Boundary Value Problems — Question 10

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Question 10

A positive point load acts at an unknown location a∈(0,1)a\in(0,1): −y″=Fδ(x−a),y(0)=y(1)=0,F>0.-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−)=−Fy'(a+)-y'(a-)=-F; no jump in yy is allowed.

Tasks

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

  2. Find the peak height and the total area ∫01y(x)dx\int_0^1y(x)\,dx. For fixed FF, determine which load location maximizes the height.

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

  4. Instead measure the endpoint slopes mL=y′(0)m_L=y'(0) and mR=y′(1)m_R=y'(1). Recover F,aF,a and state their admissibility conditions. Solve the case mL=2,mR=−1m_L=2,m_R=-1, and sketch it alongside the reflected load at 1−a1-a with the same FF.

Original worksheet page 1: question and worked solution for 8-1-010
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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=mxy=mx to the left and y=n(1−x)y=n(1-x) to the right. Continuity requires ma=n(1−a)ma=n(1-a), and the slope jump gives −n−m=−F-n-m=-F. Thus y(x)={F(1−a)x,0≤x≤a,Fa(1−x),a≤x≤1.\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 δ′\delta' term; the derivative jump −F-F gives exactly −y″=Fδ(x−a)-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 aa: H=Fa(1−a),∫01ydx=H/2.\boxed{H=Fa(1-a),\qquad \int_0^1y\,dx=H/2.} The area is that of a triangle with base one and height HH. For fixed FF, a(1−a)=1/4−(a−1/2)2a(1-a)=1/4-(a-1/2)^2 is maximized at a=1/2a=1/2, giving height F/4F/4.

Step 3: Test whether the measurements are independent. If the measured area differs from H/2H/2, no load fits the model. If it equals H/2H/2, every a∈(0,1)a\in(0,1) is compatible with F=H/[a(1−a)]\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 mL=F(1−a)m_L=F(1-a) and mR=−Fam_R=-Fa, F=mL−mR,a=−mRmL−mR.\boxed{F=m_L-m_R,\qquad a=\frac{-m_R}{m_L-m_R}.} The precise admissibility conditions are mL>0m_L>0 and mR<0m_R<0. For (mL,mR)=(2,−1)(m_L,m_R)=(2,-1), F=3,a=1/3F=3,a=1/3, with height 2/32/3 and area 1/31/3. The reflected load has the same height and area but different endpoint slopes.

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Original worksheet page 2: question and worked solution for 8-1-010

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