdot (2u) = 5 |u|^3 u \] So, \[ \frac{\partial L}{\partial u} = -\frac{1}{5} \cdot 5 |u|^3 u = -|u|^3 u \]

Next, let's compute $\frac{\partial L}{\partial (\frac{\partial u}{\partial x_i})}$ for $i=1, 2, 3$. \[ L = \frac{1}{2} \left( \left(\frac{\partial u}{\partial x_1}\right)^2 + \left(\frac{\partial u}{\partial x_2}\right)^2 + \left(\frac{\partial u}{\partial x_3}\right)^2 \right) - \frac{1}{5} |u|^5 \] The term $-\frac{1}{5} |u|^5$ does not depend on the derivatives of $u$. For $i=1$: \[ \frac{\partial L}{\partial (\frac{\partial u}{\partial x_1})} = \frac{\partial}{\partial (\frac{\partial u}{\partial x_1})} \left( \frac{1}{2} \left(\frac{\partial u}{\partial x_1}\right)^2 \right) = \frac{1}{2} \cdot 2 \left(\frac{\partial u}{\partial x_1}\right) = \frac{\partial u}{\partial x_1} \] Similarly, for $i=2$ and $i=3$: \[ \frac{\partial L}{\partial (\frac{\partial u}{\partial x_2})} = \frac{\partial u}{\partial x_2}, \quad \frac{\partial L}{\partial (\frac{\partial u}{\partial x_3})} = \frac{\partial u}{\partial x_3} \]

Substituting into Euler-Lagrange Equation

Now we substitute these results into the Euler-Lagrange equation: \[ \frac{\partial L}{\partial u} - \sum_{i=1}^3 \frac{\partial}{\partial x_i} \left( \frac{\partial L}{\partial (\frac{\partial u}{\partial x_i})} \right) = 0 \] \[ -|u|^3 u - \left( \frac{\partial}{\partial x_1} \left(\frac{\partial u}{\partial x_1}\right) + \frac{\partial}{\partial x_2} \left(\frac{\partial u}{\partial x_2}\right) + \frac{\partial}{\partial x_3} \left(\frac{\partial u}{\partial x_3}\right) \right) = 0 \] \[ -|u|^3 u - \left( \frac{\partial^2 u}{\partial x_1^2} + \frac{\partial^2 u}{\partial x_2^2} + \frac{\partial^2 u}{\partial x_3^2} \right) = 0 \] The sum of the second partial derivatives is the Laplacian of $u$, denoted by $\Delta u$. \[ \frac{\partial^2 u}{\partial x_1^2} + \frac{\partial^2 u}{\partial x_2^2} + \frac{\partial^2 u}{\partial x_3^2} = \Delta u \] So the equation becomes: \[ -|u|^3 u - \Delta u = 0 \] Rearranging the terms to solve for $\Delta u$: \[ \Delta u = -|u|^3 u \]