Basic Concepts — Question 7

Question 7 – Solution

Strategy. Differentiate a sum of nonnegative terms, then use its conservation both to bound a solution and to eliminate a zero-data difference.

Step 1: Derive the conserved quantity. For any solution, $$E'=2y'y''+8yy'=2y'(y''+4y)=0.$$ Thus $E(t)=E(0)$ throughout its interval. The prescribed data give $$\boxed{E=2^2+4(1)^2=8}.$$

Step 2: Obtain and attain the displacement bound. Since $(y')^2\ge 0$, $4y^2\le 8$, whence $\boxed{|y|\le\sqrt 2}$. For the supplied candidate, $$y'=-2\sin(2t)+2\cos(2t),\qquad y''=-4y.$$ It has the required data, and at $t=\pi/8$ it has $y=\sqrt 2$, $y'=0$. Therefore the bound is sharp, not just a convenient overestimate.

Step 3: Interpret the state curve. Writing $v=y'$ gives $4y^2+v^2=8$, an ellipse with intercepts $y=\pm\sqrt 2$ and $v=\pm 2\sqrt 2$. The initial point is $(1,2)$. Its velocity in these coordinates is $$(y',v')=(v,-4y)=(2,-4),$$ so it moves rightward and downward there. The axes represent state variables, not time versus displacement.

Step 4: Prove uniqueness and distinguish energy from data. The difference $w$ of two solutions with the same initial data satisfies $w''+4w=0$ and $w(0)=w'(0)=0$. Its conserved quantity is $(w')^2+4w^2=0$. Both terms are nonnegative, so $w=0$ everywhere on their common interval.

Energy alone only fixes an ellipse. For example, $\sqrt 2\cos(2t)$ and $\sqrt 2\sin(2t)$ both have energy 8 but different initial states. Initial position and velocity identify a point and its motion; the scalar energy value does not replace them.