Basic Concepts for nth Order Linear Equations — Question 10

Question 10 – Solution

Strategy. The linear initial-data map is explicit here, so its worst-case amplification can be determined exactly.

Step 1: Propagate all four initial errors. The difference satisfies $e^{(4)}=0$ and $e^{(j)}(0)=\delta_j$. Therefore $$\boxed{e(x)=\delta_0+\delta_1x+\frac{\delta_2x^2}{2}+\frac{\delta_3x^3}{6}.}$$ The factorials are required because the data are derivatives, not monomial coefficients.

Step 2: Prove and attain the uniform bound. For $|x|\le T$, the triangle inequality gives $$|e(x)|\le\varepsilon\left(1+|x|+\frac{|x|^2}{2}+\frac{|x|^3}{6}\right) \le\varepsilon\left(1+T+\frac{T^2}{2}+\frac{T^3}{6}\right).$$ Choose all four $\delta_j=\varepsilon$. At $x=T$, every term has the same nonnegative sign, attaining equality. Thus the stated worst-case constant is exact, including the trivial case $\varepsilon=0$.

Step 3: Translate the certificate into a tolerance. At $T=2$, the amplification factor is $1+2+2+8/6=19/3$. Consequently the largest admissible tolerance is $$\boxed{\varepsilon_{\max}=\frac{10^{-2}}{19/3}=\frac 3{1900}.}$$ Any larger value fails for the equal-sign error vector from Step 2. For $\varepsilon=10^{-3}$, the exact worst-case bound is $19/3000\approx 0.0063333<0.01$.

Step 4: Show why one uncontrolled component matters. Take $y=0$ and $\widetilde y=Mx^3/6$ for arbitrary real $M$. Both solve the equation and have the same value, first derivative and second derivative at zero, but their third-derivative difference is $M$. At $x=2$, the solution error is $4|M|/3$, which is unbounded as $|M|\to\infty$. Thus controlling only three components cannot give the proposed finite guarantee. For the fourth-order IVP, the complete initial derivative vector is needed for this quantitative stability bound.