Boundary Value Problems — Question 9

Question 9 – Solution

Strategy. Combine an energy identity, a strict comparison argument and a simple barrier to certify the boundary-value error.

Step 1: Establish existence and uniqueness. A homogeneous difference $w$ with zero endpoint values satisfies $0=\int_0^1wL[w]dx=\int_0^1((w')^2+qw^2)dx$, hence $w'=0$ and $w=0$. For existence, let $p$ solve $L[p]=f$ with $p(0)=p'(0)=0$, and let $h$ solve $L[h]=0$ with $h(0)=0,h'(0)=1$. Regular IVP theory supplies both. If $h(1)=0$, uniqueness just proved would force $h=0$, a contradiction. Thus $y=p-p(1)h/h(1)$ is the required solution.

Step 2: Make comparison strict before taking a limit. Put $\psi=1+x(1-x)>0$. Then $L[\psi]=2+q\psi>0$, so $L[w+\eta\psi]>0$ and its endpoint values are positive. At a nonpositive interior minimum, its second derivative is nonnegative and $q\geq 0$ would give $L[w+\eta\psi]\leq 0$, a contradiction. Thus $w+\eta\psi>0$. Letting $\eta\downarrow 0$ proves $w\geq 0$.

Step 3: Bound the error with a barrier. Since $L[\phi]=1+q\phi\geq 1$, for $e=y-v$ we have $L[\epsilon\phi\pm e]\geq 0$, with zero endpoints. Comparison yields $$\boxed{|y(x)-v(x)|\leq\epsilon\phi(x),\qquad \max_{[0,1]}|y-v|\leq\epsilon/8.}$$ The constant $1/8$ is sharp: take $q=0$, $v=0$, $f=\epsilon$, for which $y=\epsilon\phi$ attains equality at $x=1/2$.

Step 4: Certify the proposed approximation. Here $L[\phi]-1=x^2\phi=x^3(1-x)/2$. Its derivative is $x^2(3-4x)/2$, so its maximum is $\epsilon=27/512$ at $x=3/4$. Also $L[v-y]\geq 0$, hence $y\leq v$. Combining this sign with the error bound gives $$\boxed{\left(1-\frac{27}{512}\right)\phi\leq y\leq\phi, \qquad \max_{[0,1]}|y-\phi|\leq\frac{27}{4096}.}$$ The residual is not zero; these are certified bounds, not an exact formula for $y$.