Fundamental Sets of Solutions — Question 10

Question 10 – Solution

Strategy. Choose basis functions adapted to the two measurements, then inspect their positive weights.

Step 1: Build an endpoint basis. Write $r=Ae^t+Be^{-t}$. The data give $A+B=0$ and $A(e^L-e^{-L})=1$, so $A=1/(2\sinh L)$ and $B=-A$. Reflecting $r$ about the midpoint gives $\ell(t)=r(L-t)$, hence $$\boxed{\ell(t)=\frac{\sinh(L-t)}{\sinh L},\qquad r(t)=\frac{\sinh t}{\sinh L}.}$$ Both solve $y''-y=0$ and have the stated endpoint values. Their determinant at zero is $\ell(0)r'(0)-\ell'(0)r(0)=1/\sinh L\ne 0$, so they form a fundamental set on $\mathbb R$.

Step 2: Fit the measurements. In this pair the endpoint coefficients are exactly the prescribed values: $$\boxed{y=a\ell+br.}$$ Every solution has a unique representation, and evaluating at $0,L$ forces its coefficients to be $a,b$. Thus the endpoint problem has exactly one solution.

Step 3: Bound the positive weights. On $0<t<L$, both weights are positive. The hyperbolic addition identity gives $$\ell(t)+r(t)=\frac{\cosh(t-L/2)}{\cosh(L/2)}\le 1\qquad(0\le t\le L),$$ with strict inequality in the interior, since $|t-L/2|<L/2$. If $M=\max(a,b)$, then $0<y$ in the interior and $y\le M(\ell+r)\le M$ on the full closed interval. At an endpoint the solution can equal zero if the corresponding datum is zero.

Step 4: Examine equal endpoint data. For $a=b=1$, $$y=\frac{\cosh(t-L/2)}{\cosh(L/2)},\qquad \boxed{y_{\min}=\frac 1{\cosh(L/2)}\text{ at }t=L/2.}$$ The minimum is unique because $\cosh x$ has its unique minimum at zero. The weights are positive but sum to less than one inside; this is not the straight-line interpolation of equal endpoint values.