Heat Equation with Non-Zero Temperature Boundaries — Question 9

Question 9 – Solution

Strategy. Equal sine values at the two sensors let their difference cancel the transient; a second observation time separates the remaining steady and decaying parts.

Step 1: Cancel the equal transient values. At both sensor positions, $\sin(\pi x)=\sqrt 3/2$. Thus $$d:=B-A=3(y_R-y_L),\qquad m(t)=S+K e^{-\pi^2t},\quad S=(A+B)/2,\quad K=\sqrt 3 C/2.$$ The difference determines the stationary gradient at any measurement time. It must be unchanged between the two observations if the assumed model holds.

Step 2: Invert the two-time averages. Put $r=e^{-\pi^2\Delta}\in(0,1)$. From $m_0=S+K$, $m_\Delta=S+rK$, $$\boxed{K=\frac{m_0-m_\Delta}{1-r},\quad S=\frac{m_\Delta-rm_0}{1-r},\quad A=S-d/2,\ B=S+d/2,\ C=2K/\sqrt 3.}$$ The denominator is nonzero because $\Delta>0$. Equality of the observed differences is the only additional consistency condition; with it these parameters reconstruct all four readings under the stated model.

Step 3: Reconstruct the numerical example. Here $r=1/2$, $d=2$, $m_0=3$ and $m_\Delta=5/2$. Thus $S=2$, $K=1$ and $$\boxed{A=1,\quad B=3,\quad C=2/\sqrt 3,\qquad u=1+2x+\frac 2{\sqrt 3}e^{-\pi^2t}\sin(\pi x).}$$ The endpoint values are one and three. The sensor transients are one at zero and one half at $\Delta$, giving exactly the four supplied readings. The linear stationary part and the decaying sine verify $u_t=u_{xx}$.

Step 4: State the identifiability and conditioning limits. A single pair gives $d$ and $S+K$, leaving infinitely many choices of $S,K$. For two times, errors $e_0,e_\Delta$ in the averages cause $\delta S=(e_\Delta-r e_0)/(1-r)$. If $|e_0|,|e_\Delta|\le\eta$, $$|\delta S|\le\frac{1+r}{1-r}\eta\sim\frac{2\eta}{\pi^2\Delta} \quad(\Delta\downarrow 0).$$ Opposite-signed errors attain the bound. Very close observations are therefore poorly conditioned even though the noiseless parameters are unique.