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Consistent-mass P1 diffusion can leave the nonnegative cone

Statement

On the interval $[0, 2 h]$ with three uniform P1 nodes and homogeneous Neumann boundary conditions, the semidiscrete heat equation with consistent mass matrix,

M \dot{u} + ν K u = 0, ν > 0,

does not preserve nodal nonnegativity: there is a nonnegative initial vector for which the first component becomes negative immediately.

Facts & Assumptions

Given: The P1 nodal basis on two elements of length $h > 0$ , the consistent mass and Neumann stiffness matrices assembled from their defining integrals, and $ν > 0$ .

[F1]

Finite matrix exponential series and its identities: for a finite matrix $A$ , $e^{t A}$ is differentiable with derivative $A e^{t A}$ and $e^{0 A} = I$ .

[F2]

Picard-Lindelöf local existence and uniqueness for first-order systems: a locally Lipschitz finite-dimensional vector field has a unique local solution through each initial state.

[A1]

The consistent mass entries are $M_{i j} = \int_{0}^{2 h} ϕ_{i} ϕ_{j} d x$ and the stiffness entries are $K_{i j} = \int_{0}^{2 h} ϕ_{i}^{'} ϕ_{j}^{'} d x$ for the three piecewise-linear nodal basis functions.

Proof

1.1A1givenalgebra

On an element $[0, h]$ , the two affine basis functions are $ϕ_{1} (x) = 1 - x / h$ and $ϕ_{2} (x) = x / h$ . Direct integration gives $\int_{0}^{h} ϕ_{1}^{2} d x = \int_{0}^{h} ϕ_{2}^{2} d x = h / 3, \int_{0}^{h} ϕ_{1} ϕ_{2} d x = h / 6,$ and their derivatives are $- 1 / h$ and $1 / h$ , so the element stiffness entries are $1 / h, - 1 / h, - 1 / h, 1 / h$ . Assembling the two elements yields the displayed matrices: $M = \frac{h}{6} (\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 4 & 1 \\ 0 & 1 & 2 \end{array}), K = \frac{1}{h} (\begin{array}{ccc} 1 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 1 \end{array}) .$ The Neumann condition is natural in this stiffness form and adds no boundary matrix.

2.1step 1.1givenalgebra

Write $M = (h / 6) B$ and $K = C / h$ for the displayed integer matrices. Direct multiplication verifies $B^{- 1} = \frac{1}{12} (\begin{array}{ccc} 7 & - 2 & 1 \\ - 2 & 4 & - 2 \\ 1 & - 2 & 7 \end{array}), B^{- 1} C = \frac{1}{12} (\begin{array}{ccc} 9 & - 12 & 3 \\ - 6 & 12 & - 6 \\ 3 & - 12 & 9 \end{array}) .$ Thus $M^{- 1} K = 6 B^{- 1} C / h^{2}$ , so in particular $(M^{- 1} K)_{13} = 3 / (2 h^{2})$ . With $e_{3} = (0, 0, 1)^{T}$ , the ODE gives $u^{'} (0) = - ν M^{- 1} K e_{3}$ , whose first component is $- 3 ν / (2 h^{2}) < 0$ .

3.1F1F2step 2.1algebra∎

The linear field $u \mapsto - ν M^{- 1} K u$ is globally Lipschitz. By [F1], $e^{- t ν M^{- 1} K} e_{3}$ solves the IVP, and [F2] makes it the unique solution. Since $u_{1} (0) = 0$ and $u_{1}^{'} (0) < 0$ by step 2.1, the right difference quotient $u_{1} (t) / t$ is negative for all sufficiently small $t > 0$ . Hence the consistent-mass P1 semidiscrete diffusion is not positivity preserving, despite the nonnegative initial nodal vector.

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