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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

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.

Lewis Fer
on 10 Jun 2021

Hello, I am having troubles solving a system of second order nonlinear equations with boundary conditions using MATALB

Here is the equations:

f''(t)=3*f(t)*g(t) -g(t)+5*t;

g''(t)=-4f(t)*g(t)+f(t)-7*t;

the boundary conditions are: f'(0)=0 et h'(o)=5;

g(0)=3 et h'(2)=h(2)

James Tursa
on 25 Sep 2017

Edited: 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

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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