Boundary value Problem

OP

maybe i am getting confused.

for part (a)

do i just need to show that for the problem

y(0) = 0 , y(1) = 0

the only solution is that y(x) = 0.

this is easily observable, but how do you 'show' it is unique? becaue y(x) = x^2 - x seems to satify these two conditions as well.... :s-smilie:

Reply 2

OP

anyone?, lol :tongue:

Reply 3

x^{2}-x

doesn't have a zero second derivative. y must be a linear function and the initial conditions make y=0.

Reply 4

OP

AlphaNumeric

x^{2}-x

doesn't have a zero second derivative. y must be a linear function and the initial conditions make y=0.

sorry, i don't quite understand. i am new to this particular type of question, need to take it in small steps :P

Reply 5

You're looking for something which satisfies y'' = 0 and then has y(0)=y(1)=0. The function

y = x^{2}-x

satisfies y(0)=y(1)=0, but it doesn't satisfy y''=0, so it cannot be the solution.

Integrate up y'' = 0 and you get y' = A. Integrate again and you've y = Ax+B. y(0) = 0 means B=0. y(1) = 0 gives A = 0. Therefore y=0.

Reply 6

OP

so is what you have said there enough to show that y=0 is a unique and the only solution?

Reply 7

Yes. If you've a second order ODE and 2 initial conditions (I think those two are known as Neumann conditions) then you will solve it uniquely.

Alternatively, suppose there exists a function f(x) which differentiates to give y' and that isn't Cx+D (imagine some weird trig polynomial thing). Then you have the possible solution y = Ax+B and y=f(x). You know that f'(x) is the same as A (since they both differentiate to y') so you have f'(x) = A. Integrate up and you've f(x) = Ax+C. By initial conditions you have that f(x) = 0. Hence there is no other function which satifies those conditions (actually you have to do it for both y' and y'', not just y' but it's exactly the same procedure).

Reply 8

OP

right ok, that makes sense. sometimes i seem to stuggle with the most simple of things! :P

would you be able to help me further with this question, after i post what i've done so far? :smile:

for instance..

the other bit of part (a) asks to find the 'Appropriate Green's Function'

now as far as i can see

The solution is:

y(x) = \int^b_aG(x,\epsilon)f(\epsilon)d\epsilon

The generalised green's function is given by:

Unparseable latex formula:

G(x,\epsilon) = \frac{u(x)v(\epsilon)}{W[u,v](\epsilon)

for

a \leq x \leq \epsilon

Unparseable latex formula:

G(x,\epsilon) = \frac{u(\epsilon)v(x)}{W[u,v](\epsilon)

for

\epsilon \leq x \leq b

Reply 9

I'll have a look at it when I can. I'm literally heading out the door to go to the maths dept. to do some revision. I'll point Wrangler at this thread, Green's functions more his thing (and he's always hungry for rep :wink:

)

Reply 10

OP

AlphaNumeric

I'll have a look at it when I can. I'm literally heading out the door to go to the maths dept. to do some revision. I'll point Wrangler at this thread, Green's functions more his thing (and he's always hungry for rep :wink:

)

lol, ok thanks for that! :smile:

Reply 11