Solving the PDE $u_{tt}+2u_{tx}+u_{xx}=2c$

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.

I will try to shed some light, let's see if I can help:

• Q1 and Q2: Your solution given by both the 1st and 2nd methods is completely right. I have checked them both getting your solution $u(x,t) = c t^2 + t f(x-t) + g(x-t)$, where $f$ and $g$ are arbitrary functions of the characteristic(s) $x-t$. This tells you that the solution in absence of forcing (the nonhomogenous term) travels along curves given by $x - t = \xi, \ \xi \in \mathbb{R}$, where $u$ is precisely constant.

• Q3: You can check that both ways to proceed are exactly the same if you write the original PDE in canonical form. Your 2nd method just take advantage of the operator decomposition: $$(D_{xx} + 2 D_{xt} + D_{tt})u = (D_{x} + D_{t})(D_{x} + D_{t})u = (D_{x} + D_{t})^2 u, \quad D_{x_i} = \partial_{x_i}, $$ which yields to two 1st-order hyperbolic PDEs.

• Q4: The general way to solve PDEs of the form: $$A u_{xx} + B u_{xt} + C u_{xx} = P(x,t;u,u_x,u_t),$$ for some linear function $P$, is to compute first the discriminant, $\Delta = B^2 -4AC$, in order to determine whether we have a hyperbolic ($\Delta >0$), elliptic ($\Delta <0$), or parabolic ($\Delta =0$) PDE. In your case: $A=C=1, B=2$, so $\Delta = 0$, making your PDE parabolic. The next step is to compute the characteristics of the equation which are given by the equations: $$\begin{align} A \xi_{x}^2 + B \xi_x \xi_t + C\xi_t^2 = 0, \tag{1} \\ A \eta_{x}^2 + B \eta_x \eta_t + C\eta_t^2 = 0. \tag{2}\\ \end{align}$$ Solving $(1)$ for $\xi_x/\xi_t$ we arrive at: $$\frac{\xi_x}{\xi_t} = -1 = \color{blue}{-\frac{dt}{dx}}^*, \tag{3}$$ which after integration leads us to the first characteristic curve, $dx - dt = 0$, or equivalently, $x - t = c \equiv \xi$. Since the equation is parabolic, the other solution (remember $\Delta = 0$) can be arbitrarily chosen so we can make the other characteristic to be $\eta = t$ or $\eta = x$ (symmetry). Once the characteristics are computed you write the equation making the transformation $(x,t) \to (\xi,\eta)$ and using the chain rule in order to come up with the known as canonical form of the PDE which will be (prove it, taking $\eta = t$): $$ u_{\eta \eta}(\xi,\eta) = 2c, $$ which after integrating yields the desired solution: $$ u(\xi,\eta) = c \eta^2 + \eta f(\xi) + g(\xi) \Leftrightarrow u(x,t) = c t^2 + t f(x-t) +g(x-t),$$ as we expected.

• Q5: As you can see, we can only avoid this tedious work when the PDE is easily factorizable like in the present case. Otherwise, the general method always tries to put your PDE into canonical form which, by the way, is not always simpler.

I hope this helps.

Cheers!

Edit: *, the equation highlighted in blue comes from the fact that given $\xi = \xi(x,t)$, then $d\xi = \xi_x dx + \xi_t dt$, but we know that $\xi$ is a constant (characteristic), therefore $d\xi = 0$ and we have $dt/dx = -\xi_x/\xi_t$.

Edit 2: of course, we have forgotten about boundary value problems, which can be solved using completely different techniques. Here, the shape of the arbitrary functions $f$ and $g$ are either determined by the initial conditions or by any other specification of $u$ as, for example, passing through a given parametrized curve.