Solving ODE using Fourier Transform

[absolute76]

i have found the general solution which is,

u(x)= (C1 + C2x)e^ax + (1/2a)$$\int f(x-y) e^\left|y\right| dy<br /> <br /> is this correct??<br /> now, i just want your help to guide me for justifying f(x)=x^5...<br /> <br /> is that wrong if i solve the integration and just substitute the integral which is the range( infinity to negative infinity)??<br /> <br /> thank you...$$

 

[D H]

i have found the general solution which is,

u(x)= (C1 + C2x)e^ax + (1/2a)$$\int f(x-y) e^\left|y\right| dy<br /> <br /> is this correct??$$

$$That doesn't look correct. Could you show the work that led you to this result? Also, since there is no one agreed-upon definition of the Fourier transform, tell us what formalism you are using.<br /> <br /> I know just enough about Fourier transforms to be dangerous. I've asked other homework helpers who are better versed than am I in Fourier transforms to jump in and help. Until then, show some work and try to think about the second part of the question.<br /> <br /> A hint for that: Are those functions square integrable?$$

 

[absolute76]

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??

 

[absolute76]

A hint for that: Are those functions square integrable?

can you explain in detail of what you mean by that??

 

[gabbagabbahey]

You've correctly found the particular solution, $u_p(x)=\frac{1}{2a}\int_{-\infty}^{\infty}f(x-y)e^{-a|y|}dy$, (assuming $a>0$) via Fourier Transform methods. But $(c_1+c_2x)e^{ax}$ does not satisfy the homogeneous equation $-\frac{d^2 u}{dx^2}+a^2u=0$, and so it is not the correct homogeneous solution.

 

[absolute76]

You've correctly found the particular solution, $u_p(x)=\frac{1}{2a}\int_{-\infty}^{\infty}f(x-y)e^{-a|y|}dy$, (assuming $a>0$) via Fourier Transform methods. But $(c_1+c_2x)e^{ax}$ does not satisfy the homogeneous equation $-\frac{d^2 u}{dx^2}+a^2u=0$, 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??

 

[gabbagabbahey]

$$a^2=\lambda^2\implies a=\pm\lambda$$

You have two distinct roots, not one double root.

 

[absolute76]

$$a^2=\lambda^2\implies a=\pm\lambda$$

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)??

 

[gabbagabbahey]

No, $(-\lambda)^2=\lambda^2$. 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 $x^2=c$; both $x=\sqrt{c}$ and $x=-\sqrt{c}$ are solutions.

 

[absolute76]

$$a^2=\lambda^2\implies a=\pm\lambda$$

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?

 

[absolute76]

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??

 

[absolute76]

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?

 

[gabbagabbahey]

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.

 

[gabbagabbahey]

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?

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

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

 

[absolute76]

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

There is a rule that tells you how to take the FT of the product of $x^n$ 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?

 

[gabbagabbahey]

You're missing a factor of $k$ and you don't need the $1/2\pi$:

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

 

[absolute76]

You're missing a factor of $k$ and you don't need the $1/2\pi$:

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

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?

 

[gabbagabbahey]

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]<br /> <br /> You will need to use the product rule to carry out the derivative.[/tex]

 

[absolute76]

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]<br /> <br /> You will need to use the product rule to carry out the derivative.[/tex]

$$<br /> <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f615.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":confused:" title="Confused :confused:" data-smilie="5"data-shortname=":confused:" /> error?<br /> <br /> error in which part??<br /> <br /> yes i got the answer of 2k³-4k²-k/(-2k²+1)² using...(u'v-uv')/v² am i right??$$

 

[gabbagabbahey]

error?

error in which part??

$$\frac{d}{dk}\left(k\tilde{u}\right)\neq k\frac{d\tilde{u}}{dk}$$

 

[absolute76]

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]<br /> <br /> You will need to use the product rule to carry out the derivative.[/tex]

[tex] <br /> $$i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]$$<br /> <br /> owk i think i got it, is it correct...i(iu~(k)-k²u~(k))<br /> <br /> that will give me -u~(k)+ik²u~(k)---->u~(k)[-1+ik²]??<br /> <br /> owk how do i separate u~(k) so that it don't cancel each other(left and right side)??[/tex]

 

[gabbagabbahey]

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

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

No.

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

This is a basic application of the product rule.

 

[absolute76]

No.

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

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?

 

[gabbagabbahey]

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:

$$\frac{d\tilde{u}}{\tilde{u}}=2\frac{k^2-1}{k}dk$$

 

[absolute76]

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:

$$\frac{d\tilde{u}}{\tilde{u}}=2\frac{k^2-1}{k}dk$$

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?

 

[gabbagabbahey]

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 $\tilde{g}(k)=\frac{1}{k^2}$

 

[absolute76]

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 $\tilde{g}(k)=\frac{1}{k^2}$

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.

 

[gabbagabbahey]

Right, although in this case, I think your second solution comes from solving $u''(x)=0$ and [itex]xu'-u=0[/tex]. which isn't really "the homogeneous equation".[/itex]

 

[absolute76]

Right, although in this case, I think your second solution comes from solving $u''(x)=0$ and [itex]xu'-u=0[/tex]. which isn't really "the homogeneous equation".[/itex]

$<br /> owh ok..i think i get it...<br /> anyway thanks a lot for your help!<br /> i really appreciate it..$