Need help solving second-order nonlinear differential eq

WelshCorgiDude
Messages
2
Reaction score
0
Hello,

I'm a doctoral student in civil engineering. In my research I came across a differential equation for the net force acting on an object as it impacts a granular medium at low velocities.

z'' + a[ z' ]^2 + b[ z ] = c
Where a, b, and c are all constants

I believe that this equation will allow for the development of an analytical method for the determination of soil shear strength parameters in situ using rapid and inexpensive testing procedures. Unfortunately, my differential equation skills are horribly rusty, so this problem's solution has been rather ellusive. I've gotten as far as using the chain rule (?) to make the problem first-order, and developed the following:

If u = z' and u' = z'' (or u du/dz = z''), then u(du/dz) + a^2 +b[z] = c

I'm not sure where to go from here (or if I'm even going in the right direction). Textbook examples similar to this problem collapse nicely into an easily solvable problem, but not this one. I would really appreciate any guidance on this problem. Thanks in advance!

--Peter
 
Physics news on Phys.org
Sorry, I evidently turned underline on when making that second equation. Here is is again.

If u = z' and u'' = z'' (or u du/dz = z''), then u(du/dz) + a[ u ]^2 + b[ z ] = c
 
Your original DE is ##y'' + ay'^2 + by = c##. Consider the substitution ##w(y) = y'^2##. Then$$
\frac{dw}{dy} = \frac{dw}{dx}\cdot \frac{dx}{dy}= \frac{\frac{dy'^2}{dx}}{\frac{dy}{dx}}=\frac{2y'y''}{y'}=2y''$$so you have ##y''=\frac 1 2 \frac{dw}{dy}## and your DE becomes$$
\frac 1 2 \frac{dw}{dy}+aw(y)+by = c$$This is a first order constant coefficient DE with independent variable ##y##. Solve it for ##w(y)## and then solve the equation ##y'^2 = w(y)## by taking square roots and separation of variables. It should all work, at least in principle. The last integration might be interesting. Good luck.
 
Last edited:
What you could do is solve the homogeneous equation by finding the characteristic polynomial. After, use variation of parameters to find the solution to the nonhomogeneous equation. I prefer this method myself. Pauls Online Math notes has a good set of notes covering DEs
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top