Subscripted assignment dimension mismatch.

I am writing a code to solve for the variable t at various values of x in the expression bellow. k,h, and c are previously know constants. x in an input I have determined from elsewhere. diffD are the known values I am equating the expression to. The length of RHS, x2, and diffD are all 7

Ip = @(x,t) -(((5*k*t*x)*exp((c*h)/(k*t*x)))-(5*k*t*x))/((k*t*x^2)*(exp((c*h)/(k*t*x))-1));

syms t;

for i=1:length(x2)
    RHS(i)=Ip(x2(i),t);
end

for i=1:length(x2)
    eqn = RHS(i) == diffD(i);
     td(i) = solve(eqn,t);
end

Answer 1

[Walter Roberson]

Ip(x2(i),t) simplifies to -5/x2(i) which is independent of t. When you solve() for -5/x2(i)==RHS(i) for t, you get no solutions, which is length 0. You cannot write a solution of length 0 into something of length 1, td(i)

Remember, solve() is not constrained to return a single value: it can return 0, 1, or several values. Therefore unless you have proven that there is a unique solution (and that solve() is good enough to find it) you should never write the results of solve() directly into any location of fixed length.

[Ben Johannesson]

How did you simplify Ip? the solution should not be that simple

[Walter Roberson]

Look at your numerator:

-(5*k*t*x*exp(c*h/(k*t*x)) - 5*k*t*x)

clearly the 5*k*t*x factors for the terms, giving

-(5*k*t*x*( exp(c*h/(k*t*x)) - 1 ))

and the denominator has (k*t*x^2) which has k*t*x in it, so that cancels from the top and bottom leaving

-5*(exp(c*h/(k*t*x)) - 1) / (x*(exp((c*h)/(k*t*x)) - 1))

and we easily see the exp() term cancels. This gives us a net result of -5/x

Perhaps you did not want the first set of brackets that distributes the leading negative across the entire numerator: perhaps you wanted

-5*k*t*x*exp(c*h/(k*t*x)) - 5*k*t*x

That would still have the k*t*x cancel from the numerator and denominator but the (exp(c*h/(k*t*x))-1) would not cancel because in the numerator it would effectively be -5*(exp(c*h/(k*t*x))+1) with +1 instead of -1. The result would be something that could be solved for t, giving

t  = c .* h ./ (k .* x2 .* ln((RHS .* x2 - 5) ./ (RHS .* x2 + 5)))

as the general solution in vectorized form (provided that RHS and x2 are the same size and shape)