Question

f’’’-f’^2+1=0 solution of this equation with boundary conditions f’(0)=0, f’(∞)=1, f’’(∞)=0 is needed so badly

Answer 1

have you attempted this?

Answer 2

no

Answer 3

1. which grade are you in?

Answer 4

2. f'^2 means \([f'(x)]^2\) or \(f''(x)\)?

Answer 5

first one

Answer 6

square of f'

Answer 7

http://www.wolframalpha.com/input/?i=f%27%27%27%28x%29-f%27%28x%29%5E2%2B1%3D0

Answer 8

Is the first term f''' or f''

Answer 9

This is too complicated, this should not be what you'll be asked... can you double check your question?

Answer 10

question double checked :) the first one is f'''

Answer 11

still the third degree?

Answer 12

i agree, this is not a standard question

Answer 13

yes

Answer 14

Please consult your teacher.

Answer 15

this is my homework for phd lecture advanced math

Answer 16

i know it s complicated that s why searching for help

Answer 17

can you give me a day or two to come up with a solution?

Answer 18

of course that would be very nice

Answer 19

due is two days from now

Answer 20

If you find a solution please post it here too. Third order non-linear DE... I think you need to be a real maths wiz to solve this.

Answer 21

you need to solve this as differential equation from the charactreristic equation
r^3-2r+1=0 and clearly r=1 is a root for this then
(r-1)(r^2+r-1)=0 other roots will come from r^2+r-1=0
(r+1/2)^2-5/4=0 (r+1/2)=square root of 5/4
r1=-1/2+ (root5)/2 and r2=-1/2-(root 5)/2

then your solution must be in the form of y(t)=a.e^t+b.e^r1+c.e^r2
then take its first derivative and it will equal to 0 at t=0 and 1 at t=infinity from the given information

Answer 22

thanks i ll give it a shot

Answer 23

do you solve it by computer ? or handy ?
if you must solve it handy
you can use variational method like( ptero - petro galarkin method )
or if not
you want approximate solution
suppose f(x)=a+bx+cx^2+dx^3
put them in equation and find a,b,c,d
any way
another method is solving by series
(because it is not a linear ode )

Answer 24

i ve to solve it by hand, will consider recommendations

Answer 25

are you in bs level or ms level "sinan"?

Answer 26

ms level

Answer 27

amoodarya is wrong f(x) cannot be in the form of as he said because when we take its derivative and put infinity we get infinity not 1 so it is absolutely wrong

Answer 28

ok
so boundary conditions f’(∞)=1, f’’(∞)=0 are not usual boundary conditions
in ms level , because you have not learn ode course !!!
any way
by this boundary conditions f’(∞)=1, f’’(∞)=0 you need a function like exp(-x), exp(-2x) ,... so in ∞ goes to zero .

Answer 29

exactly

Answer 30

r^3-2r+1=0 is the characteristic equation for f'''-f'^2+1=0 ? ?

Answer 31

no
it is not the characteristic equation for f'''-f'^2+1=0

Answer 32

because it is square of f'

Answer 33

i couldn t even write characteristic equation

Answer 34

r^3-2r+1=0 is for f'''=2f'+f=0

Answer 35

yes

Answer 36

you are right i didnt realize that

Answer 37

i got an idea !

Answer 38

seems like a good idea, what to do after ?

Answer 39

i have been thinking about the integration of g'

Answer 40

Did you try laplace transforms? lol

Answer 41

Looking at laplace there is no easy way to do a transform for f'^2
Looking at homogenous ODE's there is no characteristic eqn for f'^2

I cant see a way forward to a solution

Answer 42

this question is a real challange :)

Answer 43

the general solution is \[f(x)=Ae ^{x}+Be ^{\frac{ -1+\sqrt{5} }{ 2 }x}+Ce ^{\frac{ -1-\sqrt{5} }{ 2 }x}\] and use the boundary condition to find A,B,C

Answer 44

thanx but how did you come to this ?

Answer 45

from charactreristic equation \[m ^{3}-2m+1=0 \] \[(m-1)(m ^{2}+m-1)=0\]

Answer 46

it's not 2m. it's f'^2 not 2f'

Answer 47

yes kc is true

Answer 48

thats the hard part f'^2

Answer 49

ah srry

Answer 50

if we let g=f' we get g''-g^2+1=0

according to my research this DE solves to an elliptic integral
In general, elliptic integrals cannot be expressed in terms of elementary functions, so there may be no solution to this particular problem

Answer 51

I was thinking of saying f'=p like yours then the equation becomes p.p''-p^2+1=0 after that if we say p''=z equation will be pz-p^2+1=0 and z= (p^2-1)/p

Answer 52

do you think i can reach somewhere from here ?

Answer 53

if you make p=f' you'll get p''-p^2+1=0
which is essentially the same, so I don't think it'll work

Answer 54

what if after getting z=(p^2-1)/p integrating both side

Answer 55

I was playing around noticing that if you plug in some values, you can find out from:

f’(0)=0, f’(∞)=1, f’’(∞)=0

f’’’(∞)-f’(∞)^2+1=0

that:
f’’’(∞)-(1)^2+1=0
f’’’(∞)=0

And you can take the derivative of the original formula

d/dx(f’’’-f’^2+1=0)
f’’’’-2f’*f’’=0
f’’’’=2f’*f’’
f’’’’(∞)=2f’(∞)*f’’(∞)
f’’’’(∞)=0

and that will continue to lead us down the road that: \[f^{(n)}(∞) =0\] for n>1.

Continuing using our differential equation we can see:
f’’’(0)-f’(0)^2+1=0
f’’’(0)=-1

but the trail stops here since we don't know f’’(0).

So from this it seems pretty fair and quite simple to guess from that:

\[f’(x)=1-e^{-x}\]

since it fulfills all our boundary conditions:

f’(0)=0, f’’’(0)=-1; f’(∞)=1, f^(n)(∞)=0 for n>1

So then, fairly satisfied, we can then integrate this to get an answer that fulfills the boundary conditions.

\[f(x)=x+e^{-x}+C\]

Answer 56

thank you :)

Answer 57

all yours :)

Answer 58

note this is a first-order ODE in \(u=f'\):$$u'+u^2+1=0$$hence $$\frac{du}{dt}=-(1+u^2)\\\frac1{1+u^2}\frac{du}{dt}=-1\\\int \frac1{1+u^2}\frac{du}{dt}dt=\int-1\ dt$$by the chain-rule the left simplifies$$\int\frac1{1+u^2}du=\int -1\ dt\\\arctan u=c_1-t\\u=\tan(c_1-t)$$

Answer 59

to solve for \(f\) we have that \(u=\frac{df}{dt}\) so$$\frac{df}{dt}=\tan(c_1-t)\\f(t)=\int \tan(c_1 -t)\ dt=-\log|c_1-t|+c_2$$

Answer 60

actually I may have misread your initial equation -- but I'm too far in to stop so:

f’’’-f’^2+1=0 solution of this equation with boundary conditions f’(0)=0, f’(∞)=1, f’’(∞)=0 is needed so badly
$$u(t)=\tan(c_1-t)\\u(0)=0\implies c_1=\pi n,n\in\mathbb{Z}\\\lim_{t\to\infty}u(t)=1\ \text{oops! u(t) permits no such limit}$$

Answer 61

Nice try Oldrin.

Kainui, if you substitute f'= 1-e^-x, f'' = e^-x, f''' = -e^-x in the original equation
then you don't get zero on the LHS, so I don't think that your valiant attempt at a solution is correct.

Answer 62

@alekos haha yeah that's kind of a problem isn't it?

Answer 63

Sinan, I'd be really interested to know if your lecturer has a solution to this very perplexing problem

Answer 64

http://www.wolframalpha.com/input/?i=y''%3D(y%2B1)(y-1)&t=crmtb01

It says there's a step-by-step solution for this on Wolfram actually.

Answer 65

something tells me elliptic functions are well past intro ODE material

Answer 66

Kainui, Sinan
I just plugged in y'''-y'^2+1=0 in the Wolfram Alpha DE solver and we get a Weierstrass Zeta function which is an elliptic function. I've never come across this type of function before, and it does say that the step by step solution is available after creating an account.

Answer 67

Just created an account but it says that there is no step by step solution available?
So we are none the wiser as to how this was arrived at, but at least we know the answer.

Answer 68

yes i know i tried everything it s not a simple question, anyway i submitted whatever i could do :) i ll ask my lecturer if he has solution, i ll keep you posted :) thanx anyway