3.8电位分布满足拉普拉斯方程
22?????2??0即??022?x?y边界条件为?x?0,0?y?by?0,0?x?a?0?0??x?a,0?y?by?b,0?x?a?U?0?分离变量,设??f(x)g(y)代入方程并且两边同时除以f(x)g(y)得f''(x)g''(y)??0f(x)g(y)f''(x)设??则f(x)g''(y)=??g(y)(1)(2)方程可写成以下形式f''(x)??f(x)?0g''(y)??g(y)?0解方程(2)并要求满足边界条件?y?0,0?x?a?0?y?b,0?x?a?0得只有??0时方程满足要求n2?2n?解得?n?g(y)?sin(y)n2bb将代入方程(1),并满足边界条件?x?0,0?y?b?0
n?x)bn?n?则?n?Ansinh(x)sin(y)bb则电位的通解为解得fn(x)?Ansinh(???n?1?n?n?Ansinh(x)sin(y)bb
代入边界条件?x?a,0?y?b?U得?n?n?a)sin(y)?Ux?abbn?1m?两边同时乘以sin(y)并对y从0到b积分,并由bbn?m?n?m时?sin(y)sin(y)dy?00bbbn?m?bn?m时?sin(y)sin(y)dy?得0bb2b?n?n?Ansinh(a)sin(y)dy?0?bbn?1?bn?n???An?sinh(a)sin(y)dy0bbn?1bm??Amsinh(a)2bbm? ??Usin(y)dy(3)0b??Ansinh(?
(1)U=U0时,由方程(3)得bbm?U0(1?cosm?)?Amsinh(a)m?2b4U01则Am?(m?1,3,5)m?sinh(m?a)b代入电位的通解求得电位为n?sinh(x)?4U0n?b???sin(y)n?sinh(n?a)bn?1,3,5b?y(2)U?U0sin时bbbm??ym??0Usin(by)dy??0U0sinbsin(by)dyb=U0(m=1)2由方程(3)得bb?U0?A1sinh(a)22bU0则A1??sinh(a)b代入电位的通解求得电位为?xsinh()?yb??U0sin()?bsinh(a)b
3.9电位分布满足拉普拉斯方程22????2???0即??022?x?y边界条件为?x?0y?0?0?U0?x?a?0?C??y??分离变量,设??f(x)g(y)代入方程并且两边同时除以f(x)g(y)得f''(x)g''(y)??0f(x)g(y)f''(x)g''(y)设???则=?f(x)g(y)方程可写成以下形式f''(x)??f(x)?0g''(y)??g(y)?0(1)(2)得解方程(1)并要求满足边界条件?x?0?0?x?a?0只有??0时方程满足要求n2?2n?解得?n?f(x)?sin(x)n2aa将代入方程(2),并满足边界条件?y???C?n?ya解得gn(y)?Anen??yn?a则?n?Ansin(x)ea
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