The goal is to couple system of two nonlinear differential equations by applying appropriate space and time discretization and Newton-Raphson scheme.

The equations system is $$\left[ \begin{array}{c} F_{1}(u,v) \\ F_{2}(u,v) \end{array}\right] = \left[ \begin{aligned} -\frac{\partial^2 u}{\partial x^2} &= v \\ \frac{\partial v}{\partial t} &=\frac{\partial J(u,v)}{\partial x} \end{aligned} \right] $$ where $u(x,t), v(x,t)$ is function of time and space.

Notice $F_{1}$ is differential equation in space exclusively, and $F_{2}$ is partial differential equation in space and time.

The discretization in space and time are subscript $i$ and superscript $j$ respectively.

For $F_1$, discretizing in space gives $$ F_{1}(u,v)\quad\rightarrow\quad \frac{u_{i+1}-2u_{i}+u_{i-1}}{\Delta x^2}+v_{i}=0 $$ The next is discretizing $F_2$ in space. $$ F_{2}(u,v)\quad\rightarrow\quad \frac{\partial v}{\partial t} = \frac{J_{i+1}-J_{i}}{\Delta x} $$

The assumption is made; $J_{i+1}$ is function of $u_{i}$, $u_{i+1}$, $v_{i}$ and $v_{i+1}$ and $J_{i}$ is function of $u_{i-1}$, $u_{i}$, $v_{i-1}$ and $v_{i}$.

In discretizing $F_{2}$ in time, the method TR-BDF2(Trapezoidal Rule - Second-order Backward Differentiation Formula) is applied, which is $$ \frac{\partial q(z)}{\partial t} = f(z,t) \quad\rightarrow\quad \frac{q^{j+1}-q^{j}}{\Delta t}= \frac{f^{j+1}+f^{j}}{2} \quad\rightarrow\quad 2q^{j+1}-\Delta t f^{j+1} -2q^{j}-\Delta t f^{j} =0 $$

Once again, superscript $j$ and $j+1$ denotes discretization in time, not power.

Applying TR-BDF2 method to space discretized $F_{2}$ gives $$ F_{2}(u,v)\quad\rightarrow\quad 2v_{i}^{j+1}- \frac{\Delta t}{\Delta x}(J_{i+1}^{j+1}-J_{i}^{j+1}) -2v_{i}^{j}- \frac{\Delta t}{\Delta x}(J_{i+1}^{j}-J_{i}^{j}) =0 $$

Now, there are two equations, one in only discretized in space and the other discretized in both time and space. How do these two equations be coupled in Newton-Raphson scheme?

Does applying TR-BDF2 to $F_{1}$ valid? Since $F_{1}$ has to remain invariant through time...

Anyway, $q$ is zero in $F_{1}$, so $f^{j+1}+ f^{j}=0$ (??) Then there are two equations both in time and space discretized. $$ F_{1}(u,v)\quad\rightarrow\quad \frac{u_{i+1}^{j+1}-2u_{i}^{j+1}+u_{i-1}^{j+1}}{\Delta x^2}+v_{i}^{j+1} +\frac{u_{i+1}^{j+1}-2u_{i}^{j}+u_{i-1}^{j}}{\Delta x^2}+v_{i}^{j} =0 $$ $$ F_{2}(u,v)\quad\rightarrow\quad 2v_{i}^{j+1}- \frac{\Delta t}{\Delta x}(J_{i+1}^{j+1}-J_{i}^{j+1}) -2v_{i}^{j}- \frac{\Delta t}{\Delta x}(J_{i+1}^{j}-J_{i}^{j}) =0 $$ Applying Newton-Raphson Scheme method to this system is, $$ u^{k+1}=u^{k}+\delta u^{k}, \qquad\qquad \left[ \mathbf{J}(u^{k}) \right]\delta u^{k} = - \left[ \mathbf{F}(u^{k}) \right] $$

where Jacobian and $\mathbf{F}$ is

$$ \left[ \begin{array}{*{8}c} U_{1D} & U_{1U} & & & V_{1D} & & & \\ U_{1L} & U_{1D} & U_{1U} & & &V_{1D}& & \\ & U_{1L} & \ddots & U_{1U}& & &\ddots& \\ & & U_{1L} & U_{1D}& & & &V_{1D}\\ U_{2D} & U_{2U} & & & V_{2D} &V_{2U}& & \\ U_{2L} & U_{2D} & U_{2U} & & V_{2L} &V_{2D}&V_{2U}& \\ & U_{2L} & \ddots & U_{2U}& &V_{2L}&\ddots&V_{2U} \\ & & U_{2L} & U_{2D}& & &V_{2L}&V_{2D}\\ \end{array} \right] \left[ \begin{array}{c} \delta u_2\\ \vdots \\ \delta u_{n-1}\\ \delta v_2\\ \vdots \\ \delta v_{n-1} \end{array} \right] = - \left[ \begin{array}{c} F_1(i=2)\\ \vdots\\ F_1(i=n-1)\\ F_2(i=2)\\ \vdots\\ F_2(i=n-1) \end{array} \right] $$

where $F_{1,2}(i)$ vector is inserting current $i$ value for $j$ terms and guess value $i$ for $j+1$ (next time) terms.

and each components are (D=main diagonal, U=first upper diagonal, L=first lower diagonal)

$$ \begin{aligned} U_{1D} &= \frac{d}{du_i^{j+1}}F_{1}^{j+1} =-\frac{2}{\Delta x^2} & (2\leq i\leq n-1) \\ U_{1U} &= \frac{d}{du_{i+1}^{j+1}}F_{1}^{j+1} =\frac{1}{\Delta x^2} & (2\leq i\leq n-2) \\ U_{1L} &= \frac{d}{du_{i-1}^{j+1}}F_{1}^{j+1} =\frac{1}{\Delta x^2} & (3\leq i\leq n-1) \\ V_{1D} &= \frac{d}{dv_{i}^{j+1}}F_{1}^{j+1} =1 & (2\leq i\leq n-1) \\ U_{2D} &= \frac{d}{du_i^{j+1}}F_{2}^{j+1} =-\frac{\Delta t}{\Delta x^2}\left( \frac{dj_{i+1}^{j+1}}{du_{i}^{j+1}} - \frac{dj_{i}^{j+1}}{du_{i}^{j+1}} \right) & (2\leq i\leq n-1) \\ U_{2U} &= \frac{d}{du_{i+1}^{j+1}}F_{2}^{j+1} =-\frac{\Delta t}{\Delta x^2}\left( \frac{dj_{i+1}^{j+1}}{du_{i+1}^{j+1}} \right) & (2\leq i\leq n-2) \\ U_{2L} &= \frac{d}{du_{i-1}^{j+1}}F_{2}^{j+1} =\frac{\Delta t}{\Delta x^2}\left( \frac{dj_{i-1}^{j+1}}{du_{i-1}^{j+1}} \right) & (3\leq i\leq n-1) \\ V_{2D} &= \frac{d}{dv_i^{j+1}}F_{2}^{j+1} =2-\frac{\Delta t}{\Delta x^2}\left( \frac{dj_{i+1}^{j+1}}{dv_{i}^{j+1}} - \frac{dj_{i}^{j+1}}{dv_{i}^{j+1}} \right) & (2\leq i\leq n-1) \\ V_{2U} &= \frac{d}{dv_{i+1}^{j+1}}F_{2}^{j+1} =-\frac{\Delta t}{\Delta x^2}\left( \frac{dj_{i+1}^{j+1}}{dv_{i+1}^{j+1}} \right) & (2\leq i\leq n-2) \\ V_{2L} &= \frac{d}{dv_{i-1}^{j+1}}F_{2}^{j+1} =\frac{\Delta t}{\Delta x^2}\left( \frac{dj_{i-1}^{j+1}}{dv_{i-1}^{j+1}} \right) & (3\leq i\leq n-1) \end{aligned} $$

Again, $J_{i+1}$ is function of $u_{i}$, $u_{i+1}$, $v_{i}$ and $v_{i+1}$ and $J_{i}$ is function of $u_{i-1}$, $u_{i}$, $v_{i-1}$ and $v_{i}$.

Is this correct way?



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