Please solve the following IBVP:
u t t = u x x on 0 < x < p i , t > 0. u_{tt} = u_{xx} \text{ on } 0 < x < pi, t > 0. u tt = u xx on 0 < x < p i , t > 0. u ( 0 , t ) = 0 , t ≥ 0. u(0,t) = 0, t \geq 0. u ( 0 , t ) = 0 , t ≥ 0. u ( p i , t ) = 0 , t ≥ 0. u(pi,t) = 0, t \geq 0. u ( p i , t ) = 0 , t ≥ 0. u ( x , 0 ) = sin 3 ( x ) 0 ≤ x ≤ p i . u(x,0) = \sin^3(x) \quad 0 \leq x \leq pi. u ( x , 0 ) = sin 3 ( x ) 0 ≤ x ≤ p i . u t ( x , 0 ) = sin ( 2 x ) 0 ≤ x ≤ p i . u_t(x,0) = \sin(2x) \quad 0 \leq x \leq pi. u t ( x , 0 ) = sin ( 2 x ) 0 ≤ x ≤ p i .
Solution:
The general solution of this IBVP is
u ( x , t ) = ∑ k = 1 ∞ ( A k cos ( k t ) + B k sin ( k t ) ) sin ( k x ) u(x,t) = \sum_{k=1}^{\infty} \left(A_k \cos(kt) + B_k \sin(kt)\right) \sin(kx) u ( x , t ) = k = 1 ∑ ∞ ( A k cos ( k t ) + B k sin ( k t ) ) sin ( k x )
where
A k = 2 π ∫ 0 π φ ( x ) sin ( k x ) d x , B k = 2 k π ∫ 0 π ψ ( x ) sin ( k x ) d x , φ ( x ) = sin 3 ( x ) , ψ ( x ) = sin ( 2 x ) . A_k = \frac{2}{\pi} \int_0^\pi \varphi(x) \sin(kx) \, dx, \quad B_k = \frac{2}{k\pi} \int_0^\pi \psi(x) \sin(kx) \, dx, \quad \varphi(x) = \sin^3(x), \quad \psi(x) = \sin(2x). A k = π 2 ∫ 0 π φ ( x ) sin ( k x ) d x , B k = kπ 2 ∫ 0 π ψ ( x ) sin ( k x ) d x , φ ( x ) = sin 3 ( x ) , ψ ( x ) = sin ( 2 x ) .
Then we have
B k = 2 k π ∫ 0 π sin ( 2 x ) ⋅ sin ( k x ) d x = { 1 2 , k = 2 ; 0 , k ≠ 2. B_k = \frac{2}{k\pi} \int_0^\pi \sin(2x) \cdot \sin(kx) \, dx = \begin{cases} \frac{1}{2}, & k = 2; \\ 0, & k \neq 2. \end{cases} B k = kπ 2 ∫ 0 π sin ( 2 x ) ⋅ sin ( k x ) d x = { 2 1 , 0 , k = 2 ; k = 2. A k = 2 π ∫ 0 π sin 3 ( x ) sin ( k x ) d x A_k = \frac{2}{\pi} \int_0^\pi \sin^3(x) \sin(kx) \, dx A k = π 2 ∫ 0 π sin 3 ( x ) sin ( k x ) d x A 1 = 2 π ∫ 0 π sin 4 ( x ) d x = 2 π ( − 3 8 sin ( x ) cos ( x ) − 1 4 sin 3 ( x ) cos ( x ) + 3 8 x ) ∣ 0 π = 3 4 ; A_1 = \frac{2}{\pi} \int_0^\pi \sin^4(x) \, dx = \frac{2}{\pi} \left( -\frac{3}{8} \sin(x) \cos(x) - \frac{1}{4} \sin^3(x) \cos(x) + \frac{3}{8} x \right) \Bigg|_0^\pi = \frac{3}{4}; A 1 = π 2 ∫ 0 π sin 4 ( x ) d x = π 2 ( − 8 3 sin ( x ) cos ( x ) − 4 1 sin 3 ( x ) cos ( x ) + 8 3 x ) ∣ ∣ 0 π = 4 3 ; A 2 = 2 π ∫ 0 π sin 3 ( x ) sin ( 2 x ) d x = 4 π ∫ 0 π sin 4 ( x ) cos ( x ) d x = A_2 = \frac{2}{\pi} \int_0^\pi \sin^3(x) \sin(2x) \, dx = \frac{4}{\pi} \int_0^\pi \sin^4(x) \cos(x) \, dx = A 2 = π 2 ∫ 0 π sin 3 ( x ) sin ( 2 x ) d x = π 4 ∫ 0 π sin 4 ( x ) cos ( x ) d x = = 4 π ∫ 0 π sin 4 ( x ) d ( sin ( x ) ) = 4 5 π ( sin 5 ( x ) ) ∣ 0 π = 0 ; = \frac{4}{\pi} \int_0^\pi \sin^4(x) \, d(\sin(x)) = \frac{4}{5\pi} \left( \sin^5(x) \right) \Bigg|_0^\pi = 0; = π 4 ∫ 0 π sin 4 ( x ) d ( sin ( x )) = 5 π 4 ( sin 5 ( x ) ) ∣ ∣ 0 π = 0 ; A 3 = 2 π ∫ 0 π sin 3 ( x ) sin ( 3 x ) d x = 2 π ∫ 0 π sin 3 ( x ) ( 3 sin ( x ) − 4 sin 3 ( x ) ) d x = A_3 = \frac{2}{\pi} \int_0^\pi \sin^3(x) \sin(3x) \, dx = \frac{2}{\pi} \int_0^\pi \sin^3(x) \left( 3 \sin(x) - 4 \sin^3(x) \right) \, dx = A 3 = π 2 ∫ 0 π sin 3 ( x ) sin ( 3 x ) d x = π 2 ∫ 0 π sin 3 ( x ) ( 3 sin ( x ) − 4 sin 3 ( x ) ) d x = = 6 π ∫ 0 π sin 4 ( x ) d x − 8 π ∫ 0 π sin 6 ( x ) d x = 6 π ⋅ 3 8 π − = \frac{6}{\pi} \int_0^\pi \sin^4(x) \, dx - \frac{8}{\pi} \int_0^\pi \sin^6(x) \, dx = \frac{6}{\pi} \cdot \frac{3}{8} \pi - = π 6 ∫ 0 π sin 4 ( x ) d x − π 8 ∫ 0 π sin 6 ( x ) d x = π 6 ⋅ 8 3 π − − 8 π ( − 1 6 sin 5 ( x ) cos ( x ) − 5 16 sin ( x ) cos ( x ) − 5 24 sin 3 ( x ) cos ( x ) + 5 16 x ) ∣ 0 π = = 9 4 − 8 π ⋅ 5 16 π = 9 4 − 5 2 = − 1 4 ; A k = 2 π ∫ 0 π sin 3 ( x ) sin ( k x ) d x = 0 if k > 3. \begin{array}{l}
- \frac{8}{\pi} \left(- \frac{1}{6} \sin^{5}(x) \cos(x) - \frac{5}{16} \sin(x) \cos(x) - \frac{5}{24} \sin^{3}(x) \cos(x) + \frac{5}{16} x\right) \Bigg|_{0}^{\pi} = \\
= \frac{9}{4} - \frac{8}{\pi} \cdot \frac{5}{16} \pi = \frac{9}{4} - \frac{5}{2} = -\frac{1}{4}; \\
A_{k} = \frac{2}{\pi} \int_{0}^{\pi} \sin^{3}(x) \sin(kx) \, dx = 0 \text{ if } k > 3.
\end{array} − π 8 ( − 6 1 sin 5 ( x ) cos ( x ) − 16 5 sin ( x ) cos ( x ) − 24 5 sin 3 ( x ) cos ( x ) + 16 5 x ) ∣ ∣ 0 π = = 4 9 − π 8 ⋅ 16 5 π = 4 9 − 2 5 = − 4 1 ; A k = π 2 ∫ 0 π sin 3 ( x ) sin ( k x ) d x = 0 if k > 3.
So we have the next solution
u ( x , t ) = ∑ k = 1 ∞ ( A k cos ( k t ) + B k sin ( k t ) ) sin ( k x ) = = 1 2 sin ( 2 t ) sin ( 2 x ) + 3 4 cos ( t ) sin ( x ) − 1 4 cos ( 3 t ) sin ( 3 x ) . \begin{array}{l}
u(x, t) = \sum_{k=1}^{\infty} \left(A_{k} \cos(kt) + B_{k} \sin(kt)\right) \sin(kx) = \\
= \frac{1}{2} \sin(2t) \sin(2x) + \frac{3}{4} \cos(t) \sin(x) - \frac{1}{4} \cos(3t) \sin(3x).
\end{array} u ( x , t ) = ∑ k = 1 ∞ ( A k cos ( k t ) + B k sin ( k t ) ) sin ( k x ) = = 2 1 sin ( 2 t ) sin ( 2 x ) + 4 3 cos ( t ) sin ( x ) − 4 1 cos ( 3 t ) sin ( 3 x ) . 
#339914 on Dec 2023