Description
Let $w(x, \ y)$ be a function that satisfies laplace's equation
$$\nabla^2 w = 0 \ \ \ \ \ \ \text{on} \ \Omega$$ $$\dfrac{\partial w}{\partial n} = \langle \mathbf{p}, \ \mathbf{t}\rangle \ \ \ \ \ \ \text{on} \ \partial \Omega$$
$\mathbf{p}(u)$ parametrizes the closed curve $\partial \Omega$ in the plane, $\mathbf{t}$ is the tangent vector
$$\mathbf{p}(u) = \left(x(u), \ y(u)\right) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{with} \ u \in \left[0, \ 1 \right]$$ $$\mathbf{t} = \dfrac{1}{\|\mathbf{p}'\|} \cdot \mathbf{p}'$$
And set
$$w(\mathbf{p}(0)) = 0$$
My objetive is finding $J$
$$J = \int_{0}^{1} w \cdot \langle \mathbf{p}, \ \mathbf{p}'\rangle \ du$$
Question
Is it possible to compute $J$ without solving the PDE ?
I mean, I don't actually need the values of $w$, only $J$. So I wondered if there's some way to do it
If it's not possible, any hint to help computing $J$ ?
More
I tried to find some relation for $w$ on the boundary, but I don't know how I can proceed, or even if it's correct:
$$\nabla^2 w = \dfrac{\partial^2 w}{\partial x^2} + \dfrac{\partial^2 w}{\partial y^2} = \dfrac{\partial^2 w}{\partial t^2} + \dfrac{\partial^2 w}{\partial n^2} = 0$$ $$\dfrac{\partial^2 w}{\partial t^2} = -\dfrac{\partial}{\partial n} \left(\dfrac{ \langle\mathbf{p}, \ \mathbf{p}'\rangle}{\|\mathbf{p}'\|}\right)$$
The motivation is to compute the torsion constant $\mathbb{J}$ of a beam:
$$\mathbb{J} = \int_{\Omega} x^2 + y^2 \ dx dy - J$$