The Heat Equation — Question 9

Question 9 – Solution

Strategy. Match the dimensionless boundary conditions as well as the interior diffusion time scale.

Step 1: Identify the dimensionless problem. The chain rule gives $\theta_\eta=\theta_{\xi\xi}$. The outward convection laws become $$\boxed{\theta_\xi(0,\eta)=\mathrm{Bi}\,\theta(0,\eta),\qquad -\theta_\xi(1,\eta)=\mathrm{Bi}\,\theta(1,\eta),\quad \mathrm{Bi}=\frac{hL}{k}.}$$ This dimensionless coefficient is the Biot number for the chosen length scale. The interior time scale is $L^2/\alpha$, but the boundary parameter also affects the dimensionless response.

Step 2: Verify the rescaled interior field. For $u_2(x,t)=u_1(x/2,t/4)$, $$(u_2)_t=\tfrac 14(u_1)_t,\qquad (u_2)_{xx}=\tfrac 14(u_1)_{xx}.$$ Thus the same diffusivity equation holds. At $t=0$, relative positions agree: $x/(2L)=(x/2)/L$, so the stated initial-shape requirement is met.

Step 3: Match both convection conditions. At the left endpoint, $$k(u_2)_x(0,t)=\tfrac 12 k(u_1)_x(0,t/4) =\frac h2\bigl(u_2(0,t)-T_a\bigr).$$ The right outward derivative scales by the same factor. Therefore the coefficient making this a rescaling of the full problem is $$\boxed{h_2=h/2,\qquad h_2(2L)/k=hL/k.}$$ Both rods then have the same dimensionless boundary problem.

Step 4: Qualify the length-squared claim. If $h_2=h$, the longer rod instead has Biot number $2hL/k$. The proposed field generally fails the boundary law; the full response is not obtained merely by multiplying time by four. A special zero-excess equilibrium cannot establish a general scaling rule. The factor-four rescaling is valid when convection is also adjusted to $h/2$ and the initial shapes match. If both rods are perfectly insulated ($h=0$), the zero-flux conditions are preserved automatically. Interior dimensional analysis alone does not justify ignoring changed boundary parameters.