Let a ∈ ℝ and b > 0 be fixed. Find all functions f : [0, ∞) → ℝ satisfying the differential equation f'(x) = -a + b/f(x) for x > 0 and f(0) = 0.
Thank you Omran, I removed the incorrect statements from my latest file. I ultimately combined both Kåre's and your approach, and came up with the file attached here.
I suggest at first replacing f'(x) by df(x)/dx. Now, multiplying both sides by f(x) leads to
fdf=(-af+b)dx
by integrating both sides, it is obtained that
1/2(f^2)=-aF+bx+c
I guess f can be of the linear form. I am not sure it is unique.
Yes, and yes^h^h^h no.
The inverse function satisfies dx/df = f/(b-af). I.e.
x(f) = ∫0f ydy/(b-ay) = −∫f0 ydy/(b-ay),
where the first formula looks most natural for f>0 and the latter for f0. For a>0 the function x(f) maps [0, b/a) to [0, ∞) in a unique monotonous manner, otherwise it maps [0,∞) to [0,∞) in a unique monotonous manner. Hence, in either case its unique inverse f(x) exist for all x in [0,∞).
For a=0 one finds y(x) = (2bx)1/2. Otherwise y(x) can be expressed in terms of the Lambert-W function.
Added: I have not carefully checked the sanity of this result, but a bit of plain algebra for a≠0 seems to lead to the expression f(x) = (b/a)[1 + W(-e-(1+a^2 x/b))], where W is the Lambert-W function: If z= u eu, then u = W(z). However, there are actually (infinity) many branches of this inversion. One must carefully choose the correct branch; it cannot be the same for a0.
Added 26.01.2021: The two functions discussed above are respectively
f+(x) = (b/a)[1+W0(−e−(1+a^2 x/b))] (non-negative for a>0), and
f-(x) = (b/a)[1+W-1(−e−(1+a^2 x/b))] (non-negative for a0. This is a mistake: Both solutions above are valid for both signs of a. This was first given correctly in the post by Omran Kouba just below.
There are exactly two solutions.
- The case a=0 is easy and simple integration yields the two solutions f+(x)=(2bx)1/2 and f-(x)=-(2bx)1/2.
- In tge case of nonzero a we note that any solution f to the proposed problem yields a solution g to the simplified problem (P) : ” g’=1+1/g with g(0)=0 “, given by g(x)=-(a/b)f(bx/a2). Conversely, any solution to (P) yields a solution to the original problem given by f(x)=-(b/a)g(a2 x/b). We only need to consider (P).
- The differential equation y’=1+1/y has a constant solution given by y(x)=-1 for all x. So, the graph of any other solution must not intersect the lines y=0, or y=-1. So, any solution g to (P) is either positive on (0,+∞) or takes its values in (-1,0). It also satisfies (g-ln(1+g))’=1. That is g(x)-ln(g(x)+1)=x+k for some constant k. The continuity of g at 0 with g(0)=0 yields k=0, so any solution g of (P) satisfies g(x)-ln(g(x)+1)=x for all x>0.
- Now, h1:(-1,0]→[0,+∞), h1(t)=t-ln(1+t) is a decreasing homeomorphism, and similarly, h2:[0,+∞)→[0,+∞), h2(t)=t-ln(1+t) is an increasing homeomorphism. Thus if g is a solution of (P) that takes its values in [0,+∞) then from h2(g(x))=x we see that g(x)=h2-1(x), for nonnegative x, and if g is a solution of (P) that takes its values in (-1,0] then from h1(g(x))=x we see that g(x)=h1-1(x), for nonnegative x. Conversely, it is straightforward to check that h1-1 and h2-1 are solutions to (P). Thus, the proposed problem has exactly two solutions for every (a,b) with real a and positive b.
I put together the excellent suggestions so far in the attached pdf file. Although I added some of my own, I have a feeling that I am still missing something... Please check the details for me and with me.
Dear George Stoica ,
As I proved there are two solutions, only one of them tends to a finite limit as x tends to infinity the other one tends to -sgn(a) times infinity as x tends to infinity.
Thank you Omran, I removed the incorrect statements from my latest file. I ultimately combined both Kåre's and your approach, and came up with the file attached here.
Dear George Stoica ,
You only need to consider the case a=0 separately because your solution does not cover this easy case. Thanks
Dear George Stoica
In addition to the helpful answers, I would like to add:
It is an elementary separable ordinary differential equation.
It has the form ( a is not zero).
(y/(b-ay))dy = dx ⇛ (-aydy)/(b-ay))= -adx
⇛ (-ay+b-b)/(b-ay))dy = -adx ⇛ (1 - b/b-ay)dy = -adx
Direct integration yeilds the general solution
y+(b/a)ln(b - ay)= -ax+c
Solve for c, we obtain: c = (b/a)ln(b)
Therefore the general solution is given by
y+(b/a)ln(b-ay)=-ax+(b/a)ln(b)⇛
(b/a)ln(1-(a/b)y) +ax + y=0 (G.S.)
In particular, if a =0,
((dy)/(dx))=b/y.y(0)=0,b>0
Then ydy=bdx and y²=2bx+c,
y(0)=0, implies the parabola. y²=2bx.
Best regards
- Badges
- Science topic
- Similar topics
- Mathematics