Boundary Conditions and Solutions to Differential Equations

[cytochrome]

Homework Statement

Show that f(x) = A exp(σx) + B exp(-σx) is a solution to the
following differential equation:

f''(x) = (σ^2)f(x)

where A, B, and σ are constants. What if a boundary condition is
included that f(-∞) = 0?

Homework Equations

differential equation: f''(x) = (σ^2)f(x)
solution: f(x) = A exp(σx) + B exp(-σx)

The Attempt at a Solution

Proof -
Plugging exp(σx) and exp(-σx) into the equation f''(x) = (σ^2)f(x) gives an equality, therefore any linear combination of exp(σx) and exp(-σx) is a solution.


If the boundary condition f(-∞) = 0 is introduced, would only exp(σx) be a solution since exp(-∞)=0?

 

[haruspex]

If the boundary condition f(-∞) = 0 is introduced, would only exp(σx) be a solution since exp(-∞)=0?

That certainly allows B=0 to be a solution, but show also that B must be zero.
(Of course, this is assuming sigma > 0.)

 

[cytochrome]

That certainly allows B=0 to be a solution, but show also that B must be zero.
(Of course, this is assuming sigma > 0.)

I'm confused about how to do this?

B must be zero because Bexp^(-∞) = 0... is that sufficient?

 

[SammyS]

I'm confused about how to do this?

B must be zero because Bexp^(-∞) = 0... is that sufficient?

No. e^-σx → +∞ as x → -∞ , if σ > 0 .

 

[cytochrome]

So if the boundary condition f(-inf) is involved, then is the answer Aexp()?

Can someone please explain boundary conditions?

 

[tiny-tim]

hi cytochrome!

(try using the X² button just above the Reply box )

So if the boundary condition f(-inf) is involved, then is the answer Aexp()?

yes, the general solution is y = Ae^σx

Can someone please explain boundary conditions?

to completely solve a differential equation, you need as many boundary conditions as there are constants

here, there were two constants but only one boundary condition, so your solution still has one unknown constant

 

[cytochrome]

hi cytochrome!

(try using the X² button just above the Reply box )


yes, the general solution is y = Ae^σx


to completely solve a differential equation, you need as many boundary conditions as there are constants

here, there were two constants but only one boundary condition, so your solution still has one unknown constant

Thanks!

So the question is for the boundary condition f(-∞), therefore the answer would just be the general solution?

 

[haruspex]

Your very first solution was correct, it's just that the reasoning you offered was insufficient. A exp(σx) is the solution that satisfies the boundary condition. You just have to prove that B=0. What would happen to f(x) at -∞ if B were nonzero?

 

[cytochrome]

Your very first solution was correct, it's just that the reasoning you offered was insufficient. A exp(σx) is the solution that satisfies the boundary condition. You just have to prove that B=0. What would happen to f(x) at -∞ if B were nonzero?

If B ≠ 0 then f(-∞) = 0 doesn't make sense since this only works if f(x) = Aexp(σx).

Is my reasoning correct?

 

[HallsofIvy]

If B ≠ 0 then f(-∞) = 0 doesn't make sense since this only works if f(x) = Aexp(σx).

Is my reasoning correct?

I don't see any "reasoning" here, simply the assertion that "if B ≠ 0 then f(-∞) = 0 doesn't make sense". WHY does it not make any sense?

 

[cytochrome]

I don't see any "reasoning" here, simply the assertion that "if B ≠ 0 then f(-∞) = 0 doesn't make sense". WHY does it not make any sense?

Well f(-∞) = 0 implies that Aexp(x) is the only possible answer. Bexp(-x) doesn't work because as x → -∞ then Bexp(-x) → 0

Is that a good way to show it?

 

[haruspex]

as x → -∞ then Bexp(-x) → 0

No it doesn't! Get that right and you're home.