# Help solving a first order non-linear differential equation derived from the navier-stokes equation

I am an engineer studying an unsteady-state flow through a pipe. The transient Bernoulli equation of this system, which I picked up from here (http://higheredbcs.wiley.com/legacy/college/fox/0471742996/webpdf/ch06.pdf) yields this differential equation: $$\frac{dv(t)}{dt}+av^2(t)+bP(t)=c$$ This is a first order non-linear differential equation. Unfortunately, I have no experience solving non-linear differential equations. From the research I have done, this type of equation looks similar to the Ricatti equation. Is there a closed form solution to the above equation? How can I solve this equation? I'm interested in getting a function that shows how the pressure or velocity of the system decays with time.

FYI: v(t) is velocity, P(t) is pressure, (a,b,c) are constants. Thank You all.

• i think only a numerical method can help here – Dr. Sonnhard Graubner May 18 '15 at 16:19
• What are your boundary conditions? – Spencer May 18 '15 at 16:36
• You might think about this: how will you get a closed form solution without explicitly specifying $P(t)$? Off hand, itt seems to me some form of quadrature involving $P(t)$ is the best to expect in the general case. Cheers! – Robert Lewis May 18 '15 at 16:41
• The only boundary condition that I know of is that the exit pressure at the end of the pipe is atmospheric. I also know the initial pressure and velocity. Thank you for all the feedback – H3G3moKnight May 18 '15 at 17:46

It is in fact a Ricatti equation. To find an explicit solution you need first a particular solution. This can be difficult to find even when an explicit form for $P$ is known.

$$\frac{dv(t)}{dt}+av^2(t)+bP(t)=c$$ Let $v(t)=\frac{y'(t)}{ay(t)}$

$v'=\frac{y''}{ay}-\frac{y'^2}{ay^2}=-a\left(\frac{y'}{ay} \right)^2-bP(t)+c$

$\frac{y''}{ay}=-bP(t)+c$

$$y''+a(bP(t)-c)y=0$$

Since the form of the function $P(t)$ is not defined, the second order ODE is on a general form. No general solution can be provided.

So, one can expect to solve it only if the fonction $P(t)$ is known. What is more, expecting doesn't mean to be sure to solve it. Solutions are known only in some particular cases of functions $P(t)$.

As a consequence, in the general case, numerical methods are recommended.

Note :

if you intend to try to solve some equations of the kind : $$y''+F(t)y=0$$ there are a few guidelines :

$F(t)=c \: \to \: y(t)$ involves sinusoïdal or hyperbolic functions.

$F(t)=a+bt \: \to \: y(t)$ involves Airy functions.

$F(t)=a+bt+ct^2 \: \to \: y(t)$ involves parabolic cylinder functions.

$F(t)=a+\frac{b}{t} \: \to \: y(t)$ involves Bessel functions.

$F(t)=a+\frac{b}{t^2} \: \to \: y(t)$ involves Bessel functions.

$F(t)=at^p \: \to \: y(t)$ involves Bessel functions.

$F(t)=ae^{bt}+c \: \to \: y(t)$ involves Bessel functions.

$F(t)=a+\frac{b}{t}+ \frac{c}{t^2} \: \to \: y(t)$ involves Kummer or Whittaker or Coulomb wave functions.

Etc. These are only rough indications about the kind of functions. Depending on particular values of the parameters, a function might become a function of lower level.

• Wow, this is very helpful thank you. Given this second order linear differential equation form, I will try specifying a linear, quadratic and exponential form for P(t) and see if the results make physical sense. Thank You all. – H3G3moKnight May 18 '15 at 17:44
• Complementary information is added to my first answer, that could help you to try some specific forms of functions. – JJacquelin May 18 '15 at 21:59
• @JJacquelin Why do you use $x$ instead of $t$ in the expressions of $P(t)$? I think it is a typo. – Navaro Aug 31 '17 at 13:42
• Typo corrected. Thanks you for the remark. – JJacquelin Aug 31 '17 at 14:54