Syntax for specifying boundary condition using dsolve.

I am working on a beam bending/deflection problem. The boundary conditions I need to satisfy are...

• Displacement at x = 0 is 0
• Slope at x = 0 is 0
• Displacement at x = L is 0
• Moment at x = L is 0

The code I have is below. The issue I am having is specifying the moment to be zero at x=L (syntax).

syms I E L f x u0(x)
% initial equation
eq1 = diff(E*I*diff(u0,x,2),x,2) + f == 0

% gen. sol.    
eq2 = dsolve(eq1)

% part. sol.
Du0 = diff(u0,x,1); DDu0 = diff(u0,x,2);
bc1 = u0(0) == 0                      % Displacement at x = 0 is 0
bc2 = Du0(0) == 0                    % Slope at x = 0 is 0
bc3 = u0(L) == 0                      % Displacement at x = L is 0
bc4 = DDu0(x==L) == 0           % Moment at x = L is 0  
eq3 = dsolve( eq1 , [ bc1, bc2 , bc3 , bc4 ] )

[Torsten]

It's strange how the corrected form of bc4 is displayed. Why is L the differentiation variable and not x as in bc2 ?

syms I E L f x u0(x)
% initial equation
eq1 = diff(E*I*diff(u0,x,2),x,2) + f == 0;
% gen. sol.    
eq2 = dsolve(eq1);
% part. sol.
Du0 = diff(u0,x,1); DDu0 = diff(u0,x,2);
bc1 = u0(0) == 0    ;                 % Displacement at x = 0 is 0
bc2 = Du0(0) == 0                    % Slope at x = 0 is 0

bc2 = 

bc3 = u0(L) == 0    ;                 % Displacement at x = L is 0
bc4 = DDu0(L) == 0                   % Moment at x = L is 0  

bc4 = 

eq3 = dsolve( eq1 , [ bc1, bc2 , bc3 , bc4 ] );

[Jonathan]

Yes that is what initially confused me. But it yields the equivalent solution.

For instance, if you specifiy L = 1 and solve both ways you get the same result...

syms I E L f x u0(x)

% initial equation

eq1 = diff(E*I*diff(u0,x,2),x,2) + fv == 0

% gen. sol.

eq2 = dsolve(eq1)

% part. sol.

Du0 = diff(u0,x,1); DDu0 = diff(u0,x,2);

bc1 = u0(0) == 0 % Displacement at x = 0 is 0

bc2 = Du0(0) == 0 % Slope at x = 0 is 0

bc3 = u0(L) == 0 % Displacement at x = L is 0

bc4 = DDu0(L) == 0 % Moment at x = L is 0

eq3 = dsolve(eq1,bc1 , bc2 , bc3 , bc4)

% solve at x=0

u0_midpoint = vpa( subs( eq3 , [ x I L] , [ 0.5 Iv Lv ] ) , 3 )

% part. sol.

Du0 = diff(u0,x,1); DDu0 = diff(u0,x,2);

bc1 = u0(0) == 0 % Displacement at x = 0 is 0

bc2 = Du0(0) == 0 % Slope at x = 0 is 0

bc3 = u0(1) == 0 % Displacement at x = L is 0

bc4 = DDu0(1) == 0 % Moment at x = L is 0

eq3 = dsolve(eq1,bc1 , bc2 , bc3 , bc4)

% solve at x=0

u0_midpoint = vpa( subs( eq3 , [ x I] , [ 0.5 Iv ] ) , 3 )

Answer 1

[John D'Errico]

Your beam is fixed in position at the left end, as is the slope at that point.

At the right end, you have fixed the location, but you want a zero bending moment. And you already know to do that, by setting the second derivative to zero there.

syms I E L f x u0(x)
% initial equation
eq1 = diff(E*I*diff(u0,x,2),x,2) + f == 0

eq1(x) = 

An order 4 ODE, so we need 4 conditions to solve.

Du0 = diff(u0,x,1); DDu0 = diff(u0,x,2);

xsol(x) = dsolve(eq1,u0(0) == 0, Du0(0) == 0, u0(L) == 0, DDu0(L) == 0)

xsol = 

This beam is one with a clamp at 0, the left end, and pinned at X==L, so it will have a zero bending moment there. This solution should have those properties. But that looks like effectively what you wrote...

Ah, looking at your code, you wrote this:

bc4 = DDu0(x==L) == 0 % Moment at x = L is 0

and that is clearly wrong. This would have worked instead:

bc4 = DDu0(L) == 0 % Moment at x = L is 0

And you actually knew how to do that! DDu0 is just a function of x, like u0 and Du0.

[Jonathan]

Thanks very much. Yep it was just a syntax issue for me.