GATE 2023 CE (CE1) – Question 52
The differential equation, $\frac{du}{dt} + 2tu^2 = 1$, is solved by employing a backward difference scheme within the finite difference framework. The value of $u$ at the $(n - 1)^{th}$ time-step, for some $n$, is 1.75. The corresponding time ($t$) is 3.14 s. Each time step is 0.01 s long. Then, the value of $(u_n - u_{n-1})$ is _____ (round off to three decimal places).
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Correct answer: -0.161 to -0.141
Explanation
The backward scheme evaluates the right-hand side at the new step: $\frac{u_n - u_{n-1}}{h} = 1 - 2t_nu_n^2$, with $t_n = 3.14 + 0.01 = 3.15$ s and $h = 0.01$. Let $\Delta = u_n - u_{n-1}$, so $\Delta = 0.01\left[1 - 6.3(1.75 + \Delta)^2\right]$. Solving this by iteration gives $\Delta = -0.151$.