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

Asked by Patrick Guarente on 25 Sep 2017

Latest activity Answered by Teja Muppirala on 26 Sep 2017

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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

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Patrick Guarente on 25 Sep 2017

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 on 25 Sep 2017

"... 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 on 25 Sep 2017

Yes the analytical solution in terms of A and B.

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Answer by Teja Muppirala on 26 Sep 2017

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.

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Answer by James Tursa on 25 Sep 2017

Edited by James Tursa on 25 Sep 2017

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

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Patrick Guarente on 25 Sep 2017

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

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Answer by Torsten on 26 Sep 2017

According to MATHEMATICA, the analytical solution is

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

Best wishes

Torsten.

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