Non linear boundary value problem with infinity.How to solve?

dx/dphi= ((b*cos(phi))/((c*((z/b)-L))-(b*(sin(phi)/x))));

dz/dphi= ((b*sin(phi))/((c*((z/b)-L))-(b*(sin(phi)/x))));

Boundary conditions:

dz/dx=tan(124.119) at (xc,zc) xc=0.39047; zc=0.26333;

z=L=0.144017750497892 at x=infinity

Answer 1

[Torsten]

Use dz/dx = dz/dt / dx/dt and the initial condition z(0.39047)=0.26333 to solve your system from above. The condition at x=infinity will either be satisfied or not - you cannot prescribe it.

Best wishes

Torsten.

[Purush otham]

Are you suggesting the use of ode45? If so, I have already used z(0.39047)=0.26333 but I do not understand where will I be able to apply the dz/dx.

Thanks

[Torsten]

You can't apply this condition.

If by chance,

(((b*sin(phi))/((c*((z/b)-L))-(b*(sin(phi)/x)))))/((b*cos(phi))/((c*((z/b)-L))-(b*(sin(phi)/x)))) = tan(124.119)

for (x,z)=(0.39047;0.26333), you are in luck.

Best wishes

Torsten.

[Purush otham]

Hi attached image for original problem, I need to solve it. What method should I use?

regards

phic=124.119 at xc=0.39047; zc=0.26333;

dx/dphi= ((b*cos(phi))/((c*((z/b)-L))-(b*(sin(phi)/x))));

dz/dphi= ((b*sin(phi))/((c*((z/b)-L))-(b*(sin(phi)/x))));

[Torsten]

Please state the complete problem without your own modifications.

[Purush otham]

Hi Torsten, I apologize for the inconvenience but as far as I see all the above data is all that is present and needed to solve and definitely accurate.

[Torsten]

What are b,c,R,rho1,rho2 ?

[Purush otham]

Hi,

b,R and c are constants.

I have edited the original comment. I need to integrate dx/dphi and dz/dphi which is actually obtained from a single non linear 2nd ode.

[Torsten]

Does

((b*sin(phi))/((c*((z/b)-L))-(b*(sin(phi)/x))))/((b*cos(phi))/((c*((z/b)-L))-(b*(sin(phi)/x))))

at xc,zc,phic equal tan(phic) ?

[Purush otham]

yes it actually does.

[Torsten]

ok, so the 1st boundary condition is satisfied (dz/dx=tan(phic) at (x,z)=(xc,zc)).

And what do you get if you use ODE45 to integrate the two first order ODEs for x and z ? Does x-> Inf ? Does z-> L simultaneously ?

[Purush otham]

When I integrate the above the obtained result does not satisfy the conditions. I did not understand exactly what you mean by the 1st boundary condition satisfied. If by satisfied you mean by substitution, then yes.If by ode45 or bvp, then no. Wait I have attached the code:

format long
xini=0.39047;
zini=0.26333;
phiini=124.119;
stepsize=1e-4;
R=((xini/sin(deg2rad(phiini))))
for beta=45
for b=1
  c=(beta/(b^2));
  L=((2/c)*((2/R)-(1/b)))
for phifin=180;

 %code begins
 phi1=deg2rad(phiini);
 phi2=deg2rad(phifin);
 f=@(phi,x)[((b*cos(phi))/((c*((x(2)/b)-L))-(b*(sin(phi)/x(1)))));((b*sin(phi))/((c*((x(2)/b)-L))-(b*(sin(phi)/x(1)))))];
 [phi,xa]=ode45(f,[phi1:stepsize:phi2],[xini zini]);

 X=xa(:,1);
 Z=xa(:,2);
     plot(X,Z,'or') %xa(:,1)=xvalues & xa(:,2)=zvalues
     ylabel('z axis')
     xlabel('x axis')
     title('tails')
     end
 end  
   end

[Torsten]

When I integrate the above the obtained result does not satisfy the conditions. I did not understand exactly what you mean by the 1st boundary condition satisfied. If by satisfied you mean by substitution, then yes.If by ode45 or bvp, then no.

But you said that

(b*sind(phiini)/(c*(zini/b-L)-b*sind(phiini)/xini))/(b*cosd(phiini)/(c*(zini/b-L)-b*sind(phiini)/xini))-tand(phiini)=0

so dz/dx = tan(phic) at (xc,zc) holds.

I don't understand what you mean by "the obtained result does not satisfy the conditions".

But the condition z=L at x=Inf is irritating. There must be a second-order ODE that you did not yet mention for which two boundary conditions have to be imposed.

Best wishes

Torsten.