The wave equation as a system of first-order PDE's

Question

My question is related to Wave equation written as a system of first order PDEs

Writing the wave equation $$u_{tt}-u_{xx}=0$$ as a system of first-order partial differential equations for $$\phi(x,t)=(\phi^1(x,t),\phi^2(x,t),\phi^3(x,t))=(u,u_x,u_t),$$ we get $$\left\{ \begin{array}{cc} \frac{\partial \phi^1(x,t)}{\partial t}=\phi^3(x,t) \\ \frac{\partial \phi^1(x,t)}{\partial x}=\phi^2(x,t) \\ \frac{\partial \phi^2(x,t)}{\partial t}=\frac{\partial \phi^3(x,t)}{\partial x} \\ \frac{\partial \phi^3(x,t)}{\partial t}=\frac{\partial \phi^2(x,t)}{\partial x} \\ \end{array} \right.$$ I want to prove that the system cannot be expressed in the form $$\left\{ \begin{array}{cc} \frac{\partial \phi^\alpha(x,t)}{\partial x}=f^\alpha (x,t,\phi) \\ \frac{\partial \phi^\alpha(x,t)}{\partial t}=g^\alpha (x,t,\phi) \end{array} \right.$$ for $\alpha=1,2,3$ and $f^\alpha(x,t,\phi)$ and $g^\alpha(x,t,\phi)$ are functions. I think that the unique relations I can get are written in the previous system, but I don't know how to prove formally that there are no more.

Any help would be appreciate.

Answer 1

I’m not much of a PDE person, but it seems this can be done by the definition of what it means to express a quantity as a function of other varying quantities. Suppose it was possible to find a relation of the form

$$u_{xx}=f^2 (x,t,u,u_x,u_t).$$

(For specificity, I just picked one of the proposed equations with $\alpha=2$ and rewrote it in terms of $u$ instead of $\phi$.)

This relation would by assumption hold over the entire solution space of the wave equation. So, imagine we have two solutions $u$ and $v$ that happen to satisfy

$$u(1,2) = v(1,2) = 3, \\ u_x(1,2) = v_x(1,2) = 4, \\ u_t(1,2) = v_t(1,2) = 5.$$

According to our hypothetical relation, we must have

$$u_{xx}(1,2) = f^2(1,2,3,4,5) = v_{xx}(1,2).$$

In other words, we may not know much about $f^2$ but we know that it’s a function, which means it has to produce the same output for the same inputs. Therefore, if in fact $u_{xx}(1,2) \ne v_{xx}(1,2)$, we contradict the existence of $f^2$.

So let’s apply this strategy now with a very simple family of solutions to the wave equation, such as:

$$ u(x,t) = A(x^2 + t^2),$$

where $A$ is any constant. Note that regardless of $A$, we have

$$u(0,0) = u_x(0,0) = u_t(0,0) = 0,$$

but on the other hand $u_{xx}(0,0) = A$. This contradicts the assumption that $u_{xx}$ can be written as a function of $x,t,u,u_x,u_t$.

[Note: My admittedly fuzzy understanding is that it’s common in PDEs to allow the equations to not hold at certain singularities. However our functions are perfectly smooth so the only singularities might be in the definition of $f$. However it is easily to generalize these examples to target any given tuple $(x,t,u,u_x,u_t) = (a,b,c,d,e)$ rather than $(0,0,0,0,0)$, so we can account for this as well. I leave the details to you if needed.]

Answer 2

Not exactly what you asked for, but maybe a sensible alternative (I hope so).
Employing Operator Calculus we indeed have sort of a splitting into first order equations: $$ u_{xx}-u_{tt} = \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial t^2} = \left[ \left(\frac{\partial}{\partial x}\right)^2 - \left(\frac{\partial}{\partial t}\right)^2 \right] u = 0 \quad \Longleftrightarrow \\ \left(\frac{\partial}{\partial x} + \frac{\partial}{\partial t}\right) \left(\frac{\partial}{\partial x} - \frac{\partial}{\partial t}\right) u = 0 \\ \left(\frac{\partial}{\partial x} - \frac{\partial}{\partial t}\right) \left(\frac{\partial}{\partial x} + \frac{\partial}{\partial t}\right) u = 0 $$ Let $u(x,t) = f(x-t)$ , then: $$ \left(\frac{\partial}{\partial x} + \frac{\partial}{\partial t}\right) f(x-t) = \frac{\partial f}{\partial (x-t)}\frac{\partial (x-t)}{\partial x} + \frac{\partial f}{\partial (x-t)}\frac{\partial (x-t)}{\partial t} = \\ = \frac{\partial f}{\partial (x-t)}(+1) + \frac{\partial f}{\partial (x-t)}(-1) = 0 $$ Let $u(x,t) = g(x+t)$ , then: $$ \left(\frac{\partial}{\partial x} - \frac{\partial}{\partial t}\right) g(x+t) = \frac{\partial g}{\partial (x+t)}\frac{\partial (x+t)}{\partial x} - \frac{\partial g}{\partial (x+t)}\frac{\partial (x+t)}{\partial t} = \\ = \frac{\partial g}{\partial (x+t)}(+1) - \frac{\partial g}{\partial (x+t)}(+1) = 0 $$ It is concluded that the general solution is given by any linear combination of $f$ and $g$ as : $$ u(x,t) = \lambda\,f(x-t) + \mu\,g(x+t) $$ Update. Writing the wave equation as a system of first-order partial differential equations, using the above: $$ \left(\frac{\partial}{\partial x} - \frac{\partial}{\partial t}\right) u = v \quad ; \quad \left(\frac{\partial}{\partial x} + \frac{\partial}{\partial t}\right) v = 0 \\ \left(\frac{\partial}{\partial x} + \frac{\partial}{\partial t}\right) u = v \quad ; \quad \left(\frac{\partial}{\partial x} - \frac{\partial}{\partial t}\right) v = 0 $$ Or: $$ \frac{\partial u}{\partial x} - \frac{\partial u}{\partial t} = v \quad ; \quad \frac{\partial v}{\partial x} + \frac{\partial v}{\partial t} = 0 \\ \frac{\partial u}{\partial x} + \frac{\partial u}{\partial t} = v \quad ; \quad \frac{\partial v}{\partial x} - \frac{\partial v}{\partial t} = 0 $$