IVP’s With Step Functions — Question 6

Question 6 – Solution

Strategy. Exact rest requires two terminal state conditions. Use the step response to convert them into two equations for the two force levels.

Step 1: Transform with the initial displacement. The level changes give $f=A+(B-A)H_a(t)-BH_T(t)$, so $$(s^2+1)Y-s=\frac{A+(B-A)e^{-as}-Be^{-Ts}}s.$$ Writing $r(t)=1-\cos t$, the inverse is $$\boxed{y=\cos t+A r(t)+(B-A)H_a(t) r(t-a)-BH_T(t) r(t-T).}$$ The unforced term $\cos t$ supplies the initial data. Since $r(0)=r'(0)=0$, each added response preserves $y,y'$ at activation, and $r''+r=1$ verifies the forcing on each open interval.

Step 2: Impose both terminal conditions. Set $c=\cos a=\sin a=1/\sqrt 2$. At $T=\pi/2$ the last response and its derivative still vanish. The terminal values are $$y(T)=cA+(1-c)B,\qquad y'(T)=-1+(1-c)A+cB.$$ Thus rest requires $$cA+(1-c)B=0,\qquad (1-c)A+cB=1.$$

Step 3: Solve and prove complete rest. The determinant is $c^2-(1-c)^2=2c-1=\sqrt 2-1>0$. Consequently $$A=-\frac{1-c}{2c-1}=\boxed{-\frac 1{\sqrt 2}},\qquad B=\frac c{2c-1}=\boxed{1+\frac 1{\sqrt 2}}.$$ These are the unique levels in the stated two-stage family. The formula satisfies the original IVP and has zero displacement and velocity at $T$. For $t>T$ the forcing is zero, so uniqueness of the homogeneous oscillator with these terminal data implies $y(t)=0$ there. Matching gives a unique solution on the entire half-line.

Step 4: Check feasibility and regularity. Both weights in $cA+(1-c)B=0$ are positive. If $A,B\ge 0$, that equation forces $A=B=0$, contradicting the velocity equation. A negative first level is therefore necessary within this family. The motion is $C^1$, while the acceleration jumps are $$[y'']_a=B-A=1+\sqrt 2,\qquad [y'']_T=-B=-1-\frac 1{\sqrt 2}.$$ After $T$ the output is identically zero, so its transform exists for $\boxed{\text{every real }s}$, despite the apparent denominators in $Y$.