∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0 Let u ( x , y ) = X ( x ) Y ( y ) X ′ ′ Y + X Y ′ ′ = 0 , X ′ ′ Y = − X Y ′ ′ X ′ ′ Y = − X Y ′ ′ = K X ′ ′ X = − Y ′ ′ Y = K X ′ ′ X = K , Y ′ ′ Y = − K If K is chosen to be positive say K = λ 2 , then X ′ ′ = λ 2 X X = A e λ x + B e − λ x Y ′ ′ = − λ 2 Y Y = C cos ( λ y ) + D sin ( λ y ) u ( x , y ) = ( A e λ x + B e − λ x ) ( C cos ( λ y ) + D sin ( λ y ) ) u ( x , 0 ) = C ( A e λ x + B e − λ x ) = sin ( π x ) u ( 0 , 0 ) = C ( A + B ) = 0 u ( 0 , y ) = ( A + B ) ( C cos ( λ y ) + D sin ( λ y ) ) = sin ( π y ) u ( 0 , 0 ) = C ( A + B ) = 0 C = 0 , A = − B u ( x , 1 ) = ( A e λ x + B e − λ x ) ( C cos ( λ ) + D sin ( λ ) ) = x 2 u ( 0 , 1 ) = ( A + B ) ( C cos ( λ ) + D sin ( λ ) ) = 0 ( A + B ) ( D sin ( λ ) ) = 0 We would not want D to be zero , so we can force λ to produce a 0 term Choosing D = 0 would make u = 0 so we must force sin ( λ ) to be zero i.e. choose λ = n π where n is a positive integer sin ( λ ) = 0 , λ = n π u ( 2 , y ) = ( A e 2 λ + B e − 2 λ ) ( C cos ( λ y ) + D sin ( λ y ) ) = y u ( 2 , 0 ) = C ( A e 2 λ + B e − 2 λ ) = 0 u ( x , y ) = ( A e λ x − A e − λ x ) ( D sin ( λ y ) ) = A D ( e n π x − e − n π x ) sin ( λ y ) = 2 A D sinh ( n π x ) sin ( n π y ) = E sinh ( n π x ) sin ( n π y ) u ( x , y ) can be superposed as; u ( x , y ) = ∑ n = 1 ∞ E n sinh ( n π x ) sin ( n π y ) u ( 2 , y ) = y u ( 2 , y ) = ∑ n = 1 ∞ E n sinh ( 2 n π ) sin ( n π y ) = ∑ n = 1 ∞ b n sin ( n π y ) Where b n = E n sinh ( 2 n π ) We have here a half-range Fourier sine series representation of a function u ( 2 , y ) defined over 0 < y < 1 , Extending u ( 2 , y ) as an even periodic function with period 2 and using standard Fourier series theory gives b n = 2 ∫ 0 1 y sin ( n π y ) d y = 2 ∫ 0 1 y d ( − cos ( n π y ) n π ) = − 2 y cos ( n π y ) n π ∣ 0 1 + 2 n π ∫ 0 1 cos ( n π y ) d y = − 2 cos ( n π ) n π + 2 n π ⋅ sin ( n π y ) n π ∣ 0 1 = − 2 cos ( n π ) n π + 2 sin ( n π ) n 2 π 2 But sin ( n π ) = 0 & cos ( n π ) = ( − 1 ) n ∀ n ∈ Z + b n = − 2 ( − 1 ) n n π E n sinh ( 2 n π ) = − 2 ( − 1 ) n n π ∴ E n = − 2 ( − 1 ) n n π ( sinh ( 2 n π ) ) ∴ u ( x , y ) = ∑ n = 1 ∞ 2 ( − 1 ) n − 1 n π ( sinh ( 2 n π ) ) sinh ( n π x ) sin ( n π y ) \displaystyle
\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\\
\textsf{Let}\,\, u(x, y) = X(x)Y(y)\\
X''Y+ XY'' = 0,\,\, X''Y= -XY''\\
X''Y= -XY'' = K\\
\frac{X''}{X} = -\frac{Y''}{Y} = K\\
\frac{X''}{X} = K,\,\, \frac{Y''}{Y} = -K\\
\textsf{If}\,\, K\,\, \textsf{is chosen to be positive}\\
\textsf{say}\,\, K = \lambda^2,\,\,\textsf{then}\,\,\\
X'' = \lambda^2 X\\
X = Ae^{\lambda x} + Be^{-\lambda x}\\
Y'' = -\lambda^2 Y\\
Y = C\cos(\lambda y) + D\sin(\lambda y)\\
u(x,y) = \left(Ae^{\lambda x} + Be^{-\lambda x}\right)\left(C\cos(\lambda y) + D\sin(\lambda y)\right)\\
u(x,0) = C\left(Ae^{\lambda x} + Be^{-\lambda x}\right) = \sin(\pi x)\\
u(0,0) = C\left(A + B\right) = 0\\
u(0,y) = \left(A + B\right)\left(C\cos(\lambda y) + D\sin(\lambda y)\right) = \sin(\pi y)\\
u(0,0) = C\left(A + B\right) = 0\\
C = 0,\,\, A = -B\\
u(x,1) = \left(Ae^{\lambda x} + Be^{-\lambda x}\right)\left(C\cos(\lambda) + D\sin(\lambda)\right) = \frac{x}{2}\\
u(0,\,1) = \left(A + B\right)\left(C\cos(\lambda) + D\sin(\lambda)\right) = 0\\
\left(A + B\right)\left(D\sin(\lambda)\right) = 0\\
\textsf{We would not want}\,\, D \,\,\textsf{to be zero},\\
\textsf{so we can force}\,\, \lambda \,\, \textsf{to produce a 0 term}\\
\textsf{Choosing}\,\, D = 0 \,\, \textsf{would make}\,\,u = 0 \\
\textsf{so we must force}\,\, \sin(\lambda)\,\, \textsf{to be zero}\\
\textsf{i.e. choose} \,\, \lambda = n\pi \\
\textsf{where}\,\, n\,\, \textsf{is a positive integer}\\
\sin(\lambda) = 0, \lambda = n\pi\\
u(2,y) = \left(Ae^{2\lambda} + Be^{-2\lambda}\right)\left(C\cos(\lambda y) + D\sin(\lambda y)\right) = y\\
u(2,0) = C\left(Ae^{2\lambda} + Be^{-2\lambda}\right) = 0\\
\begin{aligned}
u(x,y) &= \left(Ae^{\lambda x} - Ae^{-\lambda x}\right)\left(D\sin(\lambda y)\right)\\
&= AD\left(e^{n\pi x} - e^{-n\pi x}\right)\sin(\lambda y) = 2AD\sinh\left(n\pi x\right)\sin(n\pi y)
\\&= E\sinh\left(n\pi x\right)\sin(n\pi y)
\end{aligned}\\
u(x,y)\,\,\textsf{can be superposed as;}\\
\begin{aligned}
u(x,y) &=\sum_{n = 1}^{\infty} E_n\sinh\left(n\pi x\right)\sin(n\pi y)\\
\end{aligned}\\
u(2,y) =y\\
\begin{aligned}
u(2,y) &=\sum_{n = 1}^{\infty} E_n\sinh\left(2 n\pi\right)\sin(n\pi y)
\\&=\sum_{n = 1}^{\infty} b_n\sin(n\pi y)
\end{aligned}\\
\textsf{Where}\,\, b_n = E_n\sinh\left(2 n\pi\right)\\
\textsf{We have here a half-range Fourier sine}\\
\textsf{series representation of a function}\,\, u(2,y)\\
\textsf{defined over}\,\ 0< y < 1, \,\,\textsf{Extending}\,\, u(2,y)\,\, \textsf{as an}\\
\textsf{even periodic function with period}\,\, 2\,\, \textsf{and}\\
\textsf{using standard Fourier series theory gives}\\
\begin{aligned}
b_n &= 2\int_0^1 y \sin(n\pi y)\, \mathrm{d}y
\\&= 2\int_0^1 y\, \mathrm{d}\left(\frac{-\cos(n\pi y)}{n\pi}\right)
\\&= \frac{-2y\cos(n\pi y)}{n\pi}\biggr\vert_0^1 + \frac{2}{n\pi}\int_0^1 \cos(n\pi y)\,\mathrm{d}y
\\&= \frac{-2\cos(n\pi)}{n\pi}+ \frac{2}{n\pi}\cdot\frac{\sin(n\pi y)}{n\pi}\biggr\vert_0^1
\\&= \frac{-2\cos(n\pi)}{n\pi}+ \frac{2\sin(n\pi)}{n^2\pi^2}
\end{aligned}\\
\textsf{But}\,\, \sin(n\pi) = 0\,\, \&\,\, \cos(n\pi) = (-1)^n \,\,\forall n \in \mathbb{Z}^{+} \\
b_n = -\frac{2(-1)^n}{n\pi} \\
E_n\sinh\left(2 n\pi\right) = -\frac{2(-1)^n}{n\pi}\\
\therefore E_n = -\frac{2(-1)^n}{n\pi\left(\sinh\left(2 n\pi\right)\right)}\\
\begin{aligned}
\therefore u(x,y) &= \sum_{n = 1}^{\infty} \frac{2(-1)^{n - 1}}{n\pi\left(\sinh\left(2 n\pi\right)\right)}\sinh\left(n\pi x\right)\sin(n\pi y)
\end{aligned} ∂ x 2 ∂ 2 u + ∂ y 2 ∂ 2 u = 0 Let u ( x , y ) = X ( x ) Y ( y ) X ′′ Y + X Y ′′ = 0 , X ′′ Y = − X Y ′′ X ′′ Y = − X Y ′′ = K X X ′′ = − Y Y ′′ = K X X ′′ = K , Y Y ′′ = − K If K is chosen to be positive say K = λ 2 , then X ′′ = λ 2 X X = A e λ x + B e − λ x Y ′′ = − λ 2 Y Y = C cos ( λ y ) + D sin ( λ y ) u ( x , y ) = ( A e λ x + B e − λ x ) ( C cos ( λ y ) + D sin ( λ y ) ) u ( x , 0 ) = C ( A e λ x + B e − λ x ) = sin ( π x ) u ( 0 , 0 ) = C ( A + B ) = 0 u ( 0 , y ) = ( A + B ) ( C cos ( λ y ) + D sin ( λ y ) ) = sin ( π y ) u ( 0 , 0 ) = C ( A + B ) = 0 C = 0 , A = − B u ( x , 1 ) = ( A e λ x + B e − λ x ) ( C cos ( λ ) + D sin ( λ ) ) = 2 x u ( 0 , 1 ) = ( A + B ) ( C cos ( λ ) + D sin ( λ ) ) = 0 ( A + B ) ( D sin ( λ ) ) = 0 We would not want D to be zero , so we can force λ to produce a 0 term Choosing D = 0 would make u = 0 so we must force sin ( λ ) to be zero i.e. choose λ = nπ where n is a positive integer sin ( λ ) = 0 , λ = nπ u ( 2 , y ) = ( A e 2 λ + B e − 2 λ ) ( C cos ( λ y ) + D sin ( λ y ) ) = y u ( 2 , 0 ) = C ( A e 2 λ + B e − 2 λ ) = 0 u ( x , y ) = ( A e λ x − A e − λ x ) ( D sin ( λ y ) ) = A D ( e nπ x − e − nπ x ) sin ( λ y ) = 2 A D sinh ( nπ x ) sin ( nπ y ) = E sinh ( nπ x ) sin ( nπ y ) u ( x , y ) can be superposed as; u ( x , y ) = n = 1 ∑ ∞ E n sinh ( nπ x ) sin ( nπ y ) u ( 2 , y ) = y u ( 2 , y ) = n = 1 ∑ ∞ E n sinh ( 2 nπ ) sin ( nπ y ) = n = 1 ∑ ∞ b n sin ( nπ y ) Where b n = E n sinh ( 2 nπ ) We have here a half-range Fourier sine series representation of a function u ( 2 , y ) defined over 0 < y < 1 , Extending u ( 2 , y ) as an even periodic function with period 2 and using standard Fourier series theory gives b n = 2 ∫ 0 1 y sin ( nπ y ) d y = 2 ∫ 0 1 y d ( nπ − cos ( nπ y ) ) = nπ − 2 y cos ( nπ y ) ∣ ∣ 0 1 + nπ 2 ∫ 0 1 cos ( nπ y ) d y = nπ − 2 cos ( nπ ) + nπ 2 ⋅ nπ sin ( nπ y ) ∣ ∣ 0 1 = nπ − 2 cos ( nπ ) + n 2 π 2 2 sin ( nπ ) But sin ( nπ ) = 0 & cos ( nπ ) = ( − 1 ) n ∀ n ∈ Z + b n = − nπ 2 ( − 1 ) n E n sinh ( 2 nπ ) = − nπ 2 ( − 1 ) n ∴ E n = − nπ ( sinh ( 2 nπ ) ) 2 ( − 1 ) n ∴ u ( x , y ) = n = 1 ∑ ∞ nπ ( sinh ( 2 nπ ) ) 2 ( − 1 ) n − 1 sinh ( nπ x ) sin ( nπ y )
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