# solve second order non-linear partial differential equation with boundaries conditions

조회 수: 3(최근 30일)
Alessia Nannetti 2021년 1월 30일
댓글: Alessia Nannetti 2021년 2월 6일
hello everyone!
I am struggling to find a way to solve this equation
=
with those boundaries condition
where a, b, pac, pev are known constants
hope someone can help me,
thank you!
##### 댓글 수: 1표시숨기기 없음
darova 2021년 1월 31일
Its not partial equation. It's ordinary one. DId you try bvp4c?

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### 채택된 답변

Shashank Gupta 2021년 2월 2일
Hi Alessia,
There are many way to solve the above differential equation and some of them are well documented, refer to bvp4c function, it is really good way of solving boundary condition differential equation(@darova also recommend this function in the comments section). There are other ways to solve this, check out this example link.
I hope this references helps you.
Cheers.
##### 댓글 수: 1표시숨기기 없음
Alessia Nannetti 2021년 2월 6일
thank you so much for your help!
i actually solved the problem with bvp4c function.
here there is my code
clear all
close all
clc
%stationary and rje uniform
global pac pev fi beta
rsco = 0.0024; %Reference SC resistance at h0 mmHg/(Î¼l/min)
rje = 2.08; %mmHg/(Î¼l/min)
rd = 0.694; %mmHg/(Î¼l/min)
beta = rje/rsco;
pac = 15 %mmHg
pev = 9 %mmHg
n = 30; %number of CC
w = 300; %SC width Î¼m
x0 = 1;
h0 = 1;
fi = rje/rd;
osc = (2*n*w*h0)/fi,
f = 1; %Hz frequency of ocular pulse
alpha = osc*rje*f;
pi = (pac-pev)/fi;
Hs = 1-pi*(fi/(1+fi));
options = bvpset('RelTol',1e-1,'Stats','on','NMax',10000) %needed this for tollerance errors
infinity = 1;
solinit = bvpinit(linspace(0,infinity,1000),[Hs^4 0]); %changed following equation (19)
sol = bvp4c(@fsode,@fsbc,solinit,options);
x = sol.x;
y = sol.y;
H=y(1,:).^(1/4);
dHdx=(1/4)*(y(1,:).^(-3/4).*y(2,:));
% plot of dH(x)/dx
figure(1)
plot(x,dHdx,x(end),dHdx(end),'o','linewidth',2);
grid on
title('dH/dx')
xlabel('x')
ylabel('dY^(1/4)/dx')
%plot of H(x)
figure(2)
plot(x,H,x(end),H(end),'o','linewidth',2);
grid on
title('H')
xlabel('x')
ylabel('Y^(1/4)')
Q = -(beta/alpha)*H.^3.*dHdx; %dimensionless flow rate
T = -(beta/alpha)*H.*dHdx; %dimensionless shear stress
%plot of Q(x) --> dimensionless flow rate
figure(3)
plot(x,Q,'linewidth',2)
grid on
title('dimensionless flow rate')
xlabel('x')
ylabel('Q(x)')
%plot of T(x)--> dimensionless shear stress
figure(4)
plot(x,T,'linewidth',2)
grid on
title('dimensionless shear stress')
xlabel('x')
ylabel('T(x)')
function dfdeta = fsode(x,y) %Function
global beta
dfdeta = [y(2)
(4/beta)*(y(1)^(0.25)-1)];
end
function res = fsbc(ya,yb) %Boundaries Conditions, [a,b]-->is the interval
global pac pev fi beta
res = [ya(2)
-(beta/(4*fi))*yb(2)-yb(1)^(1/4)+1+pev-pac];
end
I don't know why I got this error from Matlab:
Warning: Imaginary parts of complex X and/or Y arguments ignored.
I see that I got Imaginary parts on the solution ( sol.x and sol.y) but actually I don't want them, they need to be all real. I don't know what I am doing wrong.
Many Thanks

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