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Solve $$u_{xx}-3u_{xt}-4u_{tt}=0$$ where $u(x,0)=x^2$ and $u_t(x,0)=e^x$.

My workings so far: I have factored the differential equation in the following way: $$(\delta_x-4\delta_t)(\delta_x+\delta_t)=0$$ where $\delta_x=\frac{\delta}{\delta x}$ etc. Now if we let $v$ be the solution to $(\delta_x+\delta_t)u$ then we have the following two equations: \begin{eqnarray*} (\delta_x+\delta_t)u=u_x+u_t=v\\ (\delta_x-4\delta_t)v=v_x-4v_t=0 \end{eqnarray*} Now for $v$ we simply find $$v=h(t+4x)$$ where $h$ is an arbitrary function of one variable. Now what remains is find $u$ such that $$u_x+u_t=h(t+4x)$$ I am stuck here, I thought about making a change of variables $\zeta=x+t$ and $\eta=x-t$ and thus using the product rule to show that $u_x=u_{\zeta}+u_{\eta}$ and $u_t=u_{\zeta}-u_{\eta}$ and thus $u_x+u_t=2u_{\zeta}$ and we need to solve $$u_{\zeta}=h(t+4x)$$ (I left out the factor 2 because $h$ is an arbitrary function). Do I simply integrate now and conclude $$u=f(\zeta)h(t+4x)+g(\eta)=f(x+t)h(t+4x)+g(x-t)$$ This seems wrong to me...

Some help would be greatly appreciated!

doraemonpaul
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Slugger
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2 Answers2

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Note that, since the pde is with constant coefficients, then we proceed as

$$u_{xx}-3u_{xt}-4u_{tt}=0\implies(\delta_x-4\delta_t)(\delta_x+\delta_t)u=0$$

$$\implies(\delta_xu-4\delta_tu=0)\cup(\delta_xu+\delta_tu=0)$$

$$\implies\left\{u(x,t)=F(4x+t)\right\}\cup\left\{u(x,t)=G(t-x)\right\}$$

$$\implies u(x,t)=F(4x+t)+G(t-x)$$

$$\implies u(x,t)=f\left(x+\dfrac{t}{4}\right)+g(x-t)$$

$$u(x,0)=x^2\implies f(x)+g(x)=x^2$$

$$u_t(x,0)=e^x\implies\dfrac{f'(x)}{4}-g'(x)=e^x\implies\dfrac{f(x)}{4}-g(x)=e^x+c$$

$$\therefore f(x)=\dfrac{4x^2+4e^x+4c}{5},g(x)=\dfrac{x^2-4e^x-4c}{5}$$

$$\therefore u(x,t)=\dfrac{4\left(x+\dfrac{t}{4}\right)^2+4e^{x+\frac{t}{4}}+(x-t)^2-4e^{x-t}}{5}$$

doraemonpaul
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Mhenni Benghorbal
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  • Hey thanks for your answer. I presume you meant $u(x,t)=F(4x+t)+G(t-x)$ in the final answer ($t$ and $x$ swapped around in argument of $F$). Is this the most general solution though? Is it possible that there exists a function $u$ such that $u_x-4u_t\neq 0$ and $u_x+u_t\neq 0$ but $(\delta_x-4\delta_t)(\delta_x+\delta_t)u=0$? – Slugger Feb 18 '13 at 12:29
  • Yes, the solution is $u(x,t)=F(4x+t)+G(t-x)$. It was a typo mistake. – Mhenni Benghorbal Feb 18 '13 at 22:17
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    @Mhenni Benghorbal, good answer, but you miss to find the particular solution when $u(x,0)=x^2$ and $u_t(x,0)=e^x$ . – doraemonpaul Feb 19 '13 at 13:02
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    @Mhenni Benghorbal, I have added the process of finding the particular solution when $u(x,0)=x^2$ and $u_t(x,0)=e^x$ to your answer, hope that the adding is done by you rather than done by me in next time. – doraemonpaul Feb 20 '13 at 09:13
  • @doraemonpaul: Thanks for adding the particular solution part. Good job. – Mhenni Benghorbal Dec 08 '13 at 01:48
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I solve with free CAS Maxima. We use Laplace transform method.

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Answer:

$$u=\frac{4}{5}e^{x+\frac{t}{4}}-\frac{4}{5}e^{x-t}+x^2+\frac{t^2}{4}$$

Aleksas Domarkas
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    I think the only people understanding your answer well, is the ones who know this CAS Maxima. I suggest to translate your solution into a more human readable format. It is good, that you use latex. – peterh May 29 '18 at 18:23
  • @peterh Maxima is free computer algebra system. http://maxima.sourceforge.net/ You can also solve with Maple or Mathematica. – Aleksas Domarkas May 29 '18 at 19:33
  • ...and also I like free tools (particularly if they are also open source), but it doesn't mean that I know them all. – peterh May 29 '18 at 19:34