Question

Consider the steady temperature T (2,y) in a rectangular plate that occupies 0 <<< 9 and 0 <y<5, which is heated at constant temperature 150 at 9 and 0 along its other three sides. (a) For separation solutions T(1,y) = F(x)G(y), you are given that admissible F(1) are the eigenfunctions Fn (1) = sinh(An I) for n=1,2,... and G(y) are the eigenfunctions Gn(y) = sin(Any) for n=1,2,... A for In = (b) The solution is the superposition T(z,y) = an Fn (I)Gn(y). n=1 Use the non-homogeneous boundary condition to deduce the Fourier sine series anCn sin(Any) = 150. i. Give an expression for the coefficients Cn. Cn ii. Find a 5. Note that Maple uses sinh (x) for the function sinh() 05 A7

Answer 1

just information un(x, y) + y(x,y) = 0, 0<x<I, 0 <y<b, here 1 = 9,1 = 5 u(0, y) = 0, u(), y) =150,u (x,0)= 0, u(x,b)=0 assume solution of the form u(x,y)= X( x)Y(y) x"(x)Y (y)+X(x)Y"(y) = 0 = X"(x) Y(y)=-X(x)Y"(y) X"(x)_ Y"(y) =13 x(x) Y(y) X"(x) - AX (x )= 0, Y"(y)+ AY(y) = 0 u(x,0)=0=X(x)Y(0)=0=Y(0)= 0 u(x,b)=0=X(x) Y(b)=0=Y(6)= 0 consider the de Y"(y)+ AY (y) = 0,7(0) = 0,7(b) = 0. let A = mº. Y"(y)+AY (y)=0=Y(y) = Acos(my) + B sin (my) Y(0)=0=0= A Y(6)=0=Y(6)= B sin(mb) = 0 sin (mb) = 0 = mb = n1 =m= (here b=5) ny , Y, (y) = sin n7 fill the answer = 5 F, (x) = sinh(y), G. (y) = sin(y) answer 2 (8) (1.2) - a, ảnh (47) (*) boundary conditions T(0, y) = 0, 7(9, y)=150= 9n7 nity 7 (9, y) = 150 a, sinh sin = 150 5 using fourier half sine series for f(y)=150, In a, sinh 150dy ... (1) 5 5 - плу = sin sin ** 60 9n7 sinh 5 nity 5 97 c = sinh 5 2 as 60 sin (dy) dy sinh (9)% 60 sinh (97) T 120 TT sinh (97)