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I'm interested in finding solutions a fourth order version of the standard wave equation in $d$ dimensional Minkowski spacetime $\mathcal{M}^d$. Defining $\Box := \partial_0^2 - \sum_{i = 1}^{d-1} \partial_{i}^2$, I want to find solutions to $$\Box^2\Phi(x) = 0,$$ which are not also solutions to $\Box \Phi(x) = 0$, and which transform as scalars under Lorentz transformations. I'm mostly interested in solutions for $d=4$, but I don't think that's likely to be relevant at the level of analysis I'm currently working at so I want to stick to general $d$ if possible.

I know that in solving $\Box \Phi(x) = 0$ with boundary conditions that require some kind of fast enough fall off at infinity, then the solution space can be taken to be Lorentz invariant $L^2$ integrable functions. A basis of solutions for functions with these boundary conditions can be given by plane waves $\Phi_k(x) := e^{i k\cdot x}$ where $k^2 = 0$. My understanding is that as $\Box \Phi(x) = 0$ is seperable, Sturm-Liouville theory tells us that for some choices of boundary conditions we can find a function space of solutions to the equation spanned by some basis of solutions.

(Disclaimer: I don't have a very in-depth understanding of analysis. I can see that $\Phi_k \not\in L^2(\mathcal{M}^d)$, so I understand that at some level calling this a `basis' for the space is not correct. I'm happy to call it this given that two such functions are orthogonal in the sense that $\int_{\mathcal{M}^d}dx \;\Phi_k(x)\Phi_{k'}(x) \propto \delta^d(k-k')$, and that any (I think?) function in this space can be decomposed in terms of an integral over all possible $\Phi_k$ by the Fourier transform. I appreciate that to answer my question it may be necessary to go to some more detailed level of analysis where it doesn't make sense to think of $\Phi_k$ as a basis.)

I note that, given any $f$ such that $\Box f(x) = 0,$ we can construct a solution to $\Box^2 \Phi(x) = 0$ by taking $\Phi(x) = a \cdot x f(x)$, where $a$ is a vector. For some choices of $a$ this new solution may also satisfy $\Box\left( a\cdot x f(x)\right) = 0,$ but I think it's always possible to choose $a$ such that $\Box\left( a\cdot x f(x)\right) \neq 0.$

Then my question is, given a function space of solutions corresponding to a choice of boundary conditions to $\Box \Phi(x) = 0$, is there a canonical way to extend this to a larger function space of soluitons to $\Box^2 \Phi(x) = 0$, which includes functions which satisfy $\Box^2 \Phi(x) = 0$ and $\Box \Phi(x) \neq 0$, which can be expanded in terms of a basis of functions which are orthogonal under some inner product? If not a canonical way, is there some set of ways of doing this?

So then in the case of $L^2$ functions, is it possible to do something like extend to functions which increase like $x$ as $x\rightarrow \infty$? Then I could imagine that this new space could be spanned by $\Phi_k$, and $\chi_{k,a}(x) := a \cdot x e^{i k \cdot x}$, with some inner product under which $\Phi_k, \Phi_{k'}$, $\chi_{k,a}$ and $\chi_{k',a'}$ are orthogonal to each other? I think what I've written here is probably a little naive, but I'm hoping there may be some kind of result similar to this which makes sense.

After some searching online I've found that there exists a fourth order Sturm-Liouville theory, but I wasn't able to find anything accesible for me to read, and it appears to me that $\Box^2 \Phi(x) = 0$ is anyway not seperable, at least in Cartersian coordinates. Any references to a simple introduction or review article of fourth-order Sturm-Lioville theory would be appreciated, as well as any local coordinate transformation which makes $\Box^2\Phi(x) = 0$ seperable, if you think this would be relevant to my problem.

I appreciate that you could read this question and say, 'why would you expect there to be a canonical way to extend solution spaces for $\Box \Phi(x) = 0$ to $\Box^2 \Phi(x) = 0$ with boarder boundary conditions?'. So here is some evidence that I have from solving the equation in 2D.

In 2D in lightcone coordinates $u = x^0 + x^1, v = x^0-x^1$ the wave operator factorises so that $\Box = \partial_u\partial_v$. This makes it simple to write down d'Alembert's general solution to $\Box\Phi(x) = 0$ for all possible boudary conditions, $\Phi(x) = f^+(x^0 + x^1) + f^-(x^0-x^1).$

It's also simple to solve $\Box^2\Phi(x) = 0$ in this case. Writing a seperable ansatz $\Phi(u,v) = U(u)V(v)$ the equation becomes $U'' V''=0$, which is solved non-trivially by either $U' = 0$, $V' = 0$, $U''=0$ or $V''=0$. Then the general solution is given by $\Phi(u,v) = f^+_1(u) + v f^+_2(u) + f^-_1(v) + u f^-_2(v).$ Introducing null vectors $k_+ = k(1,1)$ and $k_- = q(1,-1)$, then we see that $u \propto k_+ \cdot x$, and $v \propto k_- \cdot x.$ Then the general solution can be written as a sum over solutions of the form $$\Phi(x) = f_1(k\cdot x) + a \cdot x f_2(k \cdot x),$$ where $k^2 = 0$ and $a$ is any vector. If $a \propto k$ then the second term solves $\Box\Phi(x) = 0$, otherwise it produces new solutions that satisfy $\Box^2\Phi(x) = 0$ and $\Box\Phi(x) \neq 0$. (I took a linear combination of the lightcone coordinate solutions to produce the new solution). I believe this to be the general solution to $\Box^2\Phi(x) = 0$ in 2D for all possible boundary conditions, and I've intentionally written it in a form that extends in a natural way to general dimension $d$. So this is some kind of motivation for why I think my question may have a solution; it appears to me that what I'm asking in general dimensions happens in 2D based on the form of the general solution I gave here.

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You talk about the non-separability of the $\Box^2 \phi = 0$ equation, which I don't understand. Each plane wave $e^{ik\cdot x} = \prod_{j=0}^{d-1} e^{i k_j x^j}$ is already in separated form with the components of the null vector $k=(k_j)$ playing the role of the separation constants.

But rather than get into technicalities about what does or does not constitute a separable equation or a basis of separated solutions, why not just solve the equation via the Fourier transform?

Recall first the ordinary wave equation. If $\phi(x)$ is smooth and grows no faster than polynomially in any direction, its Fourier transform $\hat{\phi}(k)$ exists as a distribution (a subclass of tempered distributions in this case). The equation $\Box \phi(x) = 0$ translates to $k^2 \hat{\phi}(k) = 0$, where of course $k^2 = k_0^2 - \sum_{j=1}^{d-1} k_j^2$. No non-vanishing continuous, let alone smooth function of $k$ can satisfy that condition. Hence, $\hat{\phi}(k)$ must be a distribution. A natural candidate is $\hat{\phi}(k) = f(k) \delta(k^2)$, with $f(k)$ an at least continuous function, because of the simple identity $u \delta(u) = 0$. The function $f(k)$ is of course not unique, any functions $f_1,f_2$ such that the difference $f_1(k)-f_2(k)$ vanishes smoothly on the locus of $k^2$ define the same $\hat{\phi}(k)$. The level of regularity (including behavior at infinity) of $f(k)$ translates in a certain way to the level of regularity of $\phi(x)$ via the Fourier transform. The above parametrization of $\hat{\phi}(k)$ is actually exhaustive when these regularity classes are appropriately fixed. To see how the data in $f(k)$ translates to initial data for $\phi(x)$, try the $d=1$ example.

Now, on to the squared wave equation. Under the Fourier transform, $\Box^2 \phi(x) = 0$ translates to $(k^2) \hat{\phi}(k) = 0$. Taking derivatives of the $\delta$-function identity, we can get $u \delta'(u) + \delta(u) = 0$ or more importantly $u^2 \delta'(u)=0$. Hence, a natural candidate for a solution is \begin{align*} \hat{\phi}(k) &= f(k) \delta(k^2) + g(k) \delta'(k^2) , \\ &= F(k) \delta(k^2) + ia\cdot\partial_k [G(k) \delta(k^2)] \end{align*} for functions $f(k), g(k)$ of appropriate regularity and $G(k) = g(k)/(2ia\cdot k)$, $F(k) = f(k) - (a\cdot\partial_k) G(k)$. To make it easier to go between $g(k)$ and $G(k)$, the vector $a$ should not be null, so that $a\cdot k$ only vanishes at $k=0$. The corresponding formula in real space is $$ \phi(x) = \phi_1(x) + (a\cdot x) \phi_2(x) , $$ where $\phi_1(x), \phi_2(x)$ are independent solutions of the wave equation, $\Box \phi_{1,2}(x)=0$. This is I think the parametrization of solutions that you were looking for. As before, it is a matter of figuring out the right regularity classes for $F(k), G(k)$ and the corresponding initial data for $\phi(x)$ to make sure that the above parametrization is exhaustive.

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  • $\begingroup$ Thanks for a great answer, I can see this is definitely a good way to think about the problem. I have two questions; 1) Suppose $\phi_1,\phi_2 \in L^2$, then I can expand them in a basis of plane waves. Can I think of $a\cdot x \phi_2$ as having an expansion in terms of some basis, (eg. $(a\cdot x) e^{i k \cdot x}$ or similar), such that these basis elements are orthogonal each other and to the plane waves? $\endgroup$
    – Jojo
    Jul 27, 2021 at 17:35
  • $\begingroup$ 2) Using that $u^3 \delta''(u) = 0$, I can also solve the equation in Fourier space by $\hat{\phi}(k) = \Box_k \left[H(k) \delta(k^2)\right]$, because the $\Box_k$ acting on the delta function gives me a factor of $k^2$ additional to the $k^4$ coming from the original PDE. It looks to me like I could potentially continue this process using $u^{n+1} \delta^{(n)}(u) = 0$ to get more and more solutions. Did you use some kind of principle to select only the solution involving $\delta'(k^2)$ and not these higher solutions? $\endgroup$
    – Jojo
    Jul 27, 2021 at 17:41
  • $\begingroup$ When I said the equation appears to me not to be separable, I meant that eg. in 2D taking $\phi(x,t) = X(x)T(t)$, we get $X'''' T - 2 X'' T'' + X T''' =0$, and I can't see how to seperate out $T$ and $X$ here. Like you say though, I think this is not important given your Fourier space analysis $\endgroup$
    – Jojo
    Jul 27, 2021 at 17:43
  • $\begingroup$ @Joe 1) I don't know what to say about orthogonality, except the usual Fourier inversion formula. If it is not what you need, you might have to formulate your question more sharply. 2) Expand all $k$-derivatives. Any term like $h(k) \delta^{(n>1)}(k^2)$ will not be annihilated by $k^4$, unless $h(k)\propto (k^2)^{n-1}$, but then $k^2\delta^{n}(k^2) = -n\delta^{(n-1)}(k^2)$, etc. So these higher order derivatives of $\delta$ are redundant. 3) About separability, the issue is just in the terminology. $\endgroup$ Jul 27, 2021 at 21:01
  • $\begingroup$ 1) ok thanks I'll try to come up with a more clear question and make a new post 2) I see, thanks very much $\endgroup$
    – Jojo
    Jul 28, 2021 at 15:02

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