Answer to Question #225543 in Differential Equations for Asmita

Question

Obtain the solution of the wave equation ∂^u/∂t^2=c^2(∂^2u/∂x^2);0⩽x⩽π,t≥0 with the conditions u(0,t)=0,u(π,t)=0,u(x,0)=0 and ut(x,0)=sin3x+(sin5x/2).

Answer 1

According to the Fourier method, the solution to the equation has the form:

$u(x,t) = \sum\limits_{n = 1}^\infty {\left( {{A_n}\cos \frac{{\pi n}}{l}ct + {B_n}\sin \frac{{\pi n}}{l}ct} \right)} \sin \frac{{\pi n}}{l}x = \sum\limits_{n = 1}^\infty {\left( {{A_n}\cos \frac{{\pi n}}{\pi }ct + {B_n}\sin \frac{{\pi n}}{\pi }ct} \right)} \sin \frac{{\pi n}}{\pi }x = \sum\limits_{n = 1}^\infty {\left( {{A_n}\cos cnt + {B_n}\sin cnt} \right)} \sin nx$

Find the coefficients

${A_n} = \frac{2}{l}\int\limits_0^l {\varphi (x)} \sin \frac{{\pi nx}}{l}dx = \frac{2}{\pi }\int\limits_0^\pi {0 \cdot } \sin nxdx = 0$

${B_n} = \frac{2}{{\pi nc}}\int\limits_0^l {\psi (x)\sin \frac{{\pi nx}}{l}} dx = \frac{2}{{\pi nc}}\int\limits_0^\pi {(\sin 3x + \sin \frac{{5x}}{2})\sin nxdx} = \frac{2}{{\pi nc}}\int\limits_0^\pi {(\sin 3x\sin nx + \sin \frac{{5x}}{2}\sin nx)dx} = \\= \frac{1}{{\pi nc}}\int\limits_0^\pi {\left( {\cos (3 - n)x - \cos (3 + n)x + \cos \left( {\frac{5}{2} - n} \right)x - \cos \left( {\frac{5}{2} + n} \right)x} \right)} dx = \\= \frac{1}{{\pi nc}}\left( {\frac{1}{{3 - n}}\sin (3 - n)\left. x \right|_0^\pi - \frac{1}{{3 + n}}\sin (3 + n)\left. x \right|_0^\pi + \frac{1}{{\frac{5}{2} - n}}\sin \left( {\frac{5}{2} - n} \right)\left. x \right|_0^\pi - \frac{1}{{\frac{5}{2} + n}}\sin \left( {\frac{5}{2} + n} \right)\left. x \right|_0^\pi } \right) = \\= \frac{1}{{\pi nc}}\left( {0 - 0 + \frac{2}{{5 - 2n}}\sin \left( {\frac{5}{2} - n} \right)\pi - \frac{2}{{5 + 2n}}\sin \left( {\frac{5}{2} + n} \right)\pi } \right)=\\\frac{1}{{\pi nc}}\left( {\frac{2}{{5 - 2n}}\sin \left( {\frac{5}{2} - n} \right)\pi - \frac{2}{{5 + 2n}}\sin \left( {\frac{5}{2} + n} \right)\pi } \right)$

if $n \ne 3$

If $n=3$ then

${B_3} = \frac{2}{{3\pi c}}\int\limits_0^\pi {({{\sin }^2}3x + \sin \frac{{5x}}{2}\sin 3x)dx} = \frac{1}{{3\pi c}}\int\limits_0^\pi {\left( {1 - \cos 6x + \cos \left( { - \frac{x}{2}} \right) - \cos \frac{{11x}}{2}} \right)} dx =\\ = \frac{1}{{3\pi c}}\left( {\left. x \right|_0^\pi - \frac{1}{6}\sin 6\left. x \right|_0^\pi - 2\sin \left. {\frac{x}{2}} \right|_0^\pi - \frac{2}{{11}}\sin \left. {\frac{{11x}}{2}} \right|_0^\pi } \right) = \\= \frac{1}{{3\pi c}}\left( {\pi - 0 - 2 + \frac{2}{{11}}} \right) = \frac{1}{{3c}} - \frac{{20}}{{33\pi c}}$

Then

$u(x,t) = \sum\limits_{n = 1}^2 {\frac{1}{{\pi nc}}\left( {\frac{2}{{5 - 2n}}\sin \left( {\frac{5}{2} - n} \right)\pi - \frac{2}{{5 + 2n}}\sin \left( {\frac{5}{2} + n} \right)\pi } \right)\sin cnt\sin nx + } \left( {\frac{1}{{3c}} - \frac{{20}}{{33\pi c}}} \right)\sin 3ct\sin 3x + \sum\limits_{n = 4}^\infty {\frac{1}{{\pi nc}}\left( {\frac{2}{{5 - 2n}}\sin \left( {\frac{5}{2} - n} \right)\pi - \frac{2}{{5 + 2n}}\sin \left( {\frac{5}{2} + n} \right)\pi } \right)\sin cnt\sin nx}$

Answer: $u(x,t) = \sum\limits_{n = 1}^2 {\frac{1}{{\pi nc}}\left( {\frac{2}{{5 - 2n}}\sin \left( {\frac{5}{2} - n} \right)\pi - \frac{2}{{5 + 2n}}\sin \left( {\frac{5}{2} + n} \right)\pi } \right)\sin cnt\sin nx + } \left( {\frac{1}{{3c}} - \frac{{20}}{{33\pi c}}} \right)\sin 3ct\sin 3x + \sum\limits_{n = 4}^\infty {\frac{1}{{\pi nc}}\left( {\frac{2}{{5 - 2n}}\sin \left( {\frac{5}{2} - n} \right)\pi - \frac{2}{{5 + 2n}}\sin \left( {\frac{5}{2} + n} \right)\pi } \right)\sin cnt\sin nx}$