Nonconstant Coefficient IVP’s — Question 5

Question 5 – Solution

Strategy. The singular equation fixes a missing initial slope. Transform differentiation removes that slope algebraically, so compatibility must be checked separately.

Step 1: Recover compatibility and transform. At zero the equation requires $\boxed{y'(0)=0}$. Since $$\mathcal L\{ty''\}=-\frac d{ds}[s^2Y-sy(0)-y'(0)] =-2sY-s^2Y'+1,$$ adding $2\mathcal L\{y'\}$ and $\mathcal L\{ty\}$ gives $$\boxed{(s^2+1)Y'=-1.}$$

Step 2: Select and identify the inverse. The condition $Y\to 0$ as $s\to\infty$ gives $$Y(s)=\int_s^\infty\frac{du}{1+u^2}=\arctan(1/s),\quad s>0.$$ The time function $\boxed{y(t)=\sin t/t\ (t>0),\quad y(0)=1}$ has representation $y=\int_0^1\cos(ut)\,du$. Transforming this integral for $s>0$ yields $\int_0^1 s/(s^2+u^2)\,du=\arctan(1/s)$. Absolute double integrability justifies the interchange.

Step 3: Verify smoothness and uniqueness. The expansion $y=1-t^2/6+O(t^4)$ supplies a smooth extension with slope zero. Set $w=ty$. Then $w''+w=ty''+2y'+ty=0$, with $w(0)=0$, $w'(0)=1$. The regular oscillator IVP has the unique solution $w=\sin t$, proving the original equation and uniqueness of its smooth extension. The term $-y'(0)$ is constant in $s$, so it disappears when differentiating the transform of $y''$; arbitrary initial slopes are nevertheless not admissible.

Step 4: Establish the boundary value. The supplied estimate makes the tails uniformly small for $s\ge 0$. On each fixed finite interval the damped integrals converge to the undamped one as $s\downarrow 0$. Taking the interval long first and then taking $s\downarrow 0$ proves $$\boxed{Y(0)=\int_0^\infty\frac{\sin t}{t}\,dt=\pi/2.}$$ The boundary integral is conditional: intervals where $|\sin t|\ge 1/2$ give a divergent harmonic lower bound for its absolute integral. For $s<0$, fixed-length positive lobe intervals have weighted integrals growing like $e^{|s|t}/t$, violating the Cauchy criterion. The exact real domain is $\boxed{[0,\infty)}$.