Differential Equation Solution incorrect with matlab compared to maple

Hello everyone, currently I am trying to solve a differential equation relating to a beam problem, the problem I am having is that is seems maple gives better results for the exact solution than matlab does, in fact, the solution i am getting from matlab is pretty unusable.

I have a feeling the issue is with my code, of course, but I cannot seem to find it.

Here is the differential equation:

EI = 10000*[2-(x/L)] , q = 40 * (x/L)^2, and L = 10

Here is the result of maple code

as well as it's plot

Here is the code I am running, as can be seen in wSol(x), it is imaginary, which is pretty contrary to what Maple provides.

syms x w(x)

L=10;
q0=40;
a0=10000;
M1=-1000;
Q1=-100;
q=q0*(x/L)^2;
EI=a0*(2-(x/L));

Dw  = diff(w,x) ;
D2w = diff(w,x,2);
D3w = diff(w,x,3);

deq = diff(EI*diff(w,x,2),x,2)==q;
cond1 = w(0)==0;
cond2 = Dw(0)==0;
cond3 = D2w(L)==M1/subs(EI,L);
cond4 = D3w(L)==(Q1-(subs(diff(EI,x),L)*M1/subs(EI,L)))/subs(EI,L);
conds=[cond1 cond2 cond3 cond4]

conds = 

wSol(x) = dsolve(deq,conds)

wSol(x) = 

Answer 1

[Paul]

The Maple solution does include imaginary terms, like 5*pi*1i/3, but we'll show that those terms cancel out on the interval of interest.

syms x real % in case this has an impact
syms w(x)
L=10;
q0=40;
a0=10000;
M1=-1000;
Q1=-100;
q=q0*(x/L)^2;
EI=a0*(2-(x/L));
Dw  = diff(w,x) ;
D2w = diff(w,x,2);
D3w = diff(w,x,3);
deq = diff(EI*diff(w,x,2),x,2)==q;
cond1 = w(0)==0;
cond2 = Dw(0)==0;
cond3 = D2w(L)==M1/subs(EI,L);
cond4 = D3w(L)==(Q1-(subs(diff(EI,x),L)*M1/subs(EI,L)))/subs(EI,L);
conds=[cond1 cond2 cond3 cond4]

conds = 

wSol(x) = dsolve(deq,conds)

wSol(x) = 

The solution contains a term of the form:

term = (x-20)*(5*sym(pi)*1i/3 - 5*log(x-20)/3)

term = 

or focusing only on the term in the parentheses after multiplying by 3/5

term = sym(pi)*1i - log(x-20)

term = 

Over the solutions interval of interest, 0 <= x <= L, the argumument to the log function is negative real. Using the rule for log of a negative number: log(x-20) = log(20-x) + i * pi. So we have

term = - log(20-x)

term = 

and the imaginary part disappears. So make that assumption explicitly and simplify the solution.

assume(0 <= x <= L) % only care about the solution over this interval
wSol(x) = simplify(expand(wSol(x))) % expand and simplify the solution using the assumption

wSol(x) = 

fplot(wSol(x),[0 L]) % looks like the Maple plot

[Ramses Young]

Wow that makes a lot of sense!!!

I can see how limiting the domain of interests affects the imaginary portion, but can you explain how using the simplify command makes it not have the imaginary part in it?

[Paul]

I edited the answer.