TITLE: Solving the PDE $u_{tt}+2u_{tx}+u_{xx}=2c$
QUESTION [6 upvotes]: Consider the second-order parabolic inhomogeneous second-order PDE
$$
u_{tt}+2u_{tx}+u_{xx}=2c
$$
I have seen two ways to solve this problem. I would like to know (1) if Solution 1 is correct (2) if Solution 2 is correct and clarification of details (3) what is the connection between Solution 1 and 2 (4) if there are any other ways to solve the PDE and (5) in more complicated second-order PDEs (e.g. hyperbolic, elliptic), could I still use these solution strategies?
$\textbf{Solution 1:}$
We can factor
$$
\left( \frac{\partial }{\partial t} + \frac{\partial }{\partial x} \right)^2 u=2c
$$
Now, we use a transformation of variables.
Let $x=y + z$ and  $t=y$ and $v(y,z)=u(t,x)$. Then,
\begin{align*}
v_{yy} &=
\left( \frac{\partial }{\partial t} + \frac{\partial }{\partial x} \right)^2 u=2c\\
v(y,z) &= cy^2+yf_1(z) + f_2(z)\\
u(t,x) &=ct^2+tf_1(x - t) + f_2(x -t)  \qquad \textit{(1)}
\end{align*}
and by symmetry,
\begin{align*}
u(t,x) &=cx^2+xg_1(t - x) + g_2(t -x) 
\end{align*}
$\textbf{Solution 2:}$
Again we can factor
$$
\left( \frac{\partial }{\partial t} + \frac{\partial }{\partial x} \right)^2 u=2c
$$
We solve the PDE in two steps,
\begin{align*}
\left( \frac{\partial }{\partial t} + \frac{\partial }{\partial x} \right) v &=2c
\qquad \textit{(2)}\\
\left( \frac{\partial }{\partial t} + \frac{\partial }{\partial x} \right) u &=v 
\qquad \textit{(3)}
\end{align*}
By the method of characteristics for $\textit{(2)}$,we have $\frac{\partial t}{\partial w}=\frac{\partial x}{\partial w}=1$ and  $\frac{\partial v}{\partial w}=2c$ so $x=w+x_0$ and $v(?,?)=2cw + \gamma_1(x_0)$.
Also by the method of characteristics for $\textit{(3)}$, we have $\frac{\partial t}{\partial s}=\frac{\partial x}{\partial s}=1$ and $\frac{\partial u}{\partial s}=v=2cs + \gamma_1(r_0)$ so $x=t+r_0$ and 
$$u(?,?)=cs^2 + s\gamma_1(r_0) + \gamma_2(r_0)
=ct^2 + t\gamma_1(x-t) + \gamma_2(x-t) \qquad \textit{(4)}$$ 
This is the same as $\textit{(1)}$.
I'm a little confused about the details of this proof. Specifically, I'm just not sure what the arguments of the functions are and it feels like I'm substituting in where convenient to get to the solution (e.g. the arguments of $u(?,?)$ are $t$ and $x$ but then they're $s$ and $r_0$ and then I just substitute in for $s$ and $r_0$).

REPLY [4 votes]: There is another way to solve this problem, but it involves applied Lie theory and not a lot of people know about it.  I picked this up from Dr. Lawrence Dresner in Oak Ridge back in the 90's.  Your pde is invariant to Lie group
$$
G(x,t,u)=(\lambda x,\lambda^\beta t,\lambda^\alpha u)\lambda_o=1
$$such that
$$
\lambda^\alpha u(x,t)=u(\lambda x,\lambda^\beta t)
$$Differentiating both sides w.r.t. $\lambda$ and setting $\lambda = \lambda_o=1$,
$$
\alpha u=xu_x+\beta tu_t
$$The characteristic equation for this expression is
$$
\frac{du}{\alpha u}=\frac{dx}{x}=\frac{dt}{\beta t}
$$Two independent integrals are 
$$
\frac{u}{t^{\frac{\alpha}{\beta}}}
$$and
$$
\frac{x}{t^{\frac{1}{\beta}}}
$$The most general solution to your PDE is to take one of these differential invariants and set it equal to a function of the other one, with the result that
$$
u=t^{\frac{\alpha}{\beta}}F\bigg(\frac{x}{t^{\frac{1}{\beta}}}\bigg)
$$Since $x'=\lambda x$, $t'=\lambda^\beta t$, and $u'=\lambda^\alpha u$,
$$
u'_{x'x'}=\lambda^{\alpha -2}u_{xx}
$$
$$
u'_{t'x'}=\lambda^{\alpha -\beta -1}u_{tx}
$$
$$
u'_{t't'}=\lambda^{\alpha -2\beta}u_{tt}
$$and
$$
2C=\lambda^0 2C
$$then for invariance to exist, $\alpha -2=\alpha -\beta -1=\alpha -2\beta =0$ which implies $\alpha =2$ and $\beta =1$.  Therefore the most general solution to your PDE is
$$
u=t^2F(\mu), \mu=\frac{x}{t}
$$If you take derivatives of this solution you get
$$
u_{xx}=F_{\mu \mu}
$$
$$
u_{tx}=F_\mu-\mu F_{\mu \mu}
$$
$$
u_{tt}=2F-2\mu F_\mu+\mu^2 F_{\mu \mu}
$$Plug this into the PDE and simplify and you get
$$
\frac{1}{2}(1-\mu)^2F_{\mu \mu}+(1-\mu)F_\mu+F=C
$$a 2nd-order, linear ODE.  You can solve this (or hand it to Wolfram Alpha) to get $F(\mu)$.  Remember that $\mu=\frac{x}{t}$ and $u=t^2F$ and the answer is yours.