Heat Equation with Non-Zero Temperature Boundaries: Question 9-6.10 & Solution

Question 10

Let $u_t=\kappa u_{xx}$ on $0<x<L$ , with fixed endpoint temperatures $A,B$ , $\kappa,L>0$ , and stationary field $S=A+(B-A)x/L$ . For smooth compatible data define the error energy $E_v=\tfrac 12\int_0^L(u-S)^2\,dx$ and the raw quadratic quantity $E_u=\tfrac 12\int_0^L u^2\,dx$ . You may use the zero-endpoint Poincare inequality $\int v_x^2\ge(\pi/L)^2\int v^2$ and the parabolic maximum principle.

Tasks

Derive the energy identities for $E_v$ and $E_u$ , retaining all boundary terms. Prove the sharp bound $E_v(t)\le e^{-2\lambda t}E_v(0)$ , where $\lambda=\kappa\pi^2/L^2$ .
For $A=B=1$ and $u(x,0)=1-\varepsilon\sin(\pi x/L)$ , $0<\varepsilon\le 1$ , construct the solution. Prove that its raw quadratic quantity increases while its error energy decreases.
If $|u(x,0)-S(x)|\le C\sin(\pi x/L)$ , $C>0$ , use comparison to prove a pointwise error bound for all later times. Give a sufficient time for uniform temperature error at most $\delta$ , where $0<\delta<C$ .
Give a sufficient time for reducing error energy to a fraction $\eta\in(0,1)$ of its initial value, and explain why neither total heat nor raw quadratic energy alone measures distance from a maintained nonzero equilibrium.

Original worksheet page 1: question and worked solution for 9-6-010

Show solutionHide solution+

Question 10 – Solution

Strategy. Dissipation applies naturally to deviations from the maintained equilibrium; the boundary reservoirs can increase the unshifted quantities.

Step 1: Identify which energy has no boundary work. For $v=u-S$ , the endpoints vanish and $v_t=\kappa v_{xx}$ . Therefore $E_v'=-\kappa\int_0^L v_x^2\,dx\le-2\lambda E_v, \qquad E_v(t)\le e^{-2\lambda t}E_v(0).$ A first-sine error attains equality, so the rate is sharp. For the raw quantity, $\boxed{E_u'=\kappa[u u_x]_0^L-\kappa\int_0^L u_x^2\,dx.}$ The first term need not vanish with nonzero maintained boundary values.

Step 2: Exhibit simultaneous raw growth and error decay. The stated initial data give $u=1-\varepsilon e^{-\lambda t}\sin(\pi x/L)$ . Then $E_u=\frac L2-\frac{2\varepsilon L}{\pi}e^{-\lambda t} +\frac{\varepsilon^2 L}{4}e^{-2\lambda t},\qquad E_v=\frac{\varepsilon^2 L}{4}e^{-2\lambda t}.$ Writing $z=\varepsilon e^{-\lambda t}\in(0,1]$ gives $E_u'=\lambda Lz(2/\pi-z/2)>0$ , since $2/\pi>1/2$ . Meanwhile $E_v'=-2\lambda E_v<0$ . Heating by the reservoirs moves this field closer to equilibrium while increasing its raw quadratic quantity.

Step 3: Establish a pointwise stopping criterion. The barriers $\pm C e^{-\lambda t}\sin(\pi x/L)$ solve the homogeneous zero-endpoint heat equation and bound the initial error. Applying comparison to their differences with $v$ on any finite time interval gives $\boxed{|u-S|\le C e^{-\lambda t}\sin(\pi x/L)\le C e^{-\lambda t}.}$ Thus $t\ge\lambda^{-1}\log(C/\delta)$ suffices for uniform error at most $\delta$ . This conclusion uses the supplied pointwise initial bound, not merely an initial $L^2$ norm.

Step 4: Distinguish the two stopping criteria. The energy inequality gives $E_v(t)/E_v(0)\le\eta$ once $t\ge(2\lambda)^{-1}\log(1/\eta)$ for nonzero initial error; zero error is already at equilibrium. The factor of two comes from squaring a decaying amplitude. Total heat can conceal spatial differences, and raw energy includes the maintained background and boundary work. Neither is a substitute for an explicit norm of $u-S$ when measuring equilibration.

Original worksheet page 2: question and worked solution for 9-6-010

Heat Equation with Non-Zero Temperature Boundaries — Question 10

Question 10

Question 10 – Solution