Please help me to run this simple code

조회 수: 53 (최근 30일)
T K
T K 2026년 2월 13일 18:18
댓글: dpb 2026년 2월 13일 22:28
proj()
rr = 1×3
0 0.3000 0.6000
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Array indices must be positive integers or logical values.

Error in solution>proj/projfun (line 48)
dy(2)=(1/((((s^2+a1*h^2-(2*x+2+1).^2)*(1-a*t))*((h^2+a1*s^2-(2*(x+1+1)).^2)*(1-a*t)))-((a2*s*h)*(1-a*t))^2))*(((a2*s*h*a*2*(x+1)+2*(x+1)*a*a2*s*h)*t-a2*s*h*a*2*(x+1)*t^2-a2*s*h*2*(x+1)+(a*a2*s*h*a(s^2+a1*h^2)*t-a2*s*h*a(s^2+a1*h^2))*dt)*du+((s^2+a1*h^2-2*(x+1))*a*(h^2+a1*s^2)-(a*2*(x+1)*a*(a4*s*h+a1*s*h))+((a*a2*s*h)*a*(a4*s*h+a1*s*h)-(a*(s^2+a1*h^2-2*(x+1)))*(a*(h^2+a1*s^2))))*dt*dv+(a2*s*h*2*a*a3*s+a3*s*a*a2*s*h+2*a*a3*h*(s^2+a1*h^2-4*(x+1+1)^2)+a*(s^2+a1*h^2-(2*(x+1+1))^2)*(a3*h))*t*dt-(a2*s*h*a3*s)*dt-((s^2+a1*h^2-(2*(x+1))^2)*a3*h)*dt-dt*t^2*(a*a2*s*h*2*a*a3*s+a*(s^2+a1*h^2-(2*(x+1))^2)*2*a*a3*h));
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Error in bvparguments (line 96)
testODE = ode(x1,y1,odeExtras{:});
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Error in bvp4c (line 119)
bvparguments(solver_name,ode,bc,solinit,options,varargin);
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Error in solution>proj (line 19)
sol= bvp4c(@projfun,@projbc,solinit,options);
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
function sol= proj
clc;clf;clear;
myLegend1 = {};
myLegend2 = {};
k0=386; ce=3.831*10^2; mu=38.6*10^9;alfat=1.78*10^-5; rho=89.54*10^2; lamda=77.6*10^9;taw=0.5;Tnot=2.93*10^2;
c0=sqrt((lamda+2*mu)/(rho)); Betanot=(3*lamda+2*mu)*alfat; a1=mu/(lamda+2*mu);a2=(mu+lamda)/(lamda+2*mu);a3=(Betanot*Tnot)/(lamda+2*mu);omega=(rho*ce)/(k0);
a4=lamda/(lamda+2*mu);a5=(k0*omega*c0^2)/(k0);a6=(rho*ce*c0^2)/(k0);
a7=(Betanot*c0^2)/(k0); a8=a6*taw; a9=a7*taw; a10=rho*ce*taw*omega*c0^4/(k0); a11=Betanot*taw*omega*c0^4/(k0);w=rho*ce/(k0);
rr = [0 0.3 0.6 ]
for i =1:numel(rr)
a= rr(i);
s=0.5;h=0.1;
y0 = [1,0,0,0,1,0,0];
options =bvpset('stats','on','RelTol',1e-4);
m = linspace(0,10);
solinit = bvpinit(m,y0);
sol= bvp4c(@projfun,@projbc,solinit,options);
figure(1)
plot(sol.x,(sol.y(5,:)))
grid on,hold on
myLegend1{i}=['alfa= ',num2str(rr(i))];
figure(2)
plot(sol.x,(sol.y(6,:)))
title('Temperature')
grid on,hold on
myLegend2{i}=['alfa= ',num2str(rr(i))];
i=i+1;
end
figure(1)
legend(myLegend1)
hold on
figure(2)
legend(myLegend2)
hold on
function dy= projfun(x,y)
dy= zeros(7,1);
v = y(1);
dv = y(2);
u=y(3);
du=y(4);
t=y(5);
dt=y(6);
ddt=y(7);
dy(1) = dv;
dy(2)=(1/((((s^2+a1*h^2-(2*x+2+1).^2)*(1-a*t))*((h^2+a1*s^2-(2*(x+1+1)).^2)*(1-a*t)))-((a2*s*h)*(1-a*t))^2))*(((a2*s*h*a*2*(x+1)+2*(x+1)*a*a2*s*h)*t-a2*s*h*a*2*(x+1)*t^2-a2*s*h*2*(x+1)+(a*a2*s*h*a(s^2+a1*h^2)*t-a2*s*h*a(s^2+a1*h^2))*dt)*du+((s^2+a1*h^2-2*(x+1))*a*(h^2+a1*s^2)-(a*2*(x+1)*a*(a4*s*h+a1*s*h))+((a*a2*s*h)*a*(a4*s*h+a1*s*h)-(a*(s^2+a1*h^2-2*(x+1)))*(a*(h^2+a1*s^2))))*dt*dv+(a2*s*h*2*a*a3*s+a3*s*a*a2*s*h+2*a*a3*h*(s^2+a1*h^2-4*(x+1+1)^2)+a*(s^2+a1*h^2-(2*(x+1+1))^2)*(a3*h))*t*dt-(a2*s*h*a3*s)*dt-((s^2+a1*h^2-(2*(x+1))^2)*a3*h)*dt-dt*t^2*(a*a2*s*h*2*a*a3*s+a*(s^2+a1*h^2-(2*(x+1))^2)*2*a*a3*h));
dy(3)=du;
dy(4) =(1/(a*(s^2+a1*h^2-(2*x+2+1).^2))-(s^2+a1*h^2-(2*x+2+1).^2))*((a2*s*h-a*a2*s*h*t)*((1/((((s^2+a1*h^2-(2*x+2+1).^2)*(1-a*t))*((h^2+a1*s^2-(2*(x+1+1)).^2)*(1-a*t)))-((a2*s*h)*(1-a*t))^2))*(((a2*s*h*a*2*(x+1)+2*(x+1)*a*a2*s*h)*t-a2*s*h*a*2*(x+1)*t^2-a2*s*h*2*(x+1)+(a*a2*s*h*a(s^2+a1*h^2)*t-a2*s*h*a(s^2+a1*h^2))*dt)*du+((s^2+a1*h^2-2*(x+1))*a*(h^2+a1*s^2)-(a*2*(x+1)*a*(a4*s*h+a1*s*h))+((a*a2*s*h)*a*(a4*s*h+a1*s*h)-(a*(s^2+a1*h^2-2*(x+1)))*(a*(h^2+a1*s^2))))*dt*dv+(a2*s*h*2*a*a3*s+a3*s*a*a2*s*h+2*a*a3*h*(s^2+a1*h^2-4*(x+1+1)^2)+a*(s^2+a1*h^2-(2*(x+1+1))^2)*(a3*h))*t*dt-(a2*s*h*a3*s)*dt-((s^2+a1*h^2-(2*(x+1))^2)*a3*h)*dt-dt*t^2*(a*a2*s*h*2*a*a3*s+a*(s^2+a1*h^2-(2*(x+1))^2)*2*a*a3*h)))+(((a*2*(x+1)*t)-2*(x+1)-(a*s^2+a*a1*h^2)*t)*du)+(2*a*a3*s*t-a3*s)*dt-(a*a4*s*h+a*a1*s*h)*dt*dv);
dy(5)=dt;
dy(6)=ddt;
dy(7)=(1/(a*(a5*s^2*2*(x+1+1)+a5*h^2*2*(x+1+1))*t-(a5*s^2*2*(x+1+1)+a5*h^2*2*(x+1)-a8*(2*(x+1+1))^3)))*((s^2+h^2+2*a5*h*2-a6*(2*(x+1)).^2-3*a8*2*(x+1+1)*2*(x+1))*ddt);
end
end
function res= projbc(ya,yb)
res= [ya(1);
ya(3);
ya(5)-1;
ya(7);
yb(1);
yb(3);
yb(5);
];
end
  댓글 수: 3
dpb
dpb 2026년 2월 13일 20:30
편집: dpb 2026년 2월 13일 20:31
dy(2)=(1/((((s^2+a1*h^2-(2*x+2+1).^2)*(1-a*t))*((h^2+a1*s^2-(2*(x+1+1)).^2)*(1-a*t)))-((a2*s*h)*(1-a*t))^2))*(((a2*s*h*a*2*(x+1)+2*(x+1)*a*a2*s*h)*t-a2*s*h*a*2*(x+1)*t^2-a2*s*h*2*(x+1)+(a*a2*s*h*a(s^2+a1*h^2)*t-a2*s*h*a(s^2+a1*h^2))*dt)*du+((s^2+a1*h^2-2*(x+1))*a*(h^2+a1*s^2)-(a*2*(x+1)*a*(a4*s*h+a1*s*h))+((a*a2*s*h)*a*(a4*s*h+a1*s*h)-(a*(s^2+a1*h^2-2*(x+1)))*(a*(h^2+a1*s^2))))*dt*dv+(a2*s*h*2*a*a3*s+a3*s*a*a2*s*h+2*a*a3*h*(s^2+a1*h^2-4*(x+1+1)^2)+a*(s^2+a1*h^2-(2*(x+1+1))^2)*(a3*h))*t*dt-(a2*s*h*a3*s)*dt-((s^2+a1*h^2-(2*(x+1))^2)*a3*h)*dt-dt*t^2*(a*a2*s*h*2*a*a3*s+a*(s^2+a1*h^2-(2*(x+1))^2)*2*a*a3*h));
Format your code by wrapping long lines and appropriate spacing, etc., so it's possible to read...there's almost no chance of being able to parse those lines as written without missing something...making shorter, temporary variables wouldn't necessarily be a bad idea, either as one way to reduce the apparent complexity.
Stephen23
Stephen23 2026년 2월 13일 20:33
And get rid of cargo-cult code like this: clc;clf;clear;
What do you think clearing an empty function workspace will do ?

댓글을 달려면 로그인하십시오.

채택된 답변

Stephen23
Stephen23 2026년 2월 13일 20:39
You're missing the multiplication operator * in several places. Here's the corrected code with the missing * operators added:
proj()
The solution was obtained on a mesh of 100 points. The maximum residual is 4.453e-06. There were 1303 calls to the ODE function. There were 25 calls to the BC function. The solution was obtained on a mesh of 100 points. The maximum residual is 3.046e-06. There were 1502 calls to the ODE function. There were 26 calls to the BC function. The solution was obtained on a mesh of 100 points. The maximum residual is 5.754e-06. There were 1701 calls to the ODE function. There were 27 calls to the BC function.
ans = struct with fields:
solver: 'bvp4c' x: [0 0.1010 0.2020 0.3030 0.4040 0.5051 0.6061 0.7071 0.8081 0.9091 1.0101 1.1111 1.2121 1.3131 1.4141 1.5152 1.6162 1.7172 1.8182 1.9192 2.0202 2.1212 … ] (1×100 double) y: [7×100 double] yp: [7×100 double] stats: [1×1 struct]
function sol= proj
myLegend1 = {};
myLegend2 = {};
k0=386; ce=3.831*10^2; mu=38.6*10^9;alfat=1.78*10^-5; rho=89.54*10^2; lamda=77.6*10^9;taw=0.5;Tnot=2.93*10^2;
c0=sqrt((lamda+2*mu)/(rho)); Betanot=(3*lamda+2*mu)*alfat; a1=mu/(lamda+2*mu);a2=(mu+lamda)/(lamda+2*mu);a3=(Betanot*Tnot)/(lamda+2*mu);omega=(rho*ce)/(k0);
a4=lamda/(lamda+2*mu);a5=(k0*omega*c0^2)/(k0);a6=(rho*ce*c0^2)/(k0);
a7=(Betanot*c0^2)/(k0); a8=a6*taw; a9=a7*taw; a10=rho*ce*taw*omega*c0^4/(k0); a11=Betanot*taw*omega*c0^4/(k0);w=rho*ce/(k0);
rr = [0 0.3 0.6 ];
for i =1:numel(rr)
a= rr(i);
s=0.5;h=0.1;
y0 = [1,0,0,0,1,0,0];
options =bvpset('stats','on','RelTol',1e-4);
m = linspace(0,10);
solinit = bvpinit(m,y0);
sol= bvp4c(@projfun,@projbc,solinit,options);
figure(1)
plot(sol.x,(sol.y(5,:)))
grid on,hold on
myLegend1{i}=['alfa= ',num2str(rr(i))];
figure(2)
plot(sol.x,(sol.y(6,:)))
title('Temperature')
grid on,hold on
myLegend2{i}=['alfa= ',num2str(rr(i))];
end
figure(1)
legend(myLegend1)
hold on
figure(2)
legend(myLegend2)
hold on
function dy = projfun(x,y)
dy = zeros(7,1);
v = y(1);
dv = y(2);
u = y(3);
du = y(4);
t = y(5);
dt = y(6);
ddt = y(7);
dy(1) = dv;
dy(2)=(1/((((s^2+a1*h^2-(2*x+2+1).^2)*(1-a*t))*((h^2+a1*s^2-(2*(x+1+1)).^2)*(1-a*t)))-((a2*s*h)*(1-a*t))^2))*(((a2*s*h*a*2*(x+1)+2*(x+1)*a*a2*s*h)*t-a2*s*h*a*2*(x+1)*t^2-a2*s*h*2*(x+1)+(a*a2*s*h*a*(s^2+a1*h^2)*t-a2*s*h*a*(s^2+a1*h^2))*dt)*du+((s^2+a1*h^2-2*(x+1))*a*(h^2+a1*s^2)-(a*2*(x+1)*a*(a4*s*h+a1*s*h))+((a*a2*s*h)*a*(a4*s*h+a1*s*h)-(a*(s^2+a1*h^2-2*(x+1)))*(a*(h^2+a1*s^2))))*dt*dv+(a2*s*h*2*a*a3*s+a3*s*a*a2*s*h+2*a*a3*h*(s^2+a1*h^2-4*(x+1+1)^2)+a*(s^2+a1*h^2-(2*(x+1+1))^2)*(a3*h))*t*dt-(a2*s*h*a3*s)*dt-((s^2+a1*h^2-(2*(x+1))^2)*a3*h)*dt-dt*t^2*(a*a2*s*h*2*a*a3*s+a*(s^2+a1*h^2-(2*(x+1))^2)*2*a*a3*h));
dy(3)=du;
dy(4) =(1/(a*(s^2+a1*h^2-(2*x+2+1).^2)-(s^2+a1*h^2-(2*x+2+1).^2)))*((a2*s*h-a*a2*s*h*t)*((1/((((s^2+a1*h^2-(2*x+2+1).^2)*(1-a*t))*((h^2+a1*s^2-(2*(x+1+1)).^2)*(1-a*t)))-((a2*s*h)*(1-a*t))^2))*(((a2*s*h*a*2*(x+1)+2*(x+1)*a*a2*s*h)*t-a2*s*h*a*2*(x+1)*t^2-a2*s*h*2*(x+1)+(a*a2*s*h*a*(s^2+a1*h^2)*t-a2*s*h*a*(s^2+a1*h^2))*dt)*du+((s^2+a1*h^2-2*(x+1))*a*(h^2+a1*s^2)-(a*2*(x+1)*a*(a4*s*h+a1*s*h))+((a*a2*s*h)*a*(a4*s*h+a1*s*h)-(a*(s^2+a1*h^2-2*(x+1)))*(a*(h^2+a1*s^2))))*dt*dv+(a2*s*h*2*a*a3*s+a3*s*a*a2*s*h+2*a*a3*h*(s^2+a1*h^2-4*(x+1+1)^2)+a*(s^2+a1*h^2-(2*(x+1+1))^2)*(a3*h))*t*dt-(a2*s*h*a3*s)*dt-((s^2+a1*h^2-(2*(x+1))^2)*a3*h)*dt-dt*t^2*(a*a2*s*h*2*a*a3*s+a*(s^2+a1*h^2-(2*(x+1))^2)*2*a*a3*h)))+(((a*2*(x+1)*t)-2*(x+1)-(a*s^2+a*a1*h^2)*t)*du)+(2*a*a3*s*t-a3*s)*dt-(a*a4*s*h+a*a1*s*h)*dt*dv);
dy(5)=dt;
dy(6)=ddt;
dy(7)=(1/(a*(a5*s^2*2*(x+1+1)+a5*h^2*2*(x+1+1))*t-(a5*s^2*2*(x+1+1)+a5*h^2*2*(x+1)-a8*(2*(x+1+1))^3)))*((s^2+h^2+2*a5*h*2-a6*(2*(x+1)).^2-3*a8*2*(x+1+1)*2*(x+1))*ddt);
end
function res = projbc(ya,yb)
res = [ya(1);
ya(3);
ya(5)-1;
ya(7);
yb(1);
yb(3);
yb(5);
];
end
end
  댓글 수: 2
T K
T K 2026년 2월 13일 22:15
Thanks alot
dpb
dpb 2026년 2월 13일 22:28
@Stephen23 has far more patience (and better eyesight, undoubtedly) than I... <vbg>

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 String Parsing에 대해 자세히 알아보기

태그

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by