Consider a wall of thickness b extending from 0 to infinity in the y-direction and from negative infinity to positive infinity in the z-direction. (See figure 11.3-1).  The surfaces at x= +/- b/2 are held at a temperature To, whereas the bottom of the wall at the surface y=0 is at T1.  Show that the imaginary part of the function (11.3-3) gives the steady state temperature distribution Theta(x,y)=(T-To)/(T1-To). After finding Theta(x,y), find the heat flux qy(x,0) through the base of the wall.

                        figure 11.3-1

> restart;

> w:=z->log((sin(Pi*z/b)-1)/(sin(Pi*z/b)+1));This is 11.3-3. Pi was excluded in this equation in order for Maple to be able to reduce and extract the imaginary part.

> assume(x,real);assume(y,real);z=x+y*I, where x and y are real.

> v:=Im(evalc(w(x+I*y)));Im extracts the imaginary part of an expression, and evalc manipulates complex valued expressions by attempting to split expressions into real and imaginary parts. Arctan(a,b) is a function that computes the principal value of a complex number a+I*b.

> Theta1:=simplify(tan(v),trig);This is the tangent of the imaginary part of w(z)*Pi.

> Theta2:=tan(2*arctan(cos(Pi*x/b)/sinh(Pi*y/b)));This is the tangent of the solution given in 11.3-4.

> simplify(%,trig);Maple rewrites tan as sin/cos

> expand(%);Maple can now expand the arctans in the above expression into cos and sinh functions.

> Theta1-%;The difference between the solution derived above and the books solution.

> simplify(%);Maple agrees with Bird, Stewart, and Lightfoot's solution 11.3-4.

> Theta:=(x,y)->2/Pi*arctan(cos(Pi*x/b)/sinh(Pi*y/b));11.3-4

> Theta(b/2,y);Theta satisfies the boundary conditions T=To, or Theta =0, at the edges of the wall, x=+/-b/2.

> limit(Theta(x,h),h=0)-1;Theta should satisfy the boundary condition that at y=0, T=T1 or Theta=1.

> limit(2/Pi*arctan(h),h=infinity);Since sinh(0)=0, the expression above reduces to 2/Pi*arctan(infinity).

> T:=(x,y)->Theta(x,y)*(T1-To)+To;From the definition of Theta(x,y), we can get T(x,y).

> simplify(T(x,y),trig);

> T:=(x,y)->(2*arctan(cos(Pi*x/b)/sinh(Pi*y/b))*T1-2*arctan(cos(Pi*x/b)/sinh(Pi*y/b))*To+To*Pi)/Pi;This is the above simplification of T(x,y).

> qy:=(x,y)->-k*D[2](T)(x,y);The heat flux in the y direction is defined as the partial derivative of T with respect to y multiplied by the negative of the thermal conductivity.

> simplify(qy(x,y));

> limit(qy(x,h),h=0);This is the heat flux through the base of the wall, y=0 which is identical to the solution 11.3-5.