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- Thread starter ssky
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HallsofIvy

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It's hard to respond without know what you already know. Do you know how to get the "characteristic equation" for a "linear, homogeneous, differential equation with constant coefficients" such as this?

For this differential equation, the characteristic equation is [itex]r^2= -\lambda_i[/itex] and has characteristic roots [itex]r= \pm\lambda_i i[/itex] (that second "i" is the imaginary unit). That, in turn means that the general solution to the differential equation is [itex]\psi_i(x)= Ae^{i\lambda_i x}+ Ce^{-i\lambda_i x}= Ccos(\lambda_i x)+ D sin(\lambda_i x)[/itex] where A, B, C, and D are constants. Do you understand how to convert from the complex exponential to sine and cosine? It is based on [itex]e^{ix}= cos(x)+ i sin(x)[/itex].

Now look at the boundary conditions. From [itex]\psi_i(x)= Ccos(\lambda_i x)+ Dsin(\lambda_i x)[/itex], we have [itex]d\psi_i/dx= -C\lambda_i sin(\lambda_i x)+ D\lambda_i cos(\lambda_i x)[/itex] so [itex]d\psi_i/dx(0)= D\lambda_i = 0[/itex].

If [itex]\lambda_i= 0[/itex] the differential equation would be just [itex]d^2\psi_i/dx= 0[/itex] which has general solution [itex]\psi_i(x)= Ax+ B[/itex] (just integrate twice) and then [itex]d\psi/dx= A= 0[/itex] in order to satisfy [itex]d\psi/dx(0)= 0[/itex]. Then \psi_i(1)= B= 0[/itex] from the other boundary condition. That gives [itex]\psi_i(x)[/itex] identically 0. That is a solution but not a very interesting or useful one! That is called the "trivial" solution. If [itex]\lambda_i\ne 0[/itex], we must have D= 0 so that [itex]\psi_i(x)= C cos(\lambda_i x)[/itex].

Now [itex]\psi_i(1)= C cos(\lambda_i)= 0[/itex]. That is satisfied by C= 0- but again, that would mean, with both C and D equal to 0, that [itex]\psi_i(x)[/itex] is identically 0. In order to have a "non trivial" solution, we must have [itex]cos(\lambda_i)= 0[/itex]. cos(x)= 0 for x an "odd multiple of [itex]\pi/2[/itex]": [itex]\pi/2[/itex], [itex]3\pi/2[/itex], etc. That means we must have [itex]\lambda_i= (2i- 1)\pi/2[/itex]. That is the reason for the subscript "i" on the functions. We now have [itex]\psi(x)= Ccos(((2i-1)\pi/2)x)[/itex].

The intial [itex]\sqrt{2}[/itex] is the "normalizing" part. That makes the integral of the square of the function, from 0 to 1, equal to 1.

For this differential equation, the characteristic equation is [itex]r^2= -\lambda_i[/itex] and has characteristic roots [itex]r= \pm\lambda_i i[/itex] (that second "i" is the imaginary unit). That, in turn means that the general solution to the differential equation is [itex]\psi_i(x)= Ae^{i\lambda_i x}+ Ce^{-i\lambda_i x}= Ccos(\lambda_i x)+ D sin(\lambda_i x)[/itex] where A, B, C, and D are constants. Do you understand how to convert from the complex exponential to sine and cosine? It is based on [itex]e^{ix}= cos(x)+ i sin(x)[/itex].

Now look at the boundary conditions. From [itex]\psi_i(x)= Ccos(\lambda_i x)+ Dsin(\lambda_i x)[/itex], we have [itex]d\psi_i/dx= -C\lambda_i sin(\lambda_i x)+ D\lambda_i cos(\lambda_i x)[/itex] so [itex]d\psi_i/dx(0)= D\lambda_i = 0[/itex].

If [itex]\lambda_i= 0[/itex] the differential equation would be just [itex]d^2\psi_i/dx= 0[/itex] which has general solution [itex]\psi_i(x)= Ax+ B[/itex] (just integrate twice) and then [itex]d\psi/dx= A= 0[/itex] in order to satisfy [itex]d\psi/dx(0)= 0[/itex]. Then \psi_i(1)= B= 0[/itex] from the other boundary condition. That gives [itex]\psi_i(x)[/itex] identically 0. That is a solution but not a very interesting or useful one! That is called the "trivial" solution. If [itex]\lambda_i\ne 0[/itex], we must have D= 0 so that [itex]\psi_i(x)= C cos(\lambda_i x)[/itex].

Now [itex]\psi_i(1)= C cos(\lambda_i)= 0[/itex]. That is satisfied by C= 0- but again, that would mean, with both C and D equal to 0, that [itex]\psi_i(x)[/itex] is identically 0. In order to have a "non trivial" solution, we must have [itex]cos(\lambda_i)= 0[/itex]. cos(x)= 0 for x an "odd multiple of [itex]\pi/2[/itex]": [itex]\pi/2[/itex], [itex]3\pi/2[/itex], etc. That means we must have [itex]\lambda_i= (2i- 1)\pi/2[/itex]. That is the reason for the subscript "i" on the functions. We now have [itex]\psi(x)= Ccos(((2i-1)\pi/2)x)[/itex].

The intial [itex]\sqrt{2}[/itex] is the "normalizing" part. That makes the integral of the square of the function, from 0 to 1, equal to 1.

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HallsofIvy, i do not know how to thank you.

thank you very very very ... very much.

but, i don't understand :shy: the last part

can you help me pleas

thank you very very very ... very much.

but, i don't understand :shy: the last part

can you help me pleas

The intial is the "normalizing" part. That makes the integral of the square of the function, from 0 to 1, equal to 1

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Cyosis

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Cyosis

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For this differential equation, the characteristic equation is [itex]r^2= -\lambda_i[/itex] and has characteristic roots [itex]r= \pm\lambda_i i[/itex] (that second "i" is the imaginary unit). That, in turn means that the general solution to the differential equation is [itex]\psi_i(x)= Ae^{i\lambda_i x}+ Ce^{-i\lambda_i x}= Ccos(\lambda_i x)+ D sin(\lambda_i x)[/itex] where A, B, C, and D are constants. Do you understand how to convert from the complex exponential to sine and cosine? It is based on [itex]e^{ix}= cos(x)+ i sin(x)[/itex].

The difference between the previous differential equation and this one is that the first one had a second derivative in it and this one a fourth derivative. So the characteristic equation changes from [itex]r^2=-\lambda_i^2[/itex] to [itex]r^4=\mu_i^4[/itex]. This has the solutions [itex]r=\{-\mu_i,\mu_i,-i \mu_i, i\mu_i \}[/itex] as a result the general solution will become [itex]\psi(x)=A e^{-\mu x}+B e^{\mu x}+C e^{- i \mu x}+D e^{i \mu x}[/itex]. The exponents with i in their powers can be written as cosine and sine again and the real exponents can be written as cosh and sinh.

- #7

HallsofIvy

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The first was [itex]d^2\psi/dx^2= -\lambda_i^2 \psi[/itex] which has characteristic equation [itex]r^2= -\lambda_i^2[/itex] with roots [itex]\pm i\lambda_i[/itex] while the second was [itex]d^4\psi/dx^4= /mu^4\psi[/itex] which has characteristic equation [itex]r^4= \mu^4[/itex]. We can solve that by first taking the square root: [itex]r^2= \pm \mu^2[/itex]. Taking the positive root, [itex]r^2= \mu^2[/itex], and taking the square root again gives [itex]r= \pm \mu[/itex] while taking the negative root, [itex]r^2= -\mu^2[/itex] and taking the square root gives [itex]r= \pm i\mu[/itex]. The four roots of the equation are [itex]\mu[/itex], [itex]-\mu[/itex], [itex]i\mu[/itex], and [itex]-i\mu[/itex].

The imaginary roots, [itex]i\mu[/itex] and [itex]-i\mu[/itex], give [itex]sin(\mu x)[/itex] and [itex]cos(\mu x)[/itex] solutions as before. The solutions corresponding to the real roots, [itex]\mu[/itex], and [itex]-\mu[/itex], can be written [itex]e^{\mu x}[/itex] and [itex]e^{-\mu x}[/itex], but because

[tex]cosh(\mu x)= \frac{e^{\mu x}+ e^{-\mu x}}{2}[/tex]

and

[tex]sinh(\mu x)= \frac{e^{\mu x}- e^{-\mu x}}{2}[/tex]

they can be written in terms of sinh and cosh also. Since cosh(0)= 1 and sinh(0)= 0, those functions are better suited for initial value problems where we are given values of the function and its derivative at x= 0.

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Cyosis

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I am so grateful for your interesting in my problem and I will be more grateful if you tell to found the constants c1, c2 , c3 and c4 because I tried but I failed.

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Cyosis

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Show us all the steps you've done so far so we can see where you failed.

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Cyosis

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thank you but I do not understand what you mean.

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Cyosis

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From the picture you linked you get the following two equations:

[tex]

\begin{align}

& -C_3 \cosh \mu_i-C_4 \sinh \mu_i+C_3 \cos \mu_i+C_4 \sin \mu_i=0

\\

& -C_3 \sinh \mu_i-C_4 \cosh \mu_i - C_3 \sin \mu_i+C_4 \cos \mu_i=0

\end{align}

[/tex]

Solving equation (1) for C3 yields:

[tex]

C_3=-C_4 \frac{\sin \mu_i -\sinh \mu_i}{\cos \mu_i-\cosh \mu_i}

[/tex]

Plug this value for C3 into equation (2) and solve for C4.

Edit: This is a pretty weird problem, because the only function that seems to meet all requirements is 0.

[tex]

\begin{align}

& -C_3 \cosh \mu_i-C_4 \sinh \mu_i+C_3 \cos \mu_i+C_4 \sin \mu_i=0

\\

& -C_3 \sinh \mu_i-C_4 \cosh \mu_i - C_3 \sin \mu_i+C_4 \cos \mu_i=0

\end{align}

[/tex]

Solving equation (1) for C3 yields:

[tex]

C_3=-C_4 \frac{\sin \mu_i -\sinh \mu_i}{\cos \mu_i-\cosh \mu_i}

[/tex]

Plug this value for C3 into equation (2) and solve for C4.

Edit: This is a pretty weird problem, because the only function that seems to meet all requirements is 0.

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- #17

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pleas pleas pleas help me

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HallsofIvy

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yes, it is the same problem as i posted in response #5 and #16

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HallsofIvy

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Okay, the general solution is

[tex]\psi(x)= C_1cosh(\mu x)+ C_2sinh(\mu x)+ C_3cos(\mu x)+ C_4sin(\mu x)[/tex]

so

[tex]\psi'(x)= \mu C_1sinh(\mu x)+ \mu C_2 cosh(\mu x)- \mu C_3sin(\mu x)+ \mu C_4cos(\mu x)[/tex]

and the conditions are

[tex]\psi(0)= d\psi(0)/dx= 0[/tex]

[tex]\psi(1)= d\psi(1)/dx= 0[/tex]

[tex]\psi(0)= C_1+ C_3= 0[/tex]

so

[tex]C_3= -C_1[/tex]

[tex]\psi'(0)= \mu C_2+ \mu C_4= 0[/tex]

so

[tex]C_4= -C_2[/tex].

That tells us that we can write

[tex]\psi(x)= C_1cosh(\mu x)+ C_2sinh(\mu x)- C_1cos(\mu x)+ C_2sin(\mu x)[/tex]

Now we use

[tex]\psi(1)= C_1cosh(\mu)+ C_2sinh(\mu)- C_1cos(\mu)+ C_2sin(\mu)= 0[/tex]

and

[tex]\psi'(1)= \mu C_1sinh(\mu)+ \mu C_2 cosh(\mu)+ \mu C_1sin(\mu)- \mu C_2cos(\mu)= 0[/tex]

We can write those as

[tex]C_1(cosh(\mu)- cos(\mu))+ C_2(sinh(\mu)- sin(\mu))= 0[/tex]

and

[tex]C_1(sinh(\mu)+ sin(\mu)+ C_2(cosh(\mu)- cos(\mu))= 0[/tex]

In order to solve for, say, [itex]C_1[/itex], we would have to eliminate [itex]C_2[/itex].

We could do that by multiplying the first equation by [itex]cosh(\mu)- cos(\mu)[/itex], the second equation by [itex]sinh(\mu)- sin(\mu)[/itex], and subtracting the second from the first.

That gives

[tex](cosh(\mu)- cos(\mu))(cosh(\mu)- cos(\mu))C_1- (sinh(\mu)- sin(\mu))(sinh(\mu)+ sin(\mu)C_1= 0[/tex]

[tex]C_1(cosh^2(\mu)- 2sinh(\mu)sin(\mu)+ cos^2(\mu)- sinh^2(\mu)+ sin^2(\mu))= 0[/tex].

But [itex]cosh^2(\mu)- sinh^2(\mu)= 1[/itex] and [itex]cos^2(\mu)+ sin^2(\mu)= 1[/itex] so this becomes

[tex](2- 2cosh(\mu)cos(\mu))C_1= 0[/tex]

Now, one obvious solution is [itex]C_1= 0[/itex] but that leads to [itex]C_2= 0[/itex] also which means [itex]\psi(x)[/itex] is identically 0- the "trivial" solution. In order to have a "non-trivial" solution, we must have [itex]C_1\ne 0[/itex] which means we must have the coefficient

[tex]2- 2cosh^2(\mu)cos(\mu)= 0[/tex]

which leads immediately to [itex]2cosh(\mu)cos(\mu)= 2[/itex] or

[tex]cosh(\mu)cos(\mu)= 1[/itex].

[tex]\psi(x)= C_1cosh(\mu x)+ C_2sinh(\mu x)+ C_3cos(\mu x)+ C_4sin(\mu x)[/tex]

so

[tex]\psi'(x)= \mu C_1sinh(\mu x)+ \mu C_2 cosh(\mu x)- \mu C_3sin(\mu x)+ \mu C_4cos(\mu x)[/tex]

and the conditions are

[tex]\psi(0)= d\psi(0)/dx= 0[/tex]

[tex]\psi(1)= d\psi(1)/dx= 0[/tex]

[tex]\psi(0)= C_1+ C_3= 0[/tex]

so

[tex]C_3= -C_1[/tex]

[tex]\psi'(0)= \mu C_2+ \mu C_4= 0[/tex]

so

[tex]C_4= -C_2[/tex].

That tells us that we can write

[tex]\psi(x)= C_1cosh(\mu x)+ C_2sinh(\mu x)- C_1cos(\mu x)+ C_2sin(\mu x)[/tex]

Now we use

[tex]\psi(1)= C_1cosh(\mu)+ C_2sinh(\mu)- C_1cos(\mu)+ C_2sin(\mu)= 0[/tex]

and

[tex]\psi'(1)= \mu C_1sinh(\mu)+ \mu C_2 cosh(\mu)+ \mu C_1sin(\mu)- \mu C_2cos(\mu)= 0[/tex]

We can write those as

[tex]C_1(cosh(\mu)- cos(\mu))+ C_2(sinh(\mu)- sin(\mu))= 0[/tex]

and

[tex]C_1(sinh(\mu)+ sin(\mu)+ C_2(cosh(\mu)- cos(\mu))= 0[/tex]

In order to solve for, say, [itex]C_1[/itex], we would have to eliminate [itex]C_2[/itex].

We could do that by multiplying the first equation by [itex]cosh(\mu)- cos(\mu)[/itex], the second equation by [itex]sinh(\mu)- sin(\mu)[/itex], and subtracting the second from the first.

That gives

[tex](cosh(\mu)- cos(\mu))(cosh(\mu)- cos(\mu))C_1- (sinh(\mu)- sin(\mu))(sinh(\mu)+ sin(\mu)C_1= 0[/tex]

[tex]C_1(cosh^2(\mu)- 2sinh(\mu)sin(\mu)+ cos^2(\mu)- sinh^2(\mu)+ sin^2(\mu))= 0[/tex].

But [itex]cosh^2(\mu)- sinh^2(\mu)= 1[/itex] and [itex]cos^2(\mu)+ sin^2(\mu)= 1[/itex] so this becomes

[tex](2- 2cosh(\mu)cos(\mu))C_1= 0[/tex]

Now, one obvious solution is [itex]C_1= 0[/itex] but that leads to [itex]C_2= 0[/itex] also which means [itex]\psi(x)[/itex] is identically 0- the "trivial" solution. In order to have a "non-trivial" solution, we must have [itex]C_1\ne 0[/itex] which means we must have the coefficient

[tex]2- 2cosh^2(\mu)cos(\mu)= 0[/tex]

which leads immediately to [itex]2cosh(\mu)cos(\mu)= 2[/itex] or

[tex]cosh(\mu)cos(\mu)= 1[/itex].

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- #21

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thanks alot,

if [itex]C_1\ne 0[/itex]

then how we find [itex]C_1[/itex] ?

if [itex]C_1\ne 0[/itex]

then how we find [itex]C_1[/itex] ?

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- #22

HallsofIvy

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We don't. If [itex]cosh(\mu)cos(\mu)= 1[/itex], so there are non-trivial solutions, there will be an infinite number of such non-trivial functions.thanks alot,

if [itex]C_1\ne 0[/itex]

then how we find [itex]C_1[/itex] ?

Basically, you are solving an "eigenvalue" problem. If there exist non-trivial solutions to the equation [itex]Av= \mu v[/itex] for A a linear operator, then [itex]\mu[/itex] is an "eigenvalue" of A and it can be shown that the set of all eigenvectors (values of v) satisfying that equation for that particular value of [itex]\mu[/itex] form a "subspace" of all possible solutions and so there are necessarily an infinite number of them.

- #23

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thank you very much... I would like to give you this gift for helping me

http://blog.doctissimo.fr/php/blog/un_avenir_heureux/images/bouquet%20de%20fleur.gif [Broken]

[/URL]
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- #24

HallsofIvy

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[/URL]thank you very much... I would like to give you this gift for helping me

http://blog.doctissimo.fr/php/blog/un_avenir_heureux/images/bouquet%20de%20fleur.gif [Broken]

Very nice! Thank you!

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(YY)xY''=C (C:Real nnumber)

- #26

HallsofIvy

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Why in the world did you add this onto a thread everyone had finished with?

Click on the "new topic" button to start a new thread.

In fact, I am going to do that for you and name it "gerechte23's question"!

Click on the "new topic" button to start a new thread.

In fact, I am going to do that for you and name it "gerechte23's question"!

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