Proof-checking a ODE solution

Hi, MATLAB online solves this:

if true
  % code
end
syms a h Y(x) g x B E
eqn = (h^2)*diff(Y,x, 2) + (2*i*h*g)*diff(Y,x) + (g^2 - E)*Y == 0;
DY = diff(Y) 
cond1 = Y(0) == 1;
cond2 = DY(0) == 0;
Y(x) = dsolve(eqn, cond)

however MATLAB on the computer, 2017 a, does not solve it, and I get :

diff(Y(x), x)

Not enough input arguments.

Error in cond (line 24) if issparse(A)

Error in PDE_sol (line 6) Y(x) = dsolve(eqn, cond)

Which one should I "believe" in?

Answer 1

[Birdman]

Replace the line

Y(x) = dsolve(eqn, cond)

with

Y(x) = dsolve(eqn, [cond1 cond2])

[Sergio Manzetti]

Thanks! Worked nicely!

[Sergio Manzetti]

Birdman, something came up:

if true
  % code
end
syms a h Y(x) g x B E
eqn = (h^2)*diff(Y,x, 2) + (2*i*h*g)*diff(Y,x) + (g^2 - E)*Y == 0;
DY = diff(Y) 
D2Y = diff(Y,x, 2)
cond1 = Y(0) == 1;
cond2 = DY(0) == 0;
cond3 = Y(2*pi) == 1;
Y(x) = dsolve(eqn, cond)

gives

Y(x) =

C5*exp(-(x*(g*1i + (- 2*g^2 + E)^(1/2)))/h) + exp((g*1i - (- 2*g^2 + E)^(1/2))/h)*exp(-(x*(g*1i - (- 2*g^2 + E)^(1/2)))/h)*(pi - C5*exp(-(g*1i + (- 2*g^2 + E)^(1/2))/h))

on MATLAB online, while on comp with alpha 2017:

Y(2*pi) == 1

Warning: Explicit solution could not be found. > In dsolve (line 201) In PDE_sol (line 7)

Y(x) =

[ empty sym ]

[Birdman]

Firstly, change the last line to

Y(x) = dsolve(eqn, [cond1 cond2 cond3])

Secondly, supplying 2 initial condition as cond1 and cond3 seems to be wrong. What are you trying to find?

[Torsten]

Where did you define "cond" ?

Best wishes

Torsten.

[Sergio Manzetti]

cond does not need to be defined, it's not there?

[Sergio Manzetti]

Torsten, I want to apply some Boundary conditions to this, in order to find C. I checked MATLAB info on dsolve, but nothing is said about BCs.

[Torsten]

As Birdman said:

if true
  % code
end
syms a h Y(x) g x B E
eqn = (h^2)*diff(Y,x, 2) + (2*i*h*g)*diff(Y,x) + (g^2 - E)*Y == 0;
DY = diff(Y) 
D2Y = diff(Y,x, 2)
cond1 = Y(0) == 1;
cond2 = DY(0) == 0;
cond = [cond1 cond2];
Y(x) = dsolve(eqn, cond)

Prescribing three boundary conditions instead of two for a second-order equation will make the problem unsolvable in general:

if true
  % code
end
syms a h Y(x) g x B E
eqn = (h^2)*diff(Y,x, 2) + (2*i*h*g)*diff(Y,x) + (g^2 - E)*Y == 0;
DY = diff(Y) 
D2Y = diff(Y,x, 2)
cond1 = Y(0) == 1;
cond2 = DY(0) == 0;
cond3 = Y(2*pi) == 1;
cond = [cond1 cond2 cond3];
Y(x) = dsolve(eqn, cond)

Best wishes

Torsten.

[Sergio Manzetti]

Thanks Torsten, I have to apply the BC's after solving the equation with the two ICs, and then extract the result.

[Sana Syed]

If I have 2 coupled 4th order non linear ODEs and 8 BCS it should take in the 8 conditions right? Its not accepting any of the boundary conditions which are non linear.

syms uy(x) txd(x) uz(x) ux(x); %Beam Dimensions Ty=0.001,Tz=0.01,L=0.1; %Constants based on dimensions from excel calculations c21=42.5015,c22=8.42*10^-4,c3=374.81,a=100,d=0.03; %Loading Conditions normalized loads fxl=0,fyl=0,myl=0,fzl=0,mzl=0,mxdl=0.3125,myl=0;%Pure twisting

%Differential Equations Input ode1=diff(uy,x,4)-fxl*diff(uy,x,2)==-myl*diff(txd,x,2)-2*fzl*diff(txd,x)+(1-x)*fzl*txd; %For pure twisting, it reduces to D4y/Dx4=0; ode2=(c21+c22*fxl)*diff(txd,x,2)-c3*diff(txd,x,4)-(myl-fzl*(1-x))*(mzl+fyl*(1-x)-fxl*(uy(L)-uy)-(myl-fzl*(1-x))*txd)==0; %ode2=(c21+c22*fxl)*diff(txd,x,2)-c3*diff(txd,x,4)==0; ode3=diff(uz,x,2)==txd*diff(uy,x,2)-(myl-fzl*(1-x))/a; %ode3=diff(uz,x,2)==txd*diff(uy,x,2)-(1/a)*(myl-fzl*(1-x)); ode4=d*(diff(ux,x)+0.5*(diff(uy,x))^2+0.5*(diff(uz,x))^2)+(((a+1)*(diff(txd,x))^2)/2)==fxl; %End of Differential Equations Input

%Define Derivative values in terms of variables Duy=diff(uy,x); D2uy=diff(uy,x,2); D3uy=diff(uy,x,3); Dtxd=diff(txd,x); D2txd=diff(txd,x,2); D3txd=diff(txd,x,3); Duz=diff(uz,x); D2uz=diff(uz,x,2); D3uz=diff(uz,x,3); %End of Derivative values in terms of variables

%Geometric Boundary Conditions cond1=ux(0)==0; cond2=uy(0)==0; cond3=uz(0)==0; cond4=txd(0)==0; cond5=Duz(0)==0; cond6=Duy(0)==0; cond7=Dtxd(0)==0; cond8=Dtxd(L)==0; %End of Geometric Boundary conditions

 %Loading Boundary conditions
    %THIS END CONDITION GIVES ERROR: cond9=-D3uy(L)-a*(2*txd(L)*Dtxd(L)*D2uy(L)+txd(L)^2*D3uy(L)-txd(L)*D3uz(L)-Dtxd(L)*D2txd(L))+fxl*Duy(L)==fyl;
    %THIS CONDITION ALSO GIVES ERROR cond10=D2uy(L)+a*(txd(L)*D2uy(L)-D2uz(L))*txd(L)==mzl
    cond11=-c3*D3txd(L)==mxdl;
%odes=[ode1 ode2 ode3 ode4]
odes=[ode1 ode2]
conds=[cond2 cond4 cond6 cond7 cond8]
%[uySol(x), txSol(x), uzSol(x), uxSol(x)] = dsolve(odes,conds)
[uySol(x), txSol(x)] = dsolve(odes,conds)