# First order non linear boundary value pb

1. Mar 30, 2012

### sbmmth

Hi there:

I am trying to solve a two points boundary value problem.
Consider a function f:[x1,x2]->[x2,x3]

x1 and x3 are knowns
x2 is an unknown parameter

f'(x) = exp( -a*x + b*f(x) )

where b>a>0

Two boundaries conditions:

f(x1) = x2
f(x2) = x3

Does anyone know how to solve it?

Thanks!!!!

2. Mar 30, 2012

### jackmell

Consider the equation:

$$f'=e^{-ax}e^{bf}$$

now, just separate variables and integrate.

3. Mar 31, 2012

### JJacquelin

I agree, solving the ODE is rather easy. But the difficulty isn't here.
The boundary conditions raise a very difficult problem:
f(x1)=x2 and f(x2)=x3 with given x1, x3 and unknown x2
In fact, x2 cannot be expressed with a combination of a finit number of usual functions of x1 and x3.
In this case, we need numerical solving in order to compute x2 and then, the unknown constant C which is necessary to fully express f(x).

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4. Apr 1, 2012

### hunt_mat

The problem is that you have a first order equation and two unknowns and that isn't possible, for a boundary value problem you need a second order equation. One of your points could be an attractor which is possible but you would need to check that to begin with.

5. Apr 2, 2012

### JJacquelin

I do not agree. A second order equation is not needed.
Apparently, there are two boundary conditions f(x1)=x2 and f(x2)=x3. But, in fact, there is only one bondary condition, which is : f(f(x1))=x3 with known x1 and x3. Since x2 is not known, only one relationship f(f(x1))=x3 is remaining.
The ODE f'(x) = exp( -a*x + b*f(x) ) and the boundary condition f(f(x1))=x3, with given a, b, x1, x3 is a complete system, needing nothing more to be fully defined.

6. Apr 2, 2012

### hunt_mat

So I misread the question, apologies, you basically have a differential equation of the form:
$$e^{-bf(x)}\frac{df}{dx}=e^{-ax}$$
So to begin with, integrate between $x_{1}$ and $x$ to obtain:
$$e^{-bf(x)}=e^{-bx_{2}}+\frac{b}{a}e^{-ax}-\frac{b}{a}e^{-ax_{1}}$$
And integrating between $x$ and $x_{2}$ to obtain:
$$e^{-bf(x)}=e^{-bx_{3}}+\frac{b}{a}e^{-ax}-\frac{b}{a}e^{-ax_{2}}$$
Now from here I see two possible ways forward, you equate that equations and obtain an equation for $x_{2}$ and use Newtons method to find that or you can solve for the unknown term and obtain a implicit expression for $f(x)$ depending on the property of b/a. It's your choice. Do you have numerical values for $x_{1}$,$x_{3}$, a and b?