Answer to Question #225542 in Differential Equations for Asmita

Question

Find the solution of the wave equation ∂^u/∂t^2=c^2(∂^2u/∂x^2) subject to u(0,t)=0,u(π,t)=0 for all t, u(x,0)=x^2(x-π),ut(x,0)=Sin3x for 0⩽x⩽π.

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

${A_n} = \frac{2}{l}\int\limits_0^l {\varphi (x)} \sin \frac{{\pi nx}}{l}dx = \frac{2}{\pi }\int\limits_0^\pi {{x^2}\left( {x - \pi } \right)} \sin nxdx = \frac{2}{\pi }\int\limits_0^\pi {\left( {{x^3} - \pi {x^2}} \right)} \sin nxdx = \left| {\begin{matrix} {u = {x^3} - \pi {x^2}}&{dv = \sin nxdx}\\ {du = (3{x^2} - 2\pi x)dx}&{v = - \frac{1}{n}\cos nx} \end{matrix}} \right| = \\= \frac{2}{\pi }\left( { - \frac{{{x^3} - \pi {x^2}}}{n}\cos n\left. x \right|_0^\pi + \frac{1}{n}\int\limits_0^\pi {(3{x^2} - 2\pi x)\cos nxdx} } \right) = \left| {\begin{matrix} {u = 3{x^2} - 2\pi x}&{dv = \cos nxdx}\\ {du = (6x - 2\pi )dx}&{v = \frac{1}{n}\sin nx} \end{matrix}} \right| = \\ = \frac{2}{\pi }\left( {0 + \frac{1}{n}\left( {\frac{{3{x^2} - 2\pi x}}{n}\sin n\left. x \right|_0^\pi - \frac{1}{n}\int\limits_0^\pi {(6x - 2\pi )\sin nxdx} } \right)} \right) = \left| {\begin{matrix} {u = 6x - 2\pi }&{dv = \sin nxdx}\\ {du = 6dx}&{v = - \frac{1}{n}\cos nx} \end{matrix}} \right| =\\ = \frac{2}{\pi }\left( {\frac{1}{n}\left( {0 - \frac{1}{n}\left( { - \frac{{6x - 2\pi }}{n}\cos n\left. x \right|_0^\pi + \frac{1}{n}\int\limits_0^\pi {6\cos nxdx} } \right)} \right)} \right) = \frac{2}{\pi }\left( {\frac{1}{n}\left( { - \frac{1}{n}\left( { - \frac{{6\pi - 2\pi }}{n}\cos \pi n + \frac{{0 - 2\pi }}{n}\cos 0 + \frac{6}{{{n^2}}}\sin n\left. x \right|_0^\pi } \right)} \right)} \right) =\\ = - \frac{2}{{\pi {n^2}}}\left( { - \frac{{4\pi }}{n}{{( - 1)}^n} - \frac{{2\pi }}{n} + 0} \right) = \frac{8}{{{n^3}}}{( - 1)^n} + \frac{4}{{{n^3}}}$

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

if $n \ne 3$

If $n=3$ then

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

Then

$u(x,t) = \sum\limits_{n = 1}^\infty {\left( {{A_n}\cos cnt + {B_n}\sin cnt} \right)} \sin nx =\\= ( - 8 + 4)\cos ct\sin x + \left( {1 + \frac{1}{2}} \right)\cos 2ct\sin 2x + \left( { - \frac{8}{{27}} + \frac{4}{{27}}} \right)\cos 3ct\sin 3x + \frac{1}{{3c}}\sin 3ct\sin 3x + \sum\limits_{n = 4}^\infty {\cos cnt\sin nx = }\\= - 4\cos ct\sin x + \frac{3}{2}\cos 2ct\sin 2x - \frac{4}{{27}}\cos 3ct\sin 3x + \frac{1}{{3c}}\sin 3ct\sin 3x + \sum\limits_{n = 4}^\infty {\cos cnt\sin nx}$

Answer: $u(x,t)=- 4\cos ct\sin x + \frac{3}{2}\cos 2ct\sin 2x - \frac{4}{{27}}\cos 3ct\sin 3x + \frac{1}{{3c}}\sin 3ct\sin 3x + \sum\limits_{n = 4}^\infty {\cos cnt\sin nx}$