KdV ODE third-order
It is necessary to find a solution that satisfies the boundary conditions:
u′′′+u′+λu=f.
initial conditions:
u′′(0)=u′(0)+u(0),
u′′(1)=u(1),
u′(1)=u(1),
How to define and in matlab?
where λ>0, ;, , and – given numbers.
≤
where C is a positive constant independent of 𝑢 and f
clear;
syms y(x) y0 lambda u
Dy = diff(y);
D2y = diff(y,2);
D2y_2 = diff(y,2);
D3y = diff(y,3);
ode = D3y + lambda*u == y(x);
ySol(x) = dsolve(ode);It is necessary to find a solution that satisfies the boundary conditions:
u′′′+u′+λu=f.
initial conditions:
u′′(0)=u′(0)+u(0),
u′′(1)=u(1),
u′(1)=u(1),
How to define and in matlab?
where λ>0, ;, , and – given numbers.
≤
where C is a positive constant independent of 𝑢 and f
clear;
syms y(x) y0 lambda u
Dy = diff(y);
D2y = diff(y,2);
D2y_2 = diff(y,2);
D3y = diff(y,3);
ode = D3y + lambda*u == y(x);
ySol(x) = dsolve(ode); It is necessary to find a solution that satisfies the boundary conditions:
u′′′+u′+λu=f.
initial conditions:
u′′(0)=u′(0)+u(0),
u′′(1)=u(1),
u′(1)=u(1),
How to define and in matlab?
where λ>0, ;, , and – given numbers.
≤
where C is a positive constant independent of 𝑢 and f
clear;
syms y(x) y0 lambda u
Dy = diff(y);
D2y = diff(y,2);
D2y_2 = diff(y,2);
D3y = diff(y,3);
ode = D3y + lambda*u == y(x);
ySol(x) = dsolve(ode); ode, kdv, third-order, sobol-dim, matlab MATLAB Answers — New Questions