Accuracy of Stable, High-order Finite Difference Methods for Hyperbolic Systems with Non-smooth Wave Speeds

Abstract

We derive analytic solutions to the scalar and vector advection equation with variable coefficients in one spatial dimension using Laplace transform methods. These solutions are used to investigate how accuracy and stability are influenced by the presence of discontinuous wave speeds when applying high-order-accurate, skew-symmetric finite difference methods designed for smooth wave speeds. The methods satisfy a summation-by-parts rule with weak enforcement of boundary conditions and formal order of accuracy equal to 2, 3, 4 and 5. We study accuracy, stability and convergence rates for linear wave speeds that are (a) constant, (b) non-constant but smooth, (c) continuous with a discontinuous derivative, and (d) constant with a jump discontinuity. Cases (a) and (b) correspond to smooth wave speeds and yield stable schemes and theoretical convergence rates. Non-smooth wave speeds [cases (c) and (d)], however, reveal reductions in theoretical convergence rates and in the latter case, the presence of an instability.

Appendices

A Detailed Analytic Solutions

Below we provide detailed analytic solutions for the scalar and vector equations considered. The code for computing these is available online at https://github.com/brittany-erickson/analytic_wave/.

1.1 A.1 Analytic Solution to the Scalar Equation

The scalar equation (1) has analytic solution given by (14) which requires the calculation of $I_a(x)$ and $\xi (t,x)$ given by (2), (15), respectively. These are provided below for the four cases of wave speeds considered in Sect. 2.3.

For case 1,

$$\begin{aligned} I_a(x)&= x, \end{aligned}$$

(82a)

$$\begin{aligned} \xi (t,x)&= x-t. \end{aligned}$$

(82b)

For case 2,

$$\begin{aligned} I_a(x)&= \frac{1}{\epsilon }\ln \left| 1 + \epsilon x\right| \end{aligned}$$

(83a)

$$\begin{aligned} \xi (t,x)&= \frac{1}{\epsilon }\left[ e^{-\epsilon t}(1 + \epsilon x) -1 \right] . \end{aligned}$$

(83b)

For case 3,

$$\begin{aligned} I_a(x)&= x H(x_0 - x) + \left( x_0 + \frac{1}{\epsilon }\ln |1-\epsilon x_0 + \epsilon x | \right) H(x - x_0), \end{aligned}$$

(84a)

$$\begin{aligned} \xi (t,x)&= d_1(t,x) H(x_0 - x)H(x_0 - d_1(t,x)) \nonumber \\&\quad +\, d_2(t,x)H(x-x_0)H(x_0 - d_2(t,x)) \nonumber \\&\quad +\,d_3(t,x)H(x-x_0)H(d_3(t,x)-x_0), \end{aligned}$$

(84b)

where

$$\begin{aligned} d_1(t,x)&= x-t, \end{aligned}$$

(85a)

$$\begin{aligned} d_2(t,x)&= x_0 + \frac{1}{\epsilon } \ln | 1 - \epsilon x_0 + \epsilon x| - t, \end{aligned}$$

(85b)

$$\begin{aligned} d_3(t,x)&= \frac{1}{\epsilon } \left[ (1 - \epsilon x_0 + \epsilon x)e^{-\epsilon t} - 1 + \epsilon x_0\right] . \end{aligned}$$

(85c)

For case 4,

$$\begin{aligned} I_a(x)&= x H(x_0 - x) + \left[ x_0 + \frac{x-x_0}{1 + \epsilon x_0}\right] H(x-x_0), \end{aligned}$$

(86a)

$$\begin{aligned} \xi (t,x)&= d_4(t,x) H(x_0 - x)H(x_0 - d_4(t,x)) \nonumber \\&\quad +\,d_5(t,x)H(x-x_0)H(x_0 - d_5(t,x)) \nonumber \\&\quad +\,d_6(t,x)H(x-x_0)H(d_6(t,x)-x_0), \end{aligned}$$

(86b)

where

$$\begin{aligned} d_4(t,x)&= x-t, \end{aligned}$$

(87a)

$$\begin{aligned} d_5(t,x)&= x_0 + \frac{x-x_0}{1 + \epsilon x_0} - t, \end{aligned}$$

(87b)

$$\begin{aligned} d_6(t,x)&= x - t(1 + \epsilon x_0). \end{aligned}$$

(87c)

1.2 A.2 Analytic Solution to the Vector Equation

The vector equation (37) with initial/boundary conditions given by (38)–(39) has analytic solution given by (55) which requires the calculation of $I_a(x), I_b(x)$ and three characteristic variables $\xi (t,x), \omega _n(t,x)$ and $\gamma _n$ given by (2) and (56) respectively. These are provided below for the four cases of wave speeds considered in Sect. 3.4.

For case 1,

$$\begin{aligned} I_a(x)&= x, \end{aligned}$$

(88a)

$$\begin{aligned} I_b(x)&= x, \end{aligned}$$

(88b)

$$\begin{aligned} \xi (t,x)&= x-t, \end{aligned}$$

(88c)

$$\begin{aligned} \omega _n(t,x)&= x + t_n - t, \end{aligned}$$

(88d)

$$\begin{aligned} \gamma _n(t,x)&= -x + t_n - t. \end{aligned}$$

(88e)

For case 2,

$$\begin{aligned} I_a(x)&= -\frac{1}{\epsilon } \ln |1 - \epsilon x|, \end{aligned}$$

(89a)

$$\begin{aligned} I_b(x)&= \frac{1}{\epsilon } \ln |1 + \epsilon x|, \end{aligned}$$

(89b)

$$\begin{aligned} \xi (t,x)&= -\frac{1}{\epsilon }\left[ (1-\epsilon x)e^{\epsilon t} - 1\right] , \end{aligned}$$

(89c)

$$\begin{aligned} \omega _n(t,x) =&-\frac{1}{\epsilon }\left[ (1-\epsilon x)e^{\epsilon (t-t_n)} - 1\right] , \end{aligned}$$

(89d)

$$\begin{aligned} \gamma _n(t,x)&= -\frac{1}{\epsilon }\left[ (1+\epsilon x)e^{\epsilon (t-t_n)} - 1\right] . \end{aligned}$$

(89e)

For case 3,

$$\begin{aligned} I_a(x)&= -\frac{1}{\epsilon } \ln |1 - \epsilon x| H(x_0 - x) \nonumber \\&\quad +\,\left[ -\frac{1}{\epsilon } \ln |1 - \epsilon x_0| + \frac{x - x_0}{1 - \epsilon x_0} \right] H(x - x_0), \end{aligned}$$

(90a)

$$\begin{aligned} I_b(x)&= \frac{1}{\epsilon } \ln |1 + \epsilon x| H(x_0 - x) \nonumber \\&\quad +\,\left[ \frac{1}{\epsilon } \ln |1 + \epsilon x_0| + \frac{x - x_0}{1 + \epsilon x_0} \right] H(x - x_0), \end{aligned}$$

(90b)

$$\begin{aligned} \xi (t,x)&= k_1^{-}(t,x) H(x_0 - x)H(x_0 - k_1^{-}(x,t)) \nonumber \\&\quad +\, k_2^{-}(t,x)H(x-x_0)H(x_0 - k_2^{-}(x,t)) \nonumber \\&\quad +\,k_3(t,x)H(x-x_0)H(k_3(t,x)-x_0), \end{aligned}$$

(90c)

$$\begin{aligned} \omega _n(t,x)&= k_1^{-}(t-t_n,x) H(x_0 - x)H(x_0 - k_1^{-}(x,t-t_n)) \nonumber \\&\quad +\,k_2^{-}(t-t_n,x)H(x-x_0)H(x_0 - k_2^{-}(t-t_n,x)) \nonumber \\&\quad +\,k_3(t-t_n,x)H(x-x_0)H(k_3(t-t_n,x)-x_0) \nonumber \\&\quad +\,k_5^-(t,x)H(x_0-x)H(k_5^-(t,x)-x_0), \end{aligned}$$

(90d)

$$\begin{aligned} \gamma _n(t,x)&= k_1^{+}(t-t_n,x) H(x_0 - x)H(x_0 - k_1^{+}(t-t_n,x)) \nonumber \\&\quad +\,k_2^{+}(t-t_n,x)H(x-x_0)H(x_0 - k_2^{+}(x,t-t_n)) \nonumber \\&\quad +\,k_4(t,x)H(x-x_0)H(k_4(t,x)-x_0) \nonumber \\&\quad +\,k_5^+(t,x)H(x_0-x)H(k_5^+(t,x)-x_0), \end{aligned}$$

(90e)

where

$$\begin{aligned} k_1^\pm (t,x)&= -\frac{1}{\epsilon }\left[ (1\pm \epsilon x)e^{\epsilon t} - 1\right] , \end{aligned}$$

(91a)

$$\begin{aligned} k_2^\pm (t,x)&= -\frac{1}{\epsilon }\left[ (1\pm \epsilon x_0)e^{\epsilon (t \pm [x-x_0]/[1\pm \epsilon x_0]} - 1\right] , \end{aligned}$$

(91b)

$$\begin{aligned} k_3(t,x)&= x - (1-\epsilon x_0)t, \end{aligned}$$

(91c)

$$\begin{aligned} k_4(t,x)&= x_0 + (1-\epsilon x_0)\left( \frac{1}{\epsilon } \ln \left| \frac{1-\epsilon x_0}{1 + \epsilon x_0}\right| - \frac{x - x_0}{1+\epsilon x_0} + t_n - t\right) , \end{aligned}$$

(91d)

$$\begin{aligned} k_5^\pm (t,x)&= x_0 + (1-\epsilon x_0)\left( \frac{1}{\epsilon } \ln \left| \frac{1 - \epsilon x_0}{1 \pm \epsilon x}\right| + t_n - t\right) . \end{aligned}$$

(91e)

For case 4,

$$\begin{aligned} I_a(x)&= x H(x_0 - x) + \left[ x_0 + \frac{x - x_0}{1 - \epsilon x_0}\right] H(x - x_0), \end{aligned}$$

(92a)

$$\begin{aligned} I_b(x)&= x H(x_0 - x) + \left[ x_0 + \frac{x - x_0}{1 + \epsilon x_0}\right] H(x - x_0), \end{aligned}$$

(92b)

$$\begin{aligned} \xi (t,x)&= (x-t)H(x_0 - x) \nonumber \\&\quad +\, k_6(t,x)H(x - x_0)H(x_0 - k_6(t,x)) \nonumber \\&\quad +\,k_7(t,x)H(x-x_0)H(k_7(t,x)-x_0) \end{aligned}$$

(92c)

$$\begin{aligned} \omega _n(t,x)&= k_8(t,x)H(x_0 - x)H(x_0 - k_8(t,x)) \nonumber \\&\quad +\,k_6(t-t_n,x)H(x - x_0)H(x_0 - k_6(t-t_n,x)) \nonumber \\&\quad +\,k_7(t-t_n,x)H(x-x_0)H(k_7(t-t_n,x)-x_0) \nonumber \\&\quad +\,k_9(t,x)H(x_0-x)H(k_9(t,x)-x_0) \end{aligned}$$

(92d)

$$\begin{aligned} \gamma _n(t,x)&=k_8(t,-x)H(x_0 - x)H(x_0 - k_8(t,-x)) \nonumber \\&\quad +\,k_9(t,-x)H(x_0 - x)H(k_9(t,-x) - x_0) \nonumber \\&\quad +\,k_{10}(t,x)H(x-x_0)H(x_0 - k_{10}(t,x)) \end{aligned}$$

(92e)

$$\begin{aligned}&\quad +\,_{11}(t,x)H(x-x_0)H(k_{11}(t,x)-x_0), \end{aligned}$$

(92f)

where

$$\begin{aligned} k_6(t,x)&= x_0 + \frac{x - x_0}{1 - \epsilon x_0} - t, \end{aligned}$$

(93a)

$$\begin{aligned} k_7(t,x)&= x - (1-\epsilon x_0)t, \end{aligned}$$

(93b)

$$\begin{aligned} k_8(t,x)&= x + t_n - t, \end{aligned}$$

(93c)

$$\begin{aligned} k_9(t,x)&= x_0 + (1-\epsilon x_0)(x-x_0 + t_n - t), \end{aligned}$$

(93d)

$$\begin{aligned} k_{10}(t,x)&= -x_0 - \frac{x-x_0}{1 + \epsilon x_0} + t_n - t, \end{aligned}$$

(93e)

$$\begin{aligned} k_{11}(t,x)&= x_0 + (1-\epsilon x_0)\left( -1 - \frac{x-x_0}{1 + \epsilon x_0} + t_n - t\right) . \end{aligned}$$

(93f)

Including an Interface

By placing an interface at $x = 1/2$, the vector equation (37) becomes

$$\begin{aligned} u_t^L + a^L(x)u_x^L&= 0, \end{aligned}$$

(94a)

$$\begin{aligned} v_t^L - b^L(x)v_x^L&= 0, \end{aligned}$$

(94b)

$$\begin{aligned} u^L(t,0)&= \alpha v^L(t,0), \end{aligned}$$

(94c)

$$\begin{aligned} v^L(t,1/2)&= v^R(t,1/2), \end{aligned}$$

(94d)

$$\begin{aligned} u^L(0,x)&= f^L(x), \end{aligned}$$

(94e)

$$\begin{aligned} v^L(0,x)&= 0 \end{aligned}$$

(94f)

and

$$\begin{aligned} u_t^R + a^R(x)u_x^R&= 0, \end{aligned}$$

(95a)

$$\begin{aligned} v_t^R - b^R(x)v_x^R&= 0, \end{aligned}$$

(95b)

$$\begin{aligned} v^R(t,1)&= \beta u^R(t,1), \end{aligned}$$

(95c)

$$\begin{aligned} u^R(t,1/2)&= u^L(t,1/2), \end{aligned}$$

(95d)

$$\begin{aligned} u^R(0,x)&= f^R(x), \end{aligned}$$

(95e)

$$\begin{aligned} v^R(0,x)&= 0, \end{aligned}$$

(95f)

where $\alpha = \sqrt{b^L(0)/a^L(0)}$, $\beta = \sqrt{a^R(1)/b^R(1)}$. The energy method applied to (94)–(95) yields

$$\begin{aligned} \frac{d \left( ||u^L||_2^2 + ||v^L||_2^2 + ||u^R||_2^2 + ||v^R||_2^2 \right) }{dt} =&\left[ a^R(1/2) - a^L(1/2)\right] u^L(t,0.5)^2 \nonumber \\&- \left[ b^R(1/2) - b^L(1/2)\right] v^R(t,1/2)^2 + \nonumber \\&+ \displaystyle \int _0^{1/2} a_x^L (u^L)^2 - b_x^L (v^L)^2 dx \nonumber \\&+\displaystyle \int _{1/2}^1 a_x^R (u^R)^2 - b_x^R (v^R)^2 dx \end{aligned}$$

(96)

and again we note that the first two terms on the right of (96) are zero if the wave speeds are continuous across the interface.

The discrete equations are given by

$$\begin{aligned} \mathbf{u}^L_t + \frac{1}{2}\left[ \mathbf{A}^L\mathbf{D} + \mathbf{DA}^L\right] \mathbf{u}^L - \frac{1}{2}{} \mathbf{U}^L\mathbf{D}{} \mathbf{a}^L&= \sigma _1\mathbf{H}^{-1}(u^L_0 - \alpha v^L_0)\mathbf{e}_0\nonumber \\&\quad + \sigma _2\mathbf{H}^{-1}(u_N^L - u_0^R)\mathbf{e}_N \end{aligned}$$

(97a)

$$\begin{aligned} \mathbf{v}^L_t - \frac{1}{2}\left[ \mathbf{B}^L\mathbf{D} + \mathbf{DB}^L\right] \mathbf{v}^L + \frac{1}{2}{} \mathbf{V}^L\mathbf{D}{} \mathbf{b}^L&= \sigma _3\mathbf{H}^{-1}(v^L_N - v^R_0)\mathbf{e}_N, \end{aligned}$$

(97b)

$$\begin{aligned} \mathbf{u}^R_t + \frac{1}{2}\left[ \mathbf{A}^R\mathbf{D} + \mathbf{DA}^R\right] \mathbf{u}^R - \frac{1}{2}{} \mathbf{U}^R\mathbf{D}{} \mathbf{a}^R&= \sigma _4\mathbf{H}^{-1}(u^R_0 - u^L_N)\mathbf{e}_0 \end{aligned}$$

(97c)

$$\begin{aligned} \mathbf{v}^R_t - \frac{1}{2}\left[ \mathbf{B}^R\mathbf{D} + \mathbf{DB}^R\right] \mathbf{v}^R + \frac{1}{2}{} \mathbf{V}^R\mathbf{D}{} \mathbf{b}^R&= \sigma _5\mathbf{H}^{-1}(v^R_0 - v^L_N)\mathbf{e}_0\nonumber \\&\quad +\, \sigma _6\mathbf{H}^{-1}(v^R_N - \beta u^R_N)\mathbf{e}_N. \end{aligned}$$

(97d)

A discrete energy estimate can be obtained as in previous sections, yielding

$$\begin{aligned} \frac{d\left( ||\mathbf{u}^L||^2_\mathbf{H} + ||\mathbf{v}^L||^2_\mathbf{H} + ||\mathbf{u}^R||^2_\mathbf{H} + ||\mathbf{v}^R||^2_\mathbf{H} \right) }{dt}&= (\mathbf{u}^L,\mathbf{U}^L \mathbf{Da}^L)_\mathbf{H} + (\mathbf{v}^L,\mathbf{V}^L \mathbf{Db}^L)_\mathbf{H} \nonumber \\&\quad +\, (\mathbf{u}^R,\mathbf{U}^R \mathbf{Da}^R)_\mathbf{H} + (\mathbf{v}^R,\mathbf{V}^R \mathbf{Db}^R)_\mathbf{H}\nonumber \\&\quad +\, \left( a_0^R - a_N^L\right) (u_N^L)^2 \nonumber \\&\quad -\,\left( b_0^R - b_N^L\right) (v_0^R)^2 \nonumber \\&\quad +\, {\mathbf{y}_0}^T \mathbf{M}_0 \mathbf{y}_0 + {\mathbf{y}_N}^T \mathbf{M}_N \mathbf{y}_N \nonumber \\&\quad +\, {\mathbf{y}_1}^T \mathbf{M}_1 \mathbf{y}_1 + {\mathbf{y}_2}^T \mathbf{M}_2 \mathbf{y}_2 \end{aligned}$$

(98)

where matrices

$$\begin{aligned} \mathbf{M}_0&= \begin{bmatrix} a_0^L + 2\sigma _1 \quad&\quad -\alpha \sigma _1\\ -\alpha \sigma _1&\quad -b_0^L \end{bmatrix}, \quad \quad \mathbf{M}_N = \begin{bmatrix} -a_N^R&-\beta \sigma _6\\ -\beta \sigma _6&\quad 2\sigma _6 + b_N^R \end{bmatrix}, \end{aligned}$$

(99a)

$$\begin{aligned} \mathbf{M}_1&= \begin{bmatrix} -a_0^R + 2\sigma _2 \quad&\quad -\sigma _2 - \sigma _4\\ -\sigma _2 - \sigma _4&\quad a_0^R + 2 \sigma _4 \end{bmatrix}, \quad \mathbf{M}_2 = \begin{bmatrix} b_N^L + 2\sigma _3 \quad&\quad -\sigma _3 - \sigma _5\\ -\sigma _3 - \sigma _5&\quad -b_N^L + 2\sigma _5 \end{bmatrix} \end{aligned}$$

(99b)

and vectors $\mathbf{y}_0^T = [u_0^L \quad v_0^L], \mathbf{y}_N^T = [u_N^R \quad v_N^R], \mathbf{y}_1^T = [u_N^L \quad u_0^R], \mathbf{y}_2^T = [v_N^L \quad v_0^R]$. The semi-discrete estimate (98) mimics the continuous estimate (96), with some additional dissipation if the matrices (99) are negative semi-definite. This can be accomplished by choosing for the boundary SAT terms $\sigma _1 = -a_0^L$, $\sigma _6 = -b_N^R$, and interface penalties corresponding to full upwinding, namely $\sigma _2 = \sigma _5 = 0$, $\sigma _3 = -b_N^L, \sigma _4 = -a_0^R$.