How to fix it ? Warning: Do not specify equations and variables as character strings. Instead, create symbolic variables with syms. > In solve>getEqns (line 455) In solve (line 227)

my code

S=dsolve('Dx1=x2,Dx2=-lam2,Dlam1=0,Dlam2=-lam1,x1(0)=1,x2(0)=2,x1(tf)=3,lam2(tf)=0')

t='tf';

eq1=subs(S.x1)-'x1tf';

eq2=subs(S.x2)-'x2tf';

eq3=S.lam1-'lam1tf';

eq4=subs(S.lam2)-'lam2tf';

eq5='lam1tf*x2tf-0.5*lam2tf^2';

S2=solve(eq1,eq2,eq3,eq4,eq5,'tf,x1tf,x2tf,lam1tf,lam2tf','lam1tf<>0')

if i run, help pls

[Johny]

how can i fix this

sdot= @(t,s) ....

[s(2);

[2*M*s(2)*s(4)*cos(s(3))*sin(s(3))+M*g*cos(s(3))*sin(s(3))-s(4)*L*cos(s(3))+k*s(1)]/(M*sin^2(s(3))+m);

s(4);

[-2*M*s(2)*s(4)*sin(s(3))[cos^2(s(3))-sin^2(s(3))]-sin(s(3))[M*g*cos^2(s(3))-2*m*s(2)*s(4)-g*m-(g+(M*sin^2(s(3)))]+s(4)*L*cos^2(s(3))-k*cos(s(3))]/(L*m+L*M*sin^2(s(3)))];

[Walter Roberson]

@Johny

What about it? That code cannot generate the error message being discussed -- not in itself, as it just defines an array.

You need to fix some portions of the code.

sin(s(3))[M*g..etc]

That is invalid.

In MATLAB, the [] operator is used only for array (or vector) building. It can never be used to index a result -- if you want M*g..etc to be used to index the resut of sin(s(3)) then you would need to define a helper function, like

IndexAt = @(EXPRESSION, INDEX) EXPRESSION(INDEX)
sdot= @(t,s) ....
    [s(2);
     [2*M*s(2)*s(4)*cos(s(3))*sin(s(3))+M*g*cos(s(3))*sin(s(3))-s(4)*L*cos(s(3))+k*s(1)]/(M*sin^2(s(3))+m);
     s(4);
    [-2*M*s(2)*s(4)*sin(s(3))[cos^2(s(3))-sin^2(s(3))]-INDEXAT(sin(s(3)),M*g*cos^2(s(3))-2*m*s(2)*s(4)-g*m-(g+(M*sin^2(s(3))))+s(4)*L*cos^2(s(3))-k*cos(s(3))]/(L*m+L*M*sin^2(s(3)))];

But it seems very unlikely that the expression given will happen to work out as exactly an integer.

    cos^2(s(3))

as far as MATLAB is concerned, that is equivalent to

    INDEXAT(cos()^2,s(3))

which would try to invoke cos with no parameters, do a matrix squaring of the result, and index that result as s(3)

If you want to square the cosine then you need to square the result of the cosine call, such as

cos(s(3)).^2

It is recommended that [] not be used to indicate prioritization. You can certainly code things like 5*[x+9,y+7] which would in a sense give priority to calculating x+9 and y+7 before doing the multiplication. But if you had 5*[x+9] then the recommendation would be to instead write 5*(x+9) . 5*[x+9] is not invalid code, but it is not as efficient as it could be, as it is prod(5, horzcat(plus(x,9))) with the matrix-building call to horzcat() present, which is not as efficient as 5*(x+9) would be prod(5, plus(x,9)) with no extra call.

Using [] for prioritization risks unexpect results due to some subtle details of what spacing means inside [] and {} operators. [] is only recommended if you are building lists or arrays.

Answer 1

[Steven Lord]

As per the warning at the top of the documentation page for the dsolve function: "Support for character vector or string inputs will be removed in a future release. Instead, use syms to declare variables and replace inputs such as dsolve('Dy = -3*y') with syms y(t); dsolve(diff(y,t) == -3*y)."

[Muhammad Giovani]

can you fix in my command, sorry i very bad for english, im not understand sir :(

[Muhammad Giovani]

syms x(t); S=dsolve('Dx1=x2,Dx2=-lam2,Dlam1=0,Dlam2=-lam1,x1(0)=1,x2(0)=2,x1(tf)=3,lam2(tf)=0')

t='tf';

eq1=subs(S.x1)-'x1tf';

eq2=subs(S.x2)-'x2tf';

eq3=S.lam1-'lam1tf';

eq4=subs(S.lam2)-'lam2tf';

eq5='lam1tf*x2tf-0.5*lam2tf^2';

S2=solve(eq1,eq2,eq3,eq4,eq5,'tf,x1tf,x2tf,lam1tf,lam2tf','lam1tf<>0')

[Steven Lord]

This part:

syms x(t); S=dsolve('Dx1=x2')

would become something like this (untested):

syms t x1(t) x2(t);
S=dsolve(diff(x1(t), t) == x2(t))

S = 

[Muhammad Giovani]

i think in dsolve not problem, i think on command solve ?

[Muhammad Giovani]

thx for help mee, can you help me again huhu :(

[Muhammad Giovani]

i give a question

wrote a code

% Example 2.14
% Langkah pertama dengan menggunakan perintah dsolve
S=dsolve('Dx1=x2,Dx2=-lam2,Dlam1=0,Dlam2=-lam1,x1(0)=1,x2(0)=2,x1(tf)=3,lam2(tf)=0')
t='tf';
eq1=subs(S.x1)-'x1tf';
eq2=subs(S.x2)-'x2tf';
eq3=S.lam1-'lam1tf';
eq4=subs(S.lam2)-'lam2tf';
eq5='lam1tf*x2tf-0.5*lam2tf^2';
syms S2
S2=solve(eq1,eq2,eq3,eq4,eq5,'tf,x1tf,x2tf,lam1tf,lam2tf','lam1tf<>0')

%% lam1tf<>0 means lam1tf is not equal to 0;
%% This is a condition derived from eq5.
%% Otherwise, without this condition in the above
%% SOLVE routine, we get two values for tf (1 and 3 in this case) 

tf=S2.tf
x1tf=S2.x1tf;
x2tf=S2.x2tf;
clear t
x1=subs(S.x1)
x2=subs(S.x2)
lam1=subs(S.lam1)
lam2=subs(S.lam2)

%% Convert the symbolic values to
%% numerical values as shown below.
j=1;
tf=double(subs(S2.tf))
%% coverts tf from symbolic to numerical 
for tp=0:0.05:tf
t=sym(tp);
%% coverts tp from numerical to symbolic 
x1p(j)=double(subs(S.x1));
%% subs substitutes S.x1 to x1p.
x2p(j)=double(subs(S.x2));
%% double converts symbolic to numeric 
up(j)=-double(subs(S.lam2));
%% optimal control u = -lambda_2 
t1(j)=tp;
j=j+1 ;
end
plot(t1,x1p,'k',t1,x2p,'k',t1,up,'k:')
xlabel('t')
gtext ('x_1(t)1')
gtext ('x_2(t) ')
gtext ('u(t)')

[Walter Roberson]

@Muhammad Giovani

Unfortunately, dsolve() cannot work with more than one initial value per function, so you have to solve without the extra initial conditions, and then make the rest of the solution fit the initial condition.

% Example 2.14
% Langkah pertama dengan menggunakan perintah dsolve
%S=dsolve('Dx1=x2,Dx2=-lam2,Dlam1=0,Dlam2=-lam1,x1(0)=1,x2(0)=2,x1(tf)=3,lam2(tf)=0')

syms x_1(t) x_2(t) lam_1(t) lam_2(t) tf
Dx1 = diff(x_1);
Dx2 = diff(x_2);
Dlam1 = diff(lam_1);
Dlam2 = diff(lam_2);
eqn1 = Dx1 == x_2;
eqn2 = Dx2 == -lam_2;
eqn3 = Dlam1 == 0;
eqn4 = Dlam2 == -lam_1;
ic1 = x_1(0)==1;
ic2 = x_2(0)==2;
ic3 = x_1(tf)==3;
ic4 = lam_2(tf)==0;
sol_someic = dsolve([eqn1, eqn2, eqn3, eqn4, ic1, ic2, ic4])

sol_someic = struct with fields:
      x_2: C1*t - (C1*t^2)/(2*tf) + 2
      x_1: - (C1*t^3)/(6*tf) + (C1*t^2)/2 + 2*t + 1
    lam_1: -C1/tf
    lam_2: (C1*t)/tf - C1

constant_var = setdiff(symvar(sol_someic.x_1), [t, tf])

constant_var = 

constant_value = solve(subs(sol_someic.x_1, t, tf) == 3, constant_var)

constant_value = 

sol = structfun(@simplify, subs(sol_someic, constant_var, constant_value), 'uniform', 0)

sol = struct with fields:
      x_2: (3*t^2*(2*tf - 2))/(2*tf^3) - (3*t*(2*tf - 2))/tf^2 + 2
      x_1: 2*t - (3*t^2*(2*tf - 2))/(2*tf^2) + (t^3*(2*tf - 2))/(2*tf^3) + 1
    lam_1: (3*(2*tf - 2))/tf^3
    lam_2: -(6*(t - tf)*(tf - 1))/tf^3

and then you would proceed from there.