程函方程（Eikonal equation) 是一个非线性偏微分方程[1] s y s := ( u ( x , t ) t ) ) 2 + ( u ( x , t ) x ) 2 − 4 = 0 {\displaystyle sys:=(u(x,t)_{t}))^{2}+(u(x,t)_{x})^{2}-4=0} Remove ads行波解 程函方程行波图 g [ 1 ] := u ( x , t ) = C 4 − ( 2 ∗ ( C 1 + C 2 ∗ x + C 3 ∗ t ) ) / C 3 2 + C 2 2 {\displaystyle g[1]:={u(x,t)=C4-(2*(C1+C2*x+C3*t))/{\sqrt {C3^{2}+C2^{2}}}}} g [ 2 ] := u ( x , t ) = C 4 + ( 2 ∗ ( C 1 + C 2 ∗ x + C 3 ∗ t ) ) / C 3 2 + C 2 2 {\displaystyle g[2]:={u(x,t)=C4+(2*(C1+C2*x+C3*t))/{\sqrt {C3^{2}+C2^{2}}}}} Remove ads参考文献Loading content...Loading related searches...Wikiwand - on Seamless Wikipedia browsing. On steroids.Remove ads
![{\displaystyle g[1]:={u(x,t)=C4-(2*(C1+C2*x+C3*t))/{\sqrt {C3^{2}+C2^{2}}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/1dafe0881c9d08a10faf60b54880ffbe77ccbda9)
![{\displaystyle g[2]:={u(x,t)=C4+(2*(C1+C2*x+C3*t))/{\sqrt {C3^{2}+C2^{2}}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/a7df919c8349f57d7a5c653c4820b876a4c8f85f)