Variation of Parameters — Question 9

Question 9 – Solution

Strategy. Evaluate the normalized forcing at the integration variable and check the kernel away from its diagonal.

Step 1: Derive the correctly normalized kernel. The equation is $y^{(4)}=x^{-4}$. For the given basis, variation of parameters gives $u'=x^{-4}(-x^3/6,x^2/2,-x,1)^T$. Integrating from $1$ and combining yields $$\boxed{K(x,t)=\frac{(x-t)^3}{6t^4},\qquad y(x)=\int_1^xK(x,t)\,dt.}$$ The denominator is evaluated at the source variable $t$. The same polynomial expression and oriented integral apply for $0<x<1$.

Step 2: Verify both kernel requirements. For fixed $t>0$, $K$ is a cubic polynomial in $x$, so $x^4K_{xxxx}=0$ for $x>t$. Its diagonal data are $$K(t,t)=K_x(t,t)=K_{xx}(t,t)=0,\qquad K_{xxx}(t,t)=1/t^4.$$ Leibniz differentiation gives $y^{(4)}(x)=1/x^4$, because the fourth kernel derivative is zero and the third diagonal derivative supplies exactly that term.

Step 3: Evaluate and check the elementary response. Expanding $(x-t)^3/t^4$ and integrating its four powers of $t$ gives $$\boxed{y=\frac{x^3}{18}-\frac{x^2}{4}+\frac{x}{2}-\frac{\ln x}{6}-\frac{11}{36}.}$$ Its derivatives are $y'=x^2/6-x/2+1/2-1/(6x)$, $y''=x/3-1/2+1/(6x^2)$, and $y'''=1/3-1/(3x^3)$. These and $y$ vanish at $1$; one further derivative gives $y^{(4)}=x^{-4}$.

Step 4: Expose the misleading diagonal agreement. The factor $(x-t)^3$ ensures that $\widehat K$ has the same three zero diagonal values and $\widehat K_{xxx}(t,t)=1/t^4$. But its proposed response is $$\widehat y=\frac 1{6x^4}\int_1^x(x-t)^3\,dt=\frac{(x-1)^4}{24x^4}.$$ It satisfies the zero initial data, yet $$x^4\widehat y^{(4)}=-\frac 4x+\frac{30}{x^2}-\frac{60}{x^3}+\frac{35}{x^4}, \qquad \left.x^4\widehat y^{(4)}\right|_{x=2}=\boxed{3/16\ne 1}.$$ The missing requirement is the homogeneous equation in $x$ away from $x=t$. Correct diagonal data alone do not make a valid response kernel.