Question

Find the solution to the heat equation in 3 dimensions on the half space z>0, with homogeneous Neumann boundary conditions at z=0.

Answer 1

^{Heat equation in 3 dimensions is}

u_t=k(u_xx+u_yy+u_zz).

After separation of variables in above heat equation

X^"+aX=0 ,Y^"+bY=0 and Z^"+cZ=0 ,T^'+k(a+b+c)T=0.

Let us assume that a=m²,b=n²,c=r²

Above three equations are in the form of D²+k²=0 and solution become c₁coskx+c₂sinkx.

So,Finally solution for above heat equation in three dimensions is

u(x,y,z,t)=(c₁cosmx+c₂sinmx)(c₃cosny+c₄sinny)(c₅cosrz+c₆sinrz)(c₇e^k(a+b+c)).

Homogeneous Newmann boundary conditions at z=0 means u(x,y,0,t)=Q(x).

Now,let us take boundary values as x varies from 0--->p ,y varies from 0--->q and z varies from 0--->l and assume u value is 0 for above conditions to find out the coefficient values

u(0,y,z,t)=c₁=0.,u(p,y,z,t)=(c₁cosmp+c₂sinmp)=0 and we will get eigen value=m=nπ/p. Where n=0,1,2--n

In this way,c₃=0,c₄=nπ/q,c₅=0 and c₆=nπ/l.and Newman boundary condition u(x,y,z,0)=Q(x) and we will get c₇=Q(x).

So,Finally by substituting above values,we will get solution as

u(x,y,z,t)=sin((nπ/p)x)sin(nπy/q)sin(nπz/l)e^{knπ/(√((1/p2)+(1/q2)+(1/l2)))}