Boundary Value Problems — Question 10

Question 10

A positive point load acts at an unknown location $a\in(0,1)$: $$-y''=F\delta(x-a),\qquad y(0)=y(1)=0,\qquad F>0.$$ Seek a continuous, piecewise linear displacement. The distributional equation requires $y'(a+)-y'(a-)=-F$; no jump in $y$ is allowed.

Tasks

1. Derive the solution on each side of $a$ using the endpoint values, continuity and derivative jump. Verify the distributional equation.

2. Find the peak height and the total area $\int_0^1y(x)\,dx$. For fixed $F$, determine which load location maximizes the height.

3. Suppose only the peak height and total area are measured, while both $F$ and $a$ are unknown. Decide whether these measurements determine the load uniquely, and characterize all compatible loads for a positive measured height $H$.

4. Instead measure the endpoint slopes $m_L=y'(0)$ and $m_R=y'(1)$. Recover $F,a$ and state their admissibility conditions. Solve the case $m_L=2,m_R=-1$, and sketch it alongside the reflected load at $1-a$ with the same $F$.

Question 10 – Solution

Strategy. Use the boundary conditions to determine two lines, then distinguish independent measurements from redundant ones.

Step 1: Match the two linear pieces. Write $y=mx$ to the left and $y=n(1-x)$ to the right. Continuity requires $ma=n(1-a)$, and the slope jump gives $-n-m=-F$. Thus $$\boxed{y(x)=\begin{cases}F(1-a)x,&0\leq x\leq a,\\Fa(1-x),&a\leq x\leq 1.\end{cases}}$$ Both endpoint values vanish. The continuous join produces no $\delta'$ term; the derivative jump $-F$ gives exactly $-y''=F\delta(x-a)$, with zero ordinary second derivative elsewhere.

Step 2: Compute the peak and area. The left slope is positive and the right slope negative, so the peak is at $a$: $$\boxed{H=Fa(1-a),\qquad \int_0^1y\,dx=H/2.}$$ The area is that of a triangle with base one and height $H$. For fixed $F$, $a(1-a)=1/4-(a-1/2)^2$ is maximized at $a=1/2$, giving height $F/4$.

Step 3: Test whether the measurements are independent. If the measured area differs from $H/2$, no load fits the model. If it equals $H/2$, every $a\in(0,1)$ is compatible with $\boxed{F=H/[a(1-a)]}$. The two measurements supply only one independent number, so there are infinitely many compatible location-strength pairs.

Step 4: Recover the load from endpoint slopes. Since $m_L=F(1-a)$ and $m_R=-Fa$, $$\boxed{F=m_L-m_R,\qquad a=\frac{-m_R}{m_L-m_R}.}$$ The precise admissibility conditions are $m_L>0$ and $m_R<0$. For $(m_L,m_R)=(2,-1)$, $F=3,a=1/3$, with height $2/3$ and area $1/3$. The reflected load has the same height and area but different endpoint slopes.