Fourier Transform of Differential Equation

[Hart]

Homework Statement

A differential equation [*] is given by:

[tex] \frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = \frac{\partial u}{\partial t} [/tex]

By first Fourier transforming the equation (*) with respect to x, show by substitution that:

[tex] u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t} [/tex]

is the Fourier transform of the solution of [*] , where A(k) is an unknown function of k.

Homework Equations

Derivative property of Fourier transforms (with respect to x):

[tex] F\left[ \frac{\partial f}{\partial x} \right] = ikF[f] [/tex]

Also know:

[tex] F(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x)e^{ikx}dx [/tex]

[tex] F(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F(k)e^{ikx}dk [/tex]

The Attempt at a Solution

.. Fourier transform [*] with respect to x, treating t as a parameter. k used as a variable.

Firstly rearrange [*] to get just [tex] \frac{\partial^{3}u}{\partial x^{3}} [/tex] on the LHS

Then the LHS:

[tex] F\left[ \frac{\partial^{3}u}{\partial x^{3}} \right] = F\left[ \frac{\partial^{2}}{\partial x^{2}}\left( \frac{\partial u}{\partial x} \right)\ right] = -k^{2}\left( \frac{\partial^{2}}{\partial x^{2}} \right)F [/tex]

Then the RHS:

RHS = [tex] \frac{\partial u}{\partial t} - 2 \left( \frac{\partial u}{\partial x} \right) [/tex]

can take the t term outside the integral as it is just a constant parameter, therefore:

[tex] F\left[ \left(\frac{\partial u}{\partial t}\right)-\left(\frac{2\partial u}{\partial x}\right) \right] = \frac{1}{\sqrt{2\pi}}\left(\frac{\partial}{\partial t} \right)\int_{-\infty}^{\infty}\frac{-2\partial u}{\partial x}e^{-ikx} dx [/tex]

.. then obviously some more steps, not sure really where to go next though.

Not sure if I'm even going about this in vaguely the right way .. so help will definitely be appreciated!

 

[Mapes]

Can you proofread the left-hand side? It's not clear what you're doing. You want to be replacing every spatial derivative with ik.

The integral definition of F(u) isn't needed in this problem.

 

[Hart]

This is the LHS calculations again, in full:

[tex] F\left[ \frac{\partial^{3}u}{\partial x^{3}} \right] = F\left[ \frac{\partial^{2}}{\partial x^{2}}\left( \frac{\partial u}{\partial x} \right)\right]

= ikF\left[\frac{\partial u^{2}}{\partial x} \right] = (ik)^{2}F\left[u\right] = -k^{2}\left( \frac{\partial^{2}u^{2}}{\partial x^{2}} \right)[/tex]

.. hopefully that's more clear now (and somewhat correct!)?!?

 

[Mapes]

This is the LHS calculations again, in full:

[tex] F\left[ \frac{\partial^{3}u}{\partial x^{3}} \right] = F\left[ \frac{\partial^{2}}{\partial x^{2}}\left( \frac{\partial u}{\partial x} \right)\right]

= ikF\left[\frac{\partial u^{2}}{\partial x} \right] = (ik)^{2}F\left[u\right] = -k^{2}\left( \frac{\partial^{2}u^{2}}{\partial x^{2}} \right)[/tex]

.. hopefully that's more clear now (and somewhat correct!)?!?

I agree up to the second equals sign. Then you seem to lose a derivative, and later lose F altogether. Why not just use the relationship for a spatial derivative three times?

 

[Hart]

The term being [tex] \frac{\partial^{3}u}{\partial x^{3}} [/tex] and not just a second deriviative has thrown me somewhat, though I know it shouldn't!

.. the relationship for spatial derivative 3 times? Sorry to sound stupid! At the moment I'm trying to adapt a second derivative example to this question, so I thought would be some errors.

I was using this relationship for a property of Fourier transform derivatives:

[tex] F\left[ \frac{\partial f}{\partial x} \right] = ikF[f] [/tex]

Hence that's where the derivative went for F (from 3rd to 2nd).

.. bit confused now with all these derivatives! :|

 

[vela]

Use the fact that

[tex]\frac{\partial^{3}u}{\partial x^{3}} = \frac{\partial}{\partial x}\left(\frac{\partial^{2}u}{\partial x^{2}}\right)[/tex]

so

[tex]F\left[\frac{\partial^{3}u}{\partial x^{3}}\right] = F\left[\frac{\partial}{\partial x}\left(\frac{\partial^{2}u}{\partial x^{2}}\right)\right] = ?[/tex]

 

[Hart]

.. :umm: nope I don't really get it

..

[tex]F\left[\frac{\partial^{3}u}{\partial x^{3}}\right] = F\left[\frac{\partial}{\partial x}\left(\frac{\partial^{2}u}{\partial x^{2}}\right)\right] = ik[/tex] ??

(since you said earlier I need to replace every spatial derivative with ik)

.. apologies for not being very clever at the moment with this!

 

[vela]

Compare what you wrote here:

[tex] F\left[ \frac{\partial f}{\partial x} \right] = ikF[f] [/tex]

with what you have here:

[tex]F\left[\frac{\partial}{\partial x}\left(\frac{\partial^{2}u}{\partial x^{2}}\right)\right] = \cdots[/tex]

What is f(x) in this case?

(since you said earlier I need to replace every spatial derivative with ik)

.. apologies for not being very clever at the moment with this!

We're backing up a bit here since you didn't see what Mapes was saying earlier.

 

[Hart]

.. I did see Mapes comment and my comment after was an attempt to show how I had got that answer, allbeit incorrect, because I wasn't sure where I'd gone wrong as weren't completely sure how to do the calculation.

Right, so as far as where we are now..

[tex]f(x) = \frac{\partial^2 u}{\partial x^{2}}[/tex]

??

 

[vela]

Right, so as far as where we are now..

[tex]f(x) = \frac{\partial^2 u}{\partial x^{2}}[/tex]

??

Yes, so you get

[tex]F\left[\frac{\partial^{3}u}{\partial x^{3}}\right] = F\left[\frac{\partial}{\partial x}\left(\frac{\partial^{2}u}{\partial x^{2}}\right)\right] = ikF\left[\frac{\partial^2 u}{\partial x^{2}}\right][/tex]

Do you see now where this is going?

 

[Hart]

[tex]F\left[\frac{\partial^{3}u}{\partial x^{3}}\right] = F\left[\frac{\partial}{\partial x}\left(\frac{\partial^{2}u}{\partial x^{2}}\right)\right] = ikF\left[\frac{\partial^2 u}{\partial x^{2}}\right][/tex]

[tex]= -k^{2}F\left[\frac{\partial u}{\partial x}\right][/tex]

??

 

[vela]

Yes, and you want to do it one more time.

 

[Hart]

..

[tex]F\left[\frac{\partial^{3}u}{\partial x^{3}}\right] = -ik^{3}\left[u\right][/tex]

??

 

[Mapes]

..

[tex]F\left[\frac{\partial^{3}u}{\partial x^{3}}\right] = -ik^{3}\left[u\right][/tex]

??

Looks good. Perhaps you can see where this is starting to match elements in the solution.

 

[vela]

And perhaps you can see what Mapes meant when suggesting you replace every spatial derivative with ik.

[tex]
F\left[\frac{\partial^{3}u}{\partial x^{3}}\right] = (ik)^{3}F
[/tex]

 

[Hart]

Yep! It's obvious now!

Righto, now know these expressions then:

[tex]F\left[\frac{\partial^{3}u}{\partial x^{3}}\right] = (ik)^{3}F[/tex]

[tex] F\left[ \frac{\partial u}{\partial x} \right] = ikF [/tex]

so presumably:

[tex] \frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = (ik)^{3}F + 2ikF[/tex]

correct??

.. and then the other side of the original equation:

[tex] F\left[ \left(\frac{\partial u}{\partial t}\right)\right] = \frac{1}{\sqrt{2\pi}}\left(\frac{\partial}{\partial t} \right)\int_{-\infty}^{\infty} u e^{-ikx} dx [/tex]

also correct??

Hopefully getting somewhere with this now!

 

[Hart]

.. had a look at this again, and I don't really get how to now 'sum up' these results in order to get through substitution the expression required.

Also, another question following on from this..

"Find [tex]A(k)[/tex] for the case where [tex]u(x,t)[/tex] at [tex]t=0[/tex] is given by [tex]u(x,0) = U_{0}\delta(x-a)[/tex] where [tex]U_{0}[/tex] is a constant."

I think this follows from the answer to the first question then, but obviously I don't know how to do this at the moment

 

[ideasrule]

Don't get confused like I was by the question. The "u" in this equation:

[tex]
\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = \frac{\partial u}{\partial t}
[/tex]

is NOT the same as the "u" in this equation:

[tex]
u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}
[/tex]

The latter "u" is meant to be the Fourier transform of the former "u".

To proceed, "substitution" just means putting in [tex]
u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}
[/tex] for F and showing that both sides of the differential equation are equal.

 

[Hart]

.. so input this:

[tex]u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t} [/tex]

into this:

[tex]\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = \frac{\partial u}{\partial t} [/tex]

??

How do I do this seeing how there are both x and t dependant differentials, and the first equation doesn't distinguish whether it it dx or dt!? Unless differentiate it with respect to x, to get a result, and then differentiate (original equation) to get a different result for dt?!

 

[ideasrule]

No, reread my last post. I said that the "u" given by [itex]
u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}[/itex] is NOT the same as the "u" given by [itex]\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = \frac{\partial u}{\partial t} [/itex].

You did a lot of work to get [itex]\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = (ik)^{3}F + 2ikF[/itex], so use it. F is actually [itex]u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}[/itex], as the question states.

 

[Hart]

sorry, I actually got confused too by that even after reading your comment, what I meant to ask was:

.. so input this:

[tex]u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t} [/tex]

into:

[tex]\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = (ik)^{3}F + 2ikF[/tex]

hence:

[tex] = \frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = (ik)^{3}[A(k)e^{-i \left( k^{3}-2k \right) t}] + 2ik\left[A(k)e^{-i \left( k^{3}-2k \right) t}\right][/tex]

then simplify:

[tex]\left(ik(k^{2} + 2)\right)\left[A(k)e^{-i \left( k^{3}-2k \right) t}\right][/tex]

.. correct?

Then need to equate this to the other side of the original equation, i.e. the partial derivative of U by t.. I don't get how to do this part in order to get the final answer.

 

[Mapes]

.. so input this:

[tex]u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t} [/tex]

into this:

[tex]\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = \frac{\partial u}{\partial t} [/tex]

??

How do I do this seeing how there are both x and t dependant differentials, and the first equation doesn't distinguish whether it it dx or dt!? Unless differentiate it with respect to x, to get a result, and then differentiate (original equation) to get a different result for dt?!

As ideasrule noted, [itex]u(k,t)[/itex] is the Fourier transform of [itex]u(x,t)[/itex] which is the variable in the original equation. [itex]u(k,t)[/itex] does not solve the original equation, but it does solve the Fourier transform of the equation. What did you get for this Fourier transform of the original equation? Note that [itex]F[\partial u(x,t)/\partial t][/itex] is just [itex]\partial u(k,t)/\partial t[/itex]. Can you put the whole thing together?

 

[Hart]

.. is my last post not correct?

 

[ideasrule]

The right side is the Fourier transform of the partial derivative of u by t (remember that you transformed both sides). That's equal to the partial derivative of the Fourier transform of u.

 

[vela]

.. is my last post not correct?

No, because

[tex]\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = (ik)^{3}F + 2ikF[/tex]

isn't correct. On the LHS, you have the original equation; on the RHS, you have its Fourier transform. If you instead had said,

[tex]
F\left[\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right)\right] = (ik)^{3}F + 2ikF
[/tex]

that would be correct.

 

[Hart]

[tex]
F\left[\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right)\right] = (ik)^{3}F + 2ikF
[/tex]

that would be correct.

.. so this is the LHS of the original equation?!

and the RHS:

[tex] F\left[ \left(\frac{\partial u}{\partial t}\right)\right] = \frac{1}{\sqrt{2\pi}}\left(\frac{\partial}{\partial t} \right)\int_{-\infty}^{\infty} u e^{-ikx} dx [/tex]

??

 

[vela]

You can rewrite the RHS slightly as

[tex]\frac{1}{\sqrt{2\pi}}\left(\frac{\partial}{\partial t} \right)\int_{-\infty}^{\infty} u e^{-ikx} dx = \frac{\partial}{\partial t} \left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} u e^{-ikx} dx\right)[/tex]

What's the thing inside the parentheses on the RHS?

 

[Hart]

um..

this:

[tex]\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} u e^{-ikx} dx\right)[/tex]

is the Fourier transform of u ?!

i.e.

[tex]F = \left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} u e^{-ikx} dx\right)[/tex]

so the RHS will be:

[tex] = \frac{\partial}{\partial t} F[/tex]

 

[Hart]

.. so then the equation will be:

[tex] (ik)^{3}F + 2ikF = \frac{\partial}{\partial t} F[/tex]

?!

 

[Mapes]

.. so then the equation will be:

[tex] (ik)^{3}F + 2ikF = \frac{\partial}{\partial t} F[/tex]

?!

Looks good. You've now obtained an ordinary differential equation (ODE) (where you used to have a partial differential equation). What's the solution to this ODE?

 

[vela]

Right. That's the equation u(k,t) satisfies since, as the others have noted, u(k,t)=F[u(x,t)].

 

[Hart]

[tex] (ik)^{3} + 2ik = \frac{\partial}{\partial t}[/tex]

 

[Mapes]

[tex] (ik)^{3} + 2ik = \frac{\partial}{\partial t}[/tex]

Unfortunately, [itex]\partial/\partial t[/itex] alone doesn't have any meaning; it's an operator that acts on [itex]u(k,t)[/itex], which can't just be divided away.

You are looking for a function whose time derivative equals the function itself multiplied by [itex]-i(k^3-2k)[/itex].

 

[Hart]

well..

[tex]e^{-i(k^3-2k)t}[/tex]

would mean:

[tex]\left(\frac{\partial}{\partial t}\right)e^{-i(k^3-2k)t} = -i(k^3-2k)e^{-i(k^3-2k)t}[/tex]

?!?

 

[Mapes]

This is what you were hoping to demonstrate at the beginning of this thread, yes?