Partiele differentiaalvergelijking

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Partiele differentiaalvergelijking

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Na homogenisering volgt:
\(\phi_1 = c_1 cos( \sqrt{ \lambda_1} x) + c_2sin( \sqrt{\lambda_1} x)\phi_1(0)=0 \rightarrow c_1=0 \rightarrow \phi_1 =c_2sin( \sqrt{ \lambda_1} x ) \\\phi_2 = c_3 cos( 2 \sqrt{ \lambda_2} x) + c_4 sin( 2 \sqrt{\lambda_2} x)\phi_2(2)=0 \rightarrow c_3cos(4 \sqrt{ \lambda_2 } ) = - c_4 sin(4 \sqrt{\lambda_2} ) \rightarrow c_3 = -c_4 tan(4 \sqrt{\lambda_2}) \\\phi_2 = -c_4 tan( 4 \sqrt{ \lambda_2 }) cos(2 \sqrt{ \lambda_2} x) +c_4 sin(2 \sqrt{\lambda_2}x) \\u_1(1,t) = u_2(1,t) \rightarrowc_2 sin ( \sqrt{ \lambda_1} ) = c_4 [ - tan( 4 \sqrt{ \lambda_2 }) cos(2 \sqrt{ \lambda_2} ) + sin(2 \sqrt{\lambda_2})] \\u_1_x(1,t) =4 u _2_x(1,t) \rightarrowc_2 \sqrt{ \lambda_1 } cos( \sqrt{ \lambda_1} ) = 4 c_4 [ tan(4 \sqrt{ \lambda_2} ) sin( \sqrt{ \lambda_2} ) + 2 \sqrt{ \lambda_2} cos( 2 \sqrt{\lambda_2})]\)
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