I have a specific PDE in 3-dimensional space + time (the right one). u(x,t) is the unknown function (values are in R^3) and F(x,t) the right hand side, mu and lambda are positive constants.
Now consider the stationary case of the right equation, that means the term with the time derivate goes zero. I have to show now that the stationary case leads to an elliptic linear PDE of 2'nd order. The problem is the term ∇(∇.u) = grad(div(u)). If i write it out, the component-functions of u appear everywhere. For exampe the first component of the PDE reads:
-μ[u1_(1,1) + u1_(2,2) + u1(3,3)] - (μ + λ)[u1_(1,1) + u2_(1,2) + u3(1,3)] = F1(x,t)
with notation: u1_(i,j) as the second order derivative of the first component-function of u in directions j and i.
So i can't write this equation in standard form to get a Matrix A, which has to be symmetric and positive definit for the PDE to be elliptic - since i can only scale the the vectors u_(i,j) with some scalars a_ij in standard form.
Does anyone has an idea what i'm supposed to do?
I would be glad for some help! Thank you!