Variation of Parameters — Question 7

Question 7 – Solution

Strategy. Exploit the triangular derivative matrix while retaining both kernel variables.

Step 1: Solve the parameter equations. We have $F(0)=F'(0)=0$, $F''=e^{x^2/2}$ and $F'''=xF''$. Thus $1,x,F$ solve the homogeneous equation and have Wronskian $e^{x^2/2}>0$. The triangular system gives $$u_2'=e^{-x^2/2},\quad u_1'=-F'e^{-x^2/2},\quad u_0'=(xF'-F)e^{-x^2/2}.$$

Step 2: Combine the parameter integrals. Multiplying by $1,x,F(x)$ and integrating from zero yields $$\boxed{K(x,t)=e^{-t^2/2}\bigl[F(x)-F(t)-(x-t)F'(t)\bigr].}$$ Taylor’s integral identity, obtained by integrating $F''$ twice, gives for $x\ge t$ $$K(x,t)=e^{-t^2/2}\int_t^x(x-s)e^{s^2/2}\,ds,\qquad y(x)=\int_0^xK(x,t)\,dt.$$ The factor involving $t$ is essential; it comes from the inverse derivative matrix evaluated at the integration variable.

Step 3: Verify the equation and inequalities. At the diagonal, $K(t,t)=K_x(t,t)=0$, while $$K_{xx}(x,t)=e^{(x^2-t^2)/2},\qquad y''(x)=e^{x^2/2}\int_0^xe^{-t^2/2}\,dt.$$ Hence $y'''=xy''+1$ and all three initial data vanish. For $x>0$, $y''>0$, proving strict convexity. Also $y'''=1+xy''>1$ there. Integrating this strict inequality three times from the zero initial vector gives $\boxed{y(x)>x^3/6}$ for every $x>0$.

Step 4: Test translation invariance. If $K(x,t)=H(x-t)$, then $K_{xx}=H''(x-t)$ would have the same value at all pairs with the same difference. But for any fixed $h>0$, $$K_{xx}(t+h,t)=e^{th+h^2/2},$$ which varies with $t$. Therefore this kernel is not a function of $x-t$ alone. The variable coefficient distinguishes the absolute source time from the elapsed time.