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Question: Problem with dsolve()

Question:Problem with dsolve()

Rouben Rostamian 8501 Maple 2025
ode dsolve differential-equations nonhomogeneous temperature steady-state
June 16 2025
2

A colleage asked me for the solution of an elementary boundary value problem for an ODE that models the steady-state temperature distribution in a nonhomogeneous rod.  I solved it in Maple and passed the solution on to her without checking the result.  She struggled for a day or two, attempting to "debug" her finite element code whose result was not agreeing with Maple's.

Upon closer inspection, it turned out that her code was correct and the solution returned by Maple was not.  Here are the details.  I can't tell whether Maple's dsolve can be improved to provide correct solutions to this and similar problems.

Problem with dsolve()

> 

restart;
> 

kernelopts(version);

`Maple 2025.0, X86 64 LINUX, Mar 24 2025, Build ID 1909157`

A boundary value problem for an ODE

 

Let's solve the following boundary value problem:

> 

de := Diff(a(x)*Diff(u(x),x),x) = -1;

Diff(a(x)*(Diff(u(x), x)), x) = -1

> 

bc := u(-1)=0, u(1)=0;

u(-1) = 0, u(1) = 0

Maple's dsolve() fails to find a solution:

> 

dsolve({de,bc});

> 

 

Solution obtained by hand

 

The solution may be obtained by hand, as follows.

 

Integrate the DE once

a(x)*(diff(u(x), x)) = c-x.

where c is the integration constant.  Therefore

diff(u(x), x) = (c-x)/a(x) and (c-x)/a(x) = c/a(x)-x/a(x).

To determine the value of the constant c, integrate the above over [-1, 1]

and apply the boundary conditions:

c*(int(1/a(x), x = -1 .. 1))+int(x/a(x), x = -1 .. 1) = u(1)-u(-1) and u(1)-u(-1) = 0

whence
c = (int(x/a(x), x = -1 .. 1))/(int(1/a(x), x = -1 .. 1)).

Having thus determined c, we integrate the expression for diff(u(x), x) obtained above, and arrive at the solution
u(x) = int((c-xi)/a(xi), xi = -1 .. x).

Remark: The solution obtained here is valid for any a(x) as long as

the integrals encountered above make sense.  Note that there is
no requirement on the differentiability or even continuity of a(x).

Technically, a(x) can be any function such that 1/a(x) is integrable.

 

> 

 

Wrong solution returned by dsolve()

 

Let us consider a special choice of the coefficient a(x):

> 

a := x -> 1 + 2*Heaviside(x);

proc (x) options operator, arrow; 1+2*Heaviside(x) end proc

According the the calculations above, we have:

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c := int(x/a(x),x=-1..1) / int(1/a(x),x=-1..1) ;

-1/4

and therefore the correct solution is

> 

int((c-xi)/a(xi), xi=-1..x):
sol := u(x) = collect(%, Heaviside);

u(x) = ((1/6)*x+(1/3)*x^2)*Heaviside(x)-(1/4)*x-(1/2)*x^2+1/4

Here is what the solution looks like:

> 

plot(rhs(sol), x=-1..1, color=blue, thickness=3);

Let's attempt to solve the same boundary value problem with Maple's dsolve().

> 

dsolve({de,bc}):
bad_sol := collect(%, Heaviside);

u(x) = ((1/3)*x^2+x/(3*exp(-2)+3)-exp(-2)*x/(3*exp(-2)+3))*Heaviside(x)-(1/2)*x^2-x/(3*exp(-2)+3)+(3*exp(-2)+1)/(6*exp(-2)+6)

That's very different from the correct solution obtained above.   To see the problem with it, let's recall that according to the DE, the expression a(x)*(diff(u(x), x))NULLshould evaluate to  c-x for some constant c,  but what we get through bad_sol is nothing like c-x;  in fact, it's not even continuous:

> 

a(x)*diff(rhs(bad_sol), x):
collect(%, Heaviside, simplify);
plot(%, x = -0.5 .. 0.5);

((4*x-2)*exp(-2)+4*x+2)*Heaviside(x)^2/(3*exp(-2)+3)+(-(4/3)*x-1/3)*Heaviside(x)-x-1/(3*exp(-2)+3)

It appears that dsolve goes astray by attempting to expand the expression diff(a(x)*(diff(u(x), x)), x).   It shouldn't.

 

 

Download dsolve-bug.mw

 
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