Order of Consistency of the Implicit Midpoint Rule

Note: This is a standard proof technique for consistency order. The key steps are:

1. Taylor expand the exact solution
2. Taylor expand the method’s right-hand side
3. Use the chain rule for derivatives: $y^{\prime \prime }(t)=f_t+f_{y}f$
4. Show cancellation up to order $h^2$

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The Implicit Midpoint rule $y_{i+1}=y_i+hf\left(t_i+\frac{h}{2},\frac{1}{2}(y_i+y_{i+1})\right)$ The increment function $\Phi (t,y,h)=f\left(t+\frac{h}{2},\frac{1}{2}(y(t)+y(t+h))\right)$

1. Local Truncation Error

From Formulary

$$\tau (t,h)=\frac{y(t+h)-y(t)}{h}-\Phi (t,y(t),h)$$ $$\tau (t,h)=\frac{y(t+h)-y(t)}{h}-f\left(t+\frac{h}{2},\frac{1}{2}(y(t)+y(t+h))\right)$$

2. Taylor Series Expansions

We use Taylor series to expand the terms around the point $(t,y(t))$. First Term: Expanding $y(t+h)$ around $t$ yields:

$$y(t+h)=y(t)+h{y}^{\prime }(t)+\frac{h^2}{2}{y}^{\prime \prime }(t)+\mathcal{O}(h^3)$$

Note: Comparing the expansion with the given rule, we get $y^{\prime }=f$ Rearranging,

$$\frac{y(t+h)-y(t)}{h}=y^{\prime }(t)+\frac{h}{2}{y}^{\prime \prime }(t)+\mathcal{O}(h^2)$$

Second Term: We first expand the spatial argument of $f$:

$$\begin{array}{rl}\frac{1}{2}(y(t)+y(t+h))& =\frac{1}{2}(y(t)+\stackrel{\text{Taylor expansion}}{\overbrace{[y(t)+h{y}^{\prime }(t)+\mathcal{O}(h^2)]}})\\ & =y(t)+\frac{h}{2}{y}^{\prime }(t)+\mathcal{O}(h^2)\end{array}$$

Now we perform a multivariate Taylor expansion of $f$ around $(t,y(t))$:

$$f(a+\Delta x,b+\Delta y)\approx f(a,b)+\Delta x\cdot \frac{\partial f}{\partial x}+\Delta y\cdot \frac{\partial f}{\partial y}+\text{ Higher-Order Terms }$$

We have

$$f\left(t+\stackrel{\Delta x}{\overbrace{\frac{h}{2}}},y(t)+\stackrel{\Delta y}{\overbrace{\frac{h}{2}{y}^{\prime }(t)+\mathcal{O}(h^2)}}\right)=f(t,y(t))+\frac{h}{2}\frac{\partial f}{\partial t}+\frac{h}{2}{y}^{\prime }(t)\frac{\partial f}{\partial y}+\mathcal{O}(h^2)$$

Since $y^{\prime }(t)=f(t,y(t))$ and, by the chain rule, $y^{\prime \prime }(t)=\frac{d}{dt}f(t,y(t))=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial y}{y}^{\prime }(t)$, the expansion simplifies to:

$$f\left(\dots \right)=f(t,y(t))+\frac{h}{2}{y}^{\prime \prime }(t)+\mathcal{O}(h^2)=y^{\prime }(t)+\frac{h}{2}{y}^{\prime \prime }(t)+\mathcal{O}(h^2)$$

3. Combine the Expansions

Substituting the expanded forms back into the truncation error formula, we get:

$$\tau (t,h)=\left[y^{\prime }(t)+\frac{h}{2}{y}^{\prime \prime }(t)+\mathcal{O}(h^2)\right]-\left[y^{\prime }(t)+\frac{h}{2}{y}^{\prime \prime }(t)+\mathcal{O}(h^2)\right]$$ $$\boxed{\tau (t,h)=\mathcal{O}(h^2)}$$

Since the local truncation error is of the order $\mathcal{O}(h^2)$, the implicit midpoint rule has an order of consistency of at least 2.