Qin et al 2019a - Revision history

Zarate: Zarate moved page Review 140106993447 to Qin et al 2019a

2019-09-12T07:21:53Z

Zarate moved page Review 140106993447 to Qin et al 2019a

Zarate: Zarate moved page Review 140106993447 to Qin et al 2019a

2019-09-12T07:21:47Z

Zarate moved page Review 140106993447 to Qin et al 2019a

Scipediacontent at 13:42, 5 July 2019

2019-07-05T13:42:08Z

Scipediacontent at 13:41, 5 July 2019

2019-07-05T13:41:01Z

Scipediacontent at 13:39, 5 July 2019

2019-07-05T13:39:59Z

Scipediacontent at 13:38, 5 July 2019

2019-07-05T13:38:47Z

Scipediacontent at 13:34, 5 July 2019

2019-07-05T13:34:43Z

Scipediacontent at 13:31, 5 July 2019

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Scipediacontent at 13:12, 5 July 2019

2019-07-05T13:12:22Z

Scipediacontent at 13:11, 5 July 2019

2019-07-05T13:11:06Z

@@ Line 522: / Line 522: @@
 <div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
-  <math>\lbrace \begin{array}{l}
+  <math>\begin{cases} \begin{array}{l}
 \frac{\partial C_A}{\partial t}=D_A\frac{{\partial }^2C_A}{\partial x^2}-V_A\frac{\partial C_A}{\partial x}-\frac{\alpha }{{\theta }_A}\left(C_A-C_B\right)+f_a,x\in {\Omega }_1,t\in \left[0,T\right],\\
 \frac{\partial C_B}{\partial t}=D_B\frac{{\partial }^2C_B}{\partial x^2}-V_B\frac{\partial C_B}{\partial x}-\frac{\alpha }{{\theta }_B}\left(C_B-C_A\right)+f_b,x\in {\Omega }_1,t\in \left[0,T\right],\\
 C_A\left(x,0\right)=C_{A0},C_B\left(x,0\right)=C_{B0},x\in {\Omega }_1,\\
 C_A\left(0,t\right)=f_{a1},C_A\left(1,t\right)=f_{a2},C_B\left(0,t\right)=f_{b1},C_B\left(1,t\right)=f_{b2},t\in \left[0,T\right].
-\end{array}</math> </div>
+\end{array} \end{cases}</math> </div>
 Let  <math>D_A=D_B=1</math> ,  <math>V_A=V_B=1</math> ,  <math>{\theta }_A={\theta }_B=2</math> ,  <math>\alpha =0.001</math> ,  <math>{\Omega }_1=\left[0,1\right]</math> . The exact solutions are  <math>C_A\left(x,t\right)=t^2x{\left(x-1\right)}^2</math> ,  <math>C_B\left(x,t\right)=t^2x^2{\left(x-1\right)}^3</math> , where  <math>C_{A0}</math> ,  <math>C_{B0}</math> ,  <math>f_{a1}</math> ,  <math>f_{a2}</math> ,  <math>f_{b1}</math> ,  <math>f_{b2}</math> are determined by exact solutions. The space step is <math>h=1/(N-1)</math> ( <math>N</math> is the number of nodes) , the time step is  <math>\Delta t=0.001</math> . The Gaussian function  <math>\phi \left(r\right)=exp\left(-c{\left(a\cdot r\right)}^2\right)</math> as basis function, where the direction of calculation is selected from a unit circle. The direction is:
@@ Line 712: / Line 713: @@
 C_A\left(x,y,t\right)=f_1,C_B\left(x,y,t\right)=f_2,\left(x,y\right)\in {\Omega }_2,t\in \left[0,T\right].
 \end{array}\end{cases}</math> </div>
 ''' '''Let  <math>D_A=D_B=1</math> ,  <math>V_A=V_B=1</math> ,  <math>{\theta }_A={\theta }_B=1</math> ,  <math>\alpha =0.001</math> ,  <math>{\Omega }_2=\left[0,1\right]\times \left[0,1\right]</math> . The exact solutions are  <math>C_A=e^{1/2\left(x+y\right)-t}</math> ,  <math>C_B=\left(t+1\right)x^2y^2</math> , where  <math>C_{A0}</math> ,  <math>C_{B0}</math> ,  <math>f_1</math> ,  <math>f_2</math> are determined by exact solutions. Space step size is <math>h=1/(N-1)</math>, time step size is  <math>\Delta t=0.001</math> . The MQ type function  <math>\phi \left(r\right)=\sqrt{c^2+{\left(a\cdot r\right)}^2}</math> as basis function, where the direction of calculation is selected from a unit circle. The direction is:

@@ Line 706: / Line 706: @@
 <div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
-  <math>\lbrace \begin{array}{l}
+  <math>\begin{cases} \begin{array}{l}
 \frac{\partial C_A}{\partial t}=D_A\Delta C_A-V_A\left(\frac{\partial C_A}{\partial x}+\frac{\partial C_A}{\partial y}\right)-\frac{\alpha }{{\theta }_A}\left(C_A-C_B\right)+f_a,\left(x,y\right)\in {\Omega }_2,t\in \left[0,T\right],\\
 \frac{\partial C_B}{\partial t}=D_B\Delta C_B-V_B\left(\frac{\partial C_B}{\partial x}+\frac{\partial C_B}{\partial y}\right)-\frac{\alpha }{{\theta }_B}\left(C_B-C_A\right)+f_b,\left(x,y\right)\in {\Omega }_2,t\in \left[0,T\right],\\
 C_A\left(x,y,0\right)=C_{A0},C_B\left(x,y,0\right)=C_{B0},\left(x,y\right)\in {\Omega }_2,\\
 C_A\left(x,y,t\right)=f_1,C_B\left(x,y,t\right)=f_2,\left(x,y\right)\in {\Omega }_2,t\in \left[0,T\right].
-\end{array}</math> </div>
+\end{array}\end{cases}</math> </div>
 ''' '''Let  <math>D_A=D_B=1</math> ,  <math>V_A=V_B=1</math> ,  <math>{\theta }_A={\theta }_B=1</math> ,  <math>\alpha =0.001</math> ,  <math>{\Omega }_2=\left[0,1\right]\times \left[0,1\right]</math> . The exact solutions are  <math>C_A=e^{1/2\left(x+y\right)-t}</math> ,  <math>C_B=\left(t+1\right)x^2y^2</math> , where  <math>C_{A0}</math> ,  <math>C_{B0}</math> ,  <math>f_1</math> ,  <math>f_2</math> are determined by exact solutions. Space step size is <math>h=1/(N-1)</math>, time step size is  <math>\Delta t=0.001</math> . The MQ type function  <math>\phi \left(r\right)=\sqrt{c^2+{\left(a\cdot r\right)}^2}</math> as basis function, where the direction of calculation is selected from a unit circle. The direction is:

@@ Line 269: / Line 269: @@
 {| style="text-align: center; margin:auto;"
 |-
-| <math>\lbrace \begin{array}{l}
+| <math>\begin{cases} \begin{array}{l}
 \sum_{i=1}^{N_i}\left[\left(1+\frac{\alpha \Delta t}{{\theta }_A}\right)\varphi \left(\Vert I_j-I_i\Vert \right)-\Delta tD_A\Delta \varphi \left(\Vert I_j-I_i\Vert \right)+\Delta tV_A\nabla \varphi \left(\Vert I_j-I_i\Vert \right)\right]{\alpha }_1{}_i^{n+1}\\
 +\sum_{i=1}^{N_b}\left[\left(1+\frac{\alpha \Delta t}{{\theta }_A}\right)\varphi \left(\Vert I_j-B_i\Vert \right)-\Delta tD_A\Delta \varphi \left(\Vert I_j-B_i\Vert \right)+\Delta tV_A\nabla \varphi \left(\Vert I_j-B_i\Vert \right)\right]{\beta }_1{}_i^{n+1}\\
@@ Line 280: / Line 280: @@
 \sum_{i=1}^{N_i}{\alpha }_1{}_i^{n+1}\varphi \left(\Vert B_j-I_i\Vert \right)+\sum_{i=1}^{N_b}{\beta }_1{}_i^{n+1}\varphi \left(\Vert B_j-B_i\Vert \right)=f_1\left(B_j\right),j=1,2,\cdots ,N_b\\
 \sum_{i=1}^{N_i}{\alpha }_2{}_i^{n+1}\varphi \left(\Vert B_j-I_i\Vert \right)+\sum_{i=1}^{N_b}{\beta }_2{}_i^{n+1}\varphi \left(\Vert B_j-B_i\Vert \right)=f_2\left(B_j\right),j=1,2,\cdots ,N_b
-\end{array}</math>
+\end{array}\end{cases}</math>
 |}
 | style="width: 5px;text-align: right;white-space: nowrap;" | (13)

@@ Line 38: / Line 38: @@
 This paper mainly considers the 2D two-flow domain model problem, and its equation is expressed
 {| class="formulaSCP" style="width: 100%; text-align: center;"
@@ Line 44: / Line 45: @@
 {| style="text-align: center; margin:auto;"
 |-
-| [[Image:Draft_Qin_902464492-image1.png|600px]]
+| <math>\begin{cases} \begin{array}{l}
 |}
 | style="width: 5px;text-align: right;white-space: nowrap;" | (1)

@@ Line 523: / Line 523: @@
 \end{array}</math> </div>
-Let  <math>D_A=D_B=1</math> ,  <math>V_A=V_B=1</math> ,  <math>{\theta }_A={\theta }_B=2</math> ,  <math>\alpha =0.001</math> ,  <math>{\Omega }_1=\left[0,1\right]</math> . The exact solutions are  <math>C_A\left(x,t\right)=t^2x{\left(x-1\right)}^2</math> ,  <math>C_B\left(x,t\right)=t^2x^2{\left(x-1\right)}^3</math> , where  <math>C_{A0}</math> ,  <math>C_{B0}</math> ,  <math>f_{a1}</math> ,  <math>f_{a2}</math> ,  <math>f_{b1}</math> ,  <math>f_{b2}</math> are determined by exact solutions. The space step is  [[Image:Draft_Qin_902464492-image188.png|600px]] ( <math>N</math> is the number of nodes) , the time step is  <math>\Delta t=0.001</math> . The Gaussian function  <math>\phi \left(r\right)=exp\left(-c{\left(a\cdot r\right)}^2\right)</math> as basis function, where the direction of calculation is selected from a unit circle. The direction is:
+Let  <math>D_A=D_B=1</math> ,  <math>V_A=V_B=1</math> ,  <math>{\theta }_A={\theta }_B=2</math> ,  <math>\alpha =0.001</math> ,  <math>{\Omega }_1=\left[0,1\right]</math> . The exact solutions are  <math>C_A\left(x,t\right)=t^2x{\left(x-1\right)}^2</math> ,  <math>C_B\left(x,t\right)=t^2x^2{\left(x-1\right)}^3</math> , where  <math>C_{A0}</math> ,  <math>C_{B0}</math> ,  <math>f_{a1}</math> ,  <math>f_{a2}</math> ,  <math>f_{b1}</math> ,  <math>f_{b2}</math> are determined by exact solutions. The space step is <math>h=1/(N-1)</math> ( <math>N</math> is the number of nodes) , the time step is  <math>\Delta t=0.001</math> . The Gaussian function  <math>\phi \left(r\right)=exp\left(-c{\left(a\cdot r\right)}^2\right)</math> as basis function, where the direction of calculation is selected from a unit circle. The direction is:
 <div class="center" style="width: auto; margin-left: auto; margin-right: auto;">
@@ Line 707: / Line 707: @@
 \end{array}</math> </div>
-''' '''Let  <math>D_A=D_B=1</math> ,  <math>V_A=V_B=1</math> ,  <math>{\theta }_A={\theta }_B=1</math> ,  <math>\alpha =0.001</math> ,  <math>{\Omega }_2=\left[0,1\right]\times \left[0,1\right]</math> . The exact solutions are  <math>C_A=e^{1/2\left(x+y\right)-t}</math> ,  <math>C_B=\left(t+1\right)x^2y^2</math> , where  <math>C_{A0}</math> ,  <math>C_{B0}</math> ,  <math>f_1</math> ,  <math>f_2</math> are determined by exact solutions. Space step size is  [[Image:Draft_Qin_902464492-image213.png|600px]] , time step size is  <math>\Delta t=0.001</math> . The MQ type function  <math>\phi \left(r\right)=\sqrt{c^2+{\left(a\cdot r\right)}^2}</math> as basis function, where the direction of calculation is selected from a unit circle. The direction is:
+''' '''Let  <math>D_A=D_B=1</math> ,  <math>V_A=V_B=1</math> ,  <math>{\theta }_A={\theta }_B=1</math> ,  <math>\alpha =0.001</math> ,  <math>{\Omega }_2=\left[0,1\right]\times \left[0,1\right]</math> . The exact solutions are  <math>C_A=e^{1/2\left(x+y\right)-t}</math> ,  <math>C_B=\left(t+1\right)x^2y^2</math> , where  <math>C_{A0}</math> ,  <math>C_{B0}</math> ,  <math>f_1</math> ,  <math>f_2</math> are determined by exact solutions. Space step size is <math>h=1/(N-1)</math>, time step size is  <math>\Delta t=0.001</math> . The MQ type function  <math>\phi \left(r\right)=\sqrt{c^2+{\left(a\cdot r\right)}^2}</math> as basis function, where the direction of calculation is selected from a unit circle. The direction is:
 <div class="center" style="width: auto; margin-left: auto; margin-right: auto;">

← Older revision	Revision as of 07:21, 12 September 2019
(No difference)

@@ Line 209: / Line 209: @@
   <math>\frac{C_A^{n+1}-C_A^n}{\Delta t}=D_A(\frac{{\partial }^2C_A^{n+1}}{\partial x^2}+</math><math>\frac{{\partial }^2C_A^{n+1}}{\partial y^2})-V_A(\frac{\partial C_A^{n+1}}{\partial x}+</math><math>\frac{\partial C_A^{n+1}}{\partial y})-\frac{\alpha }{{\theta }_A}\left(C_A^{n+1}-\right. </math><math>\left. C_B^{n+1}\right)+f_a^{n+1},</math>  <math>\frac{C_B^{n+1}-C_B^n}{\Delta t}=D_B(\frac{{\partial }^2C_B^{n+1}}{\partial x^2}+</math><math>\frac{{\partial }^2C_B^{n+1}}{\partial y^2})-V_B(\frac{\partial C_B^{n+1}}{\partial x}+</math><math>\frac{\partial C_B^{n+1}}{\partial y})-\frac{\alpha }{{\theta }_B}\left(C_B^{n+1}-\right. </math><math>\left. C_A^{n+1}\right)+f_b^{n+1}.</math> </div>
-If <math>\phi \left(\cdot \right)</math> is not a polynomial, then ridge basis functions can approximate to almost all the functions. Here, we respectively denote functions <math>C_A^h\left(X\right)</math> , [[Image:Draft_Qin_902464492-image80.png|600px]] as approximations of  <math>C_A\left(X\right)</math> , <math>C_B\left(X\right)</math> :
+If <math>\phi \left(\cdot \right)</math> is not a polynomial, then ridge basis functions can approximate to almost all the functions. Here, we respectively denote functions <math>C_A^h\left(X\right)</math> , <math>C_B^h\left(X\right)</math> as approximations of  <math>C_A\left(X\right)</math> , <math>C_B\left(X\right)</math> :
 <div class="center" style="width: auto; margin-left: auto; margin-right: auto;">

@@ Line 64: / Line 64: @@
 '''Definition 2.2 '''Give <math>n</math> different nodes <math>{\left\{x_i\right\}}_{i=1}^n</math> and <math>n</math> data points <math>{\left\{d_i\right\}}_{i=1}^n</math> in <math>R^d</math> , the function <math>F(x)</math> statisfying
-'''Definition 2.2 '''Give [[Image:Draft_Qin_902464492-image18.png|600px]] different nodes <math>{\left\{x_i\right\}}_{i=1}^n</math> and <math>n</math> data points <math>{\left\{d_i\right\}}_{i=1}^n</math> in <math>R^d</math> , the function <math>F(x)</math> statisfying
+'''Definition 2.2 '''Give <math>n</math> different nodes <math>{\left\{x_i\right\}}_{i=1}^n</math> and <math>n</math> data points <math>{\left\{d_i\right\}}_{i=1}^n</math> in <math>R^d</math> , the function <math>F(x)</math> statisfying
 {| class="formulaSCP" style="width: 100%; text-align: center;"