Question

The differential equation \(\dfrac{du}{dt} + 2tu^{2} = 1\) is solved by a backward difference scheme. At the \((n-1)\)-th time step, \(u_{n-1}=1.75\) and \(t_{n-1}=3.14\,\text{s}\). With \(\Delta t=0.01\,\text{s}\), find \(u_n-u_{n-1}\) (round off to three decimals).

Hint:

Backward Euler for \( \dot u = f(t,u) \) uses \( \dfrac{u_n-u_{n-1}}{\Delta t}=f(t_n,u_n) \). Be ready to solve a nonlinear algebraic equation in \(u_n\) when \(f\) is nonlinear.

Answer 1

Step 1: Backward difference discretization.
Backward (implicit) Euler at time \(t_n=t_{n-1}+\Delta t = 3.15\) s: \[ \frac{u_n-u_{n-1}}{\Delta t} + 2\,t_n\,u_n^2 = 1 . \]

Step 2: Substitute data.
\(u_{n-1}=1.75,\ \Delta t=0.01,\ t_n=3.15\): \[ \frac{u_n-1.75}{0.01} + 2(3.15)u_n^2 = 1 . \]

Step 3: Solve for \(u_n\).
\[ (u_n-1.75) + 0.01\big[\,1-2(3.15)u_n^2\,\big] = 0 \Rightarrow 315\,u_n^2 + 100\,u_n - 175.1 = 0 . \]
Solving the quadratic (physically relevant root): \(u_n \approx 1.598935\).

Step 4: Required difference.
\[ u_n-u_{n-1} \approx 1.598935-1.75 = -0.151065 \approx \boxed{-0.151}. \]