Converting a special case of telegrapher's equation into the Klein-Gordon equation I am trying to convert the following special case of the telegrapher's equation into to an equation that does not consist of first-order partial derivatives:
$$w_{tt}+k(t)w_t-\alpha^2w_{xx}=0\tag{1}$$
Note that $w = w(t,x)$ and that I am using the ${\partial{f(t,x)}\over{\partial{t}}} = f_t$ notation.
I found out from here that if $k(t)$ was actually a constant instead of a function and if I substututed $w(t,x) = e^{{-}{{kt}\over2}}u(t,x)$ into equation $(1)$, I would get the Klein-Gordon equation:
$$ u_{tt}-\alpha^2u_{xx}-{1\over4}k^2u = 0\tag{2} $$
The main advantage of the resulting Klein-Gordon equation over the telegrapher's equation is that it does not have first-order partial derivatives in it. However, in my case $k(t)$ is not a constant, but a function of the variable $t$. The problem occurs when I try to substitute $w(t,x) = e^{{-}{{k(t)t}\over2}}u(t,x)$ into equation $(1)$. I get the resulting equation:
$$u_{tt} - u_tk_tt+{u\over4}\Big((k_t)^2t^2 -2k_{tt}t-4k_t-k^2\Big)-\alpha^2{u_{xx}}=0\tag{3}$$
In equation $(3)$ exists the first-order partial derivative of the function $u(t,x)$, which is the main problem. The partial derivatives of $k(t)$ are not the problem because in my case they are known.
To summarize, I am trying to obtain a resulting equation that does not have first-order partial derivative of $u(t,x)$ in it. This means that $w(t,x) = e^{{-}{{k(t)t}\over2}}u(t,x)$ is not a valid substitute function.
My questions are:

*

*Is there a function that I can plug into equation $(1)$ which would yield an equation with only second-order partial derivatives of $u(t,x)$ in it?

*If the answer is yes, what is that function?

Thank you for your time.
 A: Let's introduce a multiplier $\mu(t)$ and multiply the equation with it:
$$\mu w_{tt}+ \mu k w_t-\alpha^2 \mu w_{xx}=0$$
Then we add in another term $\beta(t)w$ and subtract it:
$$\mu w_{tt}+ \mu k w_t + \beta w - \beta w -\alpha^2 \mu w_{xx}=0$$
The idea is to regroup the first three terms so as to form $(\mu w)_{tt}$. This is possible if
$$2\mu_t = k \mu \; \text{ and } \; \mu_{tt} = \beta(t)$$
Then we have that the equation can be rewritten as
$$(\mu w)_{tt} -\alpha^2 (\mu w)_{xx} - \beta w =0$$
or with $u=\mu w$
$$u_{tt} -\alpha^2 u_{xx} - \frac{\beta}{\mu} u =0$$
For $\mu$, we can see that it is formed as
$$\mu = \exp\left(\frac{1}{2}\int k(t) dt\right)$$
which means there is a degree of freedom in the choice of $\mu$.
To put it succinctly, the generalization of your substitution for time dependent $k$ is not $w(t,x) = e^{{-}{{kt}\over2}}u(t,x)$, but $w(t,x) = e^{{-}{\int k(t) dt}\over2}u(t,x)$.
I'll let you work out what happens with $\beta$, but you should recover the exact same form of the Klein-Gordon equation you had for the $k$ time-independent case.
