Guidance of 2nd order linear ODE

QTM912

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Joined
Oct 19, 2019
Messages
3
Hi all

I need some guidance on solving a very simple ODE of the form : y'' + y' = 0 , y a function of x. This will then be used to solve an inhomogeneous ODE using a Greens function but that is a later step that cannot be addressed until the below problem is resolved. And the problem is that the boundary conditions seem to suggest a trivial solution so I want to check if the problem has been correctly specified or am I missing something.

My two independent solutions are denoted y1 and y2 such that we can call y = y1 + y2 and the boundary conditions are: when x = 0 , y1 = 0 and when x=1 y2 = 0
The solution to the ODE is found to be : y = A + B exp (-x), with A and B constants to be determined from boundary conditions

So according to me, if I set y1 = A and y2 = B exp(-x)

At x = 0, y1 = 0 implies I think that A has to be zero as there is no dependence on x and this is the only way y1 =0
At x = 1 we are told that y2 = 0 , but that suggests B = 0 which makes the whole solution trivial => A=B =0

Alternatively just as a check if we call y1 = B exp (-x) and y2 = A
then at x=0, y1 = 0 which means B=0 and when x=1 y2= 0 again implying that A = 0 => A=B=0 again

This seems to trivialise the whole problem to y = 0 which is obviously a solution but not very interesting

So either I am doing something seriously wrong such as using the wrong solution to the ODE, missing something really key or the problem is mis-specified with incorrect boundary conditions - I am inclined to believe the first two possibilities and am looking for guidance not fully answers -unless you think it is an ill posed question.

Thank you.
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
19,318
Hi all

I need some guidance on solving a very simple ODE of the form : y'' + y' = 0 , y a function of x. This will then be used to solve an inhomogeneous ODE using a Greens function but that is a later step that cannot be addressed until the below problem is resolved. And the problem is that the boundary conditions seem to suggest a trivial solution so I want to check if the problem has been correctly specified or am I missing something.

My two independent solutions are denoted y1 and y2 such that we can call y = y1 + y2 and the boundary conditions are: when x = 0 , y1 = 0 and when x=1 y2 = 0
The solution to the ODE is found to be : y = A + B exp (-x), with A and B constants to be determined from boundary conditions

So according to me, if I set y1 = A and y2 = B exp(-x)

At x = 0, y1 = 0 implies I think that A has to be zero as there is no dependence on x and this is the only way y1 =0
At x = 1 we are told that y2 = 0 , but that suggests B = 0 which makes the whole solution trivial => A=B =0

Alternatively just as a check if we call y1 = B exp (-x) and y2 = A
then at x=0, y1 = 0 which means B=0 and when x=1 y2= 0 again implying that A = 0 => A=B=0 again

This seems to trivialise the whole problem to y = 0 which is obviously a solution but not very interesting

So either I am doing something seriously wrong such as using the wrong solution to the ODE, missing something really key or the problem is mis-specified with incorrect boundary conditions - I am inclined to believe the first two possibilities and am looking for guidance not fully answers -unless you think it is an ill posed question.

Thank you.
Did you mean:

"My two independent solutions are denoted y1 and y2 such that we can call y = y1 + y2 and the boundary conditions are:

when x = 0 , y = 0 and

when x = 1, y = 0"

When y = 0 \(\displaystyle \to \) y1 + y2 = 0
 

QTM912

New member
Joined
Oct 19, 2019
Messages
3
Hi Subhotosh, thanks for replying.

Actually no, it was not what I meant, at least not until your final line -

The general solution to this is expressed as a sum of the linearly independent solutions y1 and y2

The boundary conditions that are given to me relate specifically to the two solutions y1 and y2 separately - and not to y as a whole

Therefore : at x=0, the boundary condition of the y1 solution is y1 =0

While at x=1 the boundary condition of the y2 solution is y2 =0

Where as mentioned y = y1 + y2

Of course this means, as mentioned above, that the last line of your reply : "when y=0 => y1+y2 = 0 " must be correct

This is the first time I am seeing boundary conditions expressed in this way but that's what I am given.

Hope this clarifies.

Thanks.
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
19,318
Hi Subhotosh, thanks for replying.

Actually no, it was not what I meant, at least not until your final line -

The general solution to this is expressed as a sum of the linearly independent solutions y1 and y2

The boundary conditions that are given to me relate specifically to the two solutions y1 and y2 separately - and not to y as a whole

Therefore : at x=0, the boundary condition of the y1 solution is y1 =0

While at x=1 the boundary condition of the y2 solution is y2 =0

Where as mentioned y = y1 + y2

Of course this means, as mentioned above, that the last line of your reply : "when y=0 => y1+y2 = 0 " must be correct

This is the first time I am seeing boundary conditions expressed in this way but that's what I am given.

Hope this clarifies.

Thanks.
Please post the EXACT problem that was given to you - verbatim.

Wording for this "particular" problem does not affect the result - but arithmetic will be slightly different.
 

QTM912

New member
Joined
Oct 19, 2019
Messages
3
Thank you Subhotosh once again for your efforts and comments which caused me to reflect further. Having laboured over this some more I have managed to solve the underlying problem of which this was but one part - it seems to be the case that the trivial solution may be the correct conclusion and subsequent parts of the question have answers that are not trivial. I won't trouble you further but just remark that the speed of the responses received about this are very impressive and greatly appreciated by novices like me who are looking for guidance. I will find out soon enough if I actually managed to solve this correctly.
 
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