BVP4c for three boundary set of boundary conditions

Question

Komal Goyal 2023-2-5

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1906795-bvp4c-for-three-boundary-set-of-boundary-conditions

评论： Komal Goyal 2023-2-15

Respected sir/maam,

I was trying to solve my coupled non-linear ODEs with three intervals of boundary conditions. Here It is error "uncognized function or variable D1" and some minor errors, which I am not able to rectify. Can you please help. If any information is required, please feel free to ask.

Thanking You in Advance

You help will be highly appreciable

xb = 0.4;

xc = 0.8;

xmesh = [linspace(0,0.4,100),linspace(0.4,0.8,100),linspace(0.8,1,100)];

yinit = [1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1 ];

solinit = bvpinit(xmesh,yinit);

%constant

r=1;

m=1;

cp=1;

k=1;

s=1;

bt=1;

bc2=1;

r1=1;

m1=1;

cp1=1;

k1=1;

s1=1;

bt1=1;

bc1=1;

t=1;

L=1;

e=0.001;

H=1;

R=0.2;

Da1=0.5;

Da2=0.5;

Da3=0.5;

B=1;

M=1;

Gr=1;

Gc=1;

th=pi/2;

d=1;

Pr=1;

Ec=1;

a1=1;

a2=1;

a3=1;

f1=1;

Nb=1;

Nt=1;

Le=1;

g=1;

K1=0.001;

Bi=1;

n=1;

Db=1;

Dt=1;

K=1;

Pr1=21;

v=(m*r1)/(m1*r);

si=(s1/s);

mu=(m1/m);

ro=(r/r1);

btk=(bt1/bt);

btc=(bc1/bc2);

CP=(cp/cp1);

roc=(r*cp)/(r1*cp1);

k2=(k/k1);

a=(r*cp*k1)/(k*r*cp1);

H1=sqrt(v)*H;

R1=v*R;

M1=sqrt(si/mu)*M;

Gr1=btk*v*Gr;

Gc1=btc*v*Gc;

Ec1=CP*Ec;

Nt1=(Nt)/(Dt*a);

Nb1=(Nb)/(Db*a);

Le1=a*Le*Db;

g1=(v*g)/K;

K2=(v*K1)/K;

D1=-(1/Da1)-(M^2)

D1 = -3

D2=sin(th)*Gr;

D3=sin(th)*Gc;

D4=d*((R/H)^2);

D5=(1/Pr);

D6=(f1-a1)/Pr;

D7=((Ec/Da1)+((M^2)*Ec));

D8=(Nb/Pr);

D9=(Nt/Pr);

D10=Le*Pr*(H^2);

D11=Le*Pr*R;

D12=(Nt/Nb);

D13=g*Le*Pr;

D14=K1*Le*Pr;

D15=ro*((H1)^2);

D16=-(1/Da2)-((M1)^2);

D17=sin(th)*Gr1;

D18=sin(th)*Gc1;

D19=(1/Pr1);

D20=d*(((R1)/(H1))^2);

D21=((f1-2)*k2)/Pr;

D22=((Ec1/Da2)+(Ec1*((M1)^2)));

D23=(ro*CP*Nb1)/(Pr1);

D24=(ro*CP*Nt1)/(Pr1);

D25=Le1*Pr1*(H1)^2;

D26=Le1*Pr1*R1;

D27=Nt1/Nb1;

D28=g1*Le1*Pr1;

D29=K2*Le1*Pr1;

D30=-(1/Da3)-((M)^2);

D31=(f1-a3)/Pr;

D32=(Ec/Da3)+((M^2)*Ec);

hold on

for R = 0.2

p = [r m cp k s bt bc2 r1 m1 cp1 k1 s1 bt1 bc1 t L e H R Da1 Da2 Da3 B M Gr Gc th d Pr Ec a1 a2 a3 f1 Nb Nt Le g K1 Bi n Db Dt K Pr1 D1 D2 D3

D4 D5 D6 D7 D8 D9 D10 D11 D12 D13 D14 D15 D16 D17 D18 D19 D20 D21 D22 D23 D24 D25 D26 D27 D28 D29 D30 D31 D32];

sol = bvp5c(@(x,y,r) f(x,y,r,p), @(ya,yb)bc(ya,yb,p), solinit);

plot(sol.x,real(sol.y(1,:)));

end

Error using vertcat
Dimensions of arrays being concatenated are not consistent.

hold off

function dydx = f(x,y,region,p) % equations being solved

r = p(1);

m = p(2);

cp = p(3);

k = p(4);

s = p(5);

bt = p(6);

bc2 = p(7);

r1 = p(8);

m1 = p(9);

cp1 = p(10);

k1 = p(11);

s1 = p(12);

bt1 = p(13);

bc1 = p(14);

t = p(15);

L = p(16);

e = p(17);

H = p(18);

R = p(19);

Da1 = p(20);

Da2 = p(21);

Da3 = p(22);

B = p(23);

M = p(24);

Gr = p(25);

Gc = p(26);

th = p(27);

d = p(28);

Pr = p(29);

Ec = p(30);

a1 = p(31);

a2 = p(32);

a3 = p(33);

f1 = p(34);

Nb = p(35);

Nt = p(36);

Le = p(37);

g = p(38);

K1 = p(39);

Bi = p(40);

n = p(41);

Db = p(42);

Dt = p(43);

K = p(44);

Pr1 = p(45);

D1 = p(46);

D2 = p(47);

D3 = p(48);

D4 = p(49);

D5 = p(50);

D6 = p(51);

D7 = p(52);

D8 = p(53);

D9 = p(54);

D10 = p(55);

D11 = p(56);

D12 = p(57);

D13 = p(58);

D14 = p(59);

D15 = p(60);

D16 = p(61);

D17 = p(62);

D18 = p(63);

D19 = p(64);

D20 = p(65);

D21 = p(66);

D22 = p(67);

D23 = p(68);

D24 = p(69);

D25 = p(70);

D26 = p(71);

D27 = p(72);

D28 = p(73);

D29 = p(74);

D30 = p(75);

D31 = p(76);

D32 = p(77);

dydx = zeros(6,1);

switch region

case 1 % x in [0,0.4]

dydx = [y(2);

(R*y(2))-(D1*y(1))-(D2*y(5))-(D3*y(9))-(H^2);

y(4);

(R*y(4))-((D1-(1i*(H^2)))*y(3))-(D2*y(7))-(D3*y(11))-(H^2);

y(6);

(1/(D5-D4))*((R*y(6))+(D6*y(5))-(Ec*(y(2)^2))-(D7*(y(1)^2))-(D8*y(6)*y(10))-(D9*(y(6)^2)));

y(8);

(1/(D5-D4))*((R*y(8))+((D6+(1i*(H^2)))*y(7))-(2*Ec*y(2)*y(4))-(2*D7*y(1)*y(3))-(D8*((y(6)*y(12))+((y(10)*y(8)))))-(2*D9*y(6)*y*(8)))

y(10);

(D11*y(10))+(D13*y(9))-(D12*((1/(D5-D4))*((R*y(6))+(D6*y(5))-(Ec*(y(2)^2))-(D7*(y(1)^2))-(D8*y(6)*y(10))-(D9*(y(6)^2)))))+(D14)

y(12);

(D11*y(12))+((D13+(1i*D10))*y(11))-(D12*((1/(D5-D4))*((R*y(8))+((D6+(1i*(H^2)))*y(7))-(2*Ec*y(2)*y(4))-(2*D7*y(1)*y(3))-(D8*((y(6)*y(12))+((y(10)*y(8)))))-(2*D9*y(6)*y*(8)))))];

case 2 % x in [0.4,0.8]

dydx = [y(2);

(B/(B+1))*((R1*y(2))-(D16*y(1))-(D17*y(5))-(D18*y(9))-D15);

y(4);

(B/(B+1))*((R1*y(4))-((D16-(1i*(H1^2)))*y(3))-(D17*y(7))-(D18*y(11))-D15);

y(6);

(1/(D19-D20))*((R1*y(6))+(D21*y(5))-(Ec1*((B+1)/B)*(y(2)^2))-(D22*(y(1)^2))-(D23*y(10)*y(6))-(D24*(y(6)^2)));

y(8);

(1/(D19-D20))*((R1*y(8))+((D21+(1i*(H1^2)))*y(7))-(2*Ec1*((B+1)/B)*y(2)*y(4))-(2*D22*y(1)*y(3))-(D23*((y(10)*y(8))+((y(12)*y(6)))))-(2*D24*y(6)*y(8)));

y(10);

(D26*y(10))+(D28*y(9))-(D27*((1/(D19-D20))*((R1*y(6))+(D21*y(5))-(Ec1*((B+1)/B)*(y(2)^2))-(D22*(y(1)^2))-(D23*y(10)*y(6))-(D24*(y(6)^2)))))+D29;

y(12);

(D26*y(12))+(D28*y(11))-(D27*((1/(D19-D20))*((R1*y(8))+((D21+(1i*(H1^2)))*y(7))-(2*Ec1*((B+1)/B)*y(2)*y(4))-(2*D22*y(1)*y(3))-(D23*((y(10)*y(8))+((y(12)*y(6)))))-(2*D24*y(6)*y(8)))))

];

case 3 % x in [0.8,1]

dydx = [y(2);

(R*y(2))-(D30*y(1))-(D2*y(5))-(D3*y(9))-(H^2);

y(4);

(R*y(4))-((D30-(1i*(H^2)))*y(3))-(D2*y(7))-(D3*y(11))-(H^2);

y(6);

(1/(D5-D4))*((R*y(6))+(D31*y(5))-(Ec*(y(2)^2))-(D32*(y(1)^2))-(D8*y(6)*y(10))-(D9*(y(6)^2)));

y(8);

(1/(D5-D4))*((R*y(8))+((D31+(1i*(H^2)))*y(7))-(2*Ec*y(4)*y(2))-(2*D32*y(1)*y(3))-(D8*((y(10)*y(8))+((y(6)*y(12)))))-(2*D9*y(6)*y*(8)));

y(10);

(D11*y(10))+(D13*y(9))-(D12*((1/(D5-D4))*((R*y(6))+(D31*y(5))-(Ec*(y(2)^2))-(D32*(y(1)^2))-(D8*y(6)*y(10))-(D9*(y(6)^2)))))+(D14);

y(12);

(D11*y(12))+((D13+(1i*D10))*y(11))-(D12*((1/(D5-D4))*((R*y(8))+((D31+(1i*(H^2)))*y(7))-(2*Ec*y(4)*y(2))-(2*D32*y(1)*y(3))-(D8*((y(10)*y(8))+((y(6)*y(12)))))-(2*D9*y(6)*y*(8)))))

];

end

function res = bc(YL,YR,p)

B = p(23);

m = p(9)/p(2);

k2 = p(4)/p(11);

Db = p(42);

res = [YL(1,1) % u10(0)=0

YL(3,1) % u11(0)=0

YL(5,1) % T10(0)=0

YL(7,1) % T11(0)=0

YL(9,1) % Phi(0)=0

YL(11,1) % Phi(0)=0

YR(1,1)-YL(1,2) % u20(0.4)=u10(0.4)

YR(3,1)-YL(3,2) % u21(0.4)=u11(0.4)

YR(1,2)-YL(1,3) % u30(0.8)=u20(0.8)

YR(3,2)-YL(3,3) % u31(0.8)=u21(0.8)

YR(5,1)-YL(5,2) % T20(0.4)=T10(0.4)

YR(7,1)-YL(7,2) % T21(0.4)=T11(0.4)

YR(5,2)-YL(5,3) % T30(0.8)=T20(0.8)

YR(7,2)-YL(7,3) % T31(0.8)=T21(0.8)

YR(9,1)-YL(9,2) % P20(0.4)=P10(0.4)

YR(11,1)-YL(11,2) % P21(0.4)=P11(0.4)

YR(9,2)-YL(9,3) % P30(0.8)=P20(0.8)

YR(11,2)-YL(11,3) % P31(0.8)=P21(0.8)

(m*((B+1)/B))*YR(2,1)-YL(2,2) % m*(1+(1/b))*u20'(0.4)=u10'(0.4)

(m*((B+1)/B))*YR(4,1)-YL(4,2) % m*(1+(1/b))*u21'(0.4)=u11'(0.4)

YR(2,2)-((m*((B+1)/B))*YL(2,3)) % m*(1+(1/b))*u20'(0.8)=u30'(0.8)

YR(4,2)-((m*((B+1)/B))*YL(4,3)) % m*(1+(1/b))*u21'(0.8)=u31'(0.8)

YR(6,1)-(k2*YL(6,2)) % T20'[0.4]=k*T10'[0.4]

YR(8,1)-(k2*YL(8,2)) % T21'[0.4]=k*T11'[0.4]

(k2*YR(6,2))-(YL(6,3)) % k*T30'[0.8]=T20'[0.8]

(k2*YR(8,2))-(YL(8,3)) % k*T31'[0.8]=T21'[0.8]

YR(10,1)-(Db*YL(10,2)) % P20'[0.4]=Db*P10'[0.4]

YR(12,1)-(Db*YL(12,2)) % P21'[0.4]=Db*P11'[0.4]

(Db*YR(10,2))-(YL(10,3)) % Db*P30'[0.8]=P20'[0.8]

(Db*YR(12,2))-(YL(12,3)) % Db*T31'[0.8]=P21'[0.8]

YR(1,3) % u30(1)=0

YR(3,3) % u31(1)=0

YR(5,3) % T30(1)=0

YR(7,3) % T31(1)=0

YR(9,3) % Phi30(1)=0

YR(11,3) % Phi31(1)=0

];

end

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Answer 1

Torsten 2023-2-5

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1906795-bvp4c-for-three-boundary-set-of-boundary-conditions#answer_1164125

编辑：Torsten 2023-2-5

You didn't try to understand the code I provided for the similar problem, did you ?

You still supply 6 initial conditions although you have to supply 12.

You still add the equation numbers for the different regions although they have to start with 1 in every region and go up until 12. Your method gives you a numbering up to 12 + 12 + 12 = 36 (1-12, 13-24, 25-36) for the equation numbers within the regions, although the index should only run from 1-12 in every region (1-12, 1-12, 1-12).

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Torsten 2023-2-7

在 MATLAB Online 中打开

It works now, but it takes too long to run it here online.

R1, H1 and Ec1 were not defined. I set them as

R1 = R;
H1 = H;
Ec1 = Ec;

xb = 0.4;
xc = 0.8;
xmesh = [linspace(0,0.4,100),linspace(0.4,0.8,100),linspace(0.8,1,100)];
yinit = [1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1 ];
solinit = bvpinit(xmesh,yinit);
%constant
r=1;
m=1;
cp=1;
k=1;
s=1;
bt=1;
bc2=1;
r1=1;
m1=1;
cp1=1;
k1=1;
s1=1;
bt1=1;
bc1=1;
t=1;
L=1;
e=0.001;
H=1;
R=0.2;
Da1=0.5;
Da2=0.5;
Da3=0.5;
B=1;
M=1;
Gr=1;
Gc=1;
th=pi/2;
d=1;
Pr=1;
Ec=1;
a1=1;
a2=1;
a3=1;
f1=1;
Nb=1;
Nt=1;
Le=1;
g=1;
K1=0.001;
Bi=1;
n=1;
Db=1;
Dt=1;
K=1;
Pr1=21;
v=(m*r1)/(m1*r);
si=(s1/s);
mu=(m1/m);
ro=(r/r1);
btk=(bt1/bt);
btc=(bc1/bc2);
CP=(cp/cp1);
roc=(r*cp)/(r1*cp1);
k2=(k/k1);
a=(r*cp*k1)/(k*r*cp1);
H1=sqrt(v)*H;
R1=v*H;
M1=sqrt(si/mu)*M;
Gr1=btk*v*Gr;
Gc1=btc*v*Gc;
Ec1=CP*Ec;
Nt1=(Nt)/(Dt*a);
Nb1=(Nb)/(Db*a);
Le1=a*Le*Db;
g1=(v*g)/K;
K2=(v*K1)/K;
D1=-(1/Da1)-(M^2)
D2=sin(th)*Gr;
D3=sin(th)*Gc;
D4=d*((R/H)^2);
D5=(1/Pr);
D6=(f1-a1)/Pr;
D7=((Ec/Da1)+((M^2)*Ec));
D8=(Nb/Pr);
D9=(Nt/Pr);
D10=Le*Pr*(H^2);
D11=Le*Pr*R;
D12=(Nt/Nb);
D13=g*Le*Pr;
D14=K1*Le*Pr;
D15=ro*((H1)^2);
D16=-(1/Da2)-((M1)^2);
D17=sin(th)*Gr1;
D18=sin(th)*Gc1;
D19=(1/Pr1);
D20=d*(((R1)/(H1))^2);
D21=((f1-2)*k2)/Pr;
D22=((Ec1/Da2)+(Ec1*((M1)^2)));
D23=(ro*CP*Nb1)/(Pr1);
D24=(ro*CP*Nt1)/(Pr1);
D25=Le1*Pr1*(H1)^2;
D26=Le1*Pr1*R1;
D27=Nt1/Nb1;
D28=g1*Le1*Pr1;
D29=K2*Le1*Pr1;
D30=-(1/Da3)-((M)^2);
D31=(f1-a3)/Pr;
D32=(Ec/Da3)+((M^2)*Ec);
hold on
for R = 0.2
    p = [r m cp k s bt bc2 r1 m1 cp1 k1 s1 bt1 bc1 t L e H R Da1 Da2 Da3 B M Gr Gc th d Pr Ec a1 a2 a3 f1 Nb Nt Le g K1 Bi n Db Dt K Pr1 D1 D2 D3...
         D4 D5 D6 D7 D8 D9 D10 D11 D12 D13 D14 D15 D16 D17 D18 D19 D20 D21 D22 D23 D24 D25 D26 D27 D28 D29 D30 D31 D32];
    sol = bvp5c(@(x,y,r) f(x,y,r,p), @(ya,yb)bc(ya,yb,p), solinit);
    plot(sol.x,real(sol.y(1,:)));
end
hold off
function dydx = f(x,y,region,p) % equations being solved
r = p(1);
m = p(2);
cp = p(3);
k = p(4);
s = p(5);
bt = p(6);
bc2 = p(7);
r1 = p(8);
m1 = p(9);
cp1 = p(10);
k1 = p(11);
s1 = p(12);
bt1 = p(13);
bc1 = p(14);
t = p(15);
L = p(16);
e = p(17);
H = p(18);
R = p(19);
Da1 = p(20);
Da2 = p(21);
Da3 = p(22);
B = p(23);
M = p(24);
Gr = p(25);
Gc = p(26);
th = p(27);
d = p(28);
Pr = p(29);
Ec = p(30);
a1 = p(31);
a2 = p(32);
a3 = p(33);
f1 = p(34);
Nb = p(35);
Nt = p(36);
Le = p(37);
g = p(38);
K1 = p(39);
Bi = p(40);
n = p(41);
Db = p(42);
Dt = p(43);
K = p(44);
Pr1 = p(45);
D1 = p(46);
  D2 = p(47);
  D3 = p(48);
  D4 = p(49);
  D5 = p(50);
  D6 = p(51);
  D7 = p(52);
  D8 = p(53);
  D9 = p(54);
  D10 = p(55);
  D11 = p(56);
  D12 = p(57);
  D13 = p(58);
  D14 = p(59);
  D15 = p(60);
  D16 = p(61);
  D17 = p(62);
  D18 = p(63);
  D19 = p(64);
  D20 = p(65);
  D21 = p(66);
  D22 = p(67);
  D23 = p(68);
  D24 = p(69);
  D25 = p(70);
  D26 = p(71);
  D27 = p(72);
  D28 = p(73);
  D29 = p(74);
  D30 = p(75);
  D31 = p(76);
  D32 = p(77);
  R1 = R;
  H1 = H;
  Ec1 = Ec;
dydx = zeros(12,1);
switch region 
    case 1     % x in [0,0.4]
        dydx = [y(2);
               (R*y(2))-(D1*y(1))-(D2*y(5))-(D3*y(9))-(H^2);
               y(4);
               (R*y(4))-((D1-(1i*(H^2)))*y(3))-(D2*y(7))-(D3*y(11))-(H^2);
               y(6);
               (1/(D5-D4))*((R*y(6))+(D6*y(5))-(Ec*(y(2)^2))-(D7*(y(1)^2))-(D8*y(6)*y(10))-(D9*(y(6)^2)));
               y(8);
               (1/(D5-D4))*((R*y(8))+((D6+(1i*(H^2)))*y(7))-(2*Ec*y(2)*y(4))-(2*D7*y(1)*y(3))-(D8*((y(6)*y(12))+((y(10)*y(8)))))-(2*D9*y(6)*y(8)));
               y(10);
               (D11*y(10))+(D13*y(9))-(D12*((1/(D5-D4))*((R*y(6))+(D6*y(5))-(Ec*(y(2)^2))-(D7*(y(1)^2))-(D8*y(6)*y(10))-(D9*(y(6)^2)))))+(D14);
               y(12);
               (D11*y(12))+((D13+(1i*D10))*y(11))-(D12*((1/(D5-D4))*((R*y(8))+((D6+(1i*(H^2)))*y(7))-(2*Ec*y(2)*y(4))-(2*D7*y(1)*y(3))-(D8*((y(6)*y(12))+((y(10)*y(8)))))-(2*D9*y(6)*y(8)))))]
    case 2     % x in [0.4,0.8]
        dydx = [y(2);
               (B/(B+1))*((R1*y(2))-(D16*y(1))-(D17*y(5))-(D18*y(9))-D15);
               y(4);
               (B/(B+1))*((R1*y(4))-((D16-(1i*(H1^2)))*y(3))-(D17*y(7))-(D18*y(11))-D15);
               y(6);
               (1/(D19-D20))*((R1*y(6))+(D21*y(5))-(Ec1*((B+1)/B)*(y(2)^2))-(D22*(y(1)^2))-(D23*y(10)*y(6))-(D24*(y(6)^2)));
               y(8);
               (1/(D19-D20))*((R1*y(8))+((D21+(1i*(H1^2)))*y(7))-(2*Ec1*((B+1)/B)*y(2)*y(4))-(2*D22*y(1)*y(3))-(D23*((y(10)*y(8))+((y(12)*y(6)))))-(2*D24*y(6)*y(8)));
               y(10);     
               (D26*y(10))+(D28*y(9))-(D27*((1/(D19-D20))*((R1*y(6))+(D21*y(5))-(Ec1*((B+1)/B)*(y(2)^2))-(D22*(y(1)^2))-(D23*y(10)*y(6))-(D24*(y(6)^2)))))+D29;
               y(12);
               (D26*y(12))+(D28*y(11))-(D27*((1/(D19-D20))*((R1*y(8))+((D21+(1i*(H1^2)))*y(7))-(2*Ec1*((B+1)/B)*y(2)*y(4))-(2*D22*y(1)*y(3))-(D23*((y(10)*y(8))+((y(12)*y(6)))))-(2*D24*y(6)*y(8)))))];
      case 3    % x in [0.8,1]
        dydx = [y(2);
               (R*y(2))-(D30*y(1))-(D2*y(5))-(D3*y(9))-(H^2);
               y(4);
               (R*y(4))-((D30-(1i*(H^2)))*y(3))-(D2*y(7))-(D3*y(11))-(H^2);
               y(6);
               (1/(D5-D4))*((R*y(6))+(D31*y(5))-(Ec*(y(2)^2))-(D32*(y(1)^2))-(D8*y(6)*y(10))-(D9*(y(6)^2)));
               y(8);
               (1/(D5-D4))*((R*y(8))+((D31+(1i*(H^2)))*y(7))-(2*Ec*y(4)*y(2))-(2*D32*y(1)*y(3))-(D8*((y(10)*y(8))+((y(6)*y(12)))))-(2*D9*y(6)*y(8)));
               y(10);
               (D11*y(10))+(D13*y(9))-(D12*((1/(D5-D4))*((R*y(6))+(D31*y(5))-(Ec*(y(2)^2))-(D32*(y(1)^2))-(D8*y(6)*y(10))-(D9*(y(6)^2)))))+(D14);
               y(12);
               (D11*y(12))+((D13+(1i*D10))*y(11))-(D12*((1/(D5-D4))*((R*y(8))+((D31+(1i*(H^2)))*y(7))-(2*Ec*y(4)*y(2))-(2*D32*y(1)*y(3))-(D8*((y(10)*y(8))+((y(6)*y(12)))))-(2*D9*y(6)*y(8)))))];
end
end
function res = bc(YL,YR,p)
B = p(23);
m = p(9)/p(2);
k2 = p(4)/p(11);
Db = p(42);
res = [YL(1,1)            % u10(0)=0
    YL(3,1)            % u11(0)=0
    YL(5,1)            % T10(0)=0
    YL(7,1)            % T11(0)=0
    YL(9,1)            % Phi(0)=0
    YL(11,1)           % Phi(0)=0
    YR(1,1)-YL(1,2)    % u20(0.4)=u10(0.4)
    YR(3,1)-YL(3,2)    % u21(0.4)=u11(0.4)
    YR(1,2)-YL(1,3)    % u30(0.8)=u20(0.8)
    YR(3,2)-YL(3,3)    % u31(0.8)=u21(0.8)
    YR(5,1)-YL(5,2)    % T20(0.4)=T10(0.4)
    YR(7,1)-YL(7,2)    % T21(0.4)=T11(0.4)
    YR(5,2)-YL(5,3)    % T30(0.8)=T20(0.8)
    YR(7,2)-YL(7,3)    % T31(0.8)=T21(0.8)
    YR(9,1)-YL(9,2)    % P20(0.4)=P10(0.4)
    YR(11,1)-YL(11,2)  % P21(0.4)=P11(0.4)
    YR(9,2)-YL(9,3)    % P30(0.8)=P20(0.8)
    YR(11,2)-YL(11,3)              % P31(0.8)=P21(0.8)
    (m*((B+1)/B))*YR(2,1)-YL(2,2)  % m*(1+(1/b))*u20'(0.4)=u10'(0.4)
    (m*((B+1)/B))*YR(4,1)-YL(4,2)  % m*(1+(1/b))*u21'(0.4)=u11'(0.4)
    YR(2,2)-((m*((B+1)/B))*YL(2,3)) % m*(1+(1/b))*u20'(0.8)=u30'(0.8)
    YR(4,2)-((m*((B+1)/B))*YL(4,3))  % m*(1+(1/b))*u21'(0.8)=u31'(0.8)
    YR(6,1)-(k2*YL(6,2))             % T20'[0.4]=k*T10'[0.4]
    YR(8,1)-(k2*YL(8,2))             % T21'[0.4]=k*T11'[0.4]
    (k2*YR(6,2))-(YL(6,3))           % k*T30'[0.8]=T20'[0.8]
    (k2*YR(8,2))-(YL(8,3))           % k*T31'[0.8]=T21'[0.8]
    YR(10,1)-(Db*YL(10,2))           % P20'[0.4]=Db*P10'[0.4]
    YR(12,1)-(Db*YL(12,2))           % P21'[0.4]=Db*P11'[0.4]
    (Db*YR(10,2))-(YL(10,3))         % Db*P30'[0.8]=P20'[0.8]
    (Db*YR(12,2))-(YL(12,3))           % Db*T31'[0.8]=P21'[0.8]
    YR(1,3)            % u30(1)=0
    YR(3,3)            % u31(1)=0
    YR(5,3)            % T30(1)=0
    YR(7,3)            % T31(1)=0
    YR(9,3)            % Phi30(1)=0
    YR(11,3)           % Phi31(1)=0
    ];
end 

Torsten 2023-2-15

在 MATLAB Online 中打开

x = 1 does not exist in the above expression. You wanted to restrict y to the first region, and this region ends at x = 0.4.

Use "deval" to evaluate the solution at certain x-values:

x = 1;
y = deval(sol,x)

Komal Goyal 2023-2-15

Respected @Torsten sir, I dont know, whether I am able to tell you propely or not what is my doubt. Here I attached my doubt that in which region I want to find values and code which I prepare because of your help. Kindly help me in that as well.

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BVP4c for three boundary set of boundary conditions

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BVP4c for three boundary set of boundary conditions

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