Applying Boundary Conditions in Homogeneous ODE Problems

[AntSC]

Find the eigenvalues λ, and eigenfunctions u(x), associated with the following homogeneous ODE problem:

{u}''\left ( x \right )+2{u}'\left ( x \right )+\lambda u\left ( x \right )=0\; ,\; \; u\left ( 0 \right )=u\left ( 1 \right )=0

Solution:

Try u\left ( x \right )=Ae^{rx}, which gives roots r=-1\pm \sqrt{1-\lambda }. Solution is altered with \lambda <1\; ,\; \; \lambda =1\; ,\; \; \lambda >1

For the first case \lambda <1:

General solution:
u\left ( x \right )=Ae^{\left ( -1+\sqrt{1-\lambda } \right )x}+Be^{\left ( -1-\sqrt{1-\lambda } \right )x}
u\left ( x \right )=C\cosh \left ( -1+\sqrt{1-\lambda } \right )x+D\sinh \left ( -1-\sqrt{1-\lambda } \right )x

Applying boundaries: (this is where my question lies - how to correctly apply BCs)

u\left ( 0 \right )=0 \; \; \Rightarrow \; \; C+D=0 (some cases I've seen the conclusion that only C=0). Do i assume that as \cosh is never zero that C=0 and therefore it must be that D=0. Or do i only take C=0 and then apply the second BC to see what happens to D?
The latter (assuming C=0) gives D\sinh \left ( -1-\sqrt{1-\lambda } \right )=0. So either D=0 or \sinh \left ( -1-\sqrt{1-\lambda } \right )=i\pi n.

I'm confused by the rules for the BCs. Can anyone point out how to proceed? Thanks

 

[ehild]

Find the eigenvalues λ, and eigenfunctions u(x), associated with the following homogeneous ODE problem:

{u}''\left ( x \right )+2{u}'\left ( x \right )+\lambda \left ( x \right )=0\; ,\; \; u\left ( 0 \right )=u\left ( 1 \right )=0

What does (x) mean? Is it the independent variable, and λ is also function of x, or the whole equation is multiplied by x, and λ is a constant? ehild

 

[Simon Bridge]

To apply a BC u(x=a)=b, you take the general expression for u(x), and everywhere you see an "x" you write in an "a", and at the end of the line you write "=b".

In your case, for u(0)=0 you write down:
$$C\cosh (-1+\sqrt{1-\lambda})(0)+D\sinh (-1-\sqrt{1-\lambda})(0) = 0$$ ... and simplify.

You don't need to make assumptions - you know the values of the cosh and sinh parts at x=0 and if you don't you can look them up.

Then repeat for u(1)=0 ...

Note: I had the same puzzle as ehild over $\lambda(x)$ ... I assumed it was a typo considering the solution and focussed on the answer to your question.
Have you checked your general solution against the DE?

 

[AntSC]

What does (x) mean? Is it the independent variable, and λ is also function of x, or the whole equation is multiplied by x, and λ is a constant?


ehild

Sorry, that was meant to be λu(x). I've corrected the original post.

 

[AntSC]

To apply a BC u(x=a)=b, you take the general expression for u(x), and everywhere you see an "x" you write in an "a", and at the end of the line you write "=b".

In your case, for u(0)=0 you write down:
$$C\cosh (-1+\sqrt{1-\lambda})(0)+D\sinh (-1-\sqrt{1-\lambda})(0) = 0$$ ... and simplify.

You don't need to make assumptions - you know the values of the cosh and sinh parts at x=0 and if you don't you can look them up.

Then repeat for u(1)=0 ...

Note: I had the same puzzle as ehild over $\lambda(x)$ ... I assumed it was a typo considering the solution.

Thanks Simon.
With the first BC I'm seeing that the result doesn't actually tell me anything about the coefficients as they're multiplied by zero. Is that correct? So i still know nothing about C and D. Assuming i got some kind of positive result from the first BC how does that impact on use of the second?. For instance if i got C=0 then do i ditch the first term in the general solution when i use the second BC?

 

[ehild]

Find the eigenvalues λ, and eigenfunctions u(x), associated with the following homogeneous ODE problem:

{u}''\left ( x \right )+2{u}'\left ( x \right )+\lambda u\left ( x \right )=0\; ,\; \; u\left ( 0 \right )=u\left ( 1 \right )=0

Solution:

Try u\left ( x \right )=Ae^{rx}, which gives roots r=-1\pm \sqrt{1-\lambda }. Solution is altered with \lambda <1\; ,\; \; \lambda =1\; ,\; \; \lambda >1

For the first case \lambda <1:

General solution:
u\left ( x \right )=Ae^{\left ( -1+\sqrt{1-\lambda } \right )x}+Be^{\left ( -1-\sqrt{1-\lambda } \right )x}

No need to complicate the solution with cosh and sinh.

With the notation a=√(1-λ)>0, the general solution is u(x)=e^-x(Ae^ax+Be^-ax)

At x=0, u(0)=A+B=0 --->B=-A

At x=1, u(1)=e^-1(Ae^a+Be^-a)=0---> A(e^a-e^-a)=0

Is it possible if a>0??

ehild

 

[HallsofIvy]

A lot of people consider cosh and sinh to be less complicated than e^x and e^{-x}! Especially when one of the boundary conditions is at x=0. cosh(0)= 1 and sinh(0)= 0.

 

[AntSC]

No need to complicate the solution with cosh and sinh.

With the notation a=√(1-λ)>0, the general solution is u(x)=e^-x(Ae^ax+Be^-ax)

At x=0, u(0)=A+B=0 --->B=-A

At x=1, u(1)=e^-1(Ae^a+Be^-a)=0---> A(e^a-e^-a)=0

Is it possible if a>0??

ehild

Ok. I'm happy with that. So A\left ( e^{a}-e^{-a} \right )=0 gives a=i\pi n and we can get the eigenvalues from there \lambda = 1-\pi ^2 n^2

So, how do i write the general solution now, given that we haven't found anything out about the coefficients A and B. Would it be, u\left ( x \right )=Ae^{i\pi n}-Be^{-i\pi n}

You wrote

is it possible with a>0??

Did you mean a<0?

 

[ehild]

I mean a>0 which means λ<1. Is it possible to fulfil the BC-s with λ<1? As you started to discuss that case.

And yes, you can get a solution zero at both boundaries when λ>1, that is, √(1-λ) is imaginary.

By the way, your formula with the cosh and sinh is not correct. You should have written
u(x)=e^{-x}(C\cosh(\sqrt{1-\lambda}x)+Dsinh(\sqrt{1-\lambda}x)

ehild

 

[AntSC]

I mean a>0. Is it possible to fulfil the BC-s with λ<1?

ehild

Sorry, weren't we just doing the case for λ<1?

 

[pasmith]

Find the eigenvalues λ, and eigenfunctions u(x), associated with the following homogeneous ODE problem:

{u}&#039;&#039;\left ( x \right )+2{u}&#039;\left ( x \right )+\lambda u\left ( x \right )=0\; ,\; \; u\left ( 0 \right )=u\left ( 1 \right )=0

Solution:

Try u\left ( x \right )=Ae^{rx}, which gives roots r=-1\pm \sqrt{1-\lambda }. Solution is altered with \lambda &lt;1\; ,\; \; \lambda =1\; ,\; \; \lambda &gt;1

For the first case \lambda &lt;1:

General solution:
u\left ( x \right )=Ae^{\left ( -1+\sqrt{1-\lambda } \right )x}+Be^{\left ( -1-\sqrt{1-\lambda } \right )x}
u\left ( x \right )=C\cosh \left ( -1+\sqrt{1-\lambda } \right )x+D\sinh \left ( -1-\sqrt{1-\lambda } \right )x

That should be
<br /> u(x) = Ce^{-x} \cosh((1-\lambda)^{1/2}x) + De^{-x} \sinh((1 - \lambda)^{1/2}x)<br />

Applying boundaries: (this is where my question lies - how to correctly apply BCs)

u\left ( 0 \right )=0 \; \; \Rightarrow \; \; C+D=0 (some cases I've seen the conclusion that only C=0). Do i assume that as \cosh is never zero that C=0 and therefore it must be that D=0. Or do i only take C=0 and then apply the second BC to see what happens to D?
The latter (assuming C=0) gives D\sinh \left ( -1-\sqrt{1-\lambda } \right )=0. So either D=0 or \sinh \left ( -1-\sqrt{1-\lambda } \right )=i\pi n.

I'm confused by the rules for the BCs. Can anyone point out how to proceed? Thanks

The solution to the eigenvalue problem v&#039;&#039; + \mu v = 0 with v(0) = v(1) = 0 is \mu_n = n^2\pi^2 with v_n(x) = \sin(n\pi x). This is relevant to your problem, because the substitution u(x) = e^{-x} v(x) reduces your ODE to
<br /> v&#039;&#039; + (\lambda - 1)v = 0<br />
with v(0) = v(1) = 0.

 

[AntSC]

That should be
<br /> u(x) = Ce^{-x} \cosh((1-\lambda)^{1/2}x) + De^{-x} \sinh((1 - \lambda)^{1/2}x)<br />

Thanks for that.
When we apply the first boundary u(0)=0 doesn't that yield that C=D=0?

 

[ehild]

When λ<1
at the first boundary, x=0, u(0)=C\cosh(0)=C\rightarrow C=0
at the second boundary,x=1, u(1)=D\sinh(\sqrt{1-\lambda})=0 \rightarrow D=0

ehild

 

[AntSC]

When λ<1
at the first boundary, x=0, u(0)=C\cosh(0)=C\rightarrow C=0
at the second boundary,x=1, u(1)=D\sinh(\sqrt{1-\lambda})=0 \rightarrow D=0

ehild

Why is the first boundary not u(0)=C\cosh(0)+D\sinh(0)\rightarrow C=D=0?
This is what i don't understand. Someone else was also telling me that the boundaries are applied to find a result for one coefficient at a time. Why is this?

 

[pasmith]

Why is the first boundary not u(0)=C\cosh(0)+D\sinh(0)\rightarrow C=D=0?
This is what i don't understand. Someone else was also telling me that the boundaries are applied to find a result for one coefficient at a time. Why is this?

\sinh(0) = 0, so D \sinh(0) = 0 for any D.

 

[Simon Bridge]

Why is the first boundary not u(0)=C\cosh(0)+D\sinh(0)\rightarrow C=D=0? This is what i don't understand. Someone else was also telling me that the boundaries are applied to find a result for one coefficient at a time. Why is this?

Because that would be incorrect algebra. The relationship you get is not C=D.

Since sinh(0)=0, the expression collapses to u(0)=C. That's just normal algebra.
You also know that u(0)=0, therefore C=0. Notice how, by itself, this tells you nothing about the value of D. Which is why you need another relation.

If, however, the expression was more like u(x)=Ccosh(x)+D(sinh(x)-1)
then u(0)=0 would get you 0=C-D so that C=D.
But that is not what you have. Not even close.

\sinh(0) = 0, so D \sinh(0) = 0 for any D.

Exactly what he said:
Also think about how simultaneous equations work.
This is the same thing.

 

[AntSC]

Ok, just to be really clear... As a result of the boundary conditions we have
u(0)=0 \Rightarrow C\cosh(0)+D\sinh(0)=0and
u(1)=0 \Rightarrow C\cosh \left ( -1+\sqrt{1-\lambda } \right )+D\sinh \left ( -1-\sqrt{1-\lambda } \right )=0
I understand your point about solving simultaneously but how to we infer, for the first boundary, that only C=0? Given that \sinh (0)=0, surely then D=0 also. So \cosh (0) is undefined and \sinh (0)=0 so the resultant equation for u(0)=0 is 0=0. Therefore both C and D must be zero?

 

[ehild]

Ok, just to be really clear... As a result of the boundary conditions we have
u(0)=0 \Rightarrow C\cosh(0)+D\sinh(0)=0and
u(1)=0 \Rightarrow C\cosh \left ( -1+\sqrt{1-\lambda } \right )+D\sinh \left ( -1-\sqrt{1-\lambda } \right )=0

the second formula is wrong. It should be

u(x)=e^{-x}(C\cosh(\sqrt{1-\lambda}x)+Dsinh(\sqrt{1-\lambda}x))

ehild

 

[AntSC]

the second formula is wrong.

Can you show me how this is wrong? I don't get it.

 

[ehild]

See Post #11

ehild

 

[AntSC]

Just clicked with something. Don't know why i was thinking along the lines i was. I see now that from the first boundary that it is only C=0. Based on that, the result of the second boundary can be written as just the D\sinh term. Not sure I'm writing the hyperbolic form correctly now though...

 

[ehild]

Just clicked with something. Don't know why i was thinking along the lines i was. I see now that from the first boundary that it is only C=0. Based on that, the result of the second boundary can be written as just the D\sinh term. Not sure I'm writing the hyperbolic form correctly now though...

Your hyperbolic form was wrong.

ehild

 

[AntSC]

See Post #11

ehild

I don't see how pulling out an e^{-x} changes anything. Also, i am going on the lecturer's solutions where he wrote the hyperbolic form with the -1 in it also. Could be wrong. Actually, now i look at it (i took it as given before), i can't work out the conversion to hyperbolic form correctly. I just get one \cosh term. No \sinh term.
I subbed in e^x=\cosh x + \sinh x and e^{-x}=\cosh x - \sinh x. Missing something obvious?

 

[ehild]

I don't see how pulling out an e^{-x} changes anything. Also, i am going on the lecturer's solutions where he wrote the hyperbolic form with the -1 in it also. Could be wrong. Actually, now i look at it (i took it as given before), i can't work out the conversion to hyperbolic form correctly. I just get one \cosh term. No \sinh term.
I subbed in e^x=\cosh x + \sinh x and e^{-x}=\cosh x - \sinh x. Missing something obvious?

Remember the original solution was

u\left ( x \right )=Ae^{\left ( -1+\sqrt{1-\lambda } \right )x}+Be^{\left ( -1-\sqrt{1-\lambda } \right )x}

If you replace all exponentials with the correct sinh and cosh you get something different.

ehild

 

[AntSC]

If you replace all exponentials with the correct sinh and cosh you get something different.

u\left ( x \right )=Ae^{\left ( -1+\sqrt{1-\lambda } \right )x}+Be^{\left ( -1-\sqrt{1-\lambda } \right )x}
let \mu =\sqrt{1-\lambda }
u\left ( x \right )=Ae^{-x}e^{\mu x}+Be^{-x}e^{-\mu x}
u\left ( x \right )=A\left(\cosh x-\sinh x\right)\left(\cosh \mu x+\sinh \mu x \right)+B\left(\cosh x-\sinh x\right)\left(\cosh \mu x-\sinh \mu x \right)
u\left ( x \right )=A\left[\cosh x\cosh \mu x-\sinh x\sinh \mu x+\cosh x\sinh \mu x-\sinh x\cosh \mu x\right]+B\left[\cosh x\cosh \mu x+\sinh x\sinh \mu x-\cosh x\sinh \mu x-\sinh x\cosh \mu x\right]
u\left ( x \right )=A\left[-\cosh \left(x-\mu x \right )+\sinh \left(x-\mu x \right )\right]+B\left[\cosh \left(x+\mu x \right )-\sinh \left(x+\mu x \right )\right]
and then I'm back to where i started

If i write it like you originally suggested
u\left ( x \right )=Ae^{-x}e^{\mu x}+Be^{-x}e^{-\mu x}
u\left ( x \right )=Ae^{-x}\left(\cosh \mu x+\sinh \mu x \right)+Be^{-x}\left(\cosh \mu x-\sinh \mu x \right)
u\left ( x \right )=e^{-x}\left[ \left(A+B\right)\cosh \mu x+ \left(A-B\right)\sinh \mu x\right]
u\left ( x \right )=e^{-x}\left[ C\cosh \mu x+ D\sinh \mu x\right]
So it can't be written like the lecturer has suggested. The latter form must be correct then.
Thanks for pointing me in the right direction.

Now I've got that sorted, applying boundaries:
u\left ( 0 \right )=0 \Rightarrow C\cosh 0 + D\sinh 0 = 0
Ok I'm going to ask this question again. The D-term vanishes as sinh(0)=0 but doesn't the C-term also vanish as cosh(0) is undefined? Aren't we then left with this boundary not telling us anything about either coefficients?

 

[ehild]

Now I've got that sorted, applying boundaries:
u\left ( 0 \right )=0 \Rightarrow C\cosh 0 + D\sinh 0 = 0
Ok I'm going to ask this question again. The D-term vanishes as sinh(0)=0 but doesn't the C-term also vanish as cosh(0) is undefined? Aren't we then left with this boundary not telling us anything about either coefficients?

cosh(0) is defined, it is 1.

Yes, Dsinh(0)=0 as sinh(0)=0. But that does not mean that D=0. It can be anything, as anything multiplied by zero is equal to zero.

As u(0)=0 and u(0)=1*C, C must be zero.

The first boundary condition gives you the value of C: C=0. D is unknown yet. Apply the second boundary condition.

ehild

 

[AntSC]

cosh(0) is defined, it is 1.

ehild

Oh crap. Sorry, having a tough day.

Ok. Second boundary.

u(1)=0 \Rightarrow e^{-1}Dsinh(\sqrt{1-\lambda})=0
Either D=0 or sinh(\sqrt{1-\lambda})=0 \Rightarrow \sqrt{1-\lambda}=in\pi

So \lambda=1+n^2 \pi ^2

And u(x)=e^{-x}Dsinh(in\pi ) \Rightarrow u(x)=ie^{-x}Dsin(n\pi )

 

[ehild]

Oh crap. Sorry, having a tough day.

Ok. Second boundary.

u(1)=0 \Rightarrow e^{-1}Dsinh(\sqrt{1-\lambda})=0
Either D=0 or sinh(\sqrt{1-\lambda})=0 \Rightarrow \sqrt{1-\lambda}=in\pi

NO.
We assumed that λ<1. In this case, e^{-1}Dsinh(\sqrt{1-\lambda})=0 involves that D=0.

Assuming λ<1 and both u(0)=0 and u(1)=0 results in u(x)=0.

Next step: assume λ=1. What is the form of u(x) then?

ehild

 

[AntSC]

We assumed that λ<1. In this case, e^{-1}Dsinh(\sqrt{1-\lambda})=0 involves that D=0.

ehild

How is that e^{-1}Dsinh(\sqrt{1-\lambda})=0 involves that D=0 ?

 

[ehild]

How is that e^{-1}Dsinh(\sqrt{1-\lambda})=0 involves that D=0 ?

Remember you said that λ<1. In this case, 1-λ >0, √ (1-λ) is real, and different from zero. The sinh function is zero only when its argument is zero. How else can be the product equal to zero if not with D=0?

ehild

 

[AntSC]

Fantastic. Didn't see that at all. Thanks very much!

For \lambda=1 \Rightarrow u(x)=e^{-x}[Ax+B]
So u(0)=0 \Rightarrow B=0 and u(1)=0 \Rightarrow A=0 and therefore u(x)=0

For \lambda&gt;1 \Rightarrow u(x)=e^{-x}[A\cos (\sqrt{\lambda-1})x+B\sin (\sqrt{\lambda-1})x]
So u(0)=0 \Rightarrow A=0 and u(1)=0 \Rightarrow either B=0 or \sin (\sqrt{\lambda-1})=0 \Rightarrow (\sqrt{\lambda-1})=n\pi

So the eigenvalues are \lambda =1+\pi^{2}n^{2}
And the eigenfunctions are u(x)=Be^{-x}\sin (n\pi x)

All correct?

 

[ehild]

Yes, it looks correct

ehild

 

[AntSC]

Great. Thanks so much for all your help!
I'm pleased to put this one to bed and come out having cleared some things up.
Cheers!