Let ohm x y z R3x2 y2 z2 0 ux y zt 0 for x y z and t 0

Let ohm: = {(x, y, z) R^3:x^2 + y^2 + z^2 0, u(x, y, z,t) = 0 for (x, y, z) and t > 0, u(x, y, z,y,t) = 0 for (x, y, z) and t > 0, u(x, y, z,0) = sin (pi rightarrow x^2 + y^2 + Z^2) for (x, y, z). Your solution may contain unevaluated infinite sums, but not unevaluated integrals.

Solution

Solution:Given equation is

\\frac{\\partial^{2}u }{ \\partial x^{2}} + \\frac{\\partial^{2}u }{ \\partial y^{2}}

+\\frac{\\partial^{2}u }{ \\partial z^{2}} = \\frac{\\partial u }{ \\partial t} ..................(1).

Suppose that (1) has the solution of the form

u(x,y,z,t)= X(x)Y(y)Z(z)T(t)          .............................(2)

where X, Y, Z, and T are respectively functions of x, y, z and t alone.

Substituting this value of u in (1), we have

X\'\'YZT + XY\'\'ZT + XYZ\'\'T = XYZT\'

or X\'\'/X + Y\'\'/Y + Z\'\'/Z = T\'/T ..............(3)

Since x, y, z and t are independent variables, (3) is true only if each term on each side is a

constant such that

X\'\'/X = - n², Y\'\'/Y = - m², Z\'\'/Z = - l², T\'/T = - p² ....................(4)

where n² + m²+ l²= p² ...........................(5)

We have chosen constants in (4) in such a manner so that the solution u(x,y,z,t) has

the property that u tends to zero(0) and t tends to infity. This is generally satisfied

due to physical conditions of actual physical problem.

Solving equatins in (4) , we have

X_n(x) = A_ncos nx + B_n sin nx, Y_m(y) = C_mcos my + D_m sin my,

Z_l(z) = E_l cos lz +F_l sin lz, and T_p(t) = G_pe^-p2t= H_nmle^{-(n2 + m2+ l2)t}

So u_nml(x,y,z,t) = H_nml(A_ncos nx + B_n sin nx, Y_m(y) )(C_mcos my + D_m sin my)

( E_l cos lz +F_l sin lz) e^{-(n2 + m2+ l2)t} ...........................(6)

are solution of (1).

Hence the general solution of (1) is given by

u(x,y,z,t) = \\sigma_{n=1}^{\\infty}\\sigma_{m=1}^{\\infty}\\sigma_{l=1}^{\\infty}u_nml(x,y,z,t).