I'm working on the following problem in preparation for an exam.

  1. Let $L$ denote the one-dimensional wave operator defined for functions: $u(x,t)$ (with $x∈(0,L)$ and $t>0$) by: $$L(u)=\frac{∂^2 u}{∂t^2}−c^2 \frac{∂^2 u}{∂x^2}$$ where the constant $c>0$ is the speed of sound in the medium.

a) Prove that $L$ is a linear operator. If $u_1$, $u_2$ and $f$ are three functions which satisfy $L(u_1 )=f$ and $L(u_2)=2f$. Find a solution of the homogeneous wave equation $L(u)=0$.

Here's my solution so far.

Part a was easy enough, Let $L(u)=L(au+bv)$ such that $$ \frac{∂^2 u}{∂t^2}−c^2 \frac{∂^2 u}{∂x^2}= 0 \Rightarrow \frac{∂^2 (au+bv)}{∂t^2}−c^2 \frac{∂^2 (au+bv)}{∂x^2} \\ =a\frac{∂^2 u}{∂t^2}−ac^2 \frac{∂^2 u}{∂x^2} +b\frac{∂^2 v}{∂t^2}−bc^2 \frac{∂^2 v}{∂x^2}= a(\frac{∂^2 u}{∂t^2}−c^2 \frac{∂^2 u}{∂x^2})+b(\frac{∂^2 v}{∂t^2}−c^2 \frac{∂^2 v}{∂x^2})=0$$

Which is linear and homogenous.

So my question is regarding part b. My initial thought on how to start the problem was to relate $L(u_1)$ and $L(u_2)$ in the following way. $$ L(u_1)=\frac{∂^2 u_1}{∂t^2}−c^2 \frac{∂^2 u_1}{∂x^2}=f \\ L(u_2)=\frac{∂^2 u_2}{∂t^2}−c^2 \frac{∂^2 u_2}{∂x^2} =2f \\ \Rightarrow L(u_2)=2L(u_1) \Rightarrow \frac{∂^2 u_2}{∂t^2}−c^2 \frac{∂^2 u_2}{∂x^2}=2\left [ \frac{∂^2 u_1}{∂t^2}−c^2 \frac{∂^2 u_1}{∂x^2}\right ] $$ Some algebra later, I get that $$ \frac{∂^2 (u_2 - 2u_1)}{∂t^2}−c^2 \frac{∂^2 (u_2+2u_1)}{∂x^2} =0 $$

What I was expecting to get out of this was to say something like "Let $u=u_2-2u_1$" with the hopes of getting an equation with a single $u$ only. But, it didn't quite work out that way.

I'm not really sure where to go with this, any help is appreciated.

  • $\begingroup$ It seems all right except for the last equation...it should be: $u_2-2u_1$ also in the second term $\endgroup$ – MattG88 Mar 29 '17 at 20:59
  • $\begingroup$ That's what I was thinking, but the algebra has it as a +. I'll redo that part after I get off work, maybe I dropped a negative. $\endgroup$ – Kosta Mar 29 '17 at 21:09

We can write: $$ \frac{∂^2 u_2}{∂t^2}−c^2 \frac{∂^2 u_2}{∂x^2}=2\left [ \frac{∂^2 u_1}{∂t^2}−c^2 \frac{∂^2 u_1}{∂x^2}\right ] $$ $$\frac{∂^2 u_2}{∂t^2}−c^2 \frac{∂^2 u_2}{∂x^2}-2\left [ \frac{∂^2 u_1}{∂t^2}−c^2 \frac{∂^2 u_1}{∂x^2}\right ] =0$$ $$\frac{∂^2 u_2}{∂t^2}−2\frac{∂^2 u_1}{∂t^2}−c^2\left[\frac{∂^2 u_2}{∂x^2}-2\frac{∂^2 u_1}{∂x^2}\right]=0$$


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