where are the errors in my code? please help me!!

Urgent Help ,please!

I would like to Solve an equation for a variable (y) while setting another variable (r) across a range and then substituting the solution into the upper and lower limits of an integral (q). Eventually, plotting the relationship between the intgeral (q) and the variable (r) with the range.

I know that there are multiple errors! but I can not figure out where are they!!

here is the code:

r=linspace(0,0.09);
syms y(r) 
eqn= (cosh(10^10.*y./0.5)).^2 == 5./16.*r;
V1 = double(vpasolve(eqn,[y,r],[0 2]));
V2 = double(vpasolve(eqn,[y,r],[-2 0]));
fun=@(x)0.0018./((5./((cosh(10^10.*x./0.5)).^2)-r.*16).^0.5);
q = integral(fun,V2,V1)
plot(r,q)

Answer 1

[Torsten]

syms r z
eqn = ((z+1/z)/2)^2 == 5/16*r;
S = solve(eqn,z);
y = log(S)*0.5/10^10

solves the equation

(cosh(10^10.*y./0.5)).^2 == 5./16.*r

for y.

Deduce the conditions on r that V1 and V2 in the subsequent integration are real-valued.

[Abdallah Qaswal]

Hi,

Thank you ver much for response!

Here what I wrote:

syms r z
eqn = ((z+1/z)/2)^2 == 5/16*r;
S =solve(eqn,z);
y = log(S)*0.5/10^10

y = 

V1 = double(vpasolve(y,r,[0 0.09]));

Error using mupadengine/feval_internal
More equations than variables is only supported for polynomial systems.

Error in sym/vpasolve (line 172)
sol = eng.feval_internal('symobj::vpasolve',eqns,vars,X0);

V2 = double(vpasolve(y,r,[0 0.09]));
fun=@(x)0.0018./((5./((cosh(10^10.*x./0.5)).^2)-r.*16).^0.5);
q = integral(fun,V2,V1)
plot(r,q)

Maybe, I do not understand what you meant. could you please write the full code? becuase my code still gives me error!

[Torsten]

And did you find out which of the four solutions for y give a positive real number under the log(...) and what condition on r is necessary for that this is the case ?

I mean the expression sqrt(5*r-16) gives one necessary condition that your subsequent integration is senseful, namely r >= 16/5.

Thus

V2 = log(sqrt(5*r)/4 - sqrt(5*r-16)/4)/2e10
V1 = log(sqrt(5*r)/4 + sqrt(5*r-16)/4)/2e10

under the assumption

r >= 16/5

[Abdallah Qaswal]

My problem is not with solving the equation but with how to insert the solution into the integral across the range of values of r. this is the error messege that I got:

Error using mupadengine/feval (line 187)

Symbolic parameters not supported in nonpolynomial equations.

Error in sym/vpasolve (line 172)

sol = eng.feval('symobj::vpasolve',eqns,vars,X0);

Error in edfdgf (line 4)

V1 = double(vpasolve(eqn,y,[0 2]));

[Torsten]

Your problem is that you cannot integrate for the r-values you prescribe.

As said, r must be >= 16/5. You prescribe r as linspace(0:0.09) which is obviously < 16/5.

Or you must integrate over a complex line path from V2 to V1.

[Abdallah Qaswal]

there should be something wrong, r in my physical case can not be more than 16/5. It should be within the range I put!!!

[Torsten]

Well, MATLAB told you what V1 and V2 are depending on r - and these expressions are complex numbers if r < 16/5.

And that's logical since cosh(x) >= 1 for real x, thus the right-hand side of your equation

(cosh(10^10.*y./0.5)).^2 == 5./16.*r;

must be >=1, thus

5/16*r >= 1

thus

r >= 16/5

[Walter Roberson]

y = log(S)*0.5/10^10

If that is still how you are defining y then y is a vector of symbolic values. You cannot solve() something with respect to y when y is not a symbolic variable.

subs(eqn, r, y)

[Torsten]

y = log(S)*0.5/10^10

is the symbolic vector-solution of the equation

eqn= (cosh(10^10.*y./0.5)).^2 == 5./16.*r;

got over the deviation to solve

eqn = ((z+1/z)/2)^2 == 5/16*r;
S = solve(eqn,z);
y = log(S)*0.5/10^10

first for z.

Or can MATLAB directly

syms r y
eqn = (cosh(10^10.*y./0.5)).^2 == 5./16.*r;
S = solve(eqn,y)

?

At least, both results should match.

[Abdallah Qaswal]

ok, there is a minor mistake in my code, it is 5./(16.*r) not 5./16.*r! but still there is an error here in the following code:

syms r y

r=linspace(0,0.09);

eqn = (cosh(10^10.*y./0.5)).^2 == 5./(16.*r);

S1 = solve(eqn,y,[-5 0])

S2 = solve(eqn,y,[0 5])

fun=@(x)0.0018./((5./((cosh(10^10.*x./0.5)).^2)-16.*r).^0.5);

q = integral(@(x)fun(x),S2,S1)

plot(r,q)

[Abdallah Qaswal]

this is the error messege

Error using sym.getEqnsVars>checkVariables (line 92)

Second argument must be a vector of symbolic variables.

Error in sym.getEqnsVars (line 54)

checkVariables(vars);

Error in solve>getEqns (line 429)

[eqns, vars] = sym.getEqnsVars(argv{:});

Error in solve (line 226)

[eqns,vars,options] = getEqns(varargin{:});

Error in dsfdgfvbndresdfcv (line 4)

S1 = solve(eqn,y,[-5 0])

[Torsten]

I already told you that this will not work.

This is code you can use:

% Prescribe solutions V2, V1 for (cosh(10^10.*y./0.5)).^2 == 5./16.*r
V2 = @(r) log(sqrt(5*r)/4 - sqrt(5*r-16)/4)/2e10
V1 = @(r) log(sqrt(5*r)/4 + sqrt(5*r-16)/4)/2e10
% Define function to be integrated
fun = @(x,r)0.0018./((5./((cosh(10^10.*x./0.5)).^2)-r.*16).^0.5);
% Define integral
q = @(r) integral(@(x)fun(x,r),V2(r),V1(r));
% Evaluate integral
r = 16/5:1/5:20;
Q = arrayfun(q,r)
% Plot result
plot(r,[real(Q);imag(Q)])

[Abdallah Qaswal]

May I attach an image for the the problem written by hand so that you can visualize it better?

[Torsten]

Is there something incorrect with the description in your question ?

You want to get the two real-valued solutions V1 and V2 with V1 > V2 of the equation

(cosh(10^10.*y./0.5)).^2 == 5./16.*r;

and use these solutions as lower and upper limit to evaluate the integral of the function

fun=@(x)0.0018./((5./((cosh(10^10.*x./0.5)).^2)-r.*16).^0.5);

for different values of r.

If this is correct, then the code I posted solves this problem.

[Abdallah Qaswal]

Thank you! I think this will work!

It is much appreciated!

[Abdallah Qaswal]

Thank you! It works !

Again! It is much appreciated!

[Abdallah Qaswal]

[Torsten]

If you choose

r = linspace(0,0.09)

you must know that V1 and V2 are complex numbers and that you integrate over a line in the complex domain, not over an interval in the reals.

But in case this doesn't matter for you, it's now also questionable which two of the four solutions of the equation

(cosh(10^10.*y./0.5)).^2 == 5./16.*r;

to take as lower and upper limits of the integral.

I assumed taking the real solutions under the assumption that r >= 16/5.

[Abdallah Qaswal]

you may not notice my edit; it is not 5./16.*r rather it is 5./(16.*r), hence I got four new solutions and I picked the real ones:

y =

log(-(5^(1/2) + (5 - 16*r)^(1/2))/(4*r^(1/2)))/20000000000

log((5^(1/2) + (5 - 16*r)^(1/2))/(4*r^(1/2)))/20000000000

log(-(5^(1/2) - (5 - 16*r)^(1/2))/(4*r^(1/2)))/20000000000

log((5^(1/2) - (5 - 16*r)^(1/2))/(4*r^(1/2)))/20000000000

I picked the second and the fourth ones

[Abdallah Qaswal]

Accordingly, the range (0 ,0.09) results in real plotting

[Torsten]

Yes, the little things ...

Answer 2

[Walter Roberson]

Your first solve() returns multiple roots.

Your vpasolve() is then trying to process all of the roots at the same time .

Remember that solve() and vpasolve() are trying to solve simultaneous equations. If you pass in four equations in one variable, it will not attempt to solve each of the equations independently: it will try to find a single value of the variable that satisfies all of the equations at the same time. (Imagine, for example,that you had a series of trig expressions in a single variable, then you might be trying to find the right period that solves all of the equations at once.)

If you want to solve each equation independently, then use arrayfun() to run solve.

arrayfun(@(Y) vpasolve(Y, r, [0 0.09]), Y, 'uniform', 0)

The 'uniform', 0 is there because vpasolve() might decide there are no solutions.