Solve differential equation with boundary conditions using substitution

  • #1
μ[itex]^{2}[/itex][itex]\frac{d^{2}u}{dx^{2}}[/itex]+ae[itex]^{u}[/itex]=0

Boundary conditions: u(-L)=u(L)=u[itex]_{0}[/itex]

Solve by multiplying by [itex]\frac{du}{dx}[/itex] and integrating in x

I know you have to use substitution, but I keep going in circles.
 

Answers and Replies

  • #2
67
0
μ[itex]^{2}[/itex][itex]\frac{d^{2}u}{dx^{2}}[/itex]+ae[itex]^{u}[/itex]=0

Boundary conditions: u(-L)=u(L)=u[itex]_{0}[/itex]

Solve by multiplying by [itex]\frac{du}{dx}[/itex] and integrating in x

I know you have to use substitution, but I keep going in circles.

The way that I would do it would be to rewrite the equation as
[itex]\mu[/itex][itex]^2[/itex] [itex]'' = -ae^\mu[/itex]

Then you solve for a general solution which is
[itex]\mu(x) = 1 + x[/itex]

Then you go on to get your general solution from the boundaries and such.

Another method, which would probably be easier and is what I think you were trying to do is to just integrate twice and get:
[itex]\frac{\mu^4}{12} = -ae^\mu + \mu*c_1 + c_2[/itex]

That may be slightly off, but try integrating it yourself, and then use your boundary conditions to solve for the constants.
 
Last edited:
  • #3
798
34
Hi !
I think that the two methods proposed by danielu13 are both incorrect.
Better follow the advice given in the wording : multiply by u'
m² u'' +a exp(u) = 0
m² 2 u''u' +2a exp(u) u' = 0 and integrate :
m² u'² +2a exp(u) = c
m u' = (+ or -)sqrt(c-2a exp(u))
Case +sqrt :
m (du/dx) = sqrt(c-2a exp(u))
dx = m du / sqrt(c-2a exp(u))
Then integrate, wich gives x(u)
and invert it to obtain u(x).
 

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