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I need to sole this equation for L, and I have a set of values for d and p.

p = -2(d/L)^3 + 3(d/L)^2

Im wondering how to put this into Matlab. Any help would be apprecitated.

Dimitris Kalogiros
on 12 Nov 2019

Edited: Dimitris Kalogiros
on 13 Nov 2019

It is a 3rd order equation, it has only an analytical solution.

You can use the following code:

clc; clearvars;

syms p d L

p=1;

d=3;

eq= p == -2*(d/L)^3 + 3*(d/L)^2

% analytical solution

solve(eq,L)

%numerical solution

vpasolve(eq)

And in case you want to run this for a set of values , you can do the following:

clc; clearvars;

syms p d L

% values for p, d

pValues=50:10:90;

dValues=300:50:600;

% an empty buffer for the solutions, combined with p,d

SolBuffer=cell(length(pValues)*length(dValues),3);

% solution counter

k=0;

for n=1:length(pValues)

for m=1:length(dValues)

k=k+1;

p=pValues(n);

SolBuffer{k,1}=p;

d=dValues(m);

SolBuffer{k,2}=d;

eq= p == -2*(d/L)^3 + 3*(d/L)^2;

% analytical solution

solve(eq, L);

%numerical solution

SolBuffer{k,3}=vpasolve(eq).';

end

end

% An example: the third solution

disp('data p and d :')

disp(SolBuffer{3,1})

disp(SolBuffer{3,2})

disp('solutions:')

disp(SolBuffer{3,3})

Dimitris Kalogiros
on 13 Nov 2019

clc; clearvars;

syms p d L

% values for p, d

pValues=50:10:90;

dValues=300:50:600;

% an empty buffer for the solutions, combined with p,d

SolBuffer=nan(length(pValues)*length(dValues),3);

% solution counter

k=0;

for n=1:length(pValues)

for m=1:length(dValues)

k=k+1;

p=pValues(n);

d=dValues(m);

eq= p == -2*(d/L)^3 + 3*(d/L)^2;

% analytical solution

solve(eq, L);

%numerical solution

SolBuffer(k,1:3)=vpasolve(eq).';

end

end

% display solutions

SolBuffer

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John D'Errico
on 12 Nov 2019

You have many values for both d and p? Do you want to solve it for all combinations of d and p?

Anyway, first, I would recognize that d and L are alwways combined in the same form, thus, we have d/L always. So first, substitute

X = d/L

Of course, once we know the value of X, we can always recover L, as:

L = d/X

Now your problem reduces to

p = -2*X^3 + 3*X^2

You could use a loop over all values of p, getting three roots for each value of p. Before we do even that however, do a plot.

ALWAYS PLOT EVERYTHING.

fun = @(X) -2*X.^3 + 3*X.^2;

fplot(fun,[-1,2])

yline(0);

yline(1);

See that I used the .^ operator there to avoid problems.

Because this is a cubic polynomial in X, you can think of your problem as having 3 real roots for X, whenever p is betweeen 0 and 1. For p larger than 1 or smaller than 0, the problem will have one real root and two complex roots.

For example,

polycoef = @(p) [-2 3 0 -p];

roots(polycoef(0))

ans =

0

0

1.5

roots(polycoef(-.00001))

ans =

1.5 + 0i

-1.1111e-06 + 0.0018257i

-1.1111e-06 - 0.0018257i

So if p is just slightly less than zero as I predicted, we see a real root, but then always two complex conjugate complex roots. The same thing happens at p==1, where 3 real roots turn into a real and two complex roots as we go above p=1.

For p between 0 and 1 however, three real roots for X.

roots(polycoef(0.5))

ans =

1.366

0.5

-0.36603

Again, if you then know the value of d, we can easily recover the value of L, as

L = d./X

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