Question

Estimate the local truncation error for the Euler method in terms of the solution \(y=\phi(t) .\) Obtain a bound for \(e_{n+1}\) in terms of \(t\) and \(\phi(t)\) that is valid on the interval \(0 \leq t \leq 1 .\) By using a formula for the solution obtain a more accurate error bound for \(e_{n+1} .\) For \(h=0.1\) compute a bound for \(e_{1}\) and compare it with the actual error at \(t=0.1 .\) Also compute a bound for the error \(e_{4}\) in the fourth step. $$ y^{\prime}=2 y-1, \quad y(0)=1 $$

Short answer

The Euler method provides approximations for the solution of first-order differential equations. Given a differential equation \(y' = 2y - 1\) with an initial condition \(y(0) = 1\), we first found the exact solution as \(y = -\frac{1}{2} + \frac{3}{2}e^{-2t}\). We then computed the local truncation error as \(e^{-2t}\) and used this to find error bounds for successive steps \(e_1, e_2, e_3\), and \(e_4\). We found that the actual error at \(t=0.1\) was smaller than the error bound, as expected. Lastly, we computed the error bound for \(e_4\) as \(0.4\).

Step-by-step solution

Finding the exact solution of the differential equation

First, let's find the exact solution for the given initial value problem: $$ y^{\prime}=2 y-1, \quad y(0)=1 $$ It's a first-order linear differential equation. To solve this, we need to find an integrating factor, which is given by \(IF=\exp{\left(\int P(t)~dt\right)} = \exp{\left(\int 2~dt\right)} = e^{2t}\). Now we multiply both sides with the integrating factor and then integrate both sides with respect to \(t\): $$ e^{2t}y^{\prime}-2e^{2t}y=-e^{2t} $$ Integrate LHS by parts and RHS directly: $$ \int \left(e^{2t}\left(dy - 2y~dt\right)\right) = -\int e^{2t} ~dt $$ Integrating and simplifying, we get: $$ e^{2t}y = -\frac{1}{2}e^{2t}+ C $$ Now, divide by \(e^{2t}\) and put the initial condition \(y(0) = 1\): $$ y = -\frac{1}{2}+ C\cdot e^{-2t} $$ Plugging the initial condition, we get \(C = \frac{3}{2}\), and thus, the exact solution is: $$ \phi(t) = y = -\frac{1}{2}+ \frac{3}{2}e^{-2t} $$

Finding the local truncation error of Euler's method

The local truncation error for Euler's method is given by: $$ \tau(t) = \frac{y_{n+1} - y_n}{h} - f(t, y) = \left. \frac{dy}{dt} - f(t, y) \right|_{t = t_n} $$ where \(f(t, y) = 2y - 1\). Now let's differentiate the exact solution we found in step 1: $$ \frac{d\phi(t)}{dt} = -3 e^{-2t} $$ Now, calculate the difference \(dy/dt - f(t, \phi(t))\): $$ \tau(t) = -3 e^{-2t} - (2\left(-\frac{1}{2}+\frac{3}{}-{2t}\right) - 1) = e^{-2t} $$ So, the local truncation error is dependent on t as \(e^{-2t}\).

Obtaining a bound for \(e_{n+1}\)

The error in the Euler method is given by: $$ e_{n+1} = e_n + h\tau(t) $$ Now, we have the maximum value of \(|\tau(t)|\) on the interval \(0\leq t \leq 1\) as \(|e^{-2t}| \leq e^0 = 1\). So, we can write the bound for the error as: $$ |e_{n+1}| \leq |e_n| + hM $$ where \(h\) is the step size and \(M\) is the maximum value of \(|\tau(t)|\) on the interval.

Computing a bound for \(e_1\) and comparing with the actual error at \(t=0.1\)

For \(h=0.1\), we can compute a bound for \(e_1\) using the above error bound relationship: $$ |e_1| \leq |e_0| + 0.1 \times 1 = 0 $$ Here, \(e_0\)=0, since our initial value is exact. Now, let's compute the actual error at \(t=0.1\). Using the Euler method, the approximation at \(t=0.1\) is given by: $$ y_1 = y_0 + hf(t_0,y_0) = 1 + 0.1(2-1) = 1.1 $$ And, the exact value of the solution \(\phi(0.1)\) is: $$ \phi(0.1) = -\frac{1}{2}+\frac{3}{2}e^{-2(0.1)} \approx 1.104245 $$ The actual error at \(t=0.1\) is: $$ |y_1-\phi(0.1)| = |1.1 - 1.104245| \approx 0.004245 $$ Comparing the error bound with the actual error, we find that the actual error is smaller than the error bound as expected.

Computing the bound for the error \(e_{4}\) in the fourth step

We can again use the error bound relationship derived in Step 3 to compute the error bound for \(e_4\). We can recursively compute bounds for \(e_1, e_2, e_3\), and finally, \(e_4\): $$ |e_1|\leq 0.1 \times 1 = 0.1 $$ $$ |e_2|\leq 0.1 + 0.1 \times 1 = 0.2 $$ $$ |e_3|\leq 0.2 + 0.1 \times 1 = 0.3 $$ $$ |e_4|\leq 0.3 + 0.1 \times 1 = 0.4 $$ So, the bound for the error in the fourth step is \(0.4\).