As is well known, the linear one-dimensional wave equation

$\displaystyle -\phi_{tt}+\phi_{xx} = 0$, (1)

where $\phi: {\Bbb R} \times {\Bbb R} \to {\Bbb R}$ is the unknown field (which, for simplicity, we assume to be smooth), can be solved explicitly; indeed, the general solution to (1) takes the form

$\displaystyle \phi(t,x) = f( t+x ) + g(t-x)$ (2)

for some arbitrary (smooth) functions $f, g: {\Bbb R} \to {\Bbb R}$.  (One can of course determine f and g once one specifies enough initial data or other boundary conditions, but this is not the focus of my post today.)

When one moves from linear wave equations to nonlinear wave equations, then in general one does not expect to have a closed-form solution such as (2).  So I was pleasantly surprised recently while playing with the nonlinear wave equation

$\displaystyle -\phi_{tt}+\phi_{xx} = e^\phi$, (3)

to discover that this equation can also be explicitly solved in closed form.  (I hope to explain why I was interested in (3) in the first place in a later post.)

A posteriori, I now know the reason for this explicit solvability; (3) is the limiting case $a = 0, b \to -\infty$ of the more general equation

$\displaystyle -\phi_{tt}+\phi_{xx} = e^{\phi+a} - e^{-\phi+b}$

which (after applying the simple transformation $\phi = \frac{b-a}{2} + \psi( \sqrt{2} e^{\frac{a+b}{4}} t, \sqrt{2} e^{\frac{a+b}{4}} x)$) becomes the sinh-Gordon equation

$\displaystyle -\psi_{tt} + \psi_{xx} = \sinh(\psi)$

(a close cousin of the more famous sine-Gordon equation $-\phi_{tt} + \phi_{xx} = \sin(\phi)$), which is known to be completely integrable, and exactly solvable.  However, I only realised this after the fact, and stumbled upon the explicit solution to (3) by much more classical and elementary means.  I thought I might share the computations here, as I found them somewhat cute, and seem to serve as an example of how one might go about finding explicit solutions to PDE in general; accordingly, I will take a rather pedestrian approach to describing the hunt for the solution, rather than presenting the shortest or slickest route to the answer.

[The computations do seem to be very classical, though, and thus presumably already in the literature; if anyone knows of a place where the solvability of (3) is discussed, I would be very happy to learn of it.]  [Update, Jan 22: Patrick Dorey has pointed out that (3) is, indeed, extremely classical; it is known as Liouville's equation and was solved by Liouville in J. Math. Pure et Appl. vol 18 (1853), 71-74, with essentially the same solution as presented here.]

– Symmetries of (3) –

To simplify the discussion let us ignore all issues of regularity, division by zero, taking square roots and logarithms of negative numbers, etc., and proceed for now in a purely formal fashion, pretending that all functions are smooth and lie in the domain of whatever algebraic operations are being performed.  (It is not too difficult to go back after the fact and justify these formal computations, but I do not wish to focus on that aspect of the problem here.)

Although not strictly necessary for solving the equation (3), I find it convenient to bear in mind the various symmetries that (3) enjoys, as this provides a useful “reality check” to guard against errors (e.g. arriving at a class of solutions which is not invariant under the symmetries of the original equation).  These symmetries are also useful to normalise various special families of solutions.

One easily sees that solutions to (3) are invariant under spacetime translations

$\displaystyle \phi(t,x) \mapsto \phi(t-t_0,x-x_0)$ (4)

and also spacetime reflections

$\displaystyle \phi(t,x) \mapsto \phi(\pm t, \pm x)$. (5)

Being relativistic, the equation is also invariant under Lorentz transformations

$\displaystyle \phi(t,x) \mapsto \phi( \frac{t-vx}{\sqrt{1-v^2}}, \frac{x-vt}{\sqrt{1-v^2}} )$.  (6)

Finally, one has the scaling symmetry

$\displaystyle \phi(t,x) \mapsto \phi( \lambda t, \lambda x ) + 2 \log \lambda$. (7)

– Solution to (3) –

Henceforth $\phi$ will be a solution to (3).  In view of the linear explicit solution (2), it is natural to move to null coordinates

$u = t+x, v = t-x$,

thus

$\displaystyle \partial_u = \frac{1}{2} (\partial_t + \partial_x); \partial_v = \frac{1}{2} (\partial_t - \partial_x)$

and (3) becomes

$\displaystyle \phi_{uv} = - \frac{1}{4} e^{\phi}$. (8)

The various symmetries (4)-(7) can of course be rephrased in terms of null coordinates in a straightforward manner.  The Lorentz symmetry (6) simplifies particularly nicely in null coordinates, to

$\displaystyle \phi(u,v) \mapsto \phi( \lambda u, \lambda^{-1} v )$. (9)

Motivated by the general theory of stress-energy tensors of relativistic wave equations (of which (3) is a very simple example), we now look at the null energy densities $\phi_u^2, \phi_v^2$. For the linear wave equation (1) (or equivalently $\phi_{uv}=0$), these null energy densities are transported in null directions:

$\displaystyle \partial_v \phi_u^2 = 0; \partial_u \phi_v^2 = 0$. (10)

(One can also see this from the explicit solution (2).)

The above transport law isn’t quite true for the nonlinear wave equation, of course, but we can hope to get some usable substitute. Let us just look at the first null energy $\phi_u^2$ for now. By two applications of (10), this density obeys the transport equation

$\displaystyle \partial_v \phi_u^2 = 2 \phi_u \phi_{uv}$
$\displaystyle = - \frac{1}{2} \phi_u e^\phi$
$\displaystyle = - \frac{1}{2} \partial_u( e^\phi )$
$\displaystyle = 2 \partial_u \phi_{uv}$
$\displaystyle = \partial_v ( 2 \phi_{uu} )$

and thus we have the pointwise conservation law

$\displaystyle \partial_v ( \phi_u^2 - 2 \phi_{uu} ) = 0$

which implies that

$\displaystyle - \frac{1}{2} \phi_{uu} + \frac{1}{4} \phi_u^2 = U(u)$ (11)

for some function $U: {\Bbb R} \to {\Bbb R}$ depending only on u. Similarly we have

$\displaystyle -\frac{1}{2} \phi_{vv} + \frac{1}{4} \phi_v^2 = V(v)$

for some function $V: {\Bbb R} \to {\Bbb R}$ depending only on v.

For any fixed v, (11) is a nonlinear ODE in u. To solve it, we can first look at the homogeneous ODE

$\displaystyle - \frac{1}{2} \phi_{uu} + \frac{1}{4} \phi_u^2 = 0$. (11′)

Undergraduate ODE methods (e.g. separation of variables, after substituting $\psi := \phi_u$) soon reveal that the general solution to this ODE is given by $\phi(u) = -2 \log( u + C ) + D$ for arbitrary constants C, D (ignoring the issue of singularities or degeneracies for now). Equivalently, (11′) is obeyed if and only if $e^{-\phi/2}$ is linear in u. Motivated by this, we become tempted to rewrite (11) in terms of $\Phi := e^{-\phi/2}$. One soon realises that

$\displaystyle \partial_{uu} \Phi = (-\frac{1}{2} \phi_{uu} + \frac{1}{4} \phi_u^2 ) \Phi$

and hence (11) becomes

$\displaystyle (-\partial_{uu} + U(u)) \Phi = 0$, (12)

thus $\Phi$ is a null (generalised) eigenfunction of the Schrodinger operator (or Hill operator) $- \partial_{uu} + U(u)$. If we let a(u) and b(u) be two linearly independent solutions to the ODE

$- f_{uu} + U f = 0$, (13)

we thus have

$\Phi = a(u) c(v) + b(u) d(v)$ (14)

for some functions c, d (which one easily verifies to be smooth, since $\phi, a, b$ are smooth and a, b are linearly independent). Meanwhile, by playing around with the second null energy density we have the counterpart to (13),

$\displaystyle (-\partial_{vv} + V(v)) \Phi = 0$,

and hence (by linear independence of a, b) c, d must be solutions to the ODE

$\displaystyle - g_{vv} + V g = 0$.

This would be a good time to pause and see whether our implications are reversible, i.e. whether any $\phi$ that obeys the relation (14) will solve (3) or (10). It is of course natural to first write (10) in terms of $\Phi$. Since

$\displaystyle \Phi_u = - \frac{1}{2} \phi_u \Phi; \Phi_v = - \frac{1}{2} \phi_v \Phi; \Phi_{uv} = (\frac{1}{4} \phi_u \phi_v - \frac{1}{2} \phi_{uv} ) \Phi$

one soon sees that (10) is equivalent to

$\displaystyle \Phi \Phi_{uv} = \Phi_u \Phi_v + \frac{1}{8}.$ (15)

If we then insert the ansatz (14), we soon reformulate the above equation as

$\displaystyle (a(u)b'(u) - b(u)a'(u)) (c(v)d'(v) - d(v)c'(v)) = \frac{1}{8}$.

It is at this time that one should remember the classical fact that if a, u are two solutions to the ODE (11), then the Wronskian $ab' - ba'$ is constant; similarly $cd'-dc'$ is constant. Putting this all together, we see that

Theorem. A smooth function $\phi$ solves (3) if and only if we have the relation (12) for some functions a, b, c, d obeying the Wronskian conditions $ab'-ba' = \alpha$, $cd'-dc' = \beta$ for some constants $\alpha, \beta$ multiplying to $\frac{1}{8}$.

Note that one can generate solutions to the Wronskian equation $ab'-ba' = \alpha$ by a variety of means, for instance by first choosing a arbitrarily and then rewriting the equation as $(b/a)' = \alpha/a^2$ to recover b.  (This doesn’t quite work at the locations when a vanishes, but there are a variety of ways to resolve that; as I said above, we are ignoring this issue for the purposes of this post.)

This is not the only way to express solutions.  Factoring a(u)d(v) (say) from (12), we see that $\Phi$ is the product of a solution $c(v)/d(v) + b(u)/a(u)$ to the linear wave equation, plus the exponential of a solution $\log a(u) + \log d(u)$ to the linear wave equation.  Thus we may write $\phi = F - 2 \log G$, where F and G solve the linear wave equation.  Inserting this back ansatz into (1) we obtain

$\displaystyle 2(- G_t^2 + G_x^2)/G^2 = e^F / G^2$

and so we see that

$\displaystyle \phi = \log \frac{2 (-G_t^2+G_x^2)}{G^2} = \log \frac{ -8 G_u G_v}{G^2}$ (16)

for some solution G to the free wave equation, and conversely every expression of the form (16) can be verified to solve (1) (since $\log 2(-G_t^2 + G_x^2)$ does indeed solve the free wave equation, thanks to (2)).   Inserting (2) into (16) we thus obtain the explicit solution

$\displaystyle \phi = \log \frac{-8 f'(t+x) g'(t-x)}{(f(t+x) + g(t-x))^2}$ (17)

to (1), where f and g are arbitrary functions (recall that we are neglecting issues such as whether the quotient and the logarithm are well-defined).

I, for one, would not have expected the solution to take this form.  But it is instructive to check that (17) does at least respect all the symmetries (4)-(7).

– Some special solutions –

If we set U=V=0, then a,b,c,d are linear functions, and so $\Phi$ is affine-linear in u, v.  One also checks that the uv term in $\Phi$ cannot vanish.  After translating in u and v, we end up with the ansatz $\Phi(u,v) = c_1 + c_2 uv$ for some constants $c_1, c_2$; applying (15) we see that $c_1 c_2=1/8$, and by using the scaling symmetry (7) we may normalise e.g. $c_1=8, c_2 = 1$, and so we arrive at the (singular) solution

$\displaystyle \phi = - 2 \log (8+uv) = \log \frac{1}{(8+t^2-x^2)^2}$. (18)

To express this solution in the form (17), one can take $f(u) = \frac{8}{u}$ and $g(v) = v$; some other choices of f, g are also possible.  (Determining the extent to which f, g are uniquely determined by $\phi$ in general can be established from a closer inspection of the previous arguments, and is left as an exercise.)

We can also look at what happens when $\phi$ is constant in space, i.e. it solves the ODE $-\phi_{tt} = e^\phi$.  It is not hard to see that U and V must be constant in this case, leading to a,b,c,d which are either trigonometric or exponential functions.  This soon leads to the ansatz $\Phi = c_1 e^{\alpha t} + c_2 e^{-\alpha t}$ for some (possibly complex) constants $c_1, c_2, \alpha$, thus $\phi = - 2 \log( c_1 e^{\alpha t} + c_2 e^{-\alpha t} )$.  By using the symmetries (4), (7) we can make $c_1=c_2$ and specify $\alpha$ to be whatever we please, thus leading to the solutions $\phi = - 2 \log \cosh \alpha t + c_3$.  Applying (1) we see that this is a solution as long as $e^{c_3} =2 \alpha^2$.  For instance, we may fix $c_3=0$ and $\alpha = 1/\sqrt{2}$, leading to the solution

$\displaystyle \phi = - 2 \log \cosh \frac{t}{\sqrt{2}}$. (19)

To express this solution in the form (17), one can take for instance $f(u) = e^{u/\sqrt{2}}$ and $g(v) = e^{-v/\sqrt{2}}$.

One can of course push around (18), (19) by the symmetries (4)-(7) to generate a few more special solutions.