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I have discretized a PDE with continuous finite element method in spatial variable and with implicit Euler or Crank-Nicolson in temporal variable. In both cases, I have error estimates in $L_2$ norm in form $\mathcal{O}(\Delta\,t^j + h^2),$ where $j=1$ for backward Euler and $j=2$ for Crank-Nicolson. In both cases I have a CFL contition of form $\Delta\,t = \mathcal{O}(h^2).$ Therefore, when I run the code I always take $\Delta\,t = \mathcal{O}(h^2).$ However, for this choice in backward Euler I have \begin{equation} \mathcal{O}(\Delta\,t + h^2) = \mathcal{O}(h^2)=\mathcal{O}(\Delta\,t). \end{equation} Now, in Crank-Nicolson, for the same choice of $\Delta\,t$ I have \begin{equation} \mathcal{O}(\Delta\,t^2 + h^2) = \mathcal{O}(h^2)=\mathcal{O}(\Delta\,t). \end{equation}

My question : How can I conclude to second order accuracy in time for Crank-Nicolson when $\Delta\,t = \mathcal{O}(h^2)$?

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This is one of those scenarios where assessing the convergence order of your scheme is difficult, because, as you explained, your time-step is "linked" to your spatial discretization. What you could do is manufacture an analytical problem without any spatial error. For example, in finite elements, if you are using P2 polynomials, a second order polynomial solution will be represented exactly, thus have no error in space. Then you can introduce a time variation that will generate an error in time. The following analytical solution, with the appropriate BC, would lead to no spatial error, but a time discretization error if P2 polynomials are used: $$ u = (x^2 + y^2) \sin t $$ Since the time dependence is non-polynomial, you will always have a non-zero error in time, but no error in space. Thus, you can verify the order of convergence of your scheme.

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