ANSWER on Question #76480 – Math – Differential Equations
QUESTION
Obtain a solution of the wave equation
∂ 2 u ( x , t ) ∂ t 2 = 16 ⋅ ∂ 2 u ( x , t ) ∂ x 2 \frac {\partial^ {2} u (x , t)}{\partial t ^ {2}} = 16 \cdot \frac {\partial^ {2} u (x , t)}{\partial x ^ {2}} ∂ t 2 ∂ 2 u ( x , t ) = 16 ⋅ ∂ x 2 ∂ 2 u ( x , t )
for 0 ≤ x ≤ π 0 \leq x \leq \pi 0 ≤ x ≤ π t > 0 t > 0 t > 0
{ u ( 0 , t ) = 0 u ( π , t ) = 0 − boundary conditions \left\{ \begin{array}{l} u (0, t) = 0 \\ u (\pi , t) = 0 \end{array} - \text{boundary conditions} \right. { u ( 0 , t ) = 0 u ( π , t ) = 0 − boundary conditions { u ( x , 0 ) = x ( π − x ) ∂ u ( x , 0 ) ∂ t = 0 − initial conditions \left\{ \begin{array}{c} u (x, 0) = x (\pi - x) \\ \frac {\partial u (x , 0)}{\partial t} = 0 \end{array} - \text{initial conditions} \right. { u ( x , 0 ) = x ( π − x ) ∂ t ∂ u ( x , 0 ) = 0 − initial conditions SOLUTION
0 STEP: separation of variables.
Let
u ( x , t ) = X ( x ) T ( t ) → { ∂ 2 u ( x , t ) ∂ t 2 = ∂ 2 ∂ t 2 ( X ( x ) T ( t ) ) = X ( x ) ⋅ d 2 ( T ( t ) ) d t 2 = X ( x ) ⋅ T ′ ′ ( t ) ∂ 2 u ( x , t ) ∂ x 2 = ∂ 2 ∂ x 2 ( X ( x ) T ( t ) ) = T ( t ) ⋅ d 2 ( X ( x ) ) d x 2 = X ′ ′ ( x ) ⋅ T ( t ) u (x, t) = X (x) T (t) \rightarrow \left\{ \begin{array}{l} \frac {\partial^ {2} u (x , t)}{\partial t ^ {2}} = \frac {\partial^ {2}}{\partial t ^ {2}} \big (X (x) T (t) \big) = X (x) \cdot \frac {d ^ {2} \big (T (t) \big)}{d t ^ {2}} = X (x) \cdot T ^ {\prime \prime} (t) \\ \frac {\partial^ {2} u (x , t)}{\partial x ^ {2}} = \frac {\partial^ {2}}{\partial x ^ {2}} \big (X (x) T (t) \big) = T (t) \cdot \frac {d ^ {2} \big (X (x) \big)}{d x ^ {2}} = X ^ {\prime \prime} (x) \cdot T (t) \end{array} \right. u ( x , t ) = X ( x ) T ( t ) → ⎩ ⎨ ⎧ ∂ t 2 ∂ 2 u ( x , t ) = ∂ t 2 ∂ 2 ( X ( x ) T ( t ) ) = X ( x ) ⋅ d t 2 d 2 ( T ( t ) ) = X ( x ) ⋅ T ′′ ( t ) ∂ x 2 ∂ 2 u ( x , t ) = ∂ x 2 ∂ 2 ( X ( x ) T ( t ) ) = T ( t ) ⋅ d x 2 d 2 ( X ( x ) ) = X ′′ ( x ) ⋅ T ( t )
Boundary conditions:
{ u ( 0 , t ) = 0 u ( π , t ) = 0 → { u ( 0 , t ) = X ( 0 ) T ( t ) = 0 , ∀ t > 0 u ( π , t ) = X ( π ) T ( t ) = 0 , ∀ t > 0 → { X ( 0 ) = 0 } { X ( π ) = 0 \left\{ \begin{array}{l} u (0, t) = 0 \\ u (\pi , t) = 0 \end{array} \right. \to \left\{ \begin{array}{l} u (0, t) = X (0) T (t) = 0, \forall t > 0 \\ u (\pi , t) = X (\pi) T (t) = 0, \forall t > 0 \end{array} \right. \to \boxed {\{X (0) = 0 \}} \\ \left\{ \begin{array}{l} X (\pi) = 0 \end{array} \right. { u ( 0 , t ) = 0 u ( π , t ) = 0 → { u ( 0 , t ) = X ( 0 ) T ( t ) = 0 , ∀ t > 0 u ( π , t ) = X ( π ) T ( t ) = 0 , ∀ t > 0 → { X ( 0 ) = 0 } { X ( π ) = 0
Then,
∂ 2 u ( x , t ) ∂ t 2 = 16 ⋅ ∂ 2 u ( x , t ) ∂ x 2 → X ( x ) ⋅ T ′ ′ ( t ) = 16 ⋅ X ′ ′ ( x ) ⋅ T ( t ) ∣ × 1 16 X ( x ) T ( t ) → \frac {\partial^ {2} u (x , t)}{\partial t ^ {2}} = 16 \cdot \frac {\partial^ {2} u (x , t)}{\partial x ^ {2}} \rightarrow X (x) \cdot T ^ {\prime \prime} (t) = 16 \cdot X ^ {\prime \prime} (x) \cdot T (t) | \times \frac {1}{16X(x)T(t)} \rightarrow ∂ t 2 ∂ 2 u ( x , t ) = 16 ⋅ ∂ x 2 ∂ 2 u ( x , t ) → X ( x ) ⋅ T ′′ ( t ) = 16 ⋅ X ′′ ( x ) ⋅ T ( t ) ∣ × 16 X ( x ) T ( t ) 1 → X ( x ) ⋅ T ′ ′ ( t ) 16 X ( x ) T ( t ) = 16 ⋅ X ′ ′ ( x ) ⋅ T ( t ) 16 X ( x ) T ( t ) → 1 16 ⋅ T ′ ′ ( t ) T ( t ) = X ′ ′ ( x ) X ( x ) = − λ \frac {X (x) \cdot T ^ {\prime \prime} (t)}{16X(x)T(t)} = \frac {16 \cdot X ^ {\prime \prime} (x) \cdot T (t)}{16X(x)T(t)} \rightarrow \boxed {\frac {1}{16} \cdot \frac {T ^ {\prime \prime} (t)}{T (t)} = \frac {X ^ {\prime \prime} (x)}{X (x)} = - \lambda} 16 X ( x ) T ( t ) X ( x ) ⋅ T ′′ ( t ) = 16 X ( x ) T ( t ) 16 ⋅ X ′′ ( x ) ⋅ T ( t ) → 16 1 ⋅ T ( t ) T ′′ ( t ) = X ( x ) X ′′ ( x ) = − λ
1 STEP: We solve the Sturm-Liouville problem.
(More information: https://en.wikipedia.org/wiki/Sturm%E2%80%93Liouville_theory)
In our case,
{ X ′ ′ ( x ) X ( x ) = − λ X ( 0 ) = 0 X ( π ) = 0 \left\{ \begin{array}{l} \frac {X ^ {\prime \prime} (x)}{X (x)} = - \lambda \\ X (0) = 0 \\ X (\pi) = 0 \end{array} \right. ⎩ ⎨ ⎧ X ( x ) X ′′ ( x ) = − λ X ( 0 ) = 0 X ( π ) = 0 X ′ ′ ( x ) X ( x ) = − λ → X ′ ′ ( x ) = − λ X ( x ) → X ′ ′ ( x ) + λ X ( x ) = 0 \frac {X ^ {\prime \prime} (x)}{X (x)} = - \lambda \rightarrow X ^ {\prime \prime} (x) = - \lambda X (x) \rightarrow X ^ {\prime \prime} (x) + \lambda X (x) = 0 X ( x ) X ′′ ( x ) = − λ → X ′′ ( x ) = − λ X ( x ) → X ′′ ( x ) + λ X ( x ) = 0
Let us find the solutions of the given equation in the form
X ( x ) = e k x → X ′ ′ ( x ) = k 2 ⋅ e k x X (x) = e ^ {k x} \rightarrow X ^ {\prime \prime} (x) = k ^ {2} \cdot e ^ {k x} X ( x ) = e k x → X ′′ ( x ) = k 2 ⋅ e k x
Then,
X ′ ′ ( x ) + λ X ( x ) = 0 → k 2 ⋅ e k x + λ e k x = 0 → e k x ( k 2 + λ ) = 0 → k 2 = − λ X ^ {\prime \prime} (x) + \lambda X (x) = 0 \rightarrow k ^ {2} \cdot e ^ {k x} + \lambda e ^ {k x} = 0 \rightarrow e ^ {k x} (k ^ {2} + \lambda) = 0 \rightarrow k ^ {2} = - \lambda X ′′ ( x ) + λ X ( x ) = 0 → k 2 ⋅ e k x + λ e k x = 0 → e k x ( k 2 + λ ) = 0 → k 2 = − λ k 2 = − λ → [ k 1 = − λ = i λ k 2 = − − λ = − i λ k ^ {2} = - \lambda \rightarrow \left[\begin{array}{l}k _ {1} = \sqrt {- \lambda} = i \sqrt {\lambda}\\k _ {2} = - \sqrt {- \lambda} = - i \sqrt {\lambda}\end{array}\right. k 2 = − λ → [ k 1 = − λ = i λ k 2 = − − λ = − i λ
Then,
X ( x ) = C 1 e i λ x + C 2 e − i λ x ≡ A 1 cos ( λ x ) + A 2 sin ( λ x ) X (x) = C _ {1} e ^ {i \sqrt {\lambda} x} + C _ {2} e ^ {- i \sqrt {\lambda} x} \equiv A _ {1} \cos (\sqrt {\lambda} x) + A _ {2} \sin (\sqrt {\lambda} x) X ( x ) = C 1 e i λ x + C 2 e − i λ x ≡ A 1 cos ( λ x ) + A 2 sin ( λ x ) X ( x ) = A 1 cos ( λ x ) + A 2 sin ( λ x ) \boxed {X (x) = A _ {1} \cos (\sqrt {\lambda} x) + A _ {2} \sin (\sqrt {\lambda} x)} X ( x ) = A 1 cos ( λ x ) + A 2 sin ( λ x ) X ( 0 ) = 0 = A 1 cos ( λ ⋅ 0 ) + A 2 sin ( λ ⋅ 0 ) = A 1 cos ( 0 ) + A 2 sin ( 0 ) = A 1 ⋅ 1 + A 2 ⋅ 0 → X (0) = 0 = A _ {1} \cos (\sqrt {\lambda} \cdot 0) + A _ {2} \sin (\sqrt {\lambda} \cdot 0) = A _ {1} \cos (0) + A _ {2} \sin (0) = A _ {1} \cdot 1 + A _ {2} \cdot 0 \rightarrow X ( 0 ) = 0 = A 1 cos ( λ ⋅ 0 ) + A 2 sin ( λ ⋅ 0 ) = A 1 cos ( 0 ) + A 2 sin ( 0 ) = A 1 ⋅ 1 + A 2 ⋅ 0 → A 1 = 0 \boxed {A _ {1} = 0} A 1 = 0 X ( π ) = 0 = A 2 sin ( λ π ) → sin ( λ π ) = 0 → λ π = π n , n = 1 , 2 , 3 , … X (\pi) = 0 = A _ {2} \sin (\sqrt {\lambda} \pi) \rightarrow \sin (\sqrt {\lambda} \pi) = 0 \rightarrow \sqrt {\lambda} \pi = \pi n, n = 1, 2, 3, \dots X ( π ) = 0 = A 2 sin ( λ π ) → sin ( λ π ) = 0 → λ π = πn , n = 1 , 2 , 3 , … λ n = n 2 , n = 1 , 2 , 3 , … \boxed {\lambda_ {n} = n ^ {2}, n = 1, 2, 3, \dots} λ n = n 2 , n = 1 , 2 , 3 , …
Conclusion,
{ X n ( x ) = A ⋅ sin ( n x ) λ n = n 2 n = 1 , 2 , 3 , … \left\{ \begin{array}{c} X _ {n} (x) = A \cdot \sin (n x) \\ \lambda_ {n} = n ^ {2} \\ n = 1, 2, 3, \ldots \end{array} \right. ⎩ ⎨ ⎧ X n ( x ) = A ⋅ sin ( n x ) λ n = n 2 n = 1 , 2 , 3 , …
2 STEP: Finding the general solution.
1 16 ⋅ T ′ ′ ( t ) T ( t ) = X ′ ′ ( x ) X ( x ) = − λ n → 1 16 ⋅ T ′ ′ ( t ) T ( t ) = − n 2 → T ′ ′ ( t ) = − 16 n 2 T ( t ) → \frac {1}{16} \cdot \frac {T ^ {\prime \prime} (t)}{T (t)} = \frac {X ^ {\prime \prime} (x)}{X (x)} = - \lambda_ {n} \rightarrow \frac {1}{16} \cdot \frac {T ^ {\prime \prime} (t)}{T (t)} = - n ^ {2} \rightarrow T ^ {\prime \prime} (t) = - 16 n ^ {2} T (t) \rightarrow 16 1 ⋅ T ( t ) T ′′ ( t ) = X ( x ) X ′′ ( x ) = − λ n → 16 1 ⋅ T ( t ) T ′′ ( t ) = − n 2 → T ′′ ( t ) = − 16 n 2 T ( t ) → T ′ ′ ( t ) + 16 n 2 T ( t ) = 0 T ^ {\prime \prime} (t) + 16 n ^ {2} T (t) = 0 T ′′ ( t ) + 16 n 2 T ( t ) = 0
Let us find the solutions of the given equation in the form
T ( t ) = e k t → T ′ ′ ( t ) = k 2 ⋅ e k t T (t) = e ^ {k t} \rightarrow T ^ {\prime \prime} (t) = k ^ {2} \cdot e ^ {k t} T ( t ) = e k t → T ′′ ( t ) = k 2 ⋅ e k t
Then,
T ′ ′ ( t ) + 16 n 2 T ( t ) = 0 → k 2 ⋅ e k t + 16 n 2 e k t = 0 → e k t ( k 2 + 16 n 2 ) = 0 → k 2 = − 16 n 2 T ^ {\prime \prime} (t) + 16 n ^ {2} T (t) = 0 \rightarrow k ^ {2} \cdot e ^ {k t} + 16 n ^ {2} e ^ {k t} = 0 \rightarrow e ^ {k t} (k ^ {2} + 16 n ^ {2}) = 0 \rightarrow k ^ {2} = - 16 n ^ {2} T ′′ ( t ) + 16 n 2 T ( t ) = 0 → k 2 ⋅ e k t + 16 n 2 e k t = 0 → e k t ( k 2 + 16 n 2 ) = 0 → k 2 = − 16 n 2 k 2 = − 16 n 2 → [ k 1 = − 16 n 2 = 4 i n k 2 = − − 16 n 2 = − 4 i n k ^ {2} = - 16 n ^ {2} \rightarrow \left[\begin{array}{l}k _ {1} = \sqrt {- 16 n ^ {2}} = 4 i n\\k _ {2} = - \sqrt {- 16 n ^ {2}} = - 4 i n\end{array}\right. k 2 = − 16 n 2 → [ k 1 = − 16 n 2 = 4 in k 2 = − − 16 n 2 = − 4 in
Then,
T n ( x ) = C 1 e 4 i n t + C 2 e − 4 i n t ≡ A 1 cos ( 4 n t ) + A 2 sin ( 4 n t ) T _ {n} (x) = C _ {1} e ^ {4 i n t} + C _ {2} e ^ {- 4 i n t} \equiv A _ {1} \cos (4 n t) + A _ {2} \sin (4 n t) T n ( x ) = C 1 e 4 in t + C 2 e − 4 in t ≡ A 1 cos ( 4 n t ) + A 2 sin ( 4 n t ) T n ( x ) = A n ( 1 ) cos ( 4 n t ) + A n ( 1 ) sin ( 4 n t ) \boxed {T _ {n} (x) = A _ {n} ^ {(1)} \cos (4 n t) + A _ {n} ^ {(1)} \sin (4 n t)} T n ( x ) = A n ( 1 ) cos ( 4 n t ) + A n ( 1 ) sin ( 4 n t )
Then,
u n ( x , t ) = X n ( x ) ⋅ T n ( t ) = ( A ⋅ sin ( n x ) ) ⋅ ( A n ( 1 ) cos ( 4 n t ) + A n ( 1 ) sin ( 4 n t ) ) → u _ {n} (x, t) = X _ {n} (x) \cdot T _ {n} (t) = (A \cdot \sin (n x)) \cdot \left(A _ {n} ^ {(1)} \cos (4 n t) + A _ {n} ^ {(1)} \sin (4 n t)\right) \rightarrow u n ( x , t ) = X n ( x ) ⋅ T n ( t ) = ( A ⋅ sin ( n x )) ⋅ ( A n ( 1 ) cos ( 4 n t ) + A n ( 1 ) sin ( 4 n t ) ) → u n ( x , t ) = ( B n ( 1 ) cos ( 4 n t ) + B n ( 2 ) sin ( 4 n t ) ) sin ( n x ) − p a r t i c u l a r s o l u t i o n \boxed {u _ {n} (x, t) = \left(B _ {n} ^ {(1)} \cos (4 n t) + B _ {n} ^ {(2)} \sin (4 n t)\right) \sin (n x) - p a r t i c u l a r s o l u t i o n} u n ( x , t ) = ( B n ( 1 ) cos ( 4 n t ) + B n ( 2 ) sin ( 4 n t ) ) sin ( n x ) − p a r t i c u l a rso l u t i o n
Conclusion,
u ( x , t ) = ∑ n = 1 ∞ u n ( x , t ) − general solution u(x,t) = \sum_{n=1}^{\infty} u_n(x,t) - \text{general solution} u ( x , t ) = n = 1 ∑ ∞ u n ( x , t ) − general solution u ( x , t ) = ∑ n = 1 ∞ ( B n ( 1 ) cos ( 4 n t ) + B n ( 2 ) sin ( 4 n t ) ) sin ( n x ) \boxed{u(x,t) = \sum_{n=1}^{\infty} \left(B_n^{(1)} \cos(4nt) + B_n^{(2)} \sin(4nt)\right) \sin(nx)} u ( x , t ) = n = 1 ∑ ∞ ( B n ( 1 ) cos ( 4 n t ) + B n ( 2 ) sin ( 4 n t ) ) sin ( n x )
3 STEP: Determine the coefficients B 1 B_1 B 1 B 2 B_2 B 2
To do this, you must use the initial conditions.
∂ u ( x , 0 ) ∂ t = 0 → \frac{\partial u(x,0)}{\partial t} = 0 \rightarrow ∂ t ∂ u ( x , 0 ) = 0 → ∂ ∂ t ( ∑ n = 1 ∞ ( B n ( 1 ) cos ( 4 n t ) + B n ( 2 ) sin ( 4 n t ) ) sin ( n x ) ) t = 0 = \frac{\partial}{\partial t} \left(\sum_{n=1}^{\infty} \left(B_n^{(1)} \cos(4nt) + B_n^{(2)} \sin(4nt)\right) \sin(nx)\right)_{t=0} = ∂ t ∂ ( n = 1 ∑ ∞ ( B n ( 1 ) cos ( 4 n t ) + B n ( 2 ) sin ( 4 n t ) ) sin ( n x ) ) t = 0 = = ( ∑ n = 1 ∞ ( − 4 n B n ( 1 ) sin ( 4 n t ) + 4 n B n ( 2 ) cos ( 4 n t ) ) sin ( n x ) ) t = 0 = = \left(\sum_{n=1}^{\infty} \left(-4nB_n^{(1)} \sin(4nt) + 4nB_n^{(2)} \cos(4nt)\right) \sin(nx)\right)_{t=0} = = ( n = 1 ∑ ∞ ( − 4 n B n ( 1 ) sin ( 4 n t ) + 4 n B n ( 2 ) cos ( 4 n t ) ) sin ( n x ) ) t = 0 = = ∑ n = 1 ∞ ( − 4 n B n ( 1 ) sin ( 4 n ⋅ 0 ) + 4 n B n ( 2 ) cos ( 4 n ⋅ 0 ) ) sin ( n x ) = = \sum_{n=1}^{\infty} \left(-4nB_n^{(1)} \sin(4n \cdot 0) + 4nB_n^{(2)} \cos(4n \cdot 0)\right) \sin(nx) = = n = 1 ∑ ∞ ( − 4 n B n ( 1 ) sin ( 4 n ⋅ 0 ) + 4 n B n ( 2 ) cos ( 4 n ⋅ 0 ) ) sin ( n x ) = = ∑ n = 1 ∞ ( − 4 n B n ( 1 ) ⋅ 0 + 4 n B n ( 2 ) ⋅ 1 ) sin ( n x ) = ∑ n = 1 ∞ 4 n B n ( 2 ) sin ( n x ) = 0 = \sum_{n=1}^{\infty} \left(-4nB_n^{(1)} \cdot 0 + 4nB_n^{(2)} \cdot 1\right) \sin(nx) = \sum_{n=1}^{\infty} 4nB_n^{(2)} \sin(nx) = 0 = n = 1 ∑ ∞ ( − 4 n B n ( 1 ) ⋅ 0 + 4 n B n ( 2 ) ⋅ 1 ) sin ( n x ) = n = 1 ∑ ∞ 4 n B n ( 2 ) sin ( n x ) = 0
Then,
4 n B n ( 2 ) = 0 → B n ( 2 ) = 0 4nB_n^{(2)} = 0 \rightarrow \boxed{B_n^{(2)} = 0} 4 n B n ( 2 ) = 0 → B n ( 2 ) = 0
Conclusion,
u ( x , t ) = ∑ n = 1 ∞ B n ( 1 ) cos ( 4 n t ) sin ( n x ) u(x,t) = \sum_{n=1}^{\infty} B_n^{(1)} \cos(4nt) \sin(nx) u ( x , t ) = n = 1 ∑ ∞ B n ( 1 ) cos ( 4 n t ) sin ( n x ) u ( x , 0 ) = x ( π − x ) → u(x, 0) = x(\pi - x) \rightarrow u ( x , 0 ) = x ( π − x ) → u ( x , 0 ) = ( ∑ n = 1 ∞ B n ( 1 ) cos ( 4 n t ) sin ( n x ) ) t = 0 = ∑ n = 1 ∞ B n ( 1 ) cos ( 4 n ⋅ 0 ) sin ( n x ) = = ∑ n = 1 ∞ B n ( 1 ) ⋅ 1 ⋅ sin ( n x ) → ∑ n = 1 ∞ B n ( 1 ) sin ( n x ) = x ( π − x ) \begin{aligned}
u(x, 0) &= \left(\sum_{n=1}^{\infty} B_n^{(1)} \cos(4nt) \sin(nx)\right)_{t=0} = \sum_{n=1}^{\infty} B_n^{(1)} \cos(4n \cdot 0) \sin(nx) = \\
&= \sum_{n=1}^{\infty} B_n^{(1)} \cdot 1 \cdot \sin(nx) \rightarrow \sum_{n=1}^{\infty} B_n^{(1)} \sin(nx) = x(\pi - x)
\end{aligned} u ( x , 0 ) = ( n = 1 ∑ ∞ B n ( 1 ) cos ( 4 n t ) sin ( n x ) ) t = 0 = n = 1 ∑ ∞ B n ( 1 ) cos ( 4 n ⋅ 0 ) sin ( n x ) = = n = 1 ∑ ∞ B n ( 1 ) ⋅ 1 ⋅ sin ( n x ) → n = 1 ∑ ∞ B n ( 1 ) sin ( n x ) = x ( π − x )
As we know
∫ 0 π sin ( m x ) ⋅ sin ( n x ) d x = { π 2 , n = m 0 , n ≠ m \int_{0}^{\pi} \sin(mx) \cdot \sin(nx) \, dx = \begin{cases}
\dfrac{\pi}{2}, & n = m \\
0, & n \neq m
\end{cases} ∫ 0 π sin ( m x ) ⋅ sin ( n x ) d x = { 2 π , 0 , n = m n = m
In our case,
∫ 0 π × ∣ ∑ n = 1 ∞ B n ( 1 ) sin ( n x ) = x ( π − x ) ∣ × sin ( m x ) d x \int_{0}^{\pi} \times \left| \sum_{n=1}^{\infty} B_n^{(1)} \sin(nx) = x(\pi - x) \right| \times \sin(mx) \, dx ∫ 0 π × ∣ ∣ n = 1 ∑ ∞ B n ( 1 ) sin ( n x ) = x ( π − x ) ∣ ∣ × sin ( m x ) d x B m ( 1 ) = ∫ 0 π x ( π − x ) sin ( m x ) d x = ∫ 0 π ( π x − x 2 ) sin ( m x ) d x → B_m^{(1)} = \int_{0}^{\pi} x(\pi - x) \sin(mx) \, dx = \int_{0}^{\pi} (\pi x - x^2) \sin(mx) \, dx \rightarrow B m ( 1 ) = ∫ 0 π x ( π − x ) sin ( m x ) d x = ∫ 0 π ( π x − x 2 ) sin ( m x ) d x → B m ( 1 ) = ∫ 0 π π x sin ( m x ) d x − ∫ 0 π x 2 sin ( m x ) d x = I 1 − I 2 B_m^{(1)} = \int_{0}^{\pi} \pi x \sin(mx) \, dx - \int_{0}^{\pi} x^2 \sin(mx) \, dx = I_1 - I_2 B m ( 1 ) = ∫ 0 π π x sin ( m x ) d x − ∫ 0 π x 2 sin ( m x ) d x = I 1 − I 2 I 1 = ∫ 0 π π x sin ( m x ) d x = π ⋅ ∫ 0 π x ⏟ ⋅ sin ( m x ) d x ⏟ d v = [ u = x → d u = d x d v = sin ( m x ) d x v = − cos ( m x ) m ] = = π ⋅ ( − x ⋅ cos ( m x ) m ∣ 0 π − ∫ 0 π − cos ( m x ) d x m = = π ⋅ ( − π ⋅ cos ( π m ) m − ( − 0 ⋅ cos ( m ⋅ 0 ) m ) + 1 m ∫ 0 π cos ( m x ) d x ) = = π ⋅ ( − π ⋅ ( − 1 ) m m + 1 m ⋅ sin ( m x ) m ∣ 0 π = π ⋅ ( − π ⋅ ( − 1 ) m m + sin ( m ⋅ π ) m 2 − sin ( m ⋅ 0 ) m 2 ) = = π ⋅ ( − π ⋅ ( − 1 ) m m + 0 − 0 ) = − π 2 ⋅ ( − 1 ) m m \begin{array}{l}
I_{1} = \int_{0}^{\pi} \pi x \sin(mx) \, dx = \pi \cdot \int_{0}^{\pi} \underbrace{x} \cdot \underbrace{\sin(mx) \, dx}_{dv} = \begin{bmatrix} u = x \to du = dx \\ dv = \sin(mx) \, dx \\ v = \frac{-\cos(mx)}{m} \end{bmatrix} = \\
= \pi \cdot \left(- \frac{x \cdot \cos(mx)}{m}\right|_{0}^{\pi} - \int_{0}^{\pi} \frac{-\cos(mx) \, dx}{m} = \\
= \pi \cdot \left(- \frac{\pi \cdot \cos(\pi m)}{m} - \left(- \frac{0 \cdot \cos(m \cdot 0)}{m}\right) + \frac{1}{m} \int_{0}^{\pi} \cos(mx) \, dx\right) = \\
= \pi \cdot \left(- \frac{\pi \cdot (-1)^{m}}{m} + \frac{1}{m} \cdot \frac{\sin(mx)}{m}\right|_{0}^{\pi} = \pi \cdot \left(- \frac{\pi \cdot (-1)^{m}}{m} + \frac{\sin(m \cdot \pi)}{m^{2}} - \frac{\sin(m \cdot 0)}{m^{2}}\right) = \\
= \pi \cdot \left(- \frac{\pi \cdot (-1)^{m}}{m} + 0 - 0\right) = -\frac{\pi^{2} \cdot (-1)^{m}}{m}
\end{array} I 1 = ∫ 0 π π x sin ( m x ) d x = π ⋅ ∫ 0 π x ⋅ d v sin ( m x ) d x = ⎣ ⎡ u = x → d u = d x d v = sin ( m x ) d x v = m − c o s ( m x ) ⎦ ⎤ = = π ⋅ ( − m x ⋅ c o s ( m x ) ∣ ∣ 0 π − ∫ 0 π m − c o s ( m x ) d x = = π ⋅ ( − m π ⋅ c o s ( πm ) − ( − m 0 ⋅ c o s ( m ⋅ 0 ) ) + m 1 ∫ 0 π cos ( m x ) d x ) = = π ⋅ ( − m π ⋅ ( − 1 ) m + m 1 ⋅ m s i n ( m x ) ∣ ∣ 0 π = π ⋅ ( − m π ⋅ ( − 1 ) m + m 2 s i n ( m ⋅ π ) − m 2 s i n ( m ⋅ 0 ) ) = = π ⋅ ( − m π ⋅ ( − 1 ) m + 0 − 0 ) = − m π 2 ⋅ ( − 1 ) m
Conclusion,
I 1 = − π 2 ⋅ ( − 1 ) m m I_{1} = -\frac{\pi^{2} \cdot (-1)^{m}}{m} I 1 = − m π 2 ⋅ ( − 1 ) m I 2 = ∫ 0 π x ‾ 2 ⋅ sin ( m x ) d x ⏟ d v = [ u = x 2 → d u = 2 x d x d v = sin ( m x ) d x v = − cos ( m x ) m ] = = − x 2 ⋅ cos ( m x ) m ∣ 0 π − ∫ 0 π − 2 x cos ( m x ) d x m = = − π 2 ⋅ cos ( m π ) m − ( − 0 2 ⋅ cos ( m ⋅ 0 ) m ) + 2 m ⋅ ∫ 0 π x cos ( m x ) d x = = − π 2 ⋅ ( − 1 ) m m + 2 m ⋅ ∫ 0 π x ‾ ⋅ cos ( m x ) d x ⏟ d v = [ u = x → d u = d x d v = cos ( m x ) d x v = sin ( m x ) m ] = = − π 2 ⋅ ( − 1 ) m m + 2 m ⋅ ( x ⋅ sin ( m x ) m ∣ 0 π − ∫ 0 π sin ( m x ) m d x = = − π 2 ⋅ ( − 1 ) m m + 2 m ⋅ ( π ⋅ sin ( m π ) m − 0 ⋅ sin ( m ⋅ 0 ) m − 1 m ∫ 0 π sin ( m x ) d x ) = = − π 2 ⋅ ( − 1 ) m m + 2 m ⋅ ( 0 − 0 − 1 m ⋅ ( − cos ( m x ) m ∣ 0 π ) = = − π 2 ⋅ ( − 1 ) m m + 2 m 3 ⋅ ( cos ( m π ) − cos ( m ⋅ 0 ) ) = = − π 2 ⋅ ( − 1 ) m m + 2 m 3 ⋅ ( ( − 1 ) m − 1 ) \begin{array}{l}
I_{2} = \int_{0}^{\pi} \underline{x}^{2} \cdot \underbrace{\sin(mx) \, dx}_{dv} = \left[ \begin{array}{c} u = x^{2} \rightarrow du = 2x \, dx \\ dv = \sin(mx) \, dx \\ v = \frac{-\cos(mx)}{m} \end{array} \right] = \\
= -\frac{x^{2} \cdot \cos(mx)}{m} \Bigg|_{0}^{\pi} - \int_{0}^{\pi} \frac{-2x \cos(mx) \, dx}{m} = \\
= -\frac{\pi^{2} \cdot \cos(m\pi)}{m} - \left(-\frac{0^{2} \cdot \cos(m \cdot 0)}{m}\right) + \frac{2}{m} \cdot \int_{0}^{\pi} x \cos(mx) \, dx = \\
= -\frac{\pi^{2} \cdot (-1)^{m}}{m} + \frac{2}{m} \cdot \int_{0}^{\pi} \underline{x} \cdot \underbrace{\cos(mx) \, dx}_{dv} = \left[ \begin{array}{c} u = x \rightarrow du = dx \\ dv = \cos(mx) \, dx \\ v = \frac{\sin(mx)}{m} \end{array} \right] = \\
= -\frac{\pi^{2} \cdot (-1)^{m}}{m} + \frac{2}{m} \cdot \left(\frac{x \cdot \sin(mx)}{m}\right|_{0}^{\pi} - \int_{0}^{\pi} \frac{\sin(mx)}{m} \, dx = \\
= -\frac{\pi^{2} \cdot (-1)^{m}}{m} + \frac{2}{m} \cdot \left(\frac{\pi \cdot \sin(m\pi)}{m} - \frac{0 \cdot \sin(m \cdot 0)}{m} - \frac{1}{m} \int_{0}^{\pi} \sin(mx) \, dx\right) = \\
= -\frac{\pi^{2} \cdot (-1)^{m}}{m} + \frac{2}{m} \cdot \left(0 - 0 - \frac{1}{m} \cdot \left(-\frac{\cos(mx)}{m}\right|_{0}^{\pi}\right) = \\
= -\frac{\pi^{2} \cdot (-1)^{m}}{m} + \frac{2}{m^{3}} \cdot (\cos(m\pi) - \cos(m \cdot 0)) = \\
= -\frac{\pi^{2} \cdot (-1)^{m}}{m} + \frac{2}{m^{3}} \cdot ((-1)^{m} - 1) \\
\end{array} I 2 = ∫ 0 π x 2 ⋅ d v sin ( m x ) d x = ⎣ ⎡ u = x 2 → d u = 2 x d x d v = sin ( m x ) d x v = m − c o s ( m x ) ⎦ ⎤ = = − m x 2 ⋅ c o s ( m x ) ∣ ∣ 0 π − ∫ 0 π m − 2 x c o s ( m x ) d x = = − m π 2 ⋅ c o s ( mπ ) − ( − m 0 2 ⋅ c o s ( m ⋅ 0 ) ) + m 2 ⋅ ∫ 0 π x cos ( m x ) d x = = − m π 2 ⋅ ( − 1 ) m + m 2 ⋅ ∫ 0 π x ⋅ d v cos ( m x ) d x = ⎣ ⎡ u = x → d u = d x d v = cos ( m x ) d x v = m s i n ( m x ) ⎦ ⎤ = = − m π 2 ⋅ ( − 1 ) m + m 2 ⋅ ( m x ⋅ s i n ( m x ) ∣ ∣ 0 π − ∫ 0 π m s i n ( m x ) d x = = − m π 2 ⋅ ( − 1 ) m + m 2 ⋅ ( m π ⋅ s i n ( mπ ) − m 0 ⋅ s i n ( m ⋅ 0 ) − m 1 ∫ 0 π sin ( m x ) d x ) = = − m π 2 ⋅ ( − 1 ) m + m 2 ⋅ ( 0 − 0 − m 1 ⋅ ( − m c o s ( m x ) ∣ ∣ 0 π ) = = − m π 2 ⋅ ( − 1 ) m + m 3 2 ⋅ ( cos ( mπ ) − cos ( m ⋅ 0 )) = = − m π 2 ⋅ ( − 1 ) m + m 3 2 ⋅ (( − 1 ) m − 1 ) I 2 = − π 2 ⋅ ( − 1 ) m m + 2 m 3 ⋅ ( ( − 1 ) m − 1 ) I_{2} = -\frac{\pi^{2} \cdot (-1)^{m}}{m} + \frac{2}{m^{3}} \cdot ((-1)^{m} - 1) I 2 = − m π 2 ⋅ ( − 1 ) m + m 3 2 ⋅ (( − 1 ) m − 1 )
Then,
B m ( 1 ) = I 1 − I 2 = − π 2 ⋅ ( − 1 ) m m − ( − π 2 ⋅ ( − 1 ) m m + 2 m 3 ⋅ ( ( − 1 ) m − 1 ) ) → B _ {m} ^ {(1)} = I _ {1} - I _ {2} = - \frac {\pi^ {2} \cdot (- 1) ^ {m}}{m} - \left(- \frac {\pi^ {2} \cdot (- 1) ^ {m}}{m} + \frac {2}{m ^ {3}} \cdot ((- 1) ^ {m} - 1)\right)\rightarrow B m ( 1 ) = I 1 − I 2 = − m π 2 ⋅ ( − 1 ) m − ( − m π 2 ⋅ ( − 1 ) m + m 3 2 ⋅ (( − 1 ) m − 1 ) ) → B m ( 1 ) = − π 2 ⋅ ( − 1 ) m m + π 2 ⋅ ( − 1 ) m m − 2 m 3 ⋅ ( ( − 1 ) m − 1 ) → B _ {m} ^ {(1)} = - \frac {\pi^ {2} \cdot (- 1) ^ {m}}{m} + \frac {\pi^ {2} \cdot (- 1) ^ {m}}{m} - \frac {2}{m ^ {3}} \cdot ((- 1) ^ {m} - 1) \rightarrow B m ( 1 ) = − m π 2 ⋅ ( − 1 ) m + m π 2 ⋅ ( − 1 ) m − m 3 2 ⋅ (( − 1 ) m − 1 ) → B m ( 1 ) = − 2 ⋅ ( ( − 1 ) m − 1 ) m 3 \boxed {B _ {m} ^ {(1)} = \frac {- 2 \cdot ((- 1) ^ {m} - 1)}{m ^ {3}}} B m ( 1 ) = m 3 − 2 ⋅ (( − 1 ) m − 1 )
As we know
{ ( − 1 ) m − 1 = 0 , m = 2 k , k = 1 , 2 , 3 , 4 , … ( − 1 ) m − 1 = − 2 , m = 2 k − 1 , k = 1 , 2 , 3 , 4 , … → \left\{ \begin{array}{c} (- 1) ^ {m} - 1 = 0, m = 2 k, k = 1, 2, 3, 4, \ldots \\ (- 1) ^ {m} - 1 = - 2, m = 2 k - 1, k = 1, 2, 3, 4, \ldots \end{array} \right. \to { ( − 1 ) m − 1 = 0 , m = 2 k , k = 1 , 2 , 3 , 4 , … ( − 1 ) m − 1 = − 2 , m = 2 k − 1 , k = 1 , 2 , 3 , 4 , … → { B m ( 1 ) = 0 , m = 2 k , k = 1 , 2 , 3 , 4 , … B m ( 1 ) = 4 m 3 , m = 2 k − 1 , k = 1 , 2 , 3 , 4 , … \left\{ \begin{array}{c} B _ {m} ^ {(1)} = 0, m = 2 k, k = 1, 2, 3, 4, \ldots \\ B _ {m} ^ {(1)} = \frac {4}{m ^ {3}}, m = 2 k - 1, k = 1, 2, 3, 4, \ldots \end{array} \right. { B m ( 1 ) = 0 , m = 2 k , k = 1 , 2 , 3 , 4 , … B m ( 1 ) = m 3 4 , m = 2 k − 1 , k = 1 , 2 , 3 , 4 , …
Conclusion,
u ( x , t ) = ∑ n = 1 ∞ B n ( 1 ) cos ( 4 n t ) sin ( n x ) = ∑ n = 1 ∞ ( − 2 ⋅ ( ( − 1 ) n − 1 ) n 3 ) cos ( 4 n t ) sin ( n x ) u (x, t) = \sum_ {n = 1} ^ {\infty} B _ {n} ^ {(1)} \cos (4 n t) \sin (n x) = \sum_ {n = 1} ^ {\infty} \left(\frac {- 2 \cdot ((- 1) ^ {n} - 1)}{n ^ {3}}\right) \cos (4 n t) \sin (n x) u ( x , t ) = n = 1 ∑ ∞ B n ( 1 ) cos ( 4 n t ) sin ( n x ) = n = 1 ∑ ∞ ( n 3 − 2 ⋅ (( − 1 ) n − 1 ) ) cos ( 4 n t ) sin ( n x ) u ( x , t ) = ∑ k = 1 ∞ ( 4 ( 2 k − 1 ) 3 ) cos ( 4 ( 2 k − 1 ) t ) sin ( ( 2 k − 1 ) x ) \boxed {u (x, t) = \sum_ {k = 1} ^ {\infty} \left(\frac {4}{(2 k - 1) ^ {3}}\right) \cos (4 (2 k - 1) t) \sin ((2 k - 1) x)} u ( x , t ) = k = 1 ∑ ∞ ( ( 2 k − 1 ) 3 4 ) cos ( 4 ( 2 k − 1 ) t ) sin (( 2 k − 1 ) x )
ANSWER:
u ( x , t ) = ∑ k = 1 ∞ ( 4 ( 2 k − 1 ) 3 ) cos ( 4 ( 2 k − 1 ) t ) sin ( ( 2 k − 1 ) x ) u (x, t) = \sum_ {k = 1} ^ {\infty} \left(\frac {4}{(2 k - 1) ^ {3}}\right) \cos (4 (2 k - 1) t) \sin ((2 k - 1) x) u ( x , t ) = k = 1 ∑ ∞ ( ( 2 k − 1 ) 3 4 ) cos ( 4 ( 2 k − 1 ) t ) sin (( 2 k − 1 ) x )
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#339914 on Dec 2023