Question on Calculus of Variation: Euler-Lagrange equation without integral constraint 
Let $\;y:U \to \mathbb R^m\;$ where $\;U\subset \mathbb R^n\;$ and
  $\;n,m \gt 1\;$. Find the Euler-Lagrange equation of the following
  functional :
$\;J(y)=\frac{1}{2} \int_{U} {\vert Dy \vert}^2\;dx\;$
subject to the pointwise constraint: $\;{\vert y \vert}^2 = 1\;$
that is, that $\;y\;$ take values in the unit sphere $\;S^{m-1}\;$.

To begin with, I have to mention I have the solution of the above exercise. "The corresponding Euler-Lagrange system of PDE is 
$\;\Delta y ={\vert Dy \vert}^2 y\;$ where the term $\;{\vert Dy \vert}^2\;$ is the Lagrange multiplier for the constraint". 
The problem is I 'm having trouble getting my head around it. Since today I've only seen this kind of problems:

for which the Euler-Lagrange equation is given by:

1. How should I handle the above problem since the constraint is not an integral one?
 2. How could I find $\;λ\;$ by the given conditions?
I've been stuck to this so I would appreciate any help. 
Thanks in advance!
 A: If you want to work with what you know, you can convert the pointwise constraint to an integral one by constructing a positive "error function" to match: note that $|y|^2 = 1$ everywhere if and only if $\int (|y|^2 - 1)^2=0$.
It's not too hard to do this from first principles - if $y$ is a critical point of $J$ constrained to take values in the sphere, then for any perturbation $y(s)$ of $y$ (with the same constraint) we must have $$0=\frac{d}{ds}\Big|_{s=0}\frac12\int |D(y(s))|^2=\int Dy(0) \cdot Dy_s(0)=-\int\Delta y\cdot y_s,$$ where $y_s$ denotes $\partial y / \partial s$. Since a vector field $V$ can be realized as a $y_s$ if and only if $V(x)$ is tangential to the sphere at the point $y(x)$, we thus have $\int \Delta y^\perp \cdot V = 0$ for all $V$, where $\Delta y^\perp = \Delta y - (\Delta y \cdot y) y$ is the tangential part of the Laplacian. Thus we must have $\Delta y^\perp = 0;$ i.e. $\Delta y$ is parallel to $y$.
We can write this using a Lagrange multiplier as $\Delta y = \lambda y$. Now, to find $\lambda$ we need to use the constraint. From the product rule you can obtain $$\frac 1 2\Delta |y|^2 =y \cdot\Delta y +|Dy|^2,$$which must be  zero (since $|y|^2$ is constant). Since $y$ satisfies $\Delta y = \lambda y$, we thus have $$0=\lambda |y|^2 + |Dy|^2=\lambda + |Dy|^2;$$ so we can solve for $\lambda,$ obtaining the constrained Euler-Lagrange equation $$\Delta y = -|Dy|^2 y.$$ (I am reasonably sure the sign error is in your source, not my calculations.)
