Solving an equation in Matlab

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Pablo Estuardo 2020-2-12

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/505129-solving-an-equation-in-matlab

评论： Star Strider 2020-2-13

Dear, I'm trying to resolve this equation for my thesis project, u and v are the values that I try to calculate, the only parameter that change is the Tx and Ty, all the other parameters are constant. I appreciate any help

Screen Shot 2020-02-12 at 1.08.04 PM.png

tx = [0.1 0.5 0.12 0.14 0.22 0.1 0.12 0.29]
ty = [0.4 0.12 0.34 0.14 0.13 0.03 0.2 0.13]
Omega = 7.3E-5; 
latitude = -12; 
f0 = 2*Omega*sind(latitude);
r = 1./(2.*24.*3600)
Hm = 25
rho = 1.025e3
dvdt0 = -r.*v0 - f0.*u0 -ty./(rho.*Hm);
dudt0 = -r.*u0 + f0.*v0 + tx./(rho.*Hm);

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Answer 1

Star Strider 2020-2-12

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https://ww2.mathworks.cn/matlabcentral/answers/505129-solving-an-equation-in-matlab#answer_415268

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These have analytic solutions:

syms  u(t) v(t) f taux tauy Hm rho R u0 v0 
du = diff(u);
dv = diff(v);
DE = [du - f*v == taux/(Hm*rho) - R*u;  dv - f*u == taux/(Hm*rho) - R*v]
Soln = dsolve(DE, u(0)==u0, v(0)==v0)
u_Soln = Soln.u;
v_Soln = Soln.v;
u_Soln = simplify(u_Soln, 'Steps',500)
v_Soln = simplify(v_Soln, 'Steps',500)
u_fcn = matlabFunction(u_Soln)
v_fcn = matlabFunction(v_Soln)

Note that ‘u_fcn’ and ‘v_fcn’ are anonymous functions you can use outside of the Symbolic Math Toolbox.

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Pablo Estuardo 2020-2-12

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Dear Star Stride, thanks you so much for your help and time, I'm trying to understand the error in the line Soln = dsolve(DE, u(0)==u0, v(0)==v0), I will apreciate any help.

Error using mupadengine/feval (line 163)
The equations are invalid.
Error in dsolve>mupadDsolve (line 336)
T = feval(symengine,'symobj::dsolve',sys,x,options);
Error in dsolve (line 193)
sol = mupadDsolve(args, options);

clear all
close all
taux = [0.1 0.5 0.12 0.14 0.22 0.1 0.12 0.29]
tauy = [0.4 0.12 0.34 0.14 0.13 0.03 0.2 0.13]
Omega = 7.3E-5; 
latitude = -12; 
f = 2*Omega*sind(latitude);
R = 1./(2.*24.*3600)
Hm = 25
rho = 1.025e3
%----------------------------
syms  u(t) v(t) f taux tauy Hm rho R u0 v0 
du = diff(u);
dv = diff(v);
DE = [du - f*v == taux/(Hm*rho) - R*u;  dv + f*u == tauy/(Hm*rho) - R*v]
% dvdt0 = -r.*v0 - f.*u0 -taux./(rho.*H);
% dudt0 = -r.*u0 + f.*v0 +tauy./(rho.*H);
% initial conditions for analytic solution
% u0 = 0;
% v0 = 0;
Soln = dsolve(DE, u(0)==u0, v(0)==v0)
Error using mupadengine/feval (line 163)
The equations are invalid.
Error in dsolve>mupadDsolve (line 336)
T = feval(symengine,'symobj::dsolve',sys,x,options);
Error in dsolve (line 193)
sol = mupadDsolve(args, options);
u_Soln = Soln.u;
v_Soln = Soln.v;
u_Soln = simplify(u_Soln, 'Steps',500)
v_Soln = simplify(v_Soln, 'Steps',500)
u_fcn = matlabFunction(u_Soln)
v_fcn = matlabFunction(v_Soln)

Star Strider 2020-2-12

在 MATLAB Online 中打开

I have no problems with it in R2019b.

This code:

syms  u(t) v(t) f taux tauy Hm rho R u0 v0 
du = diff(u);
dv = diff(v);
DE = [du - f*v == taux/(Hm*rho) - R*u;  dv - f*u == taux/(Hm*rho) - R*v]
Soln = dsolve(DE, u(0)==u0, v(0)==v0)
u_Soln = Soln.u;
v_Soln = Soln.v;
u_Soln = simplify(u_Soln, 'Steps',500)
v_Soln = simplify(v_Soln, 'Steps',500)
u_fcn = matlabFunction(u_Soln)
v_fcn = matlabFunction(v_Soln)

produces these results:

u_Soln =
exp(-t*(R + f))*(u0/2 - v0/2) - (exp(-t*(R - f))*(2*taux - 2*taux*exp(t*(R - f)) - Hm*R*rho*u0 - Hm*R*rho*v0 + Hm*f*rho*u0 + Hm*f*rho*v0))/(2*Hm*rho*(R - f))
 
v_Soln =
- exp(-t*(R + f))*(u0/2 - v0/2) - (exp(-t*(R - f))*(2*taux - 2*taux*exp(t*(R - f)) - Hm*R*rho*u0 - Hm*R*rho*v0 + Hm*f*rho*u0 + Hm*f*rho*v0))/(2*Hm*rho*(R - f))
 
u_fcn = @(Hm,R,f,rho,t,taux,u0,v0)exp(-t.*(R+f)).*(u0./2.0-v0./2.0)-(exp(-t.*(R-f)).*(taux.*2.0-taux.*exp(t.*(R-f)).*2.0-Hm.*R.*rho.*u0-Hm.*R.*rho.*v0+Hm.*f.*rho.*u0+Hm.*f.*rho.*v0))./(Hm.*rho.*(R-f).*2.0)
v_fcn = @(Hm,R,f,rho,t,taux,u0,v0)-exp(-t.*(R+f)).*(u0./2.0-v0./2.0)-(exp(-t.*(R-f)).*(taux.*2.0-taux.*exp(t.*(R-f)).*2.0-Hm.*R.*rho.*u0-Hm.*R.*rho.*v0+Hm.*f.*rho.*u0+Hm.*f.*rho.*v0))./(Hm.*rho.*(R-f).*2.0)

I edited the output slightly to make it readable.

Star Strider 2020-2-13

在 MATLAB Online 中打开

Do not use the ‘taux’ and ‘tauy’ data (or any of the others) when calculating the solutions to the differential equations. Leave them until the functions are evaluated.

Try this (comment-out the symbolic derivations and resulting functions first):

u_fcn = @(Hm,R,f,rho,t,taux,tauy,u0,v0)exp(-t.*(R-f)).*(u0./2.0+v0./2.0-(taux+tauy)./(Hm.*rho.*(R-f).*2.0)+(exp(t.*(R-f)).*(taux+tauy))./(Hm.*rho.*(R-f).*2.0))+exp(-t.*(R+f)).*(u0./2.0-v0./2.0-(taux-tauy)./(Hm.*rho.*(R+f).*2.0)+(exp(t.*(R+f)).*(taux-tauy))./(Hm.*rho.*(R+f).*2.0));
v_fcn = @(Hm,R,f,rho,t,taux,tauy,u0,v0)exp(-t.*(R-f)).*(u0./2.0+v0./2.0-(taux+tauy)./(Hm.*rho.*(R-f).*2.0)+(exp(R.*t-f.*t).*(taux+tauy))./(Hm.*rho.*(R-f).*2.0))-exp(-t.*(R+f)).*(u0./2.0-v0./2.0-(taux-tauy)./(Hm.*rho.*(R+f).*2.0)+(exp(R.*t+f.*t).*(taux-tauy))./(Hm.*rho.*(R+f).*2.0));
taux = [0.1 0.5 0.12 0.14 0.22 0.1 0.12 0.29]
tauy = [0.4 0.12 0.34 0.14 0.13 0.03 0.2 0.13]
Omega = 7.3E-5; 
latitude = -12; 
f = 2*Omega*sind(latitude);
R = 1./(2.*24.*3600);
Hm = 25;
rho = 1.025e3;
u0 = 0;                                                 % Substitute Correct Initial Condition
v0 = 0;                                                 % Substitute Correct Initial Condition
[Taux,Tauy] = ndgrid(taux, tauy);                       % Create Matrices From The Vectors
tv = linspace(0, 100, 3);
u_fcnt = @(t) u_fcn(Hm,R,f,rho,t,Taux,Tauy,u0,v0);      % Function Only Of ‘t’ Here

v_fcnt = @(t) v_fcn(Hm,R,f,rho,t,Taux,Tauy,u0,v0);      % Function Only Of ‘t’ Here
u = u_fcnt(1);
v = v_fcnt(1);
figure
hold on
for k = 1:numel(tv)
    surf(taux, tauy, u_fcnt(tv(k)))
end
hold off
grid on
view(30, 30)
xlabel('\tau_x')
ylabel('\tau_y')
zlabel('u')
figure
hold on
for k = 1:numel(tv)
    surf(taux, tauy, v_fcnt(tv(k)))
end
hold off
grid on
view(30, 30)
xlabel('\tau_x')
ylabel('\tau_y')
zlabel('\nu')

Also, there was an error. The correct expression for ‘DE’ is:

DE = [du - f*v == taux/(Hm*rho) - R*u; dv - f*u == tauy/(Hm*rho) - R*v]

I already corrected for that in the code for ‘u_fcn’ and ‘v_fcn’ and the rest in the code in this Comment.

Note: The plots appear to be a bit strange, because ‘taux’ and ‘tauy’ have repeated elements, and they are not sorted. You may not want (or need) to plot them, hoiwever this code demonstrates the correct way to evaluate them. They are functions of three variables

, so that must be accounted for in evaluating them.

Pablo Estuardo 2020-2-13

编辑：Pablo Estuardo 2020-2-13

在 MATLAB Online 中打开

Thanks Star, you helping me a lot, now im going to try to make the dsolve works, in that way I can add more terms to try new solutions, again, thanks you so much for your time... best regards.

syms  u(t) v(t) f taux tauy Hm rho R u0 v0 
du = diff(u);
dv = diff(v);
DE = [du - f*v == taux/(Hm*rho) - R*u;  dv + f*u == tauy/(Hm*rho) - R*v]
Soln = dsolve(DE, u(0)==u0, v(0)==v0)
u_Soln = Soln.u;
v_Soln = Soln.v;
u_Soln = simplify(u_Soln, 'Steps',500)
v_Soln = simplify(v_Soln, 'Steps',500)
u_fcn = matlabFunction(u_Soln)
v_fcn = matlabFunction(v_Soln)

Star Strider 2020-2-13

As always, my pleasure!

I will help as I can.

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Solving an equation in Matlab

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