The Heat Equation — Question 10

Question 10 – Solution

Strategy. Conservation, preservation of the temperature range and decay of a quadratic norm impose different conditions on a numerical scheme.

Step 1: Balance each cell. For the first cell, inflow minus outflow gives $U_1^{n+1}=U_1^n+r(U_2^n-U_1^n)$. Applying the same balance to the others yields $$\boxed{U^{n+1}= \begin{pmatrix}1-r&r&0\\r&1-2r&r\\0&r&1-r\end{pmatrix}U^n.}$$ Every column sums to one, so $\sum_iU_i$ is conserved. Equal cell volumes and constant capacity make this proportional to total heat. Every row also sums to one, so constant profiles are fixed.

Step 2: Characterize range preservation. All entries are nonnegative exactly when $0\leq r\leq 1/2$. Each new value is then a convex combination of old values, preserving their range. For any $r>1/2$, input $(0,1,0)$ gives a middle value $1-2r<0$, outside $[0,1]$. Thus the exact range is $\boxed{0\leq r\leq 1/2}$.

Step 3: Test the quadratic energy on each mode. The stated mutually orthogonal vectors have update factors $1$, $1-r$ and $1-3r$, respectively. The mean is fixed; the remaining squared mode components do not grow exactly when $|1-r|\leq 1$ and $|1-3r|\leq 1$. For $r\geq 0$, this is $$\boxed{0\leq r\leq 2/3.}$$ Necessity follows by choosing the offending eigenvector as input. This threshold belongs to this particular three-cell insulated matrix.

Step 4: Exhibit the distinction numerically. At $r=3/5$, $$\boxed{(0,1,0)\longmapsto(3/5,-1/5,3/5).}$$ Both sums equal one. The uncentered squared norm drops from $1$ to $9/25+1/25+9/25=19/25$; since the mean is unchanged, the centered squared norm also drops. Nevertheless the middle temperature becomes negative. The scheme is quadratically stable here but fails the discrete range principle. The stem plot shows individual cell values, without suggesting a smooth interpolated temperature profile between cell centers.