Matlab code for boundary condition
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What is the matlab code for below the boundary conditon?
(i) x=0, y=0
(ii) x=0, dy/dx +dz/dx =0
(iii) t=0, x=infinity, y=1, z=c
4 Comments
Torsten
on 27 Jun 2019
Depends on the numerical program for which you try to implement these conditions.
Salai Mathiselvi Salai Mathiselvi
on 27 Jun 2019
Edited: Salai Mathiselvi Salai Mathiselvi
on 27 Jun 2019
Torsten
on 27 Jun 2019
What is the difference between y'(x,t) and y'(t) ?
Same for z.
According to which independent variable do you differntiate when you write y', y'', z', z'' ?
Salai Mathiselvi Salai Mathiselvi's comment moved here:
d^2y/dx^2-y^2*dy/dx+a*z-c*y-dy/dt=0,
d^2z/dx^2-y^2*dz/dx+a*z-c*y-dz/dt=0
Boundary condition
(i) x=0, y=0
(ii) x=0, dy/dx +dz/dx =0
(iii) t=0, x=infinity, y=1, z=b
What is the matlab code for this equation?
Answers (1)
Torsten
on 27 Jun 2019
0 votes
So you mean
at x=0, you have y = 0 and dy/dx + dz/dx
at x=infinity, you have y=1 and z = b
and at t = 0, you have
y = 1, z = b
for all x in [0;infinity] ?
If you have numerical values for a, b and c, read the documentation of "pdepe".
It will show you how to write the MATLAB code for your equations.
3 Comments
Salai Mathiselvi Salai Mathiselvi
on 27 Jun 2019
- m = 2 is wrong.
- You didn't include the terms -y^2*dy/dx and -y^2*dz/dx in the equations.
- Your boundary conditions settings are all wrong.
To include the boundary condition dy/dx + dz/dx = 0 at x=0 for "pdepe", you will have to rewrite your system of equations in terms of y and u:=y+z. This leads to
d^2y/dx^2-y^2*dy/dx+a*(u-y)-c*y-dy/dt=0,
d^2u/dx^2-y^2*du/dx-du/dt = 0
Best wishes
Torsten.
Salai Mathiselvi Salai Mathiselvi
on 30 Jun 2019
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