# First order non linear boundary value pb

• sbmmth
In summary, the conversation discusses a two points boundary value problem involving a function with known and unknown parameters. The function follows a specific equation and has two boundary conditions. The difficulty lies in finding the unknown parameter and a constant needed to fully express the function. One approach is to solve the ODE and use numerical methods to find the unknown parameter, while another is to solve for the unknown term and obtain an implicit expression for the function.
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!

sbmmth said:
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!

Consider the equation:

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

now, just separate variables and integrate.

jackmell said:
Consider the equation:

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

now, just separate variables and integrate.

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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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.

hunt_mat said:
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.

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.

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?

## 1. What is a first order non linear boundary value problem?

A first order non linear boundary value problem is a mathematical problem that involves finding a solution to a differential equation with certain boundary conditions. It is called "non linear" because the equation is not of the form y' = f(x,y), and it is called "first order" because it involves only the first derivative of the unknown function.

## 2. How is a first order non linear boundary value problem different from a first order linear boundary value problem?

The main difference between a first order non linear boundary value problem and a first order linear boundary value problem is the form of the differential equation. A first order linear boundary value problem has an equation of the form y' = f(x,y), while a non linear problem has an equation of the form y' = f(x,y,y'). This non linearity makes it more challenging to solve.

## 3. What are some common techniques used to solve first order non linear boundary value problems?

Some common techniques used to solve first order non linear boundary value problems include separation of variables, substitution methods, and numerical methods such as the Euler method or the Runge-Kutta method. These methods involve breaking down the problem into smaller, more manageable parts and finding a numerical solution.

## 4. What are the applications of first order non linear boundary value problems in science?

First order non linear boundary value problems have various applications in science, particularly in physics, engineering, and biology. They can be used to model and understand complex systems and phenomena, such as chemical reactions, population growth, and fluid dynamics.

## 5. Are there any real-world examples of first order non linear boundary value problems?

Yes, there are many real-world examples of first order non linear boundary value problems. Some examples include the Brusselator model in chemical reactions, the Lotka-Volterra model in predator-prey relationships, and the Navier-Stokes equations in fluid dynamics. These problems have practical applications in fields such as chemistry, ecology, and aeronautics.

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