# Tag Info

Accepted

### Puzzling remark about stability region of fifth-order Runge-Kutta method

van der Houwen's statement is correct, but it is not a statement about all fifth-order Runge-Kutta methods. The "Taylor polynomials" he is referring to are (as you seem to know) just the polynomials ...
Accepted

### Numerically stable approach for calculating x in Ax=b

If the solution of $Ax=b$ is unstable, the matrix is very ill-conditioned (i.e., has a very large condition number), and (paraphrasing Lanczos) no amount of mathematical trickery can make it stable. ...
Accepted

### Forcing an ODE solver to preserve the norm

The best approach is to use an ODE solver that is guaranteed to conserve the norm of the initial condition, i.e., for which $\|y_n\| = \|y_0\|$ for all $n\in\mathbb{N}$. Such solvers exist, and are ...

### How much regularization to add to make SVD stable?

The singular value decomposition for a symmetric matrix $A=A^{T}$ is one and the same as its canonical eigendecomposition (i.e. with an orthonormal matrix-of-eigenvectors), while the same thing for a ...
Accepted

### Lagrange multipliers space is too rich in a mathematical view

Each Lagrange multiplier corresponds to a constraint. So if the space of Lagrange multipliers is too large, then you have too many constraints that can no longer be all satisfied at the same time ...
Accepted

### Lagrange multipliers space is too rich in a mathematical view

The saddle point matrix you must solve takes the following form: $$\begin{bmatrix} A & B^T \\ B \end{bmatrix},$$ where $A$ is the unconstrained matrix and $B$ is the matrix that observes the ...

### Numerical stability in the product of many matrices

Orthogonal matrices are about as well-conditioned as you can get, but numerical errors still occur. One common error is loss of orthogonality. A fix for this could be to re-orthogonalize your columns ...
Accepted

### Advantage of diagonal "jitter" for numerical stability?

So, you want to invert your matrix $A=\Phi^T\Phi$. For $A$ to be invertible it must not have zero eigenvalues. We can show that $A$ is positive semi-definite as follows. Positive semi-definite means ...

### stability of a numercial scheme for a hyperbolic system?

It is worth making some additional points. What you set out is just one version of the Lax-Wendroff method. That scheme is unique in one space dimension but has several free parameters in two or three ...

### Finite Difference Method Stability

For stability you need $\omega \leq 1$, as stated here. If your diffusivity $\alpha$ is positive (that should be based on a physical basis), $\omega$ should also be positive since $\Delta x$ and \$\...