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I am solving the heat equation discretized spatially via FEM and temporally via backward Euler. I get the system $$M \dot{u} = K u +f$$ where $u$ is a vector representing the solution at spatial locations.

For piecewise linear basis functions, the mass matrix is defined as enter image description here while the stiffness matrix $K$ is enter image description here

By discretizing this in time, I get $$u_{k+1} = Au_k + By_k$$ for some matrix $B.$ Here, $A = (I-\Delta t M^{-1}K)^{-1}$ where $\Delta t$ is the time step size.

Is it possible to show that $$\|A\|_2 < 1$$ for $\Delta t$ sufficiently small? I would like an analytical solution without computing this in Matlab.

I have a hard time analyzing this because $M^{-1}K$ is not symmetric although $M,K$ are symmetric. Otherwise, I could have just used the fact that the spectral radius and the 2 norm coincide.

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For stability analysis, we can assume $f = 0$. First we write $$ M \left( u_{k+1} - u_{k} \right) = \Delta t K u_{k+1} $$ or $$ \left( M - \Delta t K \right) u_{k+1} = M u_{k} $$ Take the eigendecomposition of $(K, M)$ (there is an explicit expression in 1D), such that $$ w_i^T K w_j = \delta_{ij} \lambda_i \quad \mbox{and} \quad w_i^T M w_j = \delta_{ij} $$ Write $$ w_i^T \left( M - \Delta t K \right) u_{k+1} = w_i^T M u_{k} $$ and express $u_k$ and $u_{k+1}$ in the basis of eigenmodes. From there, you should be able to get an analytical expression to control $\Delta t$.

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  • $\begingroup$ Hi, thanks for the reply. I am confused about 2 things: 1) I'm not interested in showing stability but rather I want to show that $\|A\|_2 < 1$. Does the approach you mentioned above work for this or is it only to show that the spectral radius is less than 1? 2) What do you mean by express $u_k,u_{k+1}$ as eigenmodes? I have never heard of the method you showed above before. Do you have a reference for this? $\endgroup$ – user1237300 May 1 at 1:49
  • $\begingroup$ 1. Why do you want to show that $||A||_2 < 1$? --- 2. $u_k = \sum_{j = 1}^{n} \mu_k^j w_j$. Using the orthogonality among the eigenvectors $(w_i)$, we get an expression between $\mu_{k+1}^{j}$ and $\mu_{k}^{j}$. $\endgroup$ – user7440 May 1 at 1:56
  • $\begingroup$ Note also $|| ( I - M^{-1} K )^{-1} x ||_2 = || ( M^{-1/2} ( I - M^{-1/2} K M^{-1/2} ) M^{+1/2} )^{-1} x ||_2 = || M^{-1/2} ( I - M^{-1/2} K M^{-1/2} )^{-1} M^{+1/2} x ||_2 = || ( I - M^{-1/2} K M^{-1/2} )^{-1} M^{+1/2} x ||_{M^{-1}}$ $\endgroup$ – user7440 May 1 at 2:00
  • $\begingroup$ I'm using it for error estimates. How does the above help in finding the 2-norm of $I-\Delta t M^{-1}K$? $\endgroup$ – user1237300 May 1 at 2:08
  • $\begingroup$ When you use the expression with $M^{-1}$-norm, the operator $I - \Delta t M^{-1/2} K M^{-1/2}$ is now symmetric and its eigenvectors are the ones of $(K, M)$. So studying $|| ( I - \Delta t M^{-1} K )^{-1} x ||_2 / || x ||_{2}$ is the same as studying $|| ( I - \Delta t M^{-1/2} K M^{-1/2})^{-1} y ||_{M^{-1}} / || y ||_{M^{-1}}$ (where $y = M^{1/2} x$) $\endgroup$ – user7440 May 1 at 2:28

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