Nonhomogeneous Differential Equations — Question 9

Question 9

Let $y$ solve $y''+y=1$ with $y(0)=y'(0)=0$. On $0\le t\le 1$, consider the approximation $Y(t)=t^2/2$ and the error $E=Y-y$. You may use $0\le\sin r\le r$ for $0\le r\le 1$.

Tasks

1. Compute the residual $R=Y\prime\prime+Y-1$ and derive the differential equation and initial data satisfied by $E$.

2. Verify by differentiating twice that $\int_0^t\sin(t-s)s^2/2\,ds$ solves this error initial-value problem. Justify that it equals $E$.

3. Prove the rigorous error bound $0\le E(t)\le t^4/24$ on $[0,1]$, including the direction of the approximation error.

4. Find the exact solution $y$ by verification, compute the actual error at $1$ to six decimal places, and compare it with the bound. Why does a residual vanishing only at zero not prove exactness?

Question 9 – Solution

Strategy. Treat the approximation residual as a forcing for the error, then bound the resulting integral with a positive kernel.

Step 1: Form the error equation. Since $Y''=1$, the residual is $R=t^2/2$. Subtraction gives $$\boxed{E''+E=t^2/2,\qquad E(0)=E'(0)=0.}$$

Step 2: Verify the integral. Set $F(t)=\int_0^t\sin(t-s)s^2/2\,ds$. Leibniz differentiation gives $$F'=\int_0^t\cos(t-s)s^2/2\,ds,\qquad F''=t^2/2-\int_0^t\sin(t-s)s^2/2\,ds.$$ Thus $F''+F=t^2/2$ and $F(0)=F'(0)=0$. The continuous coefficients and forcing give uniqueness, so $F=E$.

Step 3: Bound the error. For $0\le s\le t\le 1$, the supplied sine inequality gives $$0\le E(t)\le\frac 12\int_0^t(t-s)s^2\,ds =\frac 12(t^4/3-t^4/4)=\boxed{t^4/24}.$$ Hence $Y$ overestimates the solution. The bound vanishes at zero and remains valid across the entire stated interval.

Step 4: Check against the exact response. The function $y=1-\cos t$ has the required equation and zero data. Therefore $$\boxed{E(1)=\cos 1-\tfrac 12\approx 0.040302<1/24\approx 0.041667.}$$ The residual is zero at the initial point but is positive elsewhere on $(0,1]$. A solution must satisfy its equation at every point, so agreement at one point and in the initial data does not make $Y$ exact.