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

Problem 1: Part 16: Newton iterations
Let the nonlinear equations be $$g(\Delta\gamma) = 0$$. Recall that the Newton method requires that we iterate using the formula

\Delta\gamma_{r+1} = \Delta\gamma_r - \cfrac{g(\Delta\gamma_r)}{\cfrac{dg(\Delta\gamma_r)}{d\Delta\gamma}} $$ where $$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

\cfrac{d}{d\Delta\gamma}\left[9~\mu^2~(\Delta\gamma)^2\right] = 18~\mu^2~\Delta\gamma $$ For the second term

\cfrac{d}{d\Delta\gamma}\left[ 6~\mu~\sqrt{\cfrac{3}{2}}~\Delta\gamma~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{n}_n \right]

= 6~\mu~\sqrt{\cfrac{3}{2}}~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{n}_n $$ For the fourth term

\cfrac{d}{d\Delta\gamma}\left[ \cfrac{3}{2}~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{s}_{n+1}^{\text{trial}} \right] = 0 $$ For the third term

\cfrac{d}{d\Delta\gamma}\left[ \left\{\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right\}^2 \left\{1 - \left(\cfrac{T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} \lVert\mathbf{s}_n\rVert_{} - T_0}{T_m -T_0}\right)\right\}^2 \right] = \cfrac{d}{d\Delta\gamma}[P_n~Q_n] = \cfrac{dP_n}{d\Delta\gamma}~Q_n + \cfrac{dQ_n}{d\Delta\gamma}~P_n $$ Now,

\begin{align} \cfrac{dP_n}{d\Delta\gamma} & = \cfrac{d}{d\Delta\gamma}\left[ \left\{\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right\}^2\right] \\ & = 2~\left\{\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right\} \cfrac{d}{d\Delta\gamma}\left[ \sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right] \\ & = 2~\left\{\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right\}\left[n~B~ \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^{n-1} \right] \cfrac{d}{d\Delta\gamma}\left( \alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right) \\ & = 2~\left\{\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right\}\left[n~B~ \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^{n-1} \right] \left(\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol{\varepsilon}

^p_n\rVert_{}}\right) \\ & = 2~n~B~ \left(\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol{\varepsilon}

^p_n\rVert_{}}\right) \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^{n-1} \left[\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right] \end{align} $$ Similarly,

\begin{align} \cfrac{dQ_n}{d\Delta\gamma} & = \cfrac{d}{d\Delta\gamma}\left[ \left\{1 - \left(\cfrac{T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} \lVert\mathbf{s}_n\rVert_{} - T_0}{T_m -T_0}\right)\right\}^2 \right] \\ & = 2~\left\{1 - \left(\cfrac{T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} \lVert\mathbf{s}_n\rVert_{} - T_0}{T_m -T_0}\right)\right\} \cfrac{d}{d\Delta\gamma}\left[ - \left(\cfrac{T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} \lVert\mathbf{s}_n\rVert_{} - T_0}{T_m -T_0}\right) \right] \\ & = -2~\sqrt{\cfrac{3}{2}}~\left\{1 - \left( \cfrac{T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p}~ \lVert\mathbf{s}_n\rVert_{} - T_0}{T_m -T_0}\right)\right\} ~\left(\cfrac{\chi~\lVert\mathbf{s}_n\rVert_{}}{\rho_n~C_p~(T_m-T_0)}\right)\\ & = -\sqrt{6} ~\left(\cfrac{\chi~\lVert\mathbf{s}_n\rVert_{}}{\rho_n~C_p}\right) ~\left[\cfrac{T_m - T_n - \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p}~\lVert\mathbf{s}_n\rVert_{}} {(T_m -T_0)^2}\right] \end{align} $$ Therefore, the full expression for the derivative is

{ \begin{align} \cfrac{dg(\Delta\gamma_r)}{d\Delta\gamma} & = 18~\mu^2~\Delta\gamma - 6~\mu~\sqrt{\cfrac{3}{2}}~\mathbf{s}_{n+1}^{\text{trial}}:\mathbf{n}_n \\ & \qquad - 2~n~B~ \left(\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol{\varepsilon}

^p_n\rVert_{}}\right) \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^{n-1} \left[\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right] \left[ 1 - \left(\cfrac{T_n + \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p} \lVert\mathbf{s}_n\rVert_{} - T_0}{T_m -T_0}\right)\right]^2 \\ & \qquad + \sqrt{6} ~\left(\cfrac{\chi~\lVert\mathbf{s}_n\rVert_{}}{\rho_n~C_p}\right) ~\left[\cfrac{T_m - T_n - \sqrt{\cfrac{3}{2}}~\cfrac{\chi~\Delta\gamma}{\rho_n~C_p}~\lVert\mathbf{s}_n\rVert_{}} {(T_m -T_0)^2}\right] \left[\sigma_0 + B \left(\alpha_n+ \Delta\gamma~\cfrac{\boldsymbol{\varepsilon}^p_n:\mathbf{n}_n}{\lVert\boldsymbol

{\varepsilon}^p_n\rVert_{}}\right)^n \right]^2 \end{align} } $$