# Wave Equation Solution by Factoring Operators

To solve the Wave Equation

$$u_{tt} - c^2 u_{xx} = 0$$

One method is to start with operator factorization

$$u_{tt} - c^2 u_{xx} = \bigg( \frac{\partial }{\partial t} - c \frac{\partial }{\partial x} \bigg) \bigg( \frac{\partial }{\partial t} + c \frac{\partial }{\partial x} \bigg) u = 0$$

Then it is claimed the solution is

$$u(x,t) = f(x+ct) + g(x-ct)$$

where $f$ ve $g$ are arbitrary functions of single variable.

The proof goes by letting $v = u_t + cu_x$

then

$$v_t - cv_x = 0$$

has to be true. Then simultaneously the two equations are solved

$$v_t - cv_x = 0$$

$$u_t + cu_x = v$$

We know the solution for the top equation above is

$$v(x,t) = h(x+ ct)$$

where $h$ is an arbitrary function. Now wave equation is

$$u_t + cu_x = h(x + ct)$$

Here is what I do not understand. At this point that a solution is guessed as $u(x,t) = f(x+ct)$ and the book says "it is easy to check by differentiation"

$f'(s) = h(s) / 2c$.

What do I do with $h(s) / 2c$? Is the proof complete with this conclusion? When I plug in $f(x+ct)$ for $u(x,t)$ yes, I do get this equality but how does this help?

Also, after proving $f(x+ct)$ is a solution, then magically a $g(x-ct)$ is added, where does this come from? I do understand linear independence, but why didnt we add $g(x+2ct)$ for example?

Thanks,

• There is some interesting elementary stuff here www.stanford.edu/class/math220a/handouts/waveequation1.pdf – BB_ML Nov 21 '11 at 15:41
• Properly, this is a wave equation, not heat. – paul garrett Nov 21 '11 at 21:02

It's easier to understand if you change to new variables $\xi=x+ct$ and $\eta=x-ct$. Then the PDE becomes $\partial^2 u/\partial \xi \partial \eta = 0$. Integration with respect to $\eta$ gives $\partial u/\partial \xi = f(\xi)$ with $f$ an arbitrary function (the "constant" of integration, constant here meaning "independent of $\eta$"). Integrating again, now with respect to $\xi$, gives $u=F(\xi)+g(\eta)$ with $g$ an arbitrary function and $F$ the antiderivative of $f$.

• Hi Hans, I believe you are talking about utilizing characteristic coordinates, which Strauss also shows later in his book. Could you answer my question, that how f, f'(s) is worked into this thing? Thanks. – BB_ML Nov 22 '11 at 10:29
• The condition says that in order for $f$ to satisfy the equation, it must equal $2c$ times an antiderivative of $h$. But the function $h$ was arbitrary, so $f$ is arbitrary too. And adding $g(x-ct)$ corresponds to adding the integration "constant". – Hans Lundmark Nov 22 '11 at 10:43

The solutions, functions f and g, are called characteristic curves on two first order linear PDEs. They are solved by the method of characteristics, which is often missed in engineering courses on PDEs.

Notice that the wave equation's partial derivatives below are factored into first order PDE operators like a second degree polynomial. Then, these PDE operators are applied one at a time to get the wave equation for u.

$$u_{tt} - c^2 u_{xx} = \bigg( \frac{\partial }{\partial t} - c \frac{\partial }{\partial x} \bigg) \bigg( \frac{\partial }{\partial t} + c \frac{\partial }{\partial x} \bigg) u = 0$$

Learn and apply the method of characteristics to each first order PDE below to get functions f and g. They specify the characteristic curves for the wave equation.

$$u_{t} + c u_{x} = 0$$ $$u_{t} - c u_{x} = 0$$

We first solve the Wave Equation

$$u_{tt} - c^2 u_{xx} = 0$$

by operator factorization

$$u_{tt} - c^2 u_{xx} = \bigg( \frac{\partial }{\partial t} - c \frac{\partial }{\partial x} \bigg) \bigg( \frac{\partial }{\partial t} + c \frac{\partial }{\partial x} \bigg) u = 0$$

Hence we define characteristics of the solution $$u(x,t)$$ as $$\xi = x+ct$$ and $$\eta = x-ct$$ and a new function $$V( \xi, \eta) = u(x,t)$$.

Now by chain rule we get $$c \frac{\partial u }{\partial x} = c \frac{\partial V}{\partial \xi} \frac{\partial \xi }{\partial x} + c \frac{\partial V}{\partial \eta} \frac{\partial \eta }{\partial x} = c \frac{\partial V}{\partial \xi} + c \frac{\partial V}{\partial \eta} \tag i$$

Similarly, we also obtain
$$\frac{\partial u }{\partial t} = \frac{\partial V}{\partial \xi} \frac{\partial \xi }{\partial t} + \frac{\partial V}{\partial \eta} \frac{\partial \eta }{\partial t} = c \frac{\partial V}{\partial \xi} - c \frac{\partial V}{\partial \eta} \tag {ii}$$

Now adding $$\text{(i)}$$ and $$\text{(ii)}$$ we have $$\bigg( \frac{\partial }{\partial t} + c \frac{\partial }{\partial x} \bigg)u = 2 \displaystyle c \frac{\partial V }{\partial \xi} \tag {iii}$$

Subtracting $$\text{(ii)}$$ from $$\text{(i)}$$ gives $$\bigg( \frac{\partial }{\partial t} - c \frac{\partial }{\partial x} \bigg)u =2 c \frac{\partial V}{\partial \eta} \tag {iv}$$

Operating $$\text {(iii)}$$ on $$\text{(iv)}$$ we evaluate $$2 c \frac{\partial }{\partial \eta} \left( 2 c \frac{\partial V }{\partial \xi} \right) = \bigg( \frac{\partial }{\partial t} - c \frac{\partial }{\partial x} \bigg) \bigg( \frac{\partial }{\partial t} + c \frac{\partial }{\partial x} \bigg) u =0$$

Which gives $$\frac{ \partial^2 u} { \partial \xi \partial \eta} = 0 \tag {v}$$

Integrating $$\text{(v)}$$ with respect to $$\eta$$ gives $$\frac{\partial u}{\partial \xi } = f( \xi)$$ form the properties of PDEs.

Integrating $$\frac{\partial u}{\partial \xi } = f( \xi)$$ gives
$$\int \frac{\partial u}{\partial \xi } d \xi = \int f( \xi) d \xi + g( \eta )= F( \xi) + g( \eta)\text{, where } F( \xi) = \int f( \xi) d \xi$$

Hence we have $$u(x,t)= F(x+ct)+g(x-ct)$$

• By reading the above two posts i was able to come up with this answer. I hope many wont struggle like i did – Patrick Chidzalo Aug 17 '15 at 21:06
• You may find the \tag{} latex command useful to mark equations. – Enredanrestos Aug 17 '15 at 21:07