How to solve a double complex ode using ode45 in MATLAB

The odes are:

ode(1) = diff(x,t) == (sqrt(1+x^2)*y)/(x*(t^2));
ode(2) = diff(y,t) == (x^3)*(t^2);
given x(0)=infty; y(0)=0.

I want to draw two picture to reflect x^3(u)-y relation and the u axis should be drawn by log(u) scale.

Answer 1

[Walter Roberson]

The infinite boundary condition is a problem for symbolic solution: dsolve() rejects it.

The 0 initial time is a problem: you divide by time, so you generate a numeric infinity at the initial time, which numeric solvers cannot recover from.

If you set initial time above 0, and set the boundary condition to be finite, then you get a singularity. The time of singularity depends upon the boundary condition you set -- with the failure time being approximately 1 over the boundary condition.

syms x(t) y(t)
dx = diff(x)

dx(t) = 

dy = diff(y)

dy(t) = 

odes = [dx == (sqrt(1+x^2)*y)/(x*(t^2));
        dy == (x^3)*(t^2)]

odes(t) = 

ic = [x(0) == 1e8, y(0) == 0]

ic = 

sol = dsolve(odes, ic)

Warning: Unable to find symbolic solution.

 
sol =
 
[ empty sym ]
 

[eqs,vars] = reduceDifferentialOrder(odes,[x(t), y(t)])

eqs = 

vars = 

[M,F] = massMatrixForm(eqs,vars)

M = 

F = 

f = M\F

f = 

odefun = odeFunction(f,vars)

odefun = function_handle with value:
    @(t,in2)[(1.0./t.^2.*in2(2,:).*sqrt(in2(1,:).^2+1.0))./in2(1,:);t.^2.*in2(1,:).^3]

xy0 = double(rhs(ic))

xy0 = 1×2

100000000 0

tspan = [1e-50 100]

tspan = 1×2

0.0000 100.0000

[t, xy] = ode45(odefun, tspan, xy0);

Warning: Failure at t=2.575086e-08. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (5.293956e-23) at time t.

figure
semilogy(t, xy(:,1), 'k+-')
title('x')

figure
plot(t, xy(:,2), 'k*-')
title('y')

[嘉杰 程]

your method works, thanks, but there's another question is that could you draw an asymptotic line when x(u) approaches +infty, y=?

[嘉杰 程]

and maybe you didn't notice that I want to plot y-u relation (u=x^3), can you do that?

[嘉杰 程]

this is y-x relation, and how to print y-u relation (u=x^3)

[Walter Roberson]

syms x(t) y(t)
dx = diff(x)

dx(t) = 

dy = diff(y)

dy(t) = 

odes = [dx == (sqrt(1+x^2)*y)/(x*(t^2));
        dy == (x^3)*(t^2)]

odes(t) = 

ic = [x(0) == 1e8, y(0) == 0]

ic = 

[eqs,vars] = reduceDifferentialOrder(odes,[x(t), y(t)])

eqs = 

vars = 

[M,F] = massMatrixForm(eqs,vars)

M = 

F = 

f = M\F

f = 

odefun = odeFunction(f,vars)

odefun = function_handle with value:
    @(t,in2)[(1.0./t.^2.*in2(2,:).*sqrt(in2(1,:).^2+1.0))./in2(1,:);t.^2.*in2(1,:).^3]

xy0 = double(rhs(ic))

xy0 = 1×2

100000000 0

tspan = [1e-50 100]

tspan = 1×2

0.0000 100.0000

[t, xy] = ode45(odefun, tspan, xy0);

Warning: Failure at t=2.575086e-08. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (5.293956e-23) at time t.

y = xy(:,2);
u = xy(:,1).^3;

ax = subplot(2,1,1)

ax = 
  Axes with properties:

             XLim: [0 1]
             YLim: [0 1]
           XScale: 'linear'
           YScale: 'linear'
    GridLineStyle: '-'
         Position: [0.1300 0.5838 0.7750 0.3412]
            Units: 'normalized'

  Show all properties

plot(ax, u, y);
xlabel(ax, 'u'); ylabel(ax, 'y')
ax.XScale = 'log';
ax.YScale = 'log';

ax = subplot(2,1,2)

ax = 
  Axes with properties:

             XLim: [0 1]
             YLim: [0 1]
           XScale: 'linear'
           YScale: 'linear'
    GridLineStyle: '-'
         Position: [0.1300 0.1100 0.7750 0.3412]
            Units: 'normalized'

  Show all properties

plot(ax, y, u);
xlabel(ax, 'y'); ylabel(ax, 'u')
ax.XScale = 'log';
ax.YScale = 'log';

[嘉杰 程]

thank you！

Answer 2

[Cris LaPierre]

You can solve a symbolic ODE with initial/boundary conditions using dsolve

• S = dsolve(eqn,cond)