Fourier Transform of Differential Equation

[Hart]

Homework Statement

A differential equation [*] is given by:

$$\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = \frac{\partial u}{\partial t}$$

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

$$u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}$$

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):

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

Also know:

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

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

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 $$\frac{\partial^{3}u}{\partial x^{3}}$$ 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 = $$\frac{\partial u}{\partial t} - 2 \left( \frac{\partial u}{\partial x} \right)$$

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

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

.. 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 :grumpy: .. 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:

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

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

 

[Mapes]

This is the LHS calculations again, in full:

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

.. 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 $$\frac{\partial^{3}u}{\partial x^{3}}$$ 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:

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

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

$$\frac{\partial^{3}u}{\partial x^{3}} = \frac{\partial}{\partial x}\left(\frac{\partial^{2}u}{\partial x^{2}}\right)$$

so

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

 

[Hart]

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

..

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

(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:

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

with what you have here:

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

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..

$$f(x) = \frac{\partial^2 u}{\partial x^{2}}$$

??

 

[vela]

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

$$f(x) = \frac{\partial^2 u}{\partial x^{2}}$$

??

Yes, so you get

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

Do you see now where this is going?

 

[Hart]

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

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

??

 

[vela]

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

 

[Hart]

..

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

??

 

[Mapes]

..

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

??

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:

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

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 $$A(k)$$ for the case where $$u(x,t)$$ at $$t=0$$ is given by $$u(x,0) = U_{0}\delta(x-a)$$ where $$U_{0}$$ 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:

$$\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = \frac{\partial u}{\partial t}$$

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

$$u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}$$

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

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

 

[Hart]

.. so input this:

$$u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}$$

into this:

$$\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = \frac{\partial u}{\partial t}$$

??

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 $u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}$ is NOT the same as the "u" given by $\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = \frac{\partial u}{\partial t}$.

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 $u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}$, 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:

$$u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}$$

into:

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

hence:

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

then simplify:

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

.. 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:

$$u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}$$

into this:

$$\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = \frac{\partial u}{\partial t}$$

??

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, $u(k,t)$ is the Fourier transform of $u(x,t)$ which is the variable in the original equation. $u(k,t)$ 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 $F[\partial u(x,t)/\partial t]$ is just $\partial u(k,t)/\partial t$. 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:

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

??

 

[vela]

You can rewrite the RHS slightly as

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

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

 

[Hart]

um..

this:

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

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]

$$(ik)^{3} + 2ik = \frac{\partial}{\partial t}$$

 

[Mapes]

$$(ik)^{3} + 2ik = \frac{\partial}{\partial t}$$

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

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

 

[Hart]

well..

$$e^{-i(k^3-2k)t}$$

would mean:

$$\left(\frac{\partial}{\partial t}\right)e^{-i(k^3-2k)t} = -i(k^3-2k)e^{-i(k^3-2k)t}$$

?!?

 

[Mapes]

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