Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 16

Problem 1: Part 16: Newton iterations

Let the nonlinear equations be $g (Δ γ) = 0$ . Recall that the Newton method requires that we iterate using the formula

Δ γ_{r + 1} = Δ γ_{r} - \frac{g (Δ γ_{r})}{\frac{d g (Δ γ_{r})}{d Δ γ}}

where $r$ is the Newton iteration number. Derive an expression for the derivative of $g$ that is required in the above formula.

Let us find the derivatives term by term. For the first term

\frac{d}{d Δ γ} [9 μ^{2} (Δ γ)^{2}] = 18 μ^{2} Δ γ

For the second term

\frac{d}{d Δ γ} [6 μ \sqrt{\frac{3}{2}} Δ γ 𝐬_{n + 1}^{trial} : 𝐧_{n}] = 6 μ \sqrt{\frac{3}{2}} 𝐬_{n + 1}^{trial} : 𝐧_{n}

For the fourth term

\frac{d}{d Δ γ} [\frac{3}{2} 𝐬_{n + 1}^{trial} : 𝐬_{n + 1}^{trial}] = 0

For the third term

\frac{d}{d Δ γ} [{σ_{0} + B {(α_{n} + Δ γ \frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}})}^{n}}^{2} {1 - (\frac{T_{n} + \sqrt{\frac{3}{2}} \frac{χ Δ γ}{ρ_{n} C_{p}} ‖ 𝐬_{n} ‖_{} - T_{0}}{T_{m} - T_{0}})}^{2}] = \frac{d}{d Δ γ} [P_{n} Q_{n}] = \frac{d P_{n}}{d Δ γ} Q_{n} + \frac{d Q_{n}}{d Δ γ} P_{n}

Now,

\begin{matrix} \frac{d P_{n}}{d Δ γ} & = \frac{d}{d Δ γ} [{σ_{0} + B {(α_{n} + Δ γ \frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}})}^{n}}^{2}] \\ = 2 {σ_{0} + B {(α_{n} + Δ γ \frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}})}^{n}} \frac{d}{d Δ γ} [σ_{0} + B {(α_{n} + Δ γ \frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}})}^{n}] \\ = 2 {σ_{0} + B {(α_{n} + Δ γ \frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}})}^{n}} [n B {(α_{n} + Δ γ \frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}})}^{n - 1}] \frac{d}{d Δ γ} (α_{n} + Δ γ \frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}}) \\ = 2 {σ_{0} + B {(α_{n} + Δ γ \frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}})}^{n}} [n B {(α_{n} + Δ γ \frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}})}^{n - 1}] (\frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}}) \\ = 2 n B (\frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}}) {(α_{n} + Δ γ \frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}})}^{n - 1} [σ_{0} + B {(α_{n} + Δ γ \frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}})}^{n}] \end{matrix}

Similarly,

\begin{matrix} \frac{d Q_{n}}{d Δ γ} & = \frac{d}{d Δ γ} [{1 - (\frac{T_{n} + \sqrt{\frac{3}{2}} \frac{χ Δ γ}{ρ_{n} C_{p}} ‖ 𝐬_{n} ‖_{} - T_{0}}{T_{m} - T_{0}})}^{2}] \\ = 2 {1 - (\frac{T_{n} + \sqrt{\frac{3}{2}} \frac{χ Δ γ}{ρ_{n} C_{p}} ‖ 𝐬_{n} ‖_{} - T_{0}}{T_{m} - T_{0}})} \frac{d}{d Δ γ} [- (\frac{T_{n} + \sqrt{\frac{3}{2}} \frac{χ Δ γ}{ρ_{n} C_{p}} ‖ 𝐬_{n} ‖_{} - T_{0}}{T_{m} - T_{0}})] \\ = - 2 \sqrt{\frac{3}{2}} {1 - (\frac{T_{n} + \sqrt{\frac{3}{2}} \frac{χ Δ γ}{ρ_{n} C_{p}} ‖ 𝐬_{n} ‖_{} - T_{0}}{T_{m} - T_{0}})} (\frac{χ ‖ 𝐬_{n} ‖_{}}{ρ_{n} C_{p} (T_{m} - T_{0})}) \\ = - \sqrt{6} (\frac{χ ‖ 𝐬_{n} ‖_{}}{ρ_{n} C_{p}}) [\frac{T_{m} - T_{n} - \sqrt{\frac{3}{2}} \frac{χ Δ γ}{ρ_{n} C_{p}} ‖ 𝐬_{n} ‖_{}}{(T_{m} - T_{0})^{2}}] \end{matrix}

Therefore, the full expression for the derivative is

\begin{matrix} \frac{d g (Δ γ_{r})}{d Δ γ} & = 18 μ^{2} Δ γ - 6 μ \sqrt{\frac{3}{2}} 𝐬_{n + 1}^{trial} : 𝐧_{n} \\ - 2 n B (\frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}}) {(α_{n} + Δ γ \frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}})}^{n - 1} [σ_{0} + B {(α_{n} + Δ γ \frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}})}^{n}] {[1 - (\frac{T_{n} + \sqrt{\frac{3}{2}} \frac{χ Δ γ}{ρ_{n} C_{p}} ‖ 𝐬_{n} ‖_{} - T_{0}}{T_{m} - T_{0}})]}^{2} \\ + \sqrt{6} (\frac{χ ‖ 𝐬_{n} ‖_{}}{ρ_{n} C_{p}}) [\frac{T_{m} - T_{n} - \sqrt{\frac{3}{2}} \frac{χ Δ γ}{ρ_{n} C_{p}} ‖ 𝐬_{n} ‖_{}}{(T_{m} - T_{0})^{2}}] {[σ_{0} + B {(α_{n} + Δ γ \frac{ε_{n}^{p} : 𝐧_{n}}{‖ ε_{n}^{p} ‖_{}})}^{n}]}^{2} \end{matrix}