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## Main Question or Discussion Point

Hi guys!

I'm to find the solution to

[tex]\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} [/tex]

Subject to an initial condition

[tex] u(x,0) = u_0(x) = a \exp(- \frac{x^2}{2c^2}) [/tex]

And Neumann boundary conditions

[tex] \frac{\partial u}{\partial x} (-1,t) = \frac{\partial u}{\partial x} (1,t) = 0 [/tex]

I can usually do this no problem assuming the domain is for instance [0,L], but I get stuck with this one :

Using separation of variables :

[tex] u(x,t) = f(x)g(t) [/tex]

This yields:

[tex] \frac{1}{g} \frac{dg}{dt} = \frac{1}{f} \frac{d^2f}{dx^2} = -\lambda [/tex]

Spatial Part:

[tex] \frac{1}{f} \frac{d^2f}{dx^2} = -\lambda [/tex]

[tex] \frac{d^2f}{dx^2} + \lambda f = 0 [/tex]

Therefore :

[tex] f(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x) [/tex]

And since I'm considering Neumann Boundary conditions I get the derivative of this

[tex] f \prime (x) = -A \sqrt{\lambda}\sin(\sqrt{\lambda} x) + B \sqrt{\lambda} \cos(\sqrt{\lambda} x) [/tex]

So, [tex] f \prime (-1) = 0 [/tex]

This gives:

[tex] A \sin(\sqrt{\lambda}) + B \cos(\sqrt{\lambda}) = 0[/tex]

And for [tex] f \prime (1) = 0 [/tex]

I get :

[tex] -A \sin(\sqrt{\lambda}) + B \cos(\sqrt{\lambda}) = 0[/tex]

So from these two equations I can conclude that:

Firstly by just adding the two equations

[tex] B \cos(\sqrt{\lambda}) = 0 [/tex]

So either [itex] B = 0 [/itex] or [itex] \cos(\sqrt{\lambda}) = 0 [/itex]

Now substituting [itex] B \cos(\sqrt{\lambda}) = 0 [/itex] back into [itex] A \sin(\sqrt{\lambda}) + B \cos(\sqrt{\lambda}) = 0[/itex]

I also get

[tex] A \sin(\sqrt{\lambda}) = 0 [/tex]

So either [itex] A = 0 [/itex] or [itex] \sin(\sqrt{\lambda}) = 0 [/itex]

Obviously [itex] \sin(\sqrt{\lambda}) [/itex] and [itex] \cos(\sqrt{\lambda}) [/itex] can't both equal zero, so how do I approach this...

Apologies if this is a stupid question..

Any help would be greatly appreciated

Max

I'm to find the solution to

[tex]\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} [/tex]

Subject to an initial condition

[tex] u(x,0) = u_0(x) = a \exp(- \frac{x^2}{2c^2}) [/tex]

And Neumann boundary conditions

[tex] \frac{\partial u}{\partial x} (-1,t) = \frac{\partial u}{\partial x} (1,t) = 0 [/tex]

I can usually do this no problem assuming the domain is for instance [0,L], but I get stuck with this one :

Using separation of variables :

[tex] u(x,t) = f(x)g(t) [/tex]

This yields:

[tex] \frac{1}{g} \frac{dg}{dt} = \frac{1}{f} \frac{d^2f}{dx^2} = -\lambda [/tex]

Spatial Part:

[tex] \frac{1}{f} \frac{d^2f}{dx^2} = -\lambda [/tex]

[tex] \frac{d^2f}{dx^2} + \lambda f = 0 [/tex]

Therefore :

[tex] f(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x) [/tex]

And since I'm considering Neumann Boundary conditions I get the derivative of this

[tex] f \prime (x) = -A \sqrt{\lambda}\sin(\sqrt{\lambda} x) + B \sqrt{\lambda} \cos(\sqrt{\lambda} x) [/tex]

So, [tex] f \prime (-1) = 0 [/tex]

This gives:

[tex] A \sin(\sqrt{\lambda}) + B \cos(\sqrt{\lambda}) = 0[/tex]

And for [tex] f \prime (1) = 0 [/tex]

I get :

[tex] -A \sin(\sqrt{\lambda}) + B \cos(\sqrt{\lambda}) = 0[/tex]

So from these two equations I can conclude that:

Firstly by just adding the two equations

[tex] B \cos(\sqrt{\lambda}) = 0 [/tex]

So either [itex] B = 0 [/itex] or [itex] \cos(\sqrt{\lambda}) = 0 [/itex]

Now substituting [itex] B \cos(\sqrt{\lambda}) = 0 [/itex] back into [itex] A \sin(\sqrt{\lambda}) + B \cos(\sqrt{\lambda}) = 0[/itex]

I also get

[tex] A \sin(\sqrt{\lambda}) = 0 [/tex]

So either [itex] A = 0 [/itex] or [itex] \sin(\sqrt{\lambda}) = 0 [/itex]

Obviously [itex] \sin(\sqrt{\lambda}) [/itex] and [itex] \cos(\sqrt{\lambda}) [/itex] can't both equal zero, so how do I approach this...

Apologies if this is a stupid question..

Any help would be greatly appreciated

Max