u x x + u y y = 0 , x ∈ ( 0 , a ) , y ∈ ( 0 , b ) u ( 0 , y ) = 0 , u x ( a , y ) = 0 , y ∈ [ 0 , b ] u ( x , 0 ) = f ( x ) , u ( x , b ) = 0 , x ∈ [ 0 , a ] u ( x , y ) = X ( x ) Y ( y ) u x = X ′ Y , u x x = X ′ ′ Y u y = X Y ′ , u y y = X Y ′ ′ u_{xx}+u_{yy}=0, x\in(0,a), y\in(0,b)\\
u(0,y)=0,u_x(a,y)=0,y\in[0,b]\\
u(x,0)=f(x), u(x,b)=0, x\in[0,a]\\
u(x,y)=X(x)Y(y)\\
u_x=X'Y, u_{xx}=X''Y\\
u_y=XY',u_{yy}=XY''\\ u xx + u yy = 0 , x ∈ ( 0 , a ) , y ∈ ( 0 , b ) u ( 0 , y ) = 0 , u x ( a , y ) = 0 , y ∈ [ 0 , b ] u ( x , 0 ) = f ( x ) , u ( x , b ) = 0 , x ∈ [ 0 , a ] u ( x , y ) = X ( x ) Y ( y ) u x = X ′ Y , u xx = X ′′ Y u y = X Y ′ , u yy = X Y ′′
the separation of variables and used
u ( 0 , y ) = 0 , u x ( a , y ) = 0 , y ∈ [ 0 , b ] u(0,y)=0,u_x(a,y)=0,y\in[0,b] u ( 0 , y ) = 0 , u x ( a , y ) = 0 , y ∈ [ 0 , b ]
we have
X ′ ′ ( x ) + λ X ( x ) = 0 , 0 < x < a , X ( 0 ) = X ′ ( a ) = 0 X''(x)+\lambda X(x)=0, 0<x<a, X(0)=X'(a)=0 X ′′ ( x ) + λ X ( x ) = 0 , 0 < x < a , X ( 0 ) = X ′ ( a ) = 0
eigenvalues
λ k = ( 2 k + 1 a π 2 ) 2 , k ∈ Z + \lambda_k=(\frac{2k+1}{a}\frac{\pi}{2})^2, k\in Z_+ λ k = ( a 2 k + 1 2 π ) 2 , k ∈ Z +
eigenfunctions
X k ( x ) = sin 2 k + 1 a π 2 x , k ∈ Z + X_k(x)=\sin\frac{2k+1}{a}\frac{\pi}{2}x, k\in Z_+ X k ( x ) = sin a 2 k + 1 2 π x , k ∈ Z +
Y Y Y
Y k ′ ′ − λ k Y k = 0 , k ∈ Z + Y k ( y ) = A k e λ k y + B k e − λ k y , k ∈ Z + Y''_k-\lambda_kY_k=0, k\in Z_+\\
Y_k(y)=A_ke^{\sqrt{\lambda_k}y}+B_ke^{-\sqrt{\lambda_k}y}, k\in Z_+ Y k ′′ − λ k Y k = 0 , k ∈ Z + Y k ( y ) = A k e λ k y + B k e − λ k y , k ∈ Z +
solution of problem is
u ( x , y ) = ∑ k = 0 ∞ ( A k e 2 k + 1 a π 2 y + B k e − 2 k + 1 a π 2 y ) sin 2 k + 1 a π 2 x u(x,y)=\sum\limits_{k=0}^{\infty}(A_ke^{\frac{2k+1}{a}\frac{\pi}{2}y}+B_ke^{-\frac{2k+1}{a}\frac{\pi}{2}y})\sin\frac{2k+1}{a}\frac{\pi}{2}x u ( x , y ) = k = 0 ∑ ∞ ( A k e a 2 k + 1 2 π y + B k e − a 2 k + 1 2 π y ) sin a 2 k + 1 2 π x
u ( x , y ) → u ( x , 0 ) = f ( x ) , u ( x , b ) = 0 , x ∈ [ 0 , a ] u(x,y)\to u(x,0)=f(x), u(x,b)=0, x\in[0,a] u ( x , y ) → u ( x , 0 ) = f ( x ) , u ( x , b ) = 0 , x ∈ [ 0 , a ]
{ ∑ k = 0 ∞ ( A k + B k ) sin 2 k + 1 a π 2 x = f ( x ) ∑ k = 0 ∞ ( A k e 2 k + 1 a π 2 b + B k e − 2 k + 1 a π 2 b ) sin 2 k + 1 a π 2 x = 0 , x ∈ [ 0 , a ] { A k + B k = f k A k e 2 k + 1 a π 2 b + B k e − 2 k + 1 a π 2 b = 0 , k ∈ Z + \left\{\begin{matrix}
\sum\limits_{k=0}^{\infty}(A_k+B_k)\sin\frac{2k+1}{a}\frac{\pi}{2}x=f(x)\\
\sum\limits_{k=0}^{\infty}(A_ke^{\frac{2k+1}{a}\frac{\pi}{2}b}+B_ke^{-\frac{2k+1}{a}\frac{\pi}{2}b})\sin\frac{2k+1}{a}\frac{\pi}{2}x=0, \end{matrix}\right.\\
x\in[0,a]\\
\left\{\begin{matrix}
A_k+B_k=f_k\\
A_ke^{\frac{2k+1}{a}\frac{\pi}{2}b}+B_ke^{-\frac{2k+1}{a}\frac{\pi}{2}b}=0, k\in Z_+
\end{matrix}\right. ⎩ ⎨ ⎧ k = 0 ∑ ∞ ( A k + B k ) sin a 2 k + 1 2 π x = f ( x ) k = 0 ∑ ∞ ( A k e a 2 k + 1 2 π b + B k e − a 2 k + 1 2 π b ) sin a 2 k + 1 2 π x = 0 , x ∈ [ 0 , a ] { A k + B k = f k A k e a 2 k + 1 2 π b + B k e − a 2 k + 1 2 π b = 0 , k ∈ Z +
where
f k = 2 a ∫ 0 a f ( z ) sin 2 k + 1 a π 2 z d z , k ∈ Z + f_k=\frac{2}{a}\int^{a}_{0}f(z)\sin\frac{2k+1}{a}\frac{\pi}{2}zdz, k\in Z_+ f k = a 2 ∫ 0 a f ( z ) sin a 2 k + 1 2 π z d z , k ∈ Z +
Then
A k = − f k e − 2 k + 1 a π 2 b 2 s h 2 k + 1 a π 2 b B k = f k e − 2 k + 1 a π 2 b 2 s h 2 k + 1 a π 2 b , k ∈ Z + A_k=\frac{-f_ke^{-\frac{2k+1}{a}\frac{\pi}{2}b}}{2sh\frac{2k+1}{a}\frac{\pi}{2}b}\\
B_k=\frac{f_ke^{-\frac{2k+1}{a}\frac{\pi}{2}b}}{2sh\frac{2k+1}{a}\frac{\pi}{2}b}, k\in Z_+ A k = 2 s h a 2 k + 1 2 π b − f k e − a 2 k + 1 2 π b B k = 2 s h a 2 k + 1 2 π b f k e − a 2 k + 1 2 π b , k ∈ Z +
u ( x , y ) = ∑ k = 0 ∞ f k s h 2 k + 1 a π 2 ( b − y ) s h 2 k + 1 a π 2 y sin ( 2 k + 1 ) π 2 a x , x ∈ ( 0 , a ) , y ∈ ( 0 , b ) u(x,y)=\sum ^{\infty}_{k=0}\frac{f_ksh{\frac{2k+1}{a}\frac{\pi}{2}(b-y)}}{sh\frac{2k+1}{a}\frac{\pi}{2}y}\sin\frac{(2k+1)\pi}{2a}x, \\
x\in(0,a), y\in(0,b) u ( x , y ) = ∑ k = 0 ∞ s h a 2 k + 1 2 π y f k s h a 2 k + 1 2 π ( b − y ) sin 2 a ( 2 k + 1 ) π x , x ∈ ( 0 , a ) , y ∈ ( 0 , b )
Comments