Boundary Value Problems — Question 5

Question 5 – Solution

Strategy. Determine the integration constants from the correct derivative data, then use an energy identity to check uniqueness.

Step 1: Solve both sets of endpoint equations. Starting from $y=x^4+ax^3+bx^2+cx+d$, the two solutions are $$\boxed{y_C=x^2(1-x)^2,\qquad y_S=x^4-2x^3+x.}$$ Both fourth derivatives equal 24 and both endpoint values are zero. Also $y_C'=2x(1-x)(1-2x)$ vanishes at both ends, while $y_S''=12x(x-1)$ vanishes at both ends. These verify the distinct boundary data.

Step 2: Prove uniqueness in each case. A difference $w$ has $w^{(4)}=0$ and homogeneous boundary conditions. Twice integrating by parts gives $$0=\int_0^1ww^{(4)}dx=[ww'''-w'w'']_0^1+\int_0^1(w'')^2dx.$$ The boundary term is zero for clamped data because $w=w'=0$, and for simply supported data because $w=w''=0$. Thus $w''=0$; its zero endpoint values imply $w=0$. Each design has exactly one solution.

Step 3: Locate and compare the maxima. The derivative of $y_C$ changes from positive to negative at $1/2$. For the other design, $y_S'=(2x-1)(2x^2-2x-1)$, whose second factor is negative on $[0,1]$. Thus both have their unique maximum at $1/2$, with $$\boxed{y_C(1/2)=1/16,\qquad y_S(1/2)=5/16.}$$ The derivative signs establish maxima; symmetry by itself only suggests a stationary midpoint.

Step 4: Compare the complete profiles. The difference is $y_S-y_C=x(1-x)>0$ inside the interval, and the maximum ratio is five. The same differential load is compatible with different responses because fixing slopes differs from fixing second derivatives at the boundaries.