Para se deduzir várias das expressões nesta página, é necessário perceber o método de soluções separáveis .
Já tínhamos visto a Equação do Calor anteriormente .
Nesta página apenas vão estar as expressões mais comuns para a equação do calor, que quando memorizadas permitem simplificar os cálculos.
A Equação do Calor tem a seguinte expressão:
∂ u ∂ t = k ∂ 2 u ∂ x 2 \frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2} ∂ t ∂ u = k ∂ x 2 ∂ 2 u A resolução desta equação vai depender das condições fronteira
Trata-se de um problema muito comum, que é descrito pelo seguinte sistema:
{ ∂ u ∂ t = k ∂ 2 u ∂ x 2 t > 0 , x ∈ ] 0 , L [ u ( t , 0 ) = u ( t , L ) = 0 t > 0 u ( 0 , x ) = f ( x ) x ∈ ] 0 , L [ \begin{cases}
\dfrac{\partial u}{\partial t} = k \dfrac{\partial^2 u}{\partial x^2} & t > 0, x \in \left]0, L \right[\\
u(t,0 ) = u(t,L) = 0 & t>0\\
u(0,x) = f(x) & x \in \left]0, L\right[
\end{cases} ⎩ ⎨ ⎧ ∂ t ∂ u = k ∂ x 2 ∂ 2 u u ( t , 0 ) = u ( t , L ) = 0 u ( 0 , x ) = f ( x ) t > 0 , x ∈ ] 0 , L [ t > 0 x ∈ ] 0 , L [ Após aplicar o método de separação de variáveis ,
obtemos a seguinte solução para este problema:
u ( t , x ) = ∑ n = 1 ∞ c n e − n 2 π 2 k t L 2 sin ( n π x L ) , c n ∈ R u(t,x) = \sum_{n=1}^{\infty} c_n e^{-\dfrac{n^2 \pi^2 k t}{L^2}} \sin \left(\frac{n\pi x}{L}\right) \quad, \quad c_n \in \R u ( t , x ) = n = 1 ∑ ∞ c n e − L 2 n 2 π 2 k t sin ( L nπ x ) , c n ∈ R Para determinar sucessão c n c_n c n
u ( 0 , x ) = ∑ n = 1 + ∞ c n sin ( n π x L ) = f ( x ) u(0,x) = \sum_{n=1}^{+\infty} c_n \sin \left(\frac{n\pi x}{L}\right) = f(x) u ( 0 , x ) = n = 1 ∑ + ∞ c n sin ( L nπ x ) = f ( x ) Como podemos ver, a expressão é semelhante à da série de senos .
Se f ( x ) f(x) f ( x ) c n c_n c n b n b_n b n
c n = b n = 2 L ∫ 0 L f ( x ) sin ( n π x L ) d x c_n = b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin \left(\frac{n\pi x}{L}\right) \d x c n = b n = L 2 ∫ 0 L f ( x ) sin ( L nπ x ) d x Exemplo
Assumindo L = π L = \pi L = π
f ( x ) = sin ( 2 x ) − 3 sin ( 5 x ) f(x) = \sin(2x) - 3\sin(5x) f ( x ) = sin ( 2 x ) − 3 sin ( 5 x ) resolva a equação de calor com condições fronteiras de Dirichlet homogéneas.
Então,
u ( 0 , x ) = ∑ n = 1 + ∞ c n sin ( n x ) = sin ( 2 x ) − 3 sin ( 5 x ) u(0,x) = \sum_{n=1}^{+\infty} c_n \sin (nx) = \sin(2x) - 3\sin(5x) u ( 0 , x ) = n = 1 ∑ + ∞ c n sin ( n x ) = sin ( 2 x ) − 3 sin ( 5 x ) Como f ( x ) f(x) f ( x ) c n c_n c n
c 2 = 1 c 5 = − 3 c n = 0 , ∀ n ∈ N \ { 2 , 5 } \begin{darray}{l}
c_2 = 1\\
c_5 = -3\\
c_n = 0 &, \forall n \in \N \backslash \{2,5\}
\end{darray} c 2 = 1 c 5 = − 3 c n = 0 , ∀ n ∈ N \ { 2 , 5 } Então, a solução final da equação do calor é:
u ( t , x ) = e − 4 k t sin ( 2 x ) − 3 e − 25 k t sin ( 5 x ) u(t,x) = e^{-4kt} \sin(2x) - 3 e^{-25 kt} \sin(5x) u ( t , x ) = e − 4 k t sin ( 2 x ) − 3 e − 25 k t sin ( 5 x ) Exemplo (f ( x ) f(x) f ( x ) Considerando L = π L=\pi L = π
f ( x ) = π 2 − ∣ x − π 2 ∣ = { x se 0 ≤ x ≤ π 2 π − x se π 2 < x ≤ π f(x) = \frac{\pi}{2} - \left|x - \frac{\pi}{2}\right| = \begin{cases}
x & \text{se } 0 \leq x \leq \frac{\pi}{2} \\
\pi - x & \text{se } \frac{\pi}{2} < x \leq \pi
\end{cases} f ( x ) = 2 π − x − 2 π = { x π − x se 0 ≤ x ≤ 2 π se 2 π < x ≤ π resolva a equação de calor com condições fronteiras de Dirichlet homogéneas.
Então,
u ( 0 , x ) = ∑ n = 1 ∞ c n sin ( n x ) = π 2 − ∣ π 2 − x ∣ u(0,x) = \sum_{n=1}^{\infty}c_n\sin(nx) = \frac{\pi}{2} - \left|\frac{\pi}{2} - x\right| u ( 0 , x ) = n = 1 ∑ ∞ c n sin ( n x ) = 2 π − 2 π − x Como f ( x ) f(x) f ( x )
S sin f ( x ) = ∑ n = 1 ∞ b n sin ( n x ) S_{\sin}f(x) = \sum_{n=1}^{\infty}b_n\sin(nx) S s i n f ( x ) = n = 1 ∑ ∞ b n sin ( n x ) tal que
b n = 2 π ∫ 0 π f ( x ) sin ( n x ) d x = 2 π [ ∫ 0 π 2 x sin ( n x ) d x + ∫ π 2 π ( π − x ) sin ( n x ) d x ] = 4 π n 2 sin n π 2 \begin{aligned}
b_n &= \frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)\d x\\
&= \frac{2}{\pi}\left[\int_{0}^{\frac{\pi}{2}}x\sin(nx)\d x + \int_{\frac{\pi}{2}}^{\pi}(\pi - x)\sin(nx)\d x\right]\\
&= \frac{4}{\pi n^2}\sin\frac{n\pi}{2}
\end{aligned} b n = π 2 ∫ 0 π f ( x ) sin ( n x ) d x = π 2 [ ∫ 0 2 π x sin ( n x ) d x + ∫ 2 π π ( π − x ) sin ( n x ) d x ] = π n 2 4 sin 2 nπ Determinamos assim a série de senos de f ( x ) f(x) f ( x )
f ( x ) = π 2 − ∣ x − π 2 ∣ = ∑ n = 1 ∞ 4 π n 2 sin n π 2 sin ( n x ) f(x) = \frac{\pi}{2} - \left|x - \frac{\pi}{2}\right| = \sum_{n=1}^{\infty}\frac{4}{\pi n^2}\sin\frac{n\pi}{2}\sin(nx) f ( x ) = 2 π − x − 2 π = n = 1 ∑ ∞ π n 2 4 sin 2 nπ sin ( n x ) Então, como podemos observar que c n = b n c_n = b_n c n = b n
c n = 4 π n 2 sin n π 2 c_n = \frac{4}{\pi n^2}\sin\frac{n\pi}{2} c n = π n 2 4 sin 2 nπ Assim, a solução da equação de calor é
u ( t , x ) = ∑ n = 1 ∞ 4 π n 2 sin n π 2 e − n 2 K t sin ( n x ) u(t, x) = \sum_{n=1}^{\infty}\frac{4}{\pi n^2}\sin\frac{n\pi}{2}e^{-n^2Kt}\sin(nx) u ( t , x ) = n = 1 ∑ ∞ π n 2 4 sin 2 nπ e − n 2 K t sin ( n x ) Consideremos o seguinte problema:
{ ∂ u ∂ t = k ∂ 2 u ∂ x 2 t > 0 , x ∈ ] 0 , π [ ∂ u ∂ x ( t , 0 ) = ∂ u ∂ x ( t , L ) = 0 t > 0 u ( 0 , x ) = f ( x ) x ∈ ] 0 , π [ \begin{cases}
\dfrac{\partial u}{\partial t} = k \dfrac{\partial^2 u}{\partial x^2} & t > 0, x \in \left]0, \pi \right[\\
\frac{\partial u}{\partial x}(t,0 ) = \frac{\partial u}{\partial x}(t,L) = 0 & t>0\\
u(0,x) = f(x) & x \in \left]0, \pi\right[
\end{cases} ⎩ ⎨ ⎧ ∂ t ∂ u = k ∂ x 2 ∂ 2 u ∂ x ∂ u ( t , 0 ) = ∂ x ∂ u ( t , L ) = 0 u ( 0 , x ) = f ( x ) t > 0 , x ∈ ] 0 , π [ t > 0 x ∈ ] 0 , π [ Fisicamente, este problema representa a propagação de calor numa barra de comprimento L L L
Também pelo método de separação de variáveis
é possível chegar a uma solução, apresentada abaixo:
u ( t , x ) = c 0 + ∑ n = 1 ∞ c n e − n 2 π 2 k t L 2 cos ( n π x L ) , c n ∈ R u(t,x) = c_0 + \sum_{n=1}^{\infty} c_n e^{- \dfrac{n^2 \pi^2 kt}{L^2}} \cos \left(\frac{n\pi x}{L}\right) \quad, \quad c_n \in \R u ( t , x ) = c 0 + n = 1 ∑ ∞ c n e − L 2 n 2 π 2 k t cos ( L nπ x ) , c n ∈ R Semelhantemente às as condições de Dirichlet, podemos determinar a sucessão c n c_n c n
u ( 0 , x ) = c 0 + ∑ n = 1 ∞ c n cos ( n π x L ) = f ( x ) u(0,x) = c_0 + \sum_{n=1}^{\infty} c_n \cos \left(\frac{n\pi x}{L}\right) = f(x) u ( 0 , x ) = c 0 + n = 1 ∑ ∞ c n cos ( L nπ x ) = f ( x ) que é semelhante a uma série de cossenos .
Novamente, se f ( x ) f(x) f ( x ) c n c_n c n
c 0 = a 0 2 = 1 L ∫ 0 L f ( x ) d x c n = a n = 2 L ∫ 0 L f ( x ) cos ( n π x L ) d x \begin{aligned}
c_0 &= \frac{a_0}{2} = \frac{1}{L} \int_{0}^{L} f(x) \d x\\
c_n &= a_n = \frac{2}{L} \int_{0}^{L} f(x) \cos \left(\frac{n\pi x}{L} \right) \d x
\end{aligned} c 0 c n = 2 a 0 = L 1 ∫ 0 L f ( x ) d x = a n = L 2 ∫ 0 L f ( x ) cos ( L nπ x ) d x A Equação de Laplace homogénea tem a seguinte forma:
∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0 \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 ∂ x 2 ∂ 2 u + ∂ y 2 ∂ 2 u = 0 Tal como fizemos para as Equações do Calor, podemos determinar uma solução através do método de soluções separáveis .
Pegando num problema mais completo (que provavelmente não é relevante decorar),
{ ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0 u ( x , 0 ) = f 1 ( x ) ( 1 ) u ( 0 , y ) = f 2 ( y ) ( 2 ) u ( x , b ) = f 3 ( x ) ( 3 ) u ( a , y ) = f 4 ( y ) ( 4 ) \begin{cases}
\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\\
u(x,0) = f_1(x)& (1)\\
u(0,y) = f_2(y)& (2)\\
u(x,b) = f_3(x)& (3)\\
u(a,y) = f_4(y)& (4)
\end{cases} ⎩ ⎨ ⎧ ∂ x 2 ∂ 2 u + ∂ y 2 ∂ 2 u = 0 u ( x , 0 ) = f 1 ( x ) u ( 0 , y ) = f 2 ( y ) u ( x , b ) = f 3 ( x ) u ( a , y ) = f 4 ( y ) ( 1 ) ( 2 ) ( 3 ) ( 4 ) Vamos dividir isto em 4 problemas:
Problema 1:
∂ 2 u 1 ∂ x 2 + ∂ 2 u 1 ∂ y 2 = 0 \frac{\partial^2 u_1}{\partial x^2} + \frac{\partial^2 u_1}{\partial y^2} = 0 ∂ x 2 ∂ 2 u 1 + ∂ y 2 ∂ 2 u 1 = 0 u 1 ( x , 0 ) = f 1 ( x ) u 1 ( 0 , y ) = 0 u 1 ( x , b ) = 0 u 1 ( a , y ) = 0 \begin{darray}{l}
u_1(x,0) = f_1(x)\\ % I
u_1(0,y) = 0\\ % II
u_1(x,b) = 0\\ % III
u_1(a,y) = 0 % IV
\end{darray} u 1 ( x , 0 ) = f 1 ( x ) u 1 ( 0 , y ) = 0 u 1 ( x , b ) = 0 u 1 ( a , y ) = 0 Problema 2:
∂ 2 u 2 ∂ x 2 + ∂ 2 u 2 ∂ y 2 = 0 \frac{\partial^2 u_2}{\partial x^2} + \frac{\partial^2 u_2}{\partial y^2} = 0 ∂ x 2 ∂ 2 u 2 + ∂ y 2 ∂ 2 u 2 = 0 u 2 ( x , 0 ) = 0 u 2 ( 0 , y ) = f 2 ( x ) u 2 ( x , b ) = 0 u 2 ( a , y ) = 0 \begin{darray}{l}
u_2(x,0) = 0\\ % I
u_2(0,y) = f_2(x)\\ % II
u_2(x,b) = 0\\ % III
u_2(a,y) = 0 % IV
\end{darray} u 2 ( x , 0 ) = 0 u 2 ( 0 , y ) = f 2 ( x ) u 2 ( x , b ) = 0 u 2 ( a , y ) = 0 Problema 3:
∂ 2 u 3 ∂ x 2 + ∂ 2 u 3 ∂ y 2 = 0 \frac{\partial^2 u_3}{\partial x^2} + \frac{\partial^2 u_3}{\partial y^2} = 0 ∂ x 2 ∂ 2 u 3 + ∂ y 2 ∂ 2 u 3 = 0 u 3 ( x , 0 ) = 0 u 3 ( 0 , y ) = 0 u 3 ( x , b ) = f 3 ( x ) u 3 ( a , y ) = 0 \begin{darray}{l}
u_3(x,0) = 0\\ % I
u_3(0,y) = 0\\ % II
u_3(x,b) = f_3(x)\\ % III
u_3(a,y) = 0 % IV
\end{darray} u 3 ( x , 0 ) = 0 u 3 ( 0 , y ) = 0 u 3 ( x , b ) = f 3 ( x ) u 3 ( a , y ) = 0 Problema 4:
∂ 2 u 4 ∂ x 2 + ∂ 2 u 4 ∂ y 2 = 0 \frac{\partial^2 u_4}{\partial x^2} + \frac{\partial^2 u_4}{\partial y^2} = 0 ∂ x 2 ∂ 2 u 4 + ∂ y 2 ∂ 2 u 4 = 0 u 4 ( x , 0 ) = 0 u 4 ( 0 , y ) = 0 u 4 ( x , b ) = 0 u 4 ( a , y ) = f 4 ( x ) \begin{darray}{l}
u_4(x,0) = 0\\ % I
u_4(0,y) = 0\\ % II
u_4(x,b) = 0\\ % III
u_4(a,y) = f_4(x) % IV
\end{darray} u 4 ( x , 0 ) = 0 u 4 ( 0 , y ) = 0 u 4 ( x , b ) = 0 u 4 ( a , y ) = f 4 ( x ) Nota do Autor
Abaixo mostram-se imediatamente as soluções dos problemas.
Resolvendo agora cada um dos problemas, omitindo os cálculos (mais abaixo irá estar um exemplo passo a passo):
u 1 = ∑ n = 1 + ∞ c 1 , n sin ( n π a x ) sh ( n π a ( b − y ) ) u_1 = \sum_{n=1}^{+\infty} c_{1,n} \sin \left(\frac{n\pi}{a}x \right) \sh \left(\frac{n\pi}{a}(b-y)\right) u 1 = n = 1 ∑ + ∞ c 1 , n sin ( a nπ x ) sh ( a nπ ( b − y ) ) u 2 = ∑ n = 1 + ∞ c 2 , n sh ( n π b ( a − x ) ) sin ( n π b y ) u_2 = \sum_{n=1}^{+\infty} c_{2,n} \sh \left(\frac{n\pi}{b}(a-x)\right) \sin \left(\frac{n\pi}{b}y \right) u 2 = n = 1 ∑ + ∞ c 2 , n sh ( b nπ ( a − x ) ) sin ( b nπ y ) u 3 = ∑ n = 1 + ∞ c 3 , n sin ( n π a x ) sh ( n π a y ) u_3 = \sum_{n=1}^{+\infty} c_{3,n} \sin \left(\frac{n\pi}{a}x \right) \sh \left(\frac{n\pi}{a}y\right) u 3 = n = 1 ∑ + ∞ c 3 , n sin ( a nπ x ) sh ( a nπ y ) u 4 = ∑ n = 1 + ∞ c 4 , n sh ( n π b x ) sin ( n π b y ) u_4 = \sum_{n=1}^{+\infty} c_{4,n} \sh \left(\frac{n\pi}{b}x\right) \sin \left(\frac{n\pi}{b}y \right) u 4 = n = 1 ∑ + ∞ c 4 , n sh ( b nπ x ) sin ( b nπ y ) E cada uma das sucessões:
c 1 , n = 2 a sh ( n π a b ) ∫ 0 a f 1 ( x ) sin ( n π a x ) d x c_{1,n} = \frac{2}{a \sh \left(\frac{n\pi}{a}b\right)} \int_{0}^{a} f_1(x) \sin \left(\frac{n\pi}{a}x \right) \d x c 1 , n = a sh ( a nπ b ) 2 ∫ 0 a f 1 ( x ) sin ( a nπ x ) d x c 2 , n = 2 a sh ( n π b a ) ∫ 0 a f 2 ( x ) sin ( n π b y ) d x c_{2,n} = \frac{2}{a \sh \left(\frac{n\pi}{b}a\right)} \int_{0}^{a} f_2(x) \sin \left(\frac{n\pi}{b}y \right) \d x c 2 , n = a sh ( b nπ a ) 2 ∫ 0 a f 2 ( x ) sin ( b nπ y ) d x c 3 , n = 2 a sh ( n π a b ) ∫ 0 a f 2 ( x ) sin ( n π a x ) d x c_{3,n} = \frac{2}{a \sh \left(\frac{n\pi}{a}b\right)} \int_{0}^{a} f_2(x) \sin \left(\frac{n\pi}{a}x \right) \d x c 3 , n = a sh ( a nπ b ) 2 ∫ 0 a f 2 ( x ) sin ( a nπ x ) d x c 4 , n = 2 a sh ( n π b a ) ∫ 0 a f 2 ( x ) sin ( n π b y ) d x c_{4,n} = \frac{2}{a \sh \left(\frac{n\pi}{b}a\right)} \int_{0}^{a} f_2(x) \sin \left(\frac{n\pi}{b}y \right) \d x c 4 , n = a sh ( b nπ a ) 2 ∫ 0 a f 2 ( x ) sin ( b nπ y ) d x Finalmente,
u = u 1 + u 2 + u 3 + u 4 u = u_1 + u_2 + u_3 + u_4 u = u 1 + u 2 + u 3 + u 4 Exemplo
Recorrendo ao método de separação de variáveis determine uma solução para o seguinte problema de valor na fronteira:
{ ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = u u ( x , 0 ) = u ( x , 1 ) = u ( 0 , y ) = 0 u ( 1 , y ) = y \begin{cases}
\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = u\\
u(x,0) = u(x,1) = u(0,y) = 0\\
u(1,y) = y
\end{cases} ⎩ ⎨ ⎧ ∂ x 2 ∂ 2 u + ∂ y 2 ∂ 2 u = u u ( x , 0 ) = u ( x , 1 ) = u ( 0 , y ) = 0 u ( 1 , y ) = y para 0 ≤ x ≤ 1 0 \leq x \leq 1 0 ≤ x ≤ 1 0 ≤ y ≤ 1 0 \leq y \leq 1 0 ≤ y ≤ 1
Como sempre, começamos por aplicar o método de soluções separáveis :
u ( x , y ) = X ( x ) Y ( y ) u(x,y) = X(x)Y(y) u ( x , y ) = X ( x ) Y ( y ) Já vimos anteriormente na equação de calor que é possível trabalharmos as condições fronteira para obter:
u ( x , 0 ) = 0 ⟹ X ( x ) Y ( 0 ) = 0 ⟹ X ( x ) = 0 ∨ Y ( 0 ) = 0 ⟹ Y ( 0 ) = 0 u ( x , 1 ) = 0 ⟹ X ( x ) Y ( 1 ) = 0 ⟹ X ( x ) = 0 ∨ Y ( 1 ) = 0 ⟹ Y ( 1 ) = 0 u ( 0 , y ) = 0 ⟹ X ( 0 ) Y ( y ) = 0 ⟹ X ( 0 ) = 0 ∨ Y ( y ) = 0 ⟹ X ( 0 ) = 0 \begin{aligned}
u(x,0) = 0 & \implies X(x)Y(0) = 0 \implies X(x) = 0 \lor Y(0) = 0 \implies Y(0) = 0\\
u(x,1) = 0 & \implies X(x)Y(1) = 0 \implies X(x) = 0 \lor Y(1) = 0 \implies Y(1) = 0\\
u(0,y) = 0 & \implies X(0)Y(y) = 0 \implies X(0) = 0 \lor Y(y) = 0 \implies X(0) = 0
\end{aligned} u ( x , 0 ) = 0 u ( x , 1 ) = 0 u ( 0 , y ) = 0 ⟹ X ( x ) Y ( 0 ) = 0 ⟹ X ( x ) = 0 ∨ Y ( 0 ) = 0 ⟹ Y ( 0 ) = 0 ⟹ X ( x ) Y ( 1 ) = 0 ⟹ X ( x ) = 0 ∨ Y ( 1 ) = 0 ⟹ Y ( 1 ) = 0 ⟹ X ( 0 ) Y ( y ) = 0 ⟹ X ( 0 ) = 0 ∨ Y ( y ) = 0 ⟹ X ( 0 ) = 0 Substituindo na expressão inicial e efetuando os cálculos segundo o método,
X ′ ′ ( x ) Y ( y ) + X ( x ) Y ′ ′ ( y ) = X ( x ) Y ( y ) ⇔ X ′ ′ ( x ) Y ( y ) = X ( x ) Y ( y ) − X ( x ) Y ′ ′ ( y ) ⇔ X ′ ′ ( x ) Y ( y ) X ( x ) = Y ( y ) − Y ′ ′ ( y ) ⇔ X ′ ′ ( x ) X ( x ) = 1 − Y ′ ′ ( y ) Y ( y ) ⇔ Y ′ ′ ( y ) Y ( y ) = 1 − X ′ ′ ( x ) X ( x ) = λ \begin{darray}{rl}
&X''(x) Y(y) + X(x) Y''(y) = X(x) Y(y)\\\\
\Leftrightarrow & X''(x) Y(y) = X(x) Y(y) - X(x) Y''(y)\\\\
\Leftrightarrow & \frac{X''(x) Y(y)}{X(x)} = Y(y) - Y''(y)\\\\
\Leftrightarrow & \frac{X''(x)}{X(x)} = 1 - \frac{Y''(y)}{Y(y)}\\\\
\Leftrightarrow & \frac{Y''(y)}{Y(y)} = 1 - \frac{X''(x)}{X(x)} = \lambda
\end{darray} ⇔ ⇔ ⇔ ⇔ X ′′ ( x ) Y ( y ) + X ( x ) Y ′′ ( y ) = X ( x ) Y ( y ) X ′′ ( x ) Y ( y ) = X ( x ) Y ( y ) − X ( x ) Y ′′ ( y ) X ( x ) X ′′ ( x ) Y ( y ) = Y ( y ) − Y ′′ ( y ) X ( x ) X ′′ ( x ) = 1 − Y ( y ) Y ′′ ( y ) Y ( y ) Y ′′ ( y ) = 1 − X ( x ) X ′′ ( x ) = λ Pegando em Y ( y ) Y(y) Y ( y )
Y ′ ′ ( y ) − λ Y ( y ) = 0 Y''(y) - \lambda Y(y) = 0 Y ′′ ( y ) − λY ( y ) = 0 Como vimos na Equação do Calor, a única solução não nula ocorre quando λ < 0 \lambda < 0 λ < 0
Y ( y ) = c 1 sin ( n π y ) λ = − n 2 π 2 \begin{darray}{cc}
Y(y) = c_1 \sin(n \pi y) & \lambda = - n^2 \pi^2
\end{darray} Y ( y ) = c 1 sin ( nπ y ) λ = − n 2 π 2 De seguida, pegamos em X ( x ) X(x) X ( x )
− X ′ ′ ( x ) X ( x ) = λ − 1 ⟹ X ′ ′ ( x ) + ( λ − 1 ) X ( x ) = 0 ⟹ X ′ ′ ( x ) + ( − n 2 π 2 − 1 ) X ( x ) = 0 ⟹ X ′ ′ ( x ) − ( n 2 π 2 + 1 ) X ( x ) = 0 \begin{aligned}
- \frac{X''(x)}{X(x)} = \lambda - 1 &\implies X''(x) + (\lambda - 1) X(x) = 0\\
&\implies X''(x) + (- n^2 \pi^2 - 1) X(x) = 0\\
&\implies X''(x) - (n^2 \pi^2 + 1) X(x) = 0
\end{aligned} − X ( x ) X ′′ ( x ) = λ − 1 ⟹ X ′′ ( x ) + ( λ − 1 ) X ( x ) = 0 ⟹ X ′′ ( x ) + ( − n 2 π 2 − 1 ) X ( x ) = 0 ⟹ X ′′ ( x ) − ( n 2 π 2 + 1 ) X ( x ) = 0 Logo, resolvendo a equação de ordem superior ,
X ( x ) = c 2 e 1 + n 2 π 2 x + c 3 e − 1 + n 2 π 2 x X(x) = c_2 e^{\sqrt{1 + n^2 \pi^2}x} + c_3 e^{-\sqrt{1 + n^2 \pi^2}x} X ( x ) = c 2 e 1 + n 2 π 2 x + c 3 e − 1 + n 2 π 2 x Pela condição fronteira u ( 0 , y ) = 0 u(0,y) = 0 u ( 0 , y ) = 0 X ( 0 ) = 0 X(0) = 0 X ( 0 ) = 0
X ( 0 ) = 0 ⟹ c 2 e 0 + c 3 e 0 = 0 ⟹ c 2 + c 3 = 0 ⟹ c 3 = − c 2 \begin{aligned}
X(0) = 0 &\implies c_2 e^{0} + c_3 e^{0} = 0\\
&\implies c_2 + c_3 = 0\\
&\implies c_3 = - c_2
\end{aligned} X ( 0 ) = 0 ⟹ c 2 e 0 + c 3 e 0 = 0 ⟹ c 2 + c 3 = 0 ⟹ c 3 = − c 2 Substituindo em X ( x ) X(x) X ( x )
X ( x ) = c 2 ( e 1 + n 2 π 2 x − e − 1 + n 2 π 2 x ) X(x) = c_2 \left(e^{\sqrt{1 + n^2 \pi^2}x} - e^{-\sqrt{1 + n^2 \pi^2}x} \right) X ( x ) = c 2 ( e 1 + n 2 π 2 x − e − 1 + n 2 π 2 x ) Relembrando que sinh ( a x ) = e a x − e − a x 2 \sinh(ax) = \frac{e^{ax} - e^{-ax}}{2} sinh ( a x ) = 2 e a x − e − a x
X ( x ) = c 3 sinh ( 1 + n 2 π 2 x ) X(x) = c_3 \sinh(\sqrt{1 + n^2 \pi^2}x) X ( x ) = c 3 sinh ( 1 + n 2 π 2 x ) Assim, finalmente, obtemos a expressão geral para a função u ( x , y ) u(x,y) u ( x , y )
u ( x , y ) = X ( x ) Y ( y ) = c 4 sinh ( 1 + n 2 π 2 x ) sin ( n π y ) u(x,y)=X(x)Y(y) = c_4 \sinh(\sqrt{1 + n^2 \pi^2}x)\sin(n \pi y) u ( x , y ) = X ( x ) Y ( y ) = c 4 sinh ( 1 + n 2 π 2 x ) sin ( nπ y ) E, como existem infinitas soluções,
u ( x , y ) = ∑ n = 1 + ∞ d n sinh ( 1 + n 2 π 2 x ) sin ( n π y ) u(x,y) = \sum_{n=1}^{+\infty} d_n \sinh(\sqrt{1 + n^2 \pi^2}x)\sin(n \pi y) u ( x , y ) = n = 1 ∑ + ∞ d n sinh ( 1 + n 2 π 2 x ) sin ( nπ y ) Finalmente, vamos determinar d n d_n d n u ( 1 , y ) u(1,y) u ( 1 , y )
u ( 1 , y ) = ∑ n = 1 + ∞ d n sinh ( 1 + n 2 π 2 ) sin ( n π y ) = y u(1,y) = \sum_{n=1}^{+\infty} d_n \sinh(\sqrt{1 + n^2 \pi^2})\sin(n \pi y) = y u ( 1 , y ) = n = 1 ∑ + ∞ d n sinh ( 1 + n 2 π 2 ) sin ( nπ y ) = y Novamente, isto relembra-nos uma série de senos , mas com uma pequena diferença:
d n sinh ( 1 + n 2 π 2 ) = b n = 2 1 ∫ 0 1 y sin ( n π y ) d y = 2 ( − 1 ) n + 1 n π d_n \sinh(\sqrt{1 + n^2 \pi^2}) = b_n = \frac{2}{1} \int_{0}^{1} y \sin(n\pi y) \d y = \frac{2(-1)^{n+1}}{n\pi} d n sinh ( 1 + n 2 π 2 ) = b n = 1 2 ∫ 0 1 y sin ( nπ y ) d y = nπ 2 ( − 1 ) n + 1 d n sinh ( 1 + n 2 π 2 ) = 2 ( − 1 ) n + 1 n π ⟹ d n = 2 ( − 1 ) n + 1 n π sinh ( 1 + n 2 π 2 ) d_n \sinh(\sqrt{1 + n^2 \pi^2}) = \frac{2(-1)^{n+1}}{n\pi} \implies d_n = \frac{2(-1)^{n+1}}{n\pi \sinh(\sqrt{1 + n^2 \pi^2})} d n sinh ( 1 + n 2 π 2 ) = nπ 2 ( − 1 ) n + 1 ⟹ d n = nπ sinh ( 1 + n 2 π 2 ) 2 ( − 1 ) n + 1 Assim, a solução do problema
u ( x , y ) = ∑ n = 1 + ∞ 2 ( − 1 ) n + 1 n π sinh ( 1 + n 2 π 2 ) sinh ( 1 + n 2 π 2 x ) sin ( n π y ) u(x,y) = \sum_{n=1}^{+\infty} \frac{2(-1)^{n+1}}{n\pi \sinh(\sqrt{1 + n^2 \pi^2})} \sinh(\sqrt{1 + n^2 \pi^2}x)\sin(n \pi y) u ( x , y ) = n = 1 ∑ + ∞ nπ sinh ( 1 + n 2 π 2 ) 2 ( − 1 ) n + 1 sinh ( 1 + n 2 π 2 x ) sin ( nπ y ) A equação das ondas tem a forma de
∂ 2 u ∂ t 2 = c 2 ∂ 2 u ∂ x 2 \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2} ∂ t 2 ∂ 2 u = c 2 ∂ x 2 ∂ 2 u Outra notação possível para este problema (e para outros vistos anteriormente) é a seguinte:
u t t = c 2 u x x u_{tt} = c^2 u_{xx} u tt = c 2 u xx Mais genericamente, o problema pode ser definido por
{ u x x = c 2 u t t u ( t , 0 ) = 0 u ( t , L ) = 0 u ( 0 , x ) = u 0 ( x ) ∂ u ∂ t ( 0 , x ) = v 0 ( x ) \begin{cases}
u_{xx} = c^2 u_{tt}\\
u(t,0) = 0\\
u(t,L) = 0\\
u(0,x) = u_0 (x)\\
\frac{\partial u}{\partial t} (0, x) = v_0(x)
\end{cases} ⎩ ⎨ ⎧ u xx = c 2 u tt u ( t , 0 ) = 0 u ( t , L ) = 0 u ( 0 , x ) = u 0 ( x ) ∂ t ∂ u ( 0 , x ) = v 0 ( x ) Pelo método de separação de variáveis conseguimos chegar à seguinte solução geral:
u ( t , x ) = ∑ n = 1 + ∞ ( A n cos ( c π n L t ) + B n sin ( c π n L t ) ) sin ( π n L x ) u(t,x) = \sum_{n=1}^{+\infty} \left(A_n \cos\left(\frac{c\pi n}{L}t\right) + B_n \sin\left(\frac{c\pi n}{L}t\right)\right) \sin\left(\frac{\pi n}{L}x\right) u ( t , x ) = n = 1 ∑ + ∞ ( A n cos ( L c πn t ) + B n sin ( L c πn t ) ) sin ( L πn x ) A n = 2 L ∫ 0 L u 0 ( x ) sin ( π n L x ) d x A_n = \frac{2}{L} \int_{0}^{L} u_0(x) \sin\left(\frac{\pi n}{L}x\right) \d x A n = L 2 ∫ 0 L u 0 ( x ) sin ( L πn x ) d x B n = 2 c π n ∫ 0 L v 0 ( x ) sin ( π n L x ) d x B_n = \frac{2}{c\pi n} \int_{0}^{L} v_0(x) \sin\left(\frac{\pi n}{L}x\right) \d x B n = c πn 2 ∫ 0 L v 0 ( x ) sin ( L πn x ) d x Obviamente, se B n = 0 B_n = 0 B n = 0
u ( t , x ) = ∑ n = 1 + ∞ A n cos ( c π n L t ) sin ( π n L x ) u(t,x) = \sum_{n=1}^{+\infty} A_n \cos\left(\frac{c\pi n}{L}t\right) \sin\left(\frac{\pi n}{L}x\right) u ( t , x ) = n = 1 ∑ + ∞ A n cos ( L c πn t ) sin ( L πn x ) No entanto, como podemos reparar, esta expressão é muito confusa e complicada.
Por esta razão, existe outra fórmula, que nos permite expressar a equação das ondas de uma forma mais simples.
Esta forma vem-nos simplificar a equação das ondas. Novamente, tendo o problema
{ u x x = c 2 u t t u ( 0 , x ) = f ( x ) ∂ u ∂ t ( 0 , x ) = g ( x ) \begin{cases}
u_{xx} = c^2 u_{tt}\\
u(0,x) = f(x)\\
\frac{\partial u}{\partial t} (0, x) = g(x)
\end{cases} ⎩ ⎨ ⎧ u xx = c 2 u tt u ( 0 , x ) = f ( x ) ∂ t ∂ u ( 0 , x ) = g ( x ) é possível descrever a sua solução segundo duas funções p p p q q q
u ( x , t ) = p ( x + c t ) + q ( x − c t ) u(x,t) = p(x+ct) + q(x-ct) u ( x , t ) = p ( x + c t ) + q ( x − c t ) A Fórmula de D'Alembert
u ( x , t ) = f ( x − c t ) + f ( x + c t ) 2 + 1 2 c ∫ x − c t x + c t g ( s ) d s u(x,t) = \frac{f(x-ct) + f(x+ct)}{2} + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) \d s u ( x , t ) = 2 f ( x − c t ) + f ( x + c t ) + 2 c 1 ∫ x − c t x + c t g ( s ) d s Demonstração da Fórmula de D'Alembert VIDEO
Exemplo
Resolva o seguinte problema
∂ 2 u ∂ t 2 = c 2 ∂ 2 u ∂ x 2 u ( x , 0 ) = cos ( x ) − 1 ∂ u ∂ t ( x , 0 ) = 0 , 0 ≤ x ≤ 2 π u ( 0 , t ) = 0 u ( 2 π , t ) = 0 \begin{darray}{ll}
\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}\\
u(x,0) = \cos(x) - 1 & \frac{\partial u}{\partial t} (x,0) = 0, 0 \leq x \leq 2\pi\\
u(0,t) = 0 & u(2\pi,t) = 0
\end{darray} ∂ t 2 ∂ 2 u = c 2 ∂ x 2 ∂ 2 u u ( x , 0 ) = cos ( x ) − 1 u ( 0 , t ) = 0 ∂ t ∂ u ( x , 0 ) = 0 , 0 ≤ x ≤ 2 π u ( 2 π , t ) = 0 Vamos aplicar a Fórmula de D'Alembert
f ( x ) = cos ( x ) − 1 g ( x ) = 0 \begin{darray}{cc}
f(x) = \cos(x) - 1 & g(x) = 0
\end{darray} f ( x ) = cos ( x ) − 1 g ( x ) = 0 Pelo que
u ( x , t ) = cos ( x + c t ) − 1 + cos ( x − c t ) − 1 2 = cos ( x + c t ) + cos ( x − c t ) − 2 2 \begin{aligned}
u(x,t) &= \frac{\cos(x+ct) - 1 + \cos(x-ct) - 1}{2}\\
&= \frac{\cos(x+ct) + \cos(x-ct) - 2}{2}
\end{aligned} u ( x , t ) = 2 cos ( x + c t ) − 1 + cos ( x − c t ) − 1 = 2 cos ( x + c t ) + cos ( x − c t ) − 2 Aplicando agora a condição fronteira,
u ( 0 , t ) = 0 ⇔ cos ( c t ) + cos ( c t ) − 2 2 = 0 ⇔ cos ( c t ) + cos ( c t ) = 2 ⇔ 2 cos ( c t ) = 2 ⇔ cos ( c t ) = 1 ⇔ c t = 2 π n n ∈ Z \begin{aligned}
u(0,t) = 0 & \Leftrightarrow \frac{\cos(ct) + \cos(ct) - 2}{2} = 0\\
& \Leftrightarrow \cos(ct) + \cos(ct) = 2\\
& \Leftrightarrow 2 \cos(ct) = 2\\
& \Leftrightarrow \cos(ct) = 1\\
& \Leftrightarrow ct = 2\pi n & n \in \Z
\end{aligned} u ( 0 , t ) = 0 ⇔ 2 cos ( c t ) + cos ( c t ) − 2 = 0 ⇔ cos ( c t ) + cos ( c t ) = 2 ⇔ 2 cos ( c t ) = 2 ⇔ cos ( c t ) = 1 ⇔ c t = 2 πn n ∈ Z Substituindo na solução obtida, ficamos com
u ( x , t ) = cos ( x + c t ) + cos ( x − c t ) − 2 2 = cos ( x + 2 π n ) + cos ( x − 2 π n ) − 2 2 = cos ( x ) + cos ( x ) − 2 2 = 2 cos ( x ) − 2 2 = cos ( x ) − 1 \begin{aligned}
u(x,t) &= \frac{\cos(x+ct) + \cos(x-ct) - 2}{2}\\
&= \frac{\cos(x+2 \pi n) + \cos(x-2 \pi n) - 2}{2}\\
&= \frac{\cos(x) + \cos(x) - 2}{2}\\
&= \frac{2\cos(x) - 2}{2}\\
&= \cos(x) - 1
\end{aligned} u ( x , t ) = 2 cos ( x + c t ) + cos ( x − c t ) − 2 = 2 cos ( x + 2 πn ) + cos ( x − 2 πn ) − 2 = 2 cos ( x ) + cos ( x ) − 2 = 2 2 cos ( x ) − 2 = cos ( x ) − 1 A solução do problema é u ( x , t ) = cos ( x ) − 1 u(x,t) = \cos(x) - 1 u ( x , t ) = cos ( x ) − 1