A system of 1st order nonlinear differential equations

In summary: No, an analytical solution is not impossible. However, often numerical solutions are more efficient and accurate.
  • #1
quacam09
16
0
Hello,

Can you give some suggestions to solve the following system of 1st order nonlinear differential equations?

Thank you.

[tex]

\[
\begin{array}{l}
u'(t) = Au^2 (t) + B(t)u + C(t) \\
u(t) = \left[ {\begin{array}{*{20}c}
{x_1 (t)} \\
{x_2 (t)} \\
\end{array}} \right] \\
A = \left[ {\begin{array}{*{20}c}
{a_{11} } & {a_{12} } \\
{a_{21} } & {a_{22} } \\
\end{array}} \right] \\
B(t) = \left[ {\begin{array}{*{20}c}
{f_{11} (t)} & {f_{12} (t)} \\
{f_{21} (t)} & {f_{22} (t)} \\
\end{array}} \right] \\
C(t) = \left[ {\begin{array}{*{20}c}
{g_{11} (t)} & {g_{12} (t)} \\
{g_{21} (t)} & {g_{22} (t)} \\
\end{array}} \right] \\
\end{array}
\]

[/tex]
 
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  • #2
quacam09 said:
Hello,

Can you give some suggestions to solve the following system of 1st order nonlinear differential equations?

Thank you.

[tex]

\[
\begin{array}{l}
u'(t) = Au^2 (t) + B(t)u + C(t) \\
u(t) = \left[ {\begin{array}{*{20}c}
{x_1 (t)} \\
{x_2 (t)} \\
\end{array}} \right] \\
A = \left[ {\begin{array}{*{20}c}
{a_{11} } & {a_{12} } \\
{a_{21} } & {a_{22} } \\
\end{array}} \right] \\
B(t) = \left[ {\begin{array}{*{20}c}
{f_{11} (t)} & {f_{12} (t)} \\
{f_{21} (t)} & {f_{22} (t)} \\
\end{array}} \right] \\
C(t) = \left[ {\begin{array}{*{20}c}
{g_{11} (t)} & {g_{12} (t)} \\
{g_{21} (t)} & {g_{22} (t)} \\
\end{array}} \right] \\
\end{array}
\]

[/tex]

What does u2(t) mean? Is it u(t) [itex]\cdot[/itex] u(t)?

Also, shouldn't the differential equation be
[tex]u'(t) = Au^2 (t) + B(t)u(t) + C(t) [/tex]
?
 
  • #3
quacam09 said:
Hello,

Can you give some suggestions to solve the following system of 1st order nonlinear differential equations?

Thank you.

[tex]

\[
\begin{array}{l}
u'(t) = Au^2 (t) + B(t)u + C(t) \\
As Mark44 notes, the "u^2" doesn't make sense here. If it is the dot product, then multiplying it by a two by two matrix doesn't make sense. If it is the cross product, then Both Au^2 and Bu are 2 dimensional vectors but you cannot add that to C(t), a two by two matrix.

u(t) = \left[ {\begin{array}{*{20}c}
{x_1 (t)} \\
{x_2 (t)} \\
\end{array}} \right] \\
A = \left[ {\begin{array}{*{20}c}
{a_{11} } & {a_{12} } \\
{a_{21} } & {a_{22} } \\
\end{array}} \right] \\
B(t) = \left[ {\begin{array}{*{20}c}
{f_{11} (t)} & {f_{12} (t)} \\
{f_{21} (t)} & {f_{22} (t)} \\
\end{array}} \right] \\
C(t) = \left[ {\begin{array}{*{20}c}
{g_{11} (t)} & {g_{12} (t)} \\
{g_{21} (t)} & {g_{22} (t)} \\
\end{array}} \right] \\
\end{array}
\]

[/tex]
 
  • #5
For some reason, HallsOfIvy's reply didn't render correctly. Here it is.
HallsofIvy said:
As Mark44 notes, the "u^2" doesn't make sense here. If it is the dot product, then multiplying it by a two by two matrix doesn't make sense. If it is the cross product, then Both Au^2 and Bu are 2 dimensional vectors but you cannot add that to C(t), a two by two matrix.
 
  • #6
Mark44 said:
For some reason, HallsOfIvy's reply didn't render correctly. Here it is
As Mark44 notes, the "u^2" doesn't make sense here. If it is the dot product, then multiplying it by a two by two matrix doesn't make sense. If it is the cross product, then Both Au^2 and Bu are 2 dimensional vectors but you cannot add that to C(t), a two by two matrix..

Thanks for your help and sorry for unclear things.

u^2 is a cross product. It means
[tex]


\[
u^2 (t) = \left[ {\begin{array}{*{20}c}
{x_1^2 (t)} \\
{x_2^2 (t)} \\
\end{array}} \right] \\

\]
[/tex]

And C(t)
[tex]
C(t) = \left[ {\begin{array}{*{20}c}
{g_{1} (t)} \\
{g_{2} (t)} \\
\end{array}} \right] \\

[/tex]
 
  • #7
EnumaElish said:
For a numeric solution the Runge-Kutta method seems an especially good method to use: http://www.springerlink.com/content/w080u7262137j867/

Thanks EnumaElish. Is an analytical solution impossible? If there is a method to obtain an analytical solution, can you suggest me?
 

1. What does it mean for a differential equation to be nonlinear?

A nonlinear differential equation is one in which the terms involving the dependent variable and its derivatives do not form a straight line. This means that the rate of change of the dependent variable is not directly proportional to the dependent variable itself, as is the case in linear differential equations.

2. How does a system of 1st order nonlinear differential equations differ from a single nonlinear differential equation?

A system of 1st order nonlinear differential equations is a set of multiple differential equations that are related to each other and involve the same independent and dependent variables. This is different from a single nonlinear differential equation, which only involves one dependent variable and its derivatives.

3. What makes solving a system of 1st order nonlinear differential equations challenging?

Solving a system of 1st order nonlinear differential equations can be challenging because there is no general method or formula for finding exact solutions. In most cases, numerical or approximate methods must be used to find solutions, which can be time-consuming and may not always give accurate results.

4. Can a system of 1st order nonlinear differential equations have more than one solution?

Yes, a system of 1st order nonlinear differential equations can have multiple solutions. This is because the equations can have different initial conditions or parameters, which can lead to different solutions. It is important to specify the initial conditions when solving a system of nonlinear differential equations to get a unique solution.

5. How are systems of 1st order nonlinear differential equations used in real-world applications?

Systems of 1st order nonlinear differential equations are commonly used in many fields of science and engineering to model complex systems and phenomena. For example, they can be used to model the growth of populations, the spread of diseases, and the behavior of chemical reactions. They are also used in physics to describe the motion of particles in non-uniform fields.

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