Trouble with MuPad when evaluating symfun at a symbolic variable instead of a number in MATLAB

I created a symbolic function in MATLAB R2021b using this script with the goal of solving an ODE.

syms phi(x) lambda L
assume(lambda>0)
eqn_x = diff(phi,x,2) == -lambda*phi;
dphi = diff(phi,x);
cond1 = phi(0)==0;
cond2 = dphi(1)==0;
cond = [cond1, cond2]; % this is the line where the problem starts
disp(cond)

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Addendum:

I had presumed these conditions above would work with dsolve, apparently they do not as even in that case I get the same error described below. So it might not actually be a problem with using x=1 instead of x=L but a problem with how I am inputting my conditions. See the line I have commented out below.

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------

As you can see, this script gives valid conditions to solve the ODE. However, I want to set x = L instead of a number, because I want the ability change the value of L to a number after getting a solution.

I've tried the following, and consulted the MuPad tutorial, but still have not been able to get MATLAB to solve the ODE.

cond2 = dphi(L)==0;
cond = [cond1, cond2];
disp(cond)

% phi = dsolve(eqn_x,x,cond); this is commented to stop an error in MATLAB
% when running in the browser

cond2 = dphi(x==L)==0;
cond = [cond1, cond2];
disp(cond)

same error

cond2 = dphi | x == L; % based on the MuPad tutorial see link below (page 177)
cond = [cond1, cond2];
disp(cond)

MuPad Tutorial

and finally...

cond2 = subs(dphi, x = L);
cond = [cond1, cond2];
disp(cond)

same error

Obviously, the way MATLAB is interpreting my cond2 is not correct. Can someone recommend a fix or workaround?

Answer 1

[Paul]

Hi Nathaniel,

I too could not figure out how to specify cond2 the way you want.

A workaround would be to only specify cond1, apply dsolve, and then solve for the constant in the solution based on cond2.

syms phi(x) lambda L
assume(lambda>0)
eqn_x = diff(phi,x,2) == -lambda*phi

eqn_x(x) = 

dphi = diff(phi,x);
cond1 = phi(0)==0;
cond2 = dphi(0)==0;
cond = [cond1, cond2]; % this is the line where the problem starts
disp(cond)

% solve with just cond1
sol(x) = dsolve(eqn_x, cond(1))

sol(x) = 

% differentiate the solution
dsol(x) = diff(sol,x)

dsol(x) = 

% solve for C1
syms C1
C1sol = solve(dsol(L)==0,C1)

C1sol = 

0

% sub C1 back into the solution
sol(x) = subs(sol,C1,C1sol)

sol(x) = 

0

For this problem sol(x) = 0 satisfies the differential equation and both boundary conditions.

[Nathaniel H Werner]

Interesting, I think this is probably the best option. Maybe Mathworks can address this.

[Paul]

You can always contact Tech Support and submit a bug report, if you think it is a bug, or an enhancement request otherwise.

Another option that works for this case is to convert the second order ODE to a system of first order ODEs and specify the boundary conditions there. I changed cond2 so that we get a non-zero solution.

syms phi(x) lambda L
assume(lambda>0)
eqn_x = diff(phi,x,2) == -lambda*phi

eqn_x(x) = 

dphi = diff(phi,x);
[newEqs,newVars] = reduceDifferentialOrder(eqn_x,phi)

newEqs = 

newVars = 

syms Dphit(x)
sol = dsolve(newEqs,[phi(0)==0, Dphit(L) == .5])

sol = struct with fields:
      phi: sin(lambda^(1/2)*x)/(2*lambda^(1/2)*cos(L*lambda^(1/2)))
    Dphit: cos(lambda^(1/2)*x)/(2*cos(L*lambda^(1/2)))

Verify phi satisifes the differential equation

diff(sol.phi,x,2) + lambda*sol.phi

ans = 

0

Verify Dphit is the derivative of phi

diff(sol.phi,x) - sol.Dphit

ans = 

0

Verify boundary conditions

subs(sol.phi,x,0)

ans = 

0

subs(sol.Dphit,x,L)

ans = 

[Nathaniel H Werner]

Thank you for the suggestion. It is an interesting idea, since I don't have use or time to explore further at the moment I'll leave it here for now.