# CN FEM Stability [closed]

For Crank-Nicolson FEM for solving $$u_{t}-\Delta u=f$$, how can I show that it is stable?

• Can you give more detail about what you have tried? That would make it easier to help in a way other than writing up the whole proof for you (which I'm sure you can find online elsewhere). Apr 12, 2022 at 2:11
• It will involve finding the weak form, choosing the test function equal to $\u_h^m$, followed by some rearrangement of the equation, application of the Young and Cauchy-Schwartz inequalities, and concluding with a summation over all previous time levels. Apr 12, 2022 at 9:22
• What is $Ω$, a rectangle or a general area? In the second case, what properties are guaranteed for the subdivision, so that $h$ (cell diameter?) alone is characteristic? With an irregular subdivision, calling the method CN is slightly misleading, one can use the trapezoidal method in time direction, in space direction a more general finite-element approach is needed. /// But in general you get $\dot u_h=A_hu_h$ for the space discretization and $u_h^{m+1}=(2-hA)^{-1}(2+Ah)u_h^m$ for the trapezoidal method, so it just remains to show that $A$ has eigenvalues with negative real parts. Apr 18, 2022 at 6:51
• Using a quadratic grid? Apr 18, 2022 at 8:12
• Then you can use the Fourier basis to diagonalize the matrix $A$. You get close to the eigenvalues of the Laplacian for low frequencies and $-C/h^2$ with $C>0$ for the high frequencies. What $C$ is depends on the stencil used for the Lagrangian approximation. Apr 18, 2022 at 9:00

The heat equation you give has the following weak form on some finite time interval $$[0,T]$$, which is partitioned into $$0=t^0. Let the time step be constant such that $$t^{i+1}-t^{i} = \Delta t$$ for all $$i=0,...,N-1$$.

$$\left(\frac{u_{h}^{n}-u_{h}^{n-1}}{\Delta t}, v_{h}\right)+ \left( \nabla\left(\frac{u_{h}^{n}+u_{h}^{n-1}}{2}\right) , \nabla v_{h}\right)= \left(f\left(\frac{t^{n}+t^{n-1}}{2}\right), v_{h}\right)$$

Here we have that $$(\cdot,\cdot)$$ is the usual $$L^2(\Omega)$$ inner product and later we will define $$\|\cdot\|$$ to be corresponding norm induced by the $$L^2(\Omega)$$ inner product. We may choose $$v^n_h = \frac{u^n_h + u^{n-1}_h}{2}$$.

$$\frac{\|u_h^n\|^2 - \|u_h^{n-1}\|^2}{2 \Delta t} +\left\| \nabla\left( \frac{u_{h}^{n}+u_{h}^{n-1}}{2} \right) \right\|^2 \leq \left(f\left(\frac{t^{n}+t^{n-1}}{2}\right), \frac{u_{h}^{n}+u_{h}^{n-1}}{2}\right)$$

Now, we need the notion of dual norms. Let $$\|f\|_{-1}$$ be the dual norm to $$H^1_0(\Omega)$$. A good definition of this might be

$$\|f\|_{-1} = \sup_{r_h\in V} \frac{|(f,r_h)|}{\|\nabla r_h\|}$$

Essentially it tells us how much multiplying a member of $$V_h$$ by $$f$$ and integrating can "stretch" our function input $$r_h$$. Then we just need to note that for any fixed time value $$s$$, $$f(s)$$ is just a scalar function in $$\mathbb{R}^d$$. Then, using the definition of this dual norm and Young's inequality we get

$$\frac{\|u_h^n\|^2 - \|u_h^{n-1}\|^2}{2 \Delta t} +\left\| \nabla\left( \frac{u_{h}^{n}+u_{h}^{n-1}}{2} \right) \right\|^2 \leq \frac{1}{2}\left\|f\left(\frac{t^{n}+t^{n-1}}{2}\right)\right\|_{-1}^2 + \frac{1}{2}\left\|\nabla \left(\frac{u_{h}^{n}+u_{h}^{n-1}}{2}\right)\right\|^2$$

We will combine the gradient terms on the left hand side. Now, notice that we have not specified the time level. In fact this works for any time level. In particular, this works for any time level before the current one (i.e. at $$t^i$$ for $$i). Thus we may sum over the time levels. We get

$$\sum_{k=1}^n \left(\frac{\|u_h^k\|^2 - \|u_h^{k-1}\|^2}{2 \Delta t}\right) + \frac{1}{2}\sum_{k=1}^n \left\| \nabla\left( \frac{u_{h}^{k}+u_{h}^{k-1}}{2} \right) \right\|^2 \leq \sum_{k=1}^n \frac{1}{2}\left\|f\left(\frac{t^{k}+t^{k-1}}{2}\right)\right\|_{-1}^2$$

The key observation is now the first term on the left hand side is a telescoping series. It will sum to $$\|u_h^{n}\|^2 - \|u_h^0\|^2$$. The second term can be moved to the right hand side. We might as well multiply by $$2\Delta t$$. We then have

$$\|u_h^n\|^2 + \Delta t \sum_{k=1}^n \left\| \nabla\left( \frac{u_{h}^{k}+u_{h}^{k-1}}{2} \right) \right\|^2 \leq \|u_h^0\|^2 + \Delta t \sum_{k=1}^n \left\|f\left(\frac{t^{k}+t^{k-1}}{2}\right)\right\|_{-1}^2$$

In particular we can pick $$n=N$$ to get a bound on our approximation at the final time step. This has a nice interpretation as well. The temperature is bounded uniformly in time by just the initial condition and heat sources. The gradient of temperature is bounded only in an $$L^2[0,T]$$ sense. What you see here is a discrete analog of an integral from $$0$$ to $$T$$ with respect to time.

Edit: Upon request I will provide a little more context as to how this answer should be interpreted as an answer to the question as posed.

We need to notice two things: First, by the definition of the $$L^2$$ projection, $$\|u_h^0\|^2 \leq \|u^0\|^2$$. To see this, take the definition of $$L^2$$ projection and choose the test function to be $$u_h^0$$. After applying the Cauchy-Schartz and Young's inequalities to the RHS you will arrive at this result. Since it isn't the main proof here, we will move along. Thus, in the last line of the above, we can replace the discrete initial condition with the continuous one.

This gives us

$$\|u_h^n\|^2 + \Delta t \sum_{k=1}^n \left\| \nabla\left( \frac{u_{h}^{k}+u_{h}^{k-1}}{2} \right) \right\|^2 \leq \|u^0\|^2 + \Delta t \sum_{k=1}^n \left\|f\left(\frac{t^{k}+t^{k-1}}{2}\right)\right\|_{-1}^2$$

Now, it is clear that the RHS is a constant, which depends only on the problem data (initial condition $$u^0(\mathbf{x})$$ and the source function $$f(t,\mathbf{x})$$). You ask for a very specific form of final answer, but it is already contained here. Notice that both terms on the left are nonnegative. This is because the square of any number is positive or zero (a norm is nonnegative anyway) and $$\Delta t$$ is also positive. The sum of nonnegative terms multiplied by a positive number is positive.

Finally, and this should hopefully follow intuitively, if we have $$a+b\leq C$$ and both $$a$$ and $$b$$ are positive, we can have $$a\leq C$$ and $$b\leq C$$. Thus we get:

$$\|u_h^n\|^2 \leq \|u^0\|^2 + \Delta t \sum_{k=1}^n \left\|f\left(\frac{t^{k}+t^{k-1}}{2}\right)\right\|_{-1}^2$$

Again, I see no convincing reason to report your result this way: You've simply thrown away part of your hard-earned result! Typically, the result on the gradient is regarded as more impressive (since even well-behaved, bounded functions can have erratic gradients). But nonetheless it is still accurate.

Just to emphasize, what makes this result a stability result is precisely that the right hand side can be computed before any other work has been done. Since we are given $$f$$ and $$u^0$$ we can compute the corresponding norms without issue and there is to my knowledge no better or other way to express the result here. Thus using the exact language of your question we would have

$$\|u_h^n\|^2 \leq C(u^0,f) = \|u^0\|^2 + \Delta t \sum_{k=1}^n \left\|f\left(\frac{t^{k}+t^{k-1}}{2}\right)\right\|_{-1}^2$$

• How do I get from the last step to my desired result? Apr 20, 2022 at 19:37
• That IS your desired result. On the right hand side we have a bunch of stuff which depends only on the initial condition and $f$. Both of the terms on the left are positive, so either one of them is bounded by the right. If this explanation doesn't make sense, let me know and I can add one more line. But again, what you see is what was asked for. Apr 20, 2022 at 20:57

Let me give you a sketch of a possible argument: First choose $$v_h=\frac{u_h^n+u_h^{n-1}}{2}$$, then we have $$\int_{\Omega}(u_h^n-u_h^{n-1} )(u_h^n+u_h^{n-1} ) \leq \Delta t \int_{\Omega}f (u_h^n+u_h^{n-1}) ,$$ which implies that $$\|u_h^n\|_{L^2(\Omega)} \leq \Delta t \int_{\Omega}f (u_h^n+u_h^{n-1}) +\|u_h^{n-1}\|_{L^2(\Omega)}.$$ By using Cauchy's inequality and Young's inequality, we obtain $$\|u_h^n\|_{L^2(\Omega)} \leq C_1(\Delta t )^2\|f\|^2_{L^2(\Omega)}+C_2\|u_h^{n-1}\|^2_{L^2(\Omega)},$$ where $$C_1$$ and $$C_2$$ is independent of $$u,f,h$$ and $$n$$. By repeated application, we get the desired estimate.

• This is a positive term. Note that there is a $\leq$. Apr 21, 2022 at 0:57