I need to solve the following problem $$ -\dfrac{1}{r}\left( \dfrac{d}{dr} \left( r \dfrac{du}{dr} \right) \right) = f(r), ~~ r \in (0, 1)$$ $$ u'(0) = 0, ~~~ u(1) = 0 $$ with the finite element method.
I consider the Hilbert space $H$ with the following inner-product $$ (u, v) = \int\limits_{0}^{1} u(r) v(r) r dr $$ Then I define the operator $A$ as $$ Au = -\dfrac{1}{r}\left( \dfrac{d}{dr} \left( r \dfrac{du}{dr} \right) \right) $$ on the domain $$ D_A = \{ u \in C^2(0, 1), u(1) = 0, u'(0) = 0 \} $$ I can show that $A$ is symmetric and positive definite on $D_A$ and $D_A$ is dense in $H$. Then I can make a completion of $D_A$, get the energy space $H_A$ and apply Ritz method. The inner product in $H_A$ is defined as $$ [u, v] = (Au, v) = \int\limits_{0}^{1} \left[-\dfrac{1}{r}\left( \dfrac{d}{dr} \left( r \dfrac{du}{dr} \right) \right) \right] v r dr = -\int\limits_{0}^{1} \dfrac{d}{dr}\left( r \dfrac{du}{dr} \right) v dr = \int\limits_{0}^{1} r \dfrac{du}{dr} \dfrac{dv}{dr} dr. $$
Then Ritz method formulates as find $u \in H_A$ such that \begin{equation} [u, v] = (f, v), ~~ \forall v \in H_A \end{equation}
The solution of this problem is unique. I guess that functions in $H_A$ are such that $v(1) = 0$, but I completely lose the condition $u'(0) = 0$. Usually it appears in $(Au, v)$, but here it vanishes because of the weight $r$ in the inner product.
So, my question is where am I wrong? It seems that, according to the theory of the Ritz method, everything should be solved correctly and the solution does not depend on the condition on $u'(0)$. But it's obvious that this is not the case.
Thank you in advance!
Your analysis is correct up to a certain point. The Ritz method allows you to approximate the solution of the problem by finding an approximate solution within a finite-dimensional subspace of the energy space $H_A$. However, it is important to note that the Ritz method does not automatically enforce the boundary conditions of the original problem. The boundary conditions must be imposed separately.
In your case, the condition $u'(0) = 0$ is a crucial boundary condition that must be satisfied by the approximate solution. While the weight factor $r$ in the inner product affects the form of the variational problem, it does not eliminate the need to enforce the boundary condition $u'(0) = 0$.
To incorporate the boundary condition $u'(0) = 0$ into the Ritz method, you can modify the trial space of functions used in the approximation. Instead of considering all functions in $H_A$, you can consider the subspace of functions that satisfy the boundary condition $u'(0) = 0$. This can be done by adding the appropriate constraint to the variational problem. One way to do this is by introducing a Lagrange multiplier to handle the boundary condition. Let $\lambda$ be the Lagrange multiplier associated with the boundary condition $u'(0) = 0$. Then, the modified variational problem becomes: find $u \in H_A$ and $\lambda \in \mathbb{R}$ such that \begin{align*} [u, v] + \lambda v(0) &= (f, v) \quad \forall v \in H_A, \ u'(0) &= 0. \end{align*}
By introducing the Lagrange multiplier and the additional term $\lambda v(0)$ in the variational problem, you ensure that the approximate solution satisfies the boundary condition $u'(0) = 0$. The Lagrange multiplier $\lambda$ can be determined as part of the solution process.
To correctly incorporate the boundary condition $u'(0) = 0$ into the Ritz method, you need to modify the variational problem by introducing a Lagrange multiplier associated with the boundary condition. This allows you to enforce the boundary condition in addition to the other terms in the variational problem.