solving symbolic differential equations, error when using dsolve

Hello Everyone,

i'm new to Matlab and i'm trying to solve differential equations of mechanical systems (bernoulli beams).

I think my problem can be best described with a short example:

%simple example:
syms x z
syms EI L q
syms w(x) u(x)
syms qz(x) qx(x)

%given values
qz(x) = q;
qx(x) = q;
EA = EI /L^2;

%differential equations
eq1 = diff(w,x,4)==qz/EI;
eq2 = diff(u,x,2)==qx/EA;

ddw = diff(w,x,2);
My = -EI * ddw;

%boundary conditions
cond_1 = w(0) == 0;
cond_2 = My(0) == 0;
cond_3 = My(L) == 0;
cond_4 = w(L) == 0;
cond_5 = u(0) == 0;
cond_6 = u(L) == 0;

conds1 = [cond_1, cond_2, cond_3, cond_4];
conds2 = [cond_5, cond_6];
conds3 = [cond_1, cond_2, cond_3, cond_4, cond_5, cond_6];

S1 = dsolve(eq1, conds1) %works

S1 = 

S2 = dsolve(eq2, conds2) %works

S2 = 

S3 = dsolve(eq1, eq2, conds3) %doesnt work

Error using mupadengine/feval_internal
Invalid initial conditions.

Error in dsolve>mupadDsolve (line 334)
T = feval_internal(symengine,'symobj::dsolve',sys,x,options);

Error in dsolve (line 203)
sol = mupadDsolve(args, options);

In the first very simple example it es ovious, that i can solve w(x) seperate from u(x). But in want to solve more complex systems like the following one.

%more complex mechanical system:
syms w1(x) w2(x) w3(x) u1(x) u2(x) u3(x)
syms EI EA L q

%given values:
qz1 = q;
qz2 = (q * x) / L;
qz3 = 0;

qx1 = 0;
qx2 = 0;
qx3 = (q * x) / L;

EI1 = inf;
EI2 = EI;
EI3 = EI;
EA1 = EI / L^2;
EA2 = inf;
EA3 = EI / L^2;

%equations:
eqn1 = diff(w1,x,4)==qz1/EI1;
eqn2 = diff(w2,x,4)==qz2/EI2;
eqn3 = diff(w3,x,4)==qz3/EI3;
eqn4 = diff(u1,x,2)==-qx1/EA1;
eqn5 = diff(u2,x,2)==-qx3/EA2;
eqn6 = diff(u3,x,2)==-qx3/EA3;

eqns = [eqn1, eqn2, eqn3, eqn4, eqn5, eqn6];

DDDw1 = diff(w1,x,3);
DDDw2 = diff(w2,x,3);
DDDw3 = diff(w3,x,3);

DDw1 = diff(w1,x,2);
DDw2 = diff(w2,x,2);
DDw3 = diff(w3,x,2);

Dw1 = diff(w1,x);
Dw2 = diff(w2,x);
Dw3 = diff(w3,x);

My1 = -EI1 * DDw1;
My2 = -EI2 * DDw2;
My3 = -EI3 * DDw3;

Qz1 = -EI1 * DDDw1;
Qz2 = -EI2 * DDDw2;
Qz3 = -EI3 * DDDw3;

Du1 = diff(u1,x);
Du2 = diff(u2,x);
Du3 = diff(u3,x);

Nx1 = EA1 * Du1;
Nx2 = EA2 * Du2;
Nx3 = EA3 * Du3;

%boundary conditions:
condA1 = sqrt(3)==w1(0)/u1(0);
condA2 = Qz1(0)*(sqrt(3)/2)+Nx1(0)*(1/2)==0;
condA3 = Dw1 == 0;

condB1 = My1(L)==0;
condB2 = w1(L)==w1(0);
condB3 = u2(0)==w1(L)*(sqrt(3)/2)-u1(L)*(1/2);
condB4 = w2(0)==w1(L)*(1/2)-u1(L)*(sqrt(3)/2);
condB5 = Du2 == 0;
condB6 = u2(0) == u2(L);

condC1 = Qz2(L)==Qz3(0);
condC2 = My2(L)==My3(0);
condC3 = Dw2(L)==Dw3(0);
condC4 = w2(L)==w3(0);
condC5 = Nx2(L)==Nx3(0);
condC6 = u2(L)==u3(0);

condD1 = My3(L)==0;
condD2 = w3(L)==0;
condD3 = u3(L)==0;

condsA = [condA1, condA2, condA3];
condsB = [condB1, condB2, condB3, condB4, condB5, condB6];
condsC = [condC1, condC2, condC3, condC4, condC5, condC6];
condsD = [condD1, condD2, condD3];
conds = [condsA, condsB, condsC, condsD];

%solve equation system with boundary conditions:
S = dsolve(eqns, conds) %doesnt work, i guess because of the same reason as in the first example

I get the following output:

Error using mupadengine/feval_internal
Invalid equation or initial condition.

Error in dsolve>mupadDsolve (line 334)
T = feval_internal(symengine,'symobj::dsolve',sys,x,options);

Error in dsolve (line 203)
sol = mupadDsolve(args, options);

Error in baumechanik2_hue2_v2 (line 92)
S = dsolve(eqns, conds)

Why does this not work and is how can i solve this problem?

I am using R2022b

Thank you in advance!

Answer 1

[Walter Roberson]

dsolve() always solves down to f(x) + (one) constant. One there is an initial condition, it tries to solve the == value to give a specific value for the constant; this might be non-linear as constants from different levels of derivative might be involved.

After dsolve() has solved for the (one) constant for a given derivative level, there are no more constants to resolve for that level of derivative -- so it becomes a logic error to try to solve a second boundary condition for the same level of derivative.

To handle two boundary conditions at the same derivative level, you have to first dsolve() with one of the boundary conditions, and then substitute in the independent variable value at the second boundary condition, equate to the known value there, and solve for something appropriate for your system. dsolve() cannot do that automatically because dsolve() does not know which "something" you would want to solve for at that level.

It is not uncommon to encounter situations where you are best off solving a system without any boundary conditions, and then start manually substituting in the various boundary conditions and solving. There are cases where if you provide a boundary condition, dsolve gets stuck and cannot figure out how to find the function, but if you do not provide the boundary condition then dsolve() can [sometimes] give the general form of the solution that you can then work step-by-step to satisfy the boundary conditions.