Asked by Alexander Kimbley
on 8 Apr 2019

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.

Answer by Torsten
on 10 Apr 2019

Edited by Torsten
on 10 Apr 2019

Accepted Answer

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

Alexander Kimbley
on 12 Apr 2019

Ohh I see, sorry about that! It's perfect then, thanks!

Alexander Kimbley
on 12 Apr 2019

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
on 15 Apr 2019

As before, please state the problem in a mathematical fashion (equations and boundary conditions).

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Answer by Alexander Kimbley
on 16 Apr 2019

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.

Alexander Kimbley
on 16 Apr 2019

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
on 16 Apr 2019

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
on 18 Apr 2019

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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## 4 Comments

## Torsten (view profile)

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## Alexander Kimbley (view profile)

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## Torsten (view profile)

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