How do I solve a second order non linear differential equation using matlab.

I have a fluid dynamics problem and I need to derive an equation for motion.

After applying Newtons second law to the system, and replaceing all the constants with A and B. My equation looks like this.

z'' + A(z')^2 = B

With A and B both being constants.

Initial conditions being that z(0)=0, and z'(0)=0

And I need to solve for z(t).

Thank you

[James Tursa]

What have you done so far? What specific problems are you having with your code? Have you looked at the doc for ode45 to see examples of how to use ode45 to numerically solve 2nd order ODE's?

[Patrick Guarente]

Yes I looked at ode45 page and couldnt understand the time interval that I need to put in. Because I need to find the equation for all time. And I wasn't sure if that gave me an equation as an answer or just a plot.

[James Tursa]

"... I need to find the equation for all time ..."

Are you looking for an analytical/symbolic solution? I thought that you simply wanted a numerical solution given your initial starting values.

[Patrick Guarente]

Yes the analytical solution in terms of A and B.

Answer 1

[Teja Muppirala]

syms z(t) t A B

zp = diff(z,t);
zpp = diff(z,t,2);

eqn = ( zpp  + A*zp^2 == B );
cond = [z(0)==0, zp(0)==0];
zSol = dsolve(eqn,cond,'IgnoreAnalyticConstraints',true);
zSol = unique(simplify(zSol));

This gives 3 solutions:

zSol =

 log((C15*sinh(A^(1/2)*B^(1/2)*(t + A*B^(1/2)*1i)))/B^(1/2))/A
  log(-(C18*sinh(A^(3/2)*B*1i - A^(1/2)*B^(1/2)*t))/B^(1/2))/A
                                log(cosh(A^(1/2)*B^(1/2)*t))/A

The first two look weird, but are valid solutions involving complex-valued z. The 3rd solution is real, and that's probably the one that you are looking for.

Answer 2

[James Tursa]

Define a 2-element vector y:

y(1) = z
y(2) = z'

then solve your 2nd order ODE for the highest derivative:

z'' + A(z')^2 = B       ==>
z'' = - A(z')^2 + B

then calculate the y element derivative equations, using this z derivative info:

d y(1) = d z = z' = y(2)
d y(2) = d z' = z'' = -A(z')^2 + B = -A*y(2) + B

So create a derivative function based on those two equations, using the function signature that you will find in the ode45 doc. Then call it using the outline provided in the example in the doc.

EDIT: SYMBOLIC SOLUTION

>> dsolve('D2z + A*(Dz)^2 = B')
ans =
                                                                              C29 + (B^(1/2)*t)/A^(1/2)
                                                                              C27 - (B^(1/2)*t)/A^(1/2)
 log((exp(2*A^(3/2)*B^(1/2)*(C24 + t/A)) - 1)/(2*B^(1/2)*exp(A*(C16 + A^(1/2)*B^(1/2)*(C24 + t/A)))))/A
 log((exp(2*A^(3/2)*B^(1/2)*(C20 - t/A)) - 1)/(2*B^(1/2)*exp(A*(C16 + A^(1/2)*B^(1/2)*(C20 - t/A)))))/A

[Patrick Guarente]

I'm not sure where those two equations will be implemented.

Answer 3

[Torsten]

According to MATHEMATICA, the analytical solution is

z(x) = log(cosh(sqrt(A*B)*x))/A

Best wishes

Torsten.