I have this problem where I have to find the steepest descent direction of a functional. Basically, it comes down to solving a Poisson PDE subject to a certain constraint ($||v|| = 1$). The image attached below gives a better insight of the problem.
So I have to solve equation (13) to find $v$. But note that the Lagrange multiplier $\lambda$ is also a variable to such an equation. So I'm left with one equation with two variables. I believe the other equation that must be satisfied is $||v|| = 1$; but I also believe that such a condition was already used when he introduced the Lagrange Multiplier $\lambda$ (in equation (12), where he put ($\lambda |\nabla v|^2$), so I'm not sure what can I do to find the appropriate $\lambda$ and then solve (13). Any insight on this is much appreciated.
Solving this will give me the steepest descent direction, which I want to calculate numerically. Please don't suggest different methods on solving this such as robust gradient descent algoritms; I am not particularly looking for different methods for finding the steepest descent direction, as by paragraph 2 on the image, they might not be equivalent!
If this is indeed a system of non-linear equations, what would be a good numerical method to solve for $v$? Intuitively, one would fix some $\lambda$'s and then solve for $v$ by using equation (13) - and then see for which $\lambda$ we would get $||v||$ as close to 1 as possible. But this seems like an extremely robust algorithm.
Essentially you are solving the minimization of $J(v)=a^Tv+λ(v^TBv-1)$. The derivative for $v$, or its transpose, the gradient, is $$ 0=\nabla J(v)=a+2λBv $$ which has $v=-(2λB)^{-1}a$ as solution. Now this solution has additionally to satisfy the norm condition $v^TBv=1$ which after inserting gives the value for $λ$ by $$ 1=a^T(2λB)^{-1}B(2λB)^{-1}a = (2λ)^{-2}a^TB^{-1}a\implies λ=\pm\frac12\sqrt{a^TB^{-1}a}. $$ In your case, this means in short that you need to use $λ$ to scale the solution of (13) to unit norm in $H^1(\Omega)$.