Can Nonlinear Boundary Value Problems Be Solved Analytically?

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The nonlinear boundary value problem defined by the equations x'=-y/sqrt(x^2+y^2) and y'=x/sqrt(x^2+y^2) with initial conditions x(0)=y(0)=-1 and x(pi/4)=y(pi/4)=1 can be solved analytically using polar coordinates. By rewriting the system in vector form, it is established that the radius r remains constant, leading to a solution that describes circular motion. The analytical solution is expressed as x(t)=Rcos(t/R+φ) and y(t)=Rsin(t/R+φ), where R and φ are determined by the boundary conditions.

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amr07
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Dear everybody,

I am new here. I need your suggesstion and helping for solving the next nonlinear boundary value problem:

x'=-y/sqrt(x^2+y^2)
y'=x/sqrt(x^2+y^2)
with x(0)=y(0)=-1 and x(pi/4)=y(pi/4)=1

so is it easy to get the analytical solution to the problem above or I have to solve this problem numerically.
thanks in advance

Amr
 
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Using polar coordinates, and rewriting in vector form you get the equivalent equation:

[tex]\frac{d\vec{r}}{dt}=\hat{\theta}[/tex] (1)

[tex]\frac{d\vec{r}}{dt}=\hat{r}\frac{dr}{dt}+r\frac{d\theta}{dt}\hat{\theta}[/tex]

Putting this in the equation (1) and comparing different components you get:

[tex]\frac{dr}{dt}=0 => r(t) \equiv const=R[/tex]
[tex]r\frac{d\theta}{dt}=1[/tex]

Since r is constant you get:

[tex]\theta=\frac{t}{r}+\varphi[/tex]

From this you get that:

[tex]x(t)=Rcos(\frac{t}{R}+\varphi); y(t)=Rsin(\frac{t}{R}+\varphi);[/tex]

You have a circular motion with a radius and initial phase to determine from boundary conditions.
 
Last edited:

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