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??

<br /> That doesn&#039;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&#039;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&gt;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&gt;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:

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.

 

[absolute76]

Actually, I made an error in my previous post:

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.

<br /> <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&#039;v-uv&#039;)/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:

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.

<br /> <br /> &lt;br /&gt; i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]&lt;br /&gt;<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)----&gt;u~(k)[-1+ik²]??<br /> <br /> owk how do i separate u~(k) so that it don&#039;t cancel each other(left and right side)??

 

[gabbagabbahey]

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

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&#039;&#039;(x)=0 and xu&#039;-u=0[/tex]. which isn&#039;t really &quot;the homogeneous equation&quot;.

 

[absolute76]

Right, although in this case, I think your second solution comes from solving u&#039;&#039;(x)=0 and xu&#039;-u=0[/tex]. which isn&#039;t really &quot;the homogeneous equation&quot;.

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