From 76e84c411c103c08da6040e2c70920cfd3f7df2c Mon Sep 17 00:00:00 2001 From: Krishnakumar Gopalakrishnan Date: Sun, 7 Jun 2020 00:32:20 +0100 Subject: [PATCH] fixes step-52 doc issue; the correct math expr is sin(x,pi/omega)=0 --- examples/step-52/doc/intro.dox | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/examples/step-52/doc/intro.dox b/examples/step-52/doc/intro.dox index da38d817aa..4f4caf45bd 100644 --- a/examples/step-52/doc/intro.dox +++ b/examples/step-52/doc/intro.dox @@ -59,9 +59,9 @@ S=A\left(\frac{1}{v}\omega \cos(\omega t)(bx -x^2) + \sin(\omega t) \left(\Sigma_a (bx-x^2)+2D\right) \right). @f} Because the solution is a sine in time, we know that the exact solution -satisfies $\phi\left(x,\pi\right) = 0$. -Therefore, the error at time $t=\pi$ is simply the norm of the numerical -solution, i.e., $\|e(\cdot,t=\pi)\|_{L_2} = \|\phi_h(\cdot,t=\pi)\|_{L_2}$, +satisfies $\phi\left(x,\frac{\pi}{\omega}\right) = 0$. +Therefore, the error at time $t=\frac{\pi}{\omega}$ is simply the norm of the numerical +solution, i.e., $\|e(\cdot,t=\frac{\pi}{\omega})\|_{L_2} = \|\phi_h(\cdot,t=\frac{\pi}{\omega})\|_{L_2}$, and is particularly easily evaluated. In the code, we evaluate the $l_2$ norm of the vector of nodal values of $\phi_h$ instead of the $L_2$ norm of the associated spatial function, since the former is simpler to compute; however, -- 2.39.5