# Elliptic PDE: Proving that a 2nd order accurate discretization of the 2nd derivative of the unknown is 2nd order accurate for the unknown itself

For an ODE:

$$\frac{dy}{dt}=f(y(t),t)$$

$$y_{n+1}=y_{n}+\Delta tf_n$$

and it can be easily shown with a Taylor expension that:

$$y_{n+1}=y_{n}+\Delta t \frac{dy}{dt}|_n+\Delta t^2 \frac{d^2y}{dt^2}|_n$$

and therefore by comparing the scheme with the Taylor expansion we have:

$$LTE= O(\Delta t^2)$$

i.e. the local truncation error LTE for Euler explicit scales with $$\Delta t^2$$. Since we repeat this for $$n$$ times, with $$n$$ the number of grid points, the global truncation error is $$n\Delta t^2$$. Since $$n=L/\Delta t$$, with L the domain length, then the global truncation error scales with Δt, so Euler explicit is first order accurate. For an elliptic PDE:

$$\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}=0$$

if I use the central difference scheme:

$$\frac{\partial^2\phi}{\partial x^2}= \frac{\phi_{i-1,j}-2\phi_{i,j}+\phi_{i+1,j}}{dx^2}+O(\Delta x^2)$$

the error is second order as it can be shown by Taylor expanding $$\phi_{i+1}$$ and $$\phi_{i-1}$$ in the neighborhood of $$\phi_{i}$$. Similar reasoning in $$y$$ where:

$$\frac{\partial^2\phi}{\partial y^2}= \frac{\phi_{i,j-1}-2\phi_{i,j}+\phi_{i,j+1}}{dy^2}+O(\Delta y^2)$$

And here is were I am lost. The error in the derivative is second order, but what about the error in the function $$\phi$$ I am looking for? If I replace the discretization into the equation (let's assume $$\Delta y = \Delta x$$ for now) we have:

$$\frac{\phi_{i,j-1}-2\phi_{i,j}+\phi_{i,j+1}}{dx^2}+ \frac{\phi_{i-1,j}-2\phi_{i,j}+\phi_{i+1,j}}{dx^2}+2O(\Delta x^2)=0$$

which, multiplying by $$\Delta x^2$$ reads:

$$\phi_{i,j-1}-4\phi_{i,j}+\phi_{i,j+1}+ \phi_{i,i-1,j}+\phi_{i+1,j}+2O(\Delta x^4)=0$$

So $$O(\Delta x^4)$$ shows up. From here, how can I prove that $$\phi$$ is obtained with second order accuracy? I guess the Local error for $$\phi_{j}$$ is 4th order (if all the 4 neighbours were known exactly) , but the global error is second error, I just do not know how to prove it. Any suggestion?

Consider as a simple case the 1D Poisson problem (it should be simple to extend to the 2D case you describe): $$\tag{*}\label{eq:poisson} \frac{d^2 \phi}{dx^2}(x) = f(x), \quad 0 where $$f$$ is given. We approximate $$\frac{d^2 \phi}{dx^2}(x_i) = \frac{\phi_{i-1} - 2\phi_i + \phi_{i+1}}{h^2} + \mathcal{O}(h^2).$$ Let $$\phi(x)$$ denote the exact solution to the problem $$(*)$$. Suppose $$\tilde{\phi}$$ is the numerical solution obtained by solving the system of equations $$\frac{\tilde\phi_{i-1} - 2\tilde\phi_i + \tilde\phi_{i+1}}{h^2} = f_i.$$ We now write everything in terms of linear algebra. Let $$\boldsymbol\phi$$ denote the vector of values $$\phi(x_i)$$, let $$\tilde{\boldsymbol\phi}$$ denote the vector $$\tilde\phi_i$$, and let $$\boldsymbol f$$ denote the vector $$f(x_i)$$. Finally, let $$A$$ denote the matrix corresponding to the finite-difference Laplacian. For the discrete approximate solution, we have $$A \tilde{\boldsymbol\phi} = \boldsymbol f.$$ Using the truncation error for the finite-difference approximation, we have for the exact solution $$A \boldsymbol\phi = \boldsymbol f + \boldsymbol \tau, \qquad \tau_i = \mathcal{O}(h^2).$$ We want to measure the error in our approximate solution, which is given by $$\| \tilde{\boldsymbol\phi} - \boldsymbol\phi \|$$ in some norm. We see that the difference $$\boldsymbol\varepsilon = \tilde{\boldsymbol\phi} - \boldsymbol\phi$$ satisfies $$A \boldsymbol\varepsilon = A (\tilde{\boldsymbol\phi} -\boldsymbol\phi)= -\boldsymbol \tau$$ and so $$\boldsymbol\varepsilon = -A^{-1}\boldsymbol \tau.$$ In particular $$\| \boldsymbol\varepsilon \| \leq \|A^{-1}\| \|\boldsymbol\tau\|$$ where $$\|A^{-1}\|$$ is the corresponding matrix norm. If we can bound the norm of $$A^{-1}$$ by a constant, then we can conclude that $$\|\boldsymbol\varepsilon\| = \mathcal{O}(h^2)$$. If we use the 2-norm and using that $$A$$ is symmetric, we see that $$\| A^{-1} \|_2 = \frac{1}{\min |\lambda|}$$ where $$\lambda$$ is an eigenvalue of $$A$$. Using an explicit expression for the eigenvalues of $$A$$, the smallest in magnitude is given by $$\lambda = -\pi^2 + \mathcal{O}(h^2)$$ and the result follows.