Solving ODE using Fourier Transform

In summary: That looks correct to me!In summary, the conversation discusses the process of finding the general solution for a second order homogeneous equation using Fourier transforms. It is important to consider the convergence of the function being integrated, and to use the appropriate rules for taking the Fourier transform of a product.
  • #1
absolute76
21
0
i have found the general solution which is,

u(x)= (C1 + C2x)e^ax + (1/2a)[tex]\int f(x-y) e^\left|y\right| dy

is this correct??
now, i just want your help to guide me for justifying f(x)=x^5...

is that wrong if i solve the integration and just substitute the integral which is the range( infinity to negative infinity)??

thank you...
 

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  • #2
absolute76 said:
i have found the general solution which is,

u(x)= (C1 + C2x)e^ax + (1/2a)[tex]\int f(x-y) e^\left|y\right| dy

is this correct??
That doesn't look correct. Could you show the work that led you to this result? Also, since there is no one agreed-upon definition of the Fourier transform, tell us what formalism you are using.

I know just enough about Fourier transforms to be dangerous. I've asked other homework helpers who are better versed than am I in Fourier transforms to jump in and help. Until then, show some work and try to think about the second part of the question.

A hint for that: Are those functions square integrable?
 
  • #3
I Fourier transform both sides and I get this:

u~(k) = 1/(k^2 + a^2) . f~(k)

From table,

1/(k^2 + a^2) = √(∏/2) (e^-a|x|/a)----> denotes this as g~(k)

u~(k) = g~(k)*f~(k) ----> applied convolution

After convolution, I get 1/2a ∫f(x-y) e^-a|y| dy ; range -∞<y<∞

Is it correct up until here??
 
  • #4
D H said:
A hint for that: Are those functions square integrable?

can you explain in detail of what you mean by that??
 
  • #5
You've correctly found the particular solution, [itex]u_p(x)=\frac{1}{2a}\int_{-\infty}^{\infty}f(x-y)e^{-a|y|}dy[/itex], (assuming [itex]a>0[/itex]) via Fourier Transform methods. But [itex](c_1+c_2x)e^{ax}[/itex] does not satisfy the homogeneous equation [itex]-\frac{d^2 u}{dx^2}+a^2u=0[/itex], and so it is not the correct homogeneous solution.
 
  • #6
gabbagabbahey said:
You've correctly found the particular solution, [itex]u_p(x)=\frac{1}{2a}\int_{-\infty}^{\infty}f(x-y)e^{-a|y|}dy[/itex], (assuming [itex]a>0[/itex]) via Fourier Transform methods. But [itex](c_1+c_2x)e^{ax}[/itex] does not satisfy the homogeneous equation [itex]-\frac{d^2 u}{dx^2}+a^2u=0[/itex], and so it is not the correct homogeneous solution.

-d^2 u/dx^2 + a^u = 0

2nd order homogeneous eq.

-λ^2(e^ λx)+a^2e ^λx = 0
- λ^2+a^2=0
a^2= λ^2
a= λ ------- y=(c1+c2x)e^ax

is this wrong??
 
  • #7
[tex]a^2=\lambda^2\implies a=\pm\lambda[/tex]

You have two distinct roots, not one double root.
 
  • #8
gabbagabbahey said:
[tex]a^2=\lambda^2\implies a=\pm\lambda[/tex]

You have two distinct roots, not one double root.

oh issit that way?? i thought

a=(λ^2)^1/2

and will give us a=λ because we cancel the 2x(1/2)??
 
  • #9
No, [itex](-\lambda)^2=\lambda^2[/itex]. You can't only take the positive root, and then claim that it is a repeated root. It's the exact same concept as solving the equation [itex]x^2=c[/itex]; both [itex]x=\sqrt{c}[/itex] and [itex]x=-\sqrt{c}[/itex] are solutions.
 
  • #10
gabbagabbahey said:
[tex]a^2=\lambda^2\implies a=\pm\lambda[/tex]

You have two distinct roots, not one double root.

oh sorry! my mistake!

i get it already!..thank you..

anyway I am aware that for the second part f(x)=x^5,

1/2a ∫(x-y)^5 e^-a|y| dy -∞<y<∞

right??

I am aware that i have to use convergence test..isn't it the same?

i have to integrate it and substitue the range?
 
  • #11
please correct me if I am wrong,

for f(x)=x^5

let say i take -∞<y<0 (first)

then i substitute inside---->∫(x-y)^5 e^-a|y| dy

y=0---> (x)^5 [(e^0)=1]

which give us when y=0 x^5

will this solution conclude that f(x)=x^5 is convergence??
 
  • #12
If u don’t mind, can I ask one more question..

2 d²u/dx² - x du/dx + u =0

For this question, I already Fourier transform both sides which give me,

u~(k)[-2k² - xik + 1]=0… but it is impossible if just u~(k)=0

do you have any idea of solving it?
 
  • #13
absolute76 said:
please correct me if I am wrong,

for f(x)=x^5

let say i take -∞<y<0 (first)

then i substitute inside---->∫(x-y)^5 e^-a|y| dy

y=0---> (x)^5 [(e^0)=1]

which give us when y=0 x^5

will this solution conclude that f(x)=x^5 is convergence??

That doesn't look like any convergence test I've ever seen.
 
  • #14
absolute76 said:
If u don’t mind, can I ask one more question..

2 d²u/dx² - x du/dx + u =0

For this question, I already Fourier transform both sides which give me,

u~(k)[-2k² - xik + 1]=0… but it is impossible if just u~(k)=0

do you have any idea of solving it?

[tex]F\left[x\frac{du}{dx}\right]\neq ikx\tilde{u}(k)[/tex]

There is a rule that tells you how to take the FT of the product of [itex]x^n[/itex] with any function...use that rule.
 
  • #15
gabbagabbahey said:
[tex]F\left[x\frac{du}{dx}\right]\neq ikx\tilde{u}(k)[/tex]

There is a rule that tells you how to take the FT of the product of [itex]x^n[/itex] with any function...use that rule.

u~[-2k² + 1] = i/2∏ [d/dk u~(k)]

I transfer, xu’ to the right side

It this correct??

doesnt [d/dk u~(k)] =iku~(k)

is the same?
 
Last edited:
  • #16
You're missing a factor of [itex]k[/itex] and you don't need the [itex]1/2\pi[/itex]:

[tex]F\left[x\frac{du}{dx}\right]=i\frac{d}{dk}\left(F\left[\frac{du}{dx}\right]\right)=i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]=-k\frac{d\tilde{u}}{dk}[/tex]
 
  • #17
gabbagabbahey said:
You're missing a factor of [itex]k[/itex] and you don't need the [itex]1/2\pi[/itex]:

[tex]F\left[x\frac{du}{dx}\right]=i\frac{d}{dk}\left(F\left[\frac{du}{dx}\right]\right)=i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]=-k\frac{d\tilde{u}}{dk}[/tex]

ok thank you!, so now i rearrange it,

u~(k)=(-k/-2k²+1)(du~/dk)
is this correct if i diffrentiate towards k on the right side?

that will give me u~(k)=2k³-4k²-k/(-2k²+1)²?
 
Last edited:
  • #18
Actually, I made an error in my previous post:

[tex]i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]\neq-k\frac{d\tilde{u}}{dk}[/itex]

You will need to use the product rule to carry out the derivative.
 
  • #19
gabbagabbahey said:
Actually, I made an error in my previous post:

[tex]i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]\neq-k\frac{d\tilde{u}}{dk}[/itex]

You will need to use the product rule to carry out the derivative.

:confused: error?

error in which part??

yes i got the answer of 2k³-4k²-k/(-2k²+1)² using...(u'v-uv')/v² am i right??
 
  • #20
absolute76 said:
:confused: error?

error in which part??

[tex]\frac{d}{dk}\left(k\tilde{u}\right)\neq k\frac{d\tilde{u}}{dk}[/tex]
 
  • #21
gabbagabbahey said:
Actually, I made an error in my previous post:

[tex]i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]\neq-k\frac{d\tilde{u}}{dk}[/itex]

You will need to use the product rule to carry out the derivative.

[tex]
i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]
[/tex]

owk i think i got it, is it correct...i(iu~(k)-k²u~(k))

that will give me -u~(k)+ik²u~(k)---->u~(k)[-1+ik²]??

owk how do i separate u~(k) so that it don't cancel each other(left and right side)??
 
  • #22
absolute76 said:
[tex]
i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]
[/tex]

owk i think i got it, is it correct...i(iu~(k)-k²u~(k))

No.

[tex]i\frac{d}{dk}\left[(ik)\tilde{u}\right]=-\frac{d}{dk}\left[k\tilde{u}\right]=-\left(\tilde{u}\frac{dk}{dk}+k\frac{d\tilde{u}}{dk}\right)=-\tilde{u}-k\frac{d\tilde{u}}{dk}[/tex]

This is a basic application of the product rule.
 
  • #23
gabbagabbahey said:
No.

[tex]i\frac{d}{dk}\left[(ik)\tilde{u}\right]=-\frac{d}{dk}\left[k\tilde{u}\right]=-\left(\tilde{u}\frac{dk}{dk}+k\frac{d\tilde{u}}{dk}\right)=-\tilde{u}-k\frac{d\tilde{u}}{dk}[/tex]

This is a basic application of the product rule.

Ok my mistake again,

so now the equation will be u~(k)[-2k²+2]=-k du~/dk

so u~(k)= [-k/(-2k²+2) du~/dk

Is it correct if i diffrentiate towards k on the right side?
 
  • #24
absolute76 said:
so now the equation will be u~(k)[-2k²+2]=-k du~/dk

so u~(k)= [-k/(-2k²+2) du~/dk

That's correct. So now you have a 1st order ODE to solve instead of the 2nd order one you started with. This one is separable:

[tex]\frac{d\tilde{u}}{\tilde{u}}=2\frac{k^2-1}{k}dk[/tex]
 
  • #25
gabbagabbahey said:
That's correct. So now you have a 1st order ODE to solve instead of the 2nd order one you started with. This one is separable:

[tex]\frac{d\tilde{u}}{\tilde{u}}=2\frac{k^2-1}{k}dk[/tex]

Owk now is this correct?

u~(k)=e^[(k²)-(2 ln k)]

...u~(k)=e^(k²)/e^(2 ln k)

then i transform it using the table?? right?
 
  • #26
absolute76 said:
Owk now is this correct?

u~(k)=e^[(k²)-(2 ln k)]

...u~(k)=e^(k²)/e^(2 ln k)

then i transform it using the table?? right?

Right. Although, I doubt your table will give you the inverse FT of this function, so you will probably want to use the convolution theorem with [itex]\tilde{g}(k)=\frac{1}{k^2}[/itex]
 
  • #27
gabbagabbahey said:
Right. Although, I doubt your table will give you the inverse FT of this function, so you will probably want to use the convolution theorem with [itex]\tilde{g}(k)=\frac{1}{k^2}[/itex]

ok, so if i want to find the general solution,

should i just let it be in terms of convolution plus the homogenous equation right?

that should be my final general solution.
 
  • #28
Right, although in this case, I think your second solution comes from solving [itex]u''(x)=0[/itex] and [itex]xu'-u=0[/tex]. which isn't really "the homogeneous equation".
 
  • #29
gabbagabbahey said:
Right, although in this case, I think your second solution comes from solving [itex]u''(x)=0[/itex] and [itex]xu'-u=0[/tex]. which isn't really "the homogeneous equation".

owh ok..i think i get it...
anyway thanks a lot for your help!
i really appreciate it..
 

1. What is an ODE?

An ODE, or ordinary differential equation, is a mathematical equation that describes how a function changes over time. It involves derivatives of the function and is commonly used in physics and engineering to model processes and systems.

2. What is the Fourier Transform?

The Fourier Transform is a mathematical operation that converts a function from its original domain (often time or space) to a representation in the frequency domain. It decomposes a function into a sum of sinusoidal functions of different frequencies, allowing for analysis and manipulation of signals and data.

3. How can the Fourier Transform be used to solve ODEs?

By taking the Fourier Transform of both sides of an ODE, the original differential equation can be transformed into an algebraic equation in the frequency domain. This makes it easier to solve and obtain a solution, which can then be transformed back to the original domain using the inverse Fourier Transform.

4. What are the benefits of using the Fourier Transform to solve ODEs?

The Fourier Transform allows for the solution of complex ODEs that may not have analytical solutions. It also provides a better understanding of the frequency components of the solution and can be used to analyze the stability and behavior of the system over time.

5. Are there any limitations to using the Fourier Transform to solve ODEs?

The Fourier Transform is mainly applicable to linear ODEs and may not be suitable for nonlinear systems. It also assumes that the system is time-invariant, meaning that the system's behavior does not change over time. Additionally, the Fourier Transform may not be suitable for solving boundary value problems.

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