# $L_{2}$ norm of the gradient of a vector valued function.

I have a vector valued function $U(x,y)=\Big(u_{1}(x,y),u_{2}(x,y)\Big)$. I want to find $\|\nabla U\|_{L_{2}(0,1)}$, but i could not figure how can do it. Do you have any idea?

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Do you mean divergence of $U$? I don't know how you take the gradient of a vector-valued function. – arzela Aug 4 '12 at 12:48
as i know gradient(U(x,y))=jacobi($u_{1},u_{2}$) – Brhn Aug 4 '12 at 12:50
@Brhn What is the context of the problem? Is there any more information you can provide? – Jayesh Badwaik Aug 4 '12 at 16:39

It is unusual to denote the derivative of a vector valued function $(u_1, u_2)$ by gradient, but if one does there is not much choice -- it will involve the gradient of both $u_1, u_2$ in some way.

One possible and, I guess, reasonable, defintion would be $\nabla U:= (\nabla u_1, \nabla u_2)$, which is matrix of course. One possible defintion of norm would then be $|\nabla u_1|+ |\nabla u_2|$, but for the definion of an $L_2$ - norm $(|\nabla u_1|^2+ |\nabla u_2|^2)^{1/2}$ looks more natural to me, with $|\nabla u_i| = \left[\sum_k \left( \frac{\partial u_i}{\partial x^k}\right)^2\right]^{1/2}$ as norm of choice for each $i$.

The $L_2$ - norm of this would then be defined as $$\int_\Omega (|\nabla u_1|^2+ |\nabla u_2|^2)^{1/2} dx^1d x^2$$

Note that it does not make sense to define an $L_2((0,1))$-norm for a function of two variables, the domain of definition should allow for two variables.

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I know your answer is assuming things due to lack of information from the OP, but as I guess the the "gradient" is basically a jacobian matrix between the two sets of variables and hence its norm must be the Frobenius norm or a spectral norm of a matrix. I am not sure though. – Jayesh Badwaik Aug 4 '12 at 16:39
@JayeshBadwaik $(\nabla u_1, \nabla u_2)$ is, basically, the jacobian. – user20266 Aug 4 '12 at 16:59
Yeah I meant that too, I was talking about its norm. – Jayesh Badwaik Aug 4 '12 at 17:15
@JayeshBadwaik for an $L^2$ norm one would expect the root of the integral with the sum of all squared derivatives $\partial u_k/\partial x^j$ as the integrand. This is exactly what you get with the formula I suggested. – user20266 Aug 4 '12 at 17:23
@JayeshBadwaik (if you write it down in components you actually do get the Frobenius norm) – user20266 Aug 4 '12 at 17:34