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.

  • $\begingroup$ The essential way to prove that a given quantity $A$ cannot be expressed as any function of certain variables $x,y,z$ is so find two scenarios which share the same tuple $(x,y,z)$ but different values of $A$. Seems like it should be easy to find two solutions to the wave equation that agree in the value of $u$ at a particular $x,t$ but which disagree in $u_x$ and $u_t$. $\endgroup$
    – Erick Wong
    Nov 4, 2018 at 16:17
  • $\begingroup$ Thank you for your comment. I don't understand at all what you're saying. Can you give a more detailed answer, please? $\endgroup$
    – user326159
    Nov 4, 2018 at 21:43
  • $\begingroup$ My answer depends on one ambiguity in the question. Are you trying to show that there are no suitable functions $f^a$ and $g^a$ for this particular choice of variables? Or are you trying to show there is no other change of variables that allows such a representation? $\endgroup$
    – Erick Wong
    Nov 5, 2018 at 2:39
  • $\begingroup$ For this particular choice of variables $\endgroup$
    – user326159
    Nov 5, 2018 at 8:07
  • $\begingroup$ This may help math.stackexchange.com/questions/2601371/… $\endgroup$
    – Gustave
    Nov 5, 2018 at 17:35

2 Answers 2


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.]


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: $$ \frac{\partial u}{\partial x} - \frac{\partial u}{\partial t} = v \quad ; \quad \frac{\partial v}{\partial x} + \frac{\partial v}{\partial t} = 0 \\ \frac{\partial u}{\partial x} + \frac{\partial u}{\partial t} = v \quad ; \quad \frac{\partial v}{\partial x} - \frac{\partial v}{\partial t} = 0 $$

  • 2
    $\begingroup$ Thank you for your answer. I see that you have solved the wave equation but I don't see what is the relation between this and the question I've asked $\endgroup$
    – user326159
    Nov 4, 2018 at 21:41

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