comsol实例学习muchong com求解器

1、 Copyright 2016 COMSOL. Any of the images, text, and equations here may be copied and modified for your own internal use. All trademarks are the property oftheir respective owners. See.p = 2 Nk = 4 N/mu?p = 2 Nk = 4 N/mu?f (u)usolution= 0.5f (u) = p k u = 0f (u) = 2 - 4 u = 0uf (u) = p - k uusolutio

2、n= 0.5uu0=0f (u)f (u) = p k uup，kuf (u) = p(u) k(u)u mu2 uk = exp(u) N/mp = 2 Nuf (u) = 2 exp(u)uuf (u) = 2 exp(u)uuu0=0iuif(ui)f(ui)00.0002.0001.000u1= 2uiuif(ui)f(ui)00.0002.0001.00012.000-12.77-22.167u1= 1.424uiuif(ui)f(ui)00.0002.0001.00012.000-12.77-22.16721.424-3.915-10.069u1= 1.035uiuif(ui)f(

3、ui)00.0002.0001.00012.000-12.77-22.16721.424-3.915-10.06931.035-0.914-5.729uiuif(ui)f(ui)00.0002.0001.00012.000-12.77-22.16721.424-3.915-10.06931.035-0.914-5.72940.876-0.104-4.505uiuif(ui)f(ui)00.0002.0001.00012.000-12.77-22.16721.424-3.915-10.06931.035-0.914-5.72940.876-0.104-4.50550.8530.002-4.348

4、60.8520.001-4.342ur(u)iuif(ui)|ui-1 - ui|f(ui-1) - f(ui)|00.0002.00012.000-12.772.00010.7721.424-3.9150.5768.85531.035-0.9140.3893.00140.876-0.1040.1590.81050.8530.0020.0230.10260.8520.0010.0010.001f(u0)uu0=0if: |f(u)| |f(u )|i+1if(u1)f(u0)uu0=0if: |f(u)| |f(u )|i+1ifind: u = ui+1 uif(u1)f(u0)uu0=0i

5、f: |f(u)| |f(u )|i+1ifind: u = ui+1 ui pick: 0 |f(u )|i+1if(u + u )find: u = ui+1 ui pick: 0 1 find: |f(ui + u)| repeat0- ui| 缩放因子 相对容差|ui-1Solutionpukpf (u) = 0.05 exp( u )uf (u) = 0.05 exp( u )uf (u) = 0.50 exp( u )uf (u) = 0.05 exp( u )uf (u) = 0.50 exp( u )uf (u) = 1.00 exp( u )uf (u) = 0.05 exp

6、( u )uf (u) = 0.50 exp( u )uf (u) = 1.00 exp( u )uf (u) = 2.00 exp( u )uu (p)pp0p1p2p3u (p)pp0p1p2p3p3p3puk(u)k (u)u (p)p4p3p1p2upf (u) = 2 exp(u)ukNL= exp(u)u0kLIN= exp(u0=0)k() = (1-)kLIN+kNLu0f (u) = 2 - k() u: 0 1f =2 k(=0.0)u = 2 - uf =2 k(=0.0)u f =2 k(=0.5)u f =2 k(=0.0)u f =2 k(=0.5)u f =2 k

7、(=1.0)up = 2 Nku?k = 1.5k (u)k = 1k = 0.5u1.92.1k = 1.5k (u)k = 1k = 0.5u2.11.9T=0Q=100W(1+T/20K)W/(m*K)k1W/(m*K)0.1+exp(-(T/25K)2)W/(m*K)TErroru(x)xabu(x)xabu(x)xabu(x)xabu(x)xabMore Elements：收敛速率强烈依赖于阶次、类型，以及单元分布，这种收敛速率对不同的问题也完全不同。Solution10010210410610010-210-410-6 Copyright 2016 COMSOL. Any of t

8、he images, text, and equations here may be copied and modified for your own internal use. All trademarks are the property oftheir respective owners. See.k1 = 3 N/mp = 1 Nk = 2 N/mk3 = 2 N/m2u1u2k1u2k3(u1-u2)k2u1pk3(u2-u1)f (u1) = k2u1 k3 (u1 u2)f (u2) = k1u2 k3 (u2 u1) + pf (u1) = k2u1 k3 (u1 u2)f (

9、u2) = k1u2 k3 (u2 u1) + pk + k3- k3u0f (u) =21- p- kk + k3 u2 31f(u)=Ku-busolution=K-1br(u) = f(u) f (u)u2u1f(u0) = Ku0-b f(u0) = Ku0 = 0usolution=K-1bu2u1f(u0) = Ku0-b f(u0) = Ku0 = 0u2u1Memory= A dof2 + B dof + (1GB) Copyright 2016 COMSOL. Any of the images, text, and equations here may be copied

10、and modified for your own internal use. All trademarks are the property oftheir respective owners. See.Start with uV,0 & uT,0fV (uV) = KV(uT,0)uV - bVfT (uT,uV) = KT(uT,0)uT bT (uV)uV,0 & uT,0fV (uV) = KV(uT) uV - bVfT (uT,uV) = KT(uT)uT bT (uV)uV,0 & uT,0fV (uV) = KV(uT) uV - bVfT (uT,uV) = KT(uT)uT bT (uV)uV,i , uT,i初始化:uV,i+1= uV,i - KV(uT,i, uV,i)-1fV(uT,i, uV,i)求解:uT,i+1= uT,i - KT(uT,i, uV,i+1)-1fT(uT,i, uV,i+1 )求解:i = i+1迭代:ui ui-1 & fi fi-1 容差满足条件就结束:fT (uT) = KTuT - bT (uV)fV (uV) = KVuV - bVfV (uV , uT )uV KVuV 0b