How would I find the jacobian in this case? Normally the Jacobian is calculated using partial derivatives of F with respect to each of the variables, but since we are using the centered finite difference approximation, would the jacobian itself be another approximation ?
Context: Numerical analysis. Later on, I'll have to use Newtons method to solve the system.
As you may know, once you have discretized your BVP, with the convenction $u(x_i) \approx u_i$, the equations can be written as \begin{equation}D_2 \vec{u}+\lambda e^{\vec{u}}=0\end{equation} with the boundary conditions $u_1=u_n=0$, where $D_2$ is a tridiagonal matrix which discretize the second derivative, i.e., if you do the matrix-vector product $D_2 \vec{u}$ this gives you the (second order) approximation of the second derivative at each $x_i$.
$D_2$ can be built using the fact that the secondo order approximation of the second derivative is given by \begin{equation} u''(x_i)=\frac{u_{i+1}-2u_i+u_{i-1}}{h^2} + \mathcal{O}(h^2) \end{equation} where from the indices you can clearly see it's tridiagonal.
So, you have to solve the non linear equation $F(u)=0$, where $F$ is a vector values function given by $D_2 \vec{u} + \lambda e^\vec{u}=0$. Note that the initial conditions must be imposed in the first and last rows of the matrix, but other ways can be used in order to impose them.
In order to solve the equation, a Newton's method is required, and hence, the jacobian matri, which is definied by $J_{ij}=\frac{\partial F_i}{\partial x_j}$
Hence, it is $J(u)=D_2 +\lambda \text{diag}(e^{u_i})$, where I used the fact that the jacobian of a matrix vector product like $A u$ is like the scalar case, and $\text{diag}(e^{u_i})$ is
\begin{bmatrix} e^{u_1} & 0 & 0 & 0 & \dots & 0 \\ 0& e^{u_2} & 0 & 0 & \dots & 0 \\ 0 & \ddots & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & 0 \\ 0 & \dots & 0 & 0 & e^{u_{m-1}} & 0 \\ 0 & \dots & 0 & 0 & 0 & e^{u_n} \end{bmatrix}
Now you have just to apply Newton's method, with a suitable initial guess, and you're done.