Unbound Shear Layer boundary condition problem

2019 4 8
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更新时间：2019 4 18
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Hi, I'm very new to Matlab so bare with me, I'm trying to solve the unbounded shear layer problem for the internal section, between z=1 and -1. I cannot however enter the boundary conditon at z=1 to depend on a constant. Any help would be great.
Thanks.

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Torsten
Torsten 2019-4-8
编辑：Torsten 2019-4-8
Please write out the problem in a mathematical notation:
Differential equation(s):
Boundary condition(s):
Hi, the differential equation is given by
y''+(a)^2 y=0, where -1<x<1.
The boundary conditions are y(-1)=0, y(1)=A. There is also an other condtion that must be satisfied which is y'(-1)=1. Noting that A is contant and not arbitrary.
Thanks.
Torsten
Torsten 2019-4-10
And you want to determine "a" such that all three boundary conditions are met ?
Alexander Kimbley
Alexander Kimbley 2019-4-10
You would be correct. Obivously we can solve the problem analytically but the code will allow to explore the problem further with different conditions/criteria where one would not be able to analytically.

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Torsten
Torsten 2019-4-10
编辑：Torsten 2019-4-10

2 个投票

function main
xlow = -1; xhigh = 1;
A = 1;
solinit = bvpinit(linspace(xlow,xhigh,4000),[1,1],1);
sol = bvp4c(@bvp4ode, @(ya,yb,parameters)bvp4bc(ya,yb,parameters,A), solinit);
a = sol.parameters;
xint = linspace(xlow,xhigh,2000);
Sxint = deval(sol,xint);
% Analytical solution
C1 = A/(2*sin(a));
C2 = A/(2*cos(a));
fun = @(x) C1*sin(a*x) + C2*cos(a*x);
% Plot numerical vs. analytical solution
plot(xint,Sxint(1,:),xint,fun(xint))
end
function dydx = bvp4ode(x,y,parameters)
dydx=[y(2); -parameters^2*y(1)];
end
function res = bvp4bc(ya,yb,parameters,A)
res=[ya(1); yb(1)-A; ya(2)-1];
end

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Alexander Kimbley
Alexander Kimbley 2019-4-11
Thanks! Just one thing however, I can't seem to get the value of 'a', what code would give an output for it?
Torsten
Torsten 2019-4-11
But the line
a = sol.parameters
gives the value of a.
Or what do you mean ?
Alexander Kimbley
Alexander Kimbley 2019-4-12
Ohh I see, sorry about that! It's perfect then, thanks!
Alexander Kimbley
Alexander Kimbley 2019-4-12
编辑：Alexander Kimbley 2019-4-12
I have another problem but a lot more complex, at least for my coding skills that is. I'll attach it if you want to take a look, it'd be very appreciated. I'n the attached code I've just added the relevant ODE and BC's.
It's essentially the same problem but have introduced the c the complex eigenvalue, which is not known, B a fixed constant as well as more complex boundary conditions, I'm not sure if the last of them is needed for the problem however. R is also not known.
Thanks.
Torsten
Torsten 2019-4-15
As before, please state the problem in a mathematical fashion (equations and boundary conditions).

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更多回答（1 个）

Alexander Kimbley
Alexander Kimbley 2019-4-16

0 个投票

The ODE is given by
((x-c)^2 -B^2)(y''-ay)=2(B^2)(y-(x-c)y')/((x-c)^2), -1<x<1, B constant, a wavenumber, c complex wave speed.
Boundary conditions:
y(-1)=1,
y(1)=R,
y'(-1)=(1+c-a(1+c)^2)/(B^2 -(1+c)^2),
y'(1)=R(1-c-a(1-c)^2)/((1-c)^2 -B^2).
The last boundary condition may or may not be nessesary however.
Thanks.

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Torsten
Torsten 2019-4-16
What constants are given ? What constants are to be determined ?
Alexander Kimbley
Alexander Kimbley 2019-4-16
Constants given: B
Constants to be determined: c, a, R.
If the system has two many constants to be determined however, then a is probably the best to move to constants given, where a>0.
Torsten
Torsten 2019-4-16
编辑：Torsten 2019-4-16
function main
xlow = -1; xhigh = 1;
a = 1;
B = 0.5;
c0 = 0.5;
R0 = 0.5;
y10 = 1.0;
y20 = 0.0;
solinit = bvpinit(linspace(xlow,xhigh,4000),[y10,y20],[c0,R0]);
sol = bvp4c(@(x,y,p)bvp4ode(x,y,p,a,B), @(ya,yb,p)bvp4bc(ya,yb,p,a,B), solinit);
c = sol.parameters(1)
R = sol.parameters(2)
xint = linspace(xlow,xhigh,2000);
Sxint = deval(sol,xint);
plot(xint,Sxint(1,:))
end
function dydx = bvp4ode(x,y,p,a,B)
c = p(1);
R = p(2);
dydx = [y(2); 2*B^2*(y(1)-(x-c^2)*y(2))/((x-c^2)^2*((x-c^2)^2-B^2))+a*y(1)];
end
function res = bvp4bc(ya,yb,p,a,B)
c = p(1);
R = p(2);
res = [ya(1)-1; yb(1)-R; ya(2)-((1+c^2)-a*(1+c^2)^2)/(B^2-(1+c^2)^2); yb(2)-R*((1-c^2)-a*(1-c^2)^2)/((1-c^2)^2-B^2)];
end
Alexander Kimbley
Alexander Kimbley 2019-4-18
编辑：Alexander Kimbley 2019-4-18
Thanks Torsten! There just seems to be some error however, when B=0, a=0.6392, we should get that c=0 with max error being something like O(e^-5) but the output is approx 0.01 out in both complex and real parts, which leads me to think something is not quite right. I've changed the boundary conditions slightly but these do not affect the equation when B=0 either.
Is there a way to provide an inital guess for y, would this change the output?
Or could we actually make c and R variables in the function where we set them as y(3) and y(4) respectivley?
Thanks,
Alexander.
I've attached the altered code. (only the boundary conditions have changed and an error tolerance)

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