Solving Nonlinear Diff. Equation with Boundary Conditions

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The discussion focuses on solving the nonlinear differential equation u''=K/((a-u)^2) with boundary conditions u(0)=0 and u(L)=0. The solution involves manipulating the equation by multiplying with u' and integrating, leading to the expression u'=\pm\sqrt{\frac{C_{3}-C_{4}u}{a-u}}. The integration process requires a substitution v=\sqrt{\frac{a-u}{C_{3}-C_{4}u}}, ultimately simplifying to partial fractions decomposition for an implicit solution. Special attention must be given to the conditions on K and a when applying the boundary conditions.

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astromanish
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Hi,

Can somebody guide me in getting solution for the following
non-linear differential equation?

u''=K/((a-u)^2) ...K is a constant

Where u is a function of x and the boundary conditions are
u(0)=0
u(L)=0

I need solution in form of 'u' as a function of 'x' with two constants, which are to be determined from the boundary conditions.

Thanks
 
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Multiply with u', integrate, and see if you can do something more.
 
To help you along a bit:
[tex]u'u''=\frac{Ku'}{(a-u)^{2}}\to\frac{1}{2}(u')^{2}=C_{1}+\frac{K}{(a-u)}=\frac{C_{2}-C_{1}u}{a-u},C_{2}=C_{1}a+K[/tex]
Or, we get:
[tex]u'=\pm\sqrt{\frac{C_{3}-C_{4}u}{a-u}},C_{3}=2C_{2},C_{4}=2C_{1}[/tex]

Rewriting, and integrating, we have:
[tex]\int\sqrt{\frac{a-u}{C_{3}-C_{4}u}}du=x+C_{5}[/tex]
You may crack the integral at your left-hand side using the substitution:
[tex]v=\sqrt{\frac{a-u}{C_{3}-C_{4}u}}[/tex]
since we have:
[tex]u=\frac{a-C_{3}v^{2}}{1-C_{4}v^{2}}[/tex]
Thus, the problem is resolved to partial fractions decomposition, and an implicit formula of the solution function.
You might well encounter some non-trivial conditions placed on K and a when imposing the boundary conditions.

Good luck! :smile:
 
Last edited:

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