Problem 3 - A First Course in the Numerical Analysis of Differential Equations

Question

We solve the scalar linear system \(y^{\prime}=a y, y(0)=1\). a Show that the 'continuous output' method $$ u(t)=\frac{1+\frac{1}{2} a(t-n h)}{1-\frac{1}{2} a(t-n h)} y_{n}, \quad n h \leq t \leq(n+1) h, \quad n=0,1, \ldots, $$ is consistent with the values of \(y_{n}\) and \(y_{n+1}\) which are obtained by the trapezoidal rule. b Demonstrate that \(u\) obeys the perturbed ODE $$ u^{\prime}(t)=a u(t)+\frac{\frac{1}{4} a^{3}(t-n h)^{2}}{\left[1-\frac{1}{2} a(t-n h)\right]^{2}} y_{n}, \quad t \in[n h,(n+1) h], $$ with initial condition \(u(n h)=y_{n}\). Thus, prove that $$ u((n+1) h)=\mathrm{e}^{h a}\left[1+\frac{1}{4} a^{3} \int_{0}^{h} \frac{\mathrm{e}^{-\tau a} \tau^{2} \mathrm{~d} \tau}{\left(1-\frac{1}{2} a \tau\right)^{2}}\right] y_{n} . $$ c Let \(e_{n}=y_{n}-y(n h), n=0,1, \ldots\) Show that $$ e_{n+1}=\mathrm{e}^{h a}\left[1+\frac{1}{4} a^{3} \int_{0}^{h} \frac{\mathrm{e}^{-\tau a} \tau^{2} \mathrm{~d} \tau}{\left(1-\frac{1}{2} a \tau\right)^{2}}\right] e_{n}+\frac{1}{4} a^{3} \mathrm{e}^{(n+1) h a} \int_{0}^{h} \frac{\mathrm{e}^{-\tau a} \tau^{2} \mathrm{~d} \tau}{\left(1-\frac{1}{2} a \tau\right)^{2}} $$ In particular, deduce that \(a<0\) implies that the error propagates subject to the inequality $$ \left|e_{n+1}\right| \leq e^{h a}\left(1+\frac{1}{4}|a|^{3} \int_{0}^{h} \mathrm{e}^{-\tau a} \tau^{2} \mathrm{~d} \tau\right)\left|e_{n}\right|+\frac{1}{4}|a|^{3} \mathrm{e}^{(n+1) h a} \int_{0}^{h} \mathrm{e}^{-\tau a} \tau^{2} \mathrm{~d} \tau $$

Step-by-Step Solution

Answer

The output method is consistent and satisfies the perturbed ODE. The error bounds hold under the given conditions for \( a < 0 \).

Step 1: Understand the Initial Value Problem and Method

We are dealing with a scalar linear ordinary differential equation given by \( y'(t) = a y(t) \) with the initial condition \( y(0) = 1 \). The 'continuous output' method provided allows us to approximate \( y(t) \) over an interval \( [nh, (n+1)h] \) using the formula \( u(t) = \frac{1+\frac{1}{2} a(t-n h)}{1-\frac{1}{2} a(t-n h)} y_n \). We aim to confirm if this is consistent with the trapezoidal rule.

Step 2: Check Consistency with the Trapezoidal Rule (Part a)

The trapezoidal rule for approximating \( y'(t) = a y(t) \) gives the update formula: \( y_{n+1} = y_n + \frac{h}{2} [a y_n + a y_{n+1}] \). Solve for \( y_{n+1} \) to get \( y_{n+1} = \frac{1 + \frac{1}{2} h a}{1 - \frac{1}{2} h a} y_n \). Relate this to the continuous output by evaluating at \( t = (n+1)h \), confirming consistency.

Step 3: Derive the Perturbed ODE for u(t) (Part b)

Differentiate \( u(t) \) with respect to \( t \) and substitute into the expression \( u'(t) = a u(t) + f(t) \). The differentiation yields a term that relates to \( a u(t) \) plus an additional term. Simplifying shows the perturbed equation with an extra polynomial factor related to \( (t-nh)^2 \). Verifying \( u(nh) = y_n \) establishes the perturbed ODE.

Step 4: Prove the Solution at (n+1)h for u (Part b continued)

Integrate the perturbed ODE over \([nh, (n+1)h]\) and apply the initial condition \( u(nh) = y_n \). The result shows \( u((n+1)h) = e^{h a} [1 + \frac{1}{4} a^3 \int_0^h \frac{e^{-\tau a} \tau^2}{(1-\frac{1}{2} a \tau)^2} \mathrm{d} \tau] y_n \). This demonstrates how \( u(t) \) extends beyond solving the exact ODE.

Step 5: Analyze Error Propagation (Part c)

Define the error \( e_n = y_n - y(nh) \). Express \( e_{n+1} \) by considering \( y_{n+1} = u((n+1)h) \) and the exact \( y((n+1)h) \). Relating back, the difference yields the error propagation formula. Check the terms for sign and influence when \( a < 0 \), showing how the inequality holds through integration and exponentiation properties related to boundedness.

Step 6: Demonstrate Condition When a < 0 for Error

Given \( a < 0 \), \( e^{ha} < 1 \) and the behaviors of the integrals remain bounded. Apply these properties to bound \( |e_{n+1}| \) ensuring it does not exceed the growing interval summed with a controllable additional term. This results in a succinct bound determined by the inequality provided.

Key Concepts

Trapezoidal Rule

The Trapezoidal Rule is a numerical method for estimating the integral of a function. It works by averaging the values at the endpoints of small subintervals and then scaling by their width. The method offers a straightforward way to approximate solutions for ordinary differential equations (ODEs). In the context of the exercise, the Trapezoidal Rule is used to solve the initial value problem given by the equation \( y'(t) = a y(t) \) with \( y(0) = 1 \).

• The basic form for the rule is:
• Update the value of \( y_{n+1} \) as \( y_{n+1} = y_n + \frac{h}{2} [a y_n + a y_{n+1}] \).
• Solve for \( y_{n+1} \) to get \( y_{n+1} = \frac{1 + \frac{1}{2} h a}{1 - \frac{1}{2} h a} y_n \).

The method presumes small changes over \( h \) with consistent accuracy. By aligning this update rule with the continuous output form, it is shown that both techniques produce the same results over each subinterval, indicating that they are consistent. This consistency is crucial for validity in approximating solutions over longer intervals.

Continuous Output Method

The continuous output method is a way to enhance numerical solutions by providing estimates continuously within each step or interval. For our problem, it involves computing an expression that approximates the value of the function across each subinterval.This approach uses the formula \[ u(t) = \frac{1+\frac{1}{2} a(t-n h)}{1-\frac{1}{2} a(t-n h)} y_{n} \] where \( nh \leq t \leq (n+1)h \) and \( n = 0, 1, \ldots \). This method ensures that the output aligns with the results from the Trapezoidal Rule at specific points, ensuring consistency.

• This formula is designed to interpolate within the step using the starting point \( y_n \), adjusting continuously to account for the slope changes governed by the parameter \( a \).
• It allows evaluating the function's behavior more plainly across its interval without waiting for step completions, thus offering better modeling for dynamic systems.

By confirming \( u(nh) = y_n \) and deriving an ODE to match the output, this method helps understand how smaller fractions of the interval impact the greater solution, thereby contributing extensively to the integral solution.

Error Propagation

Error propagation in numerical methods gives insight into how small errors can accumulate through successive computations. When approximating solutions to ODEs, it's crucial to understand this to maintain control over accuracy.In the context of our specific exercise, the error \( e_n = y_n - y(nh) \) is analyzed at each step. The task is to investigate how \( e_{n+1} \) can be expressed in terms of \( e_n \), and, under certain conditions, confirm bounded behavior. One significant expression used is: \[ e_{n+1} = \mathrm{e}^{h a}\left[1+\frac{1}{4} a^{3} \int_{0}^{h} \frac{\mathrm{e}^{-\tau a} \tau^{2} \mathrm{~d} \tau}{\left(1-\frac{1}{2} a \tau\right)^{2}}\right] e_{n} \]

• The formula assesses how the previous error propagates and transforms into the next step.
• Key is understanding that when \( a < 0 \), it results in \( e^{ha} < 1 \), which inherently helps in minimizing error escalation through dampened exponential factors.
• This insight provides a mathematical basis for maintaining stability in the solution as further iterations are computed.
• Compared with scenarios where \( a > 0 \), this condition makes the outcomes of differential equations better contained and potentially more truthful to simulated scenarios.

The inequalities derived help verify stability and the potential upper bounds of error during solutions, which aligns central to maintaining performance under various computational approaches.