# Heat equation "like" problem: solve $u_{tt} -u_{xx} -u_t -u_x = 0$ using differential operator

I need to find general solutions to the problem $$u_{tt} -u_{xx} -u_t -u_x=0$$

I wrote the PD operator here as $P \left( \partial_t , \partial_x \right) = {\partial_{t}}^2 -{u_x}^2 -\partial_t -\partial_x$ which in turn can be factorized as $P \left( \partial_t , \partial_x \right) = \left( \partial_t +\partial_x \right) \left( \partial_t -\partial_x -Id \right)$ where $Id$ is the identity operator.
Now, the equation $Pu=0$ can be expressed by the system: $$\cases{u_t -u_x = u+v \\ v_t + v_x =0 }$$

The second equation here has the general form of $v = \phi \left( x-t \right)$ (derived using the method of characteristics).

Then I'm left with the problem $u_t - u_x = u + \phi (x-t)$ which can be written using the method of characteristics (again) as $u_t = u + \phi(x - t)$

I realize this is some kind of ODE, that I can solve for different values of $x$: $$u = \Psi(x) e^{t} + \Phi(x-t)$$

But when calculating $Pu$ I get $$Pu = -{\mathrm{e}}^t\,\frac{\partial ^2}{\partial x^2} \Psi \left(x\right)-{\mathrm{e}}^t\,\frac{\partial }{\partial x} \Psi \left(x\right)$$

And for $Pu$ to nullify, $\Psi$ must be $C e^{-x}$. Thus $u = C e^{t-x} +\Phi \left(x - t \right)$. Surely, this solution is valid, but is this as general as it can be? How can I tell I didn't miss any other solution?

For the specific equation you provided, it seems unnecessary to use differential operators. Instead, it suffices to use the change-of-variable trick that has been adopted to deal with the standard wave equation.

Define \begin{align} t&=\xi+\eta,\\ x&=\xi-\eta, \end{align} whose inverse reads \begin{align} \xi&=\frac{t+x}{2},\\ \eta&=\frac{t-x}{2}. \end{align} With these notations, we have \begin{align} \frac{\partial}{\partial t}&=\frac{1}{2}\left(\frac{\partial}{\partial\xi}+\frac{\partial}{\partial\eta}\right),\\ \frac{\partial}{\partial x}&=\frac{1}{2}\left(\frac{\partial}{\partial\xi}-\frac{\partial}{\partial\eta}\right). \end{align} Thanks to this transform, the original equation $$u_{tt}-u_{xx}-u_t-u_x=0$$ is now equivalent to $$u_{\xi\eta}-u_{\xi}=0,$$ whose general solution is obviously in view.

Firstly, $$\frac{\partial}{\partial\eta}u_{\xi}-u_{\xi}=0$$ is an ordinary differential equation, which can be re-expressed as $$\frac{\partial}{\partial\eta}\left(e^{-\eta}u_{\xi}\right)=0.$$ Its general solution is $$e^{-\eta}u_{\xi}=\Phi'(\xi)\iff u_{\xi}=e^{\eta}\Phi'(\xi).$$

Thereafter, integrate the last equation with respect to $\xi$, and $$u=e^{\eta}\Phi(\xi)+\Psi(\eta).$$

This is the general solution you are looking for. Following the above steps, it is clear that we have not left out any possibility.

Finally, rewrite the expression by using $t$ and $x$, and the general solution reads $$u(t,x)=\exp\left(\frac{t-x}{2}\right)\Phi\biggl(\frac{t+x}{2}\biggr)+\Psi\biggl(\frac{t-x}{2}\biggr).$$

Hope this could be helpful for you.

Edit: Alternative solution using the method of partial differential operator

Following what @UriaMor proposed in the question, define a partial differential operator \begin{align} P&=\partial_t^2-\partial_x^2-\partial_t-\partial_x\\ &=\left(\partial_t+\partial_x\right)\left(\partial_t-\partial_x-I\right). \end{align} As such, the original equation $Pu=0$ could be decomposed by two equations \begin{align} \left(\partial_t-\partial_x-I\right)u&=v,\\ \left(\partial_t+\partial_x\right)v&=0. \end{align} The general solution could then be obtained by solving these two equations using the method of characteristics.

Firstly, solve the second equation with respect to $v$. Note that $$\frac{{\rm d}}{{\rm d}t}v(t,x_0+t)=\frac{\partial v}{\partial t}(t,x_0+t)+\frac{\partial v}{\partial x}(t,x_0+t)=0$$ holds for all $x_0$. Therefore, $$v(t,x_0+t)=v(0,x_0):=\Phi(x_0).$$ Replace $x_0+t$ by $x$ in this last equation, and we obtain $$v(t,x)=\Phi(x-t).$$

Secondly, substitute this last result into the first equation with respect to $u$, i.e., $$u_t(t,x)-u_x(t,x)-u(t,x)=v(t,x)=\Phi(x-t).$$ Similarly, observe that $$\frac{\rm d}{{\rm d}t}u(t,x_0-t)=\frac{\partial u}{\partial t}(t,x_0-t)-\frac{\partial u}{\partial x}(t,x_0-t),$$ for which the equation above, when considered along the characteristics $\left(t,x\right)=\left(t,x_0-t\right)$, implies that $$\frac{\rm d}{{\rm d}t}u(t,x_0-t)-u(t,x_0-t)=\Phi((x_0-t)-t)=\Phi(x_0-2t),$$ or equivalently, $$\frac{\rm d}{{\rm d}t}\left[e^{-t}u(t,x_0-t)\right]=e^{-t}\Phi(x_0-2t).$$ Therefore, we obtain \begin{align} e^{-t}u(t,x_0-t)-u(0,x_0)&=\int_0^te^{-s}\Phi(x_0-2s){\rm d}s\\ &=e^{-x_0/2}\int_0^te^{(x_0-2s)/2}\Phi(x_0-2s){\rm d}s\\ &=-\frac{1}{2}e^{-x_0/2}\int_{s=0}^{s=t}e^{(x_0-2s)/2}\Phi(x_0-2s){\rm d}\left(x_0-2s\right). \end{align} Define $$\Psi(x)=\int^xe^{s/2}\Phi(s){\rm d}s,$$ and the above equality reads $$e^{-t}u(t,x_0-t)-u(0,x_0)=-\frac{1}{2}e^{-x_0/2}\left(\Psi(x_0-2t)-\Psi(x_0)\right),$$ or equivalently, \begin{align} u(t,x_0-t)&=e^tu(0,x_0)-\frac{1}{2}e^{t-x_0/2}\left(\Psi(x_0-2t)-\Psi(x_0)\right)\\ &=e^{t-x_0/2}\left[e^{x_0/2}u(0,x_0)-\frac{1}{2}\left(\Psi(x_0-2t)-\Psi(x_0)\right)\right]. \end{align} Define \begin{align} \phi(x)&=e^{x/2}u(0,x)+\frac{1}{2}\Psi(x),\\ \psi(x)&=-\frac{1}{2}e^{-x/2}\Psi(x), \end{align} and the above result goes that $$u(t,x_0-t)=e^{t-x_0/2}\phi(x_0)+\psi(x_0-2t).$$ Finally, replace $x_0-t$ by $x$, and one eventually obtains $$u(t,x)=\exp\left(\frac{t-x}{2}\right)\phi(x+t)+\psi(x-t),$$ which agrees with the result we have figured out previously.

• It helps to see that I didn't get the most general solution... Maybe I should have put it more clearly, but the goal here was to use differential operator Apr 14 '18 at 12:25
• @UriaMor: Never mind. I am reading your PD details. I will get back to you very soon :-) Apr 14 '18 at 12:29
• @UriaMor: Seems that the problem occurs when you stated "the problem $u_t−u_x=u+\phi(x−t)$ which can be written using the method of characteristics (again) as $u_t=u+\phi(x−t)$". The idea works, but as you perform the method of characteristics again, you need to change the variable notations. That is, here in the last equation should actually be put as $\frac{\rm d}{{\rm d}t}u(t,x-t)=u(t,x-t)+\phi((x-t)-t)$. Apr 14 '18 at 12:43
• +1 for your answer...you should keep the answer it may helps someone even it dosent fit op'question entirely... Apr 14 '18 at 13:07
• @Isham: Thank you for your advice :-) Apr 14 '18 at 13:42

Solve $$u_{tt} -u_{xx} -u_t -u_x=0\qquad\qquad (1)$$

• $D_t^2-D_x^2-D_t-D_y=-\left( {D_x}-{D_t}+1\right) \, \left( {D_x}+{D_t}\right)$
• Solution of $\;u_x+u_t=0\;$ is $\;u_1=f(x-t)$
• Solution of $\;u_x-u_t+u=0\;$ is $\;u_2=e^tg(x+t)$
• Then general solution of $(1)\;$ is $$u=u_1+u_2=f(x-t)+e^tg(x+t)$$