I'm given the following PDE: $$u_{tt}-u_{xx}+u^3=0$$ My source says that the associated operator is $$L:=\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}+u^2$$ which is arrived at simply by factoring the common term $u$ and thinking of the operator as left multiplying the function.$$\frac{\partial^2}{\partial t^2}u-\frac{\partial^2}{\partial x^2}u+u^3=\left (\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}+u^2 \right )u=Lu=0$$ Is this correct? In general how do I find and write the associated operator of a PDE? Do I simply do as my source does and just factor out the $u$? For eg. is the operator in $$Lu=u_x+u_y+1$$ given by $$L= \frac{\partial}{\partial x}+\frac{\partial}{\partial y}+\frac{1}{u}$$ as given here?
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1$\begingroup$ It's not linear because of the $u^3$ term $\endgroup$– DylanMar 14, 2019 at 7:24
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$\begingroup$ @Dylan My source of confusion is how to write the operator. The linearity part I've understood and since it wasn't relevant, I've removed it from the question. $\endgroup$– ZSMJMar 14, 2019 at 7:32
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1$\begingroup$ The source you cite uses rather ambiguous notation. You cannot simply "factor" the term $\partial^2_t - \partial^2_x + u^2$, as it will still depend on the function $u$. The "operator" should be an object which is a priori independent of the function: it should be an object which takes a function $u$ and maps it to a new function $f$. $\endgroup$– charMar 26, 2019 at 21:21
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1$\begingroup$ Remaining in your context. Perhaps it is better to consider the "operator" $Lu = (\partial^2_t - \partial^2_x) u + f(u)$, where $f(x) = x^3$ for $x \in \mathbb{R}$. This allows you to see why exactly the resulting equation is nonlinear, and how you may proceed in linearizing it (as suggested in the previous comments). $\endgroup$– charMar 26, 2019 at 21:23
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1$\begingroup$ @bgsk Yes, breaking it down like this makes sense and makes it easy to see the non linearity. I was really confused by how he just factored out the $u$. Could you paste these comments as an answer so I can give you the bounty? Also, could you have a look at this and recommend some books? $\endgroup$– ZSMJMar 27, 2019 at 4:46
1 Answer
As indicated by the OP, I will regroup my comments as an answer.
The source notes you cite make use of rather ambiguous/confusing notation. Indeed, one cannot simply "factor out" the term $\partial_t^2 - \partial_x^2 + u^2$ and call it "the operator", as this object clearly depends on the function $u$. The operator should be an object which is a priori independent of the function: it takes a "function" $u$ as input and yields another "function" $g$ as output.
In relation to your actual question. I believe it is better to consider the "operator" $Nu = (\partial_t^2 - \partial_x^2)u + f(u)$, where $f(x) = x^3$ for $x \in \mathbb{R}$. It is now clear that the operator $N$ is nonlinear, and the linear and nonlinear parts are moreover explicitly split. For futher reference, the linear operator $L = \partial_t^2 - \partial_x^2$ is called the wave operator (also called the d'Alembertian), as it is the governing linear operator in the wave equation \begin{equation} \partial_t^2 u - \partial_t^2 u = 0. \end{equation} Hence the equation associated to the "operator" you were considering is the nonlinear wave equation \begin{equation} \partial_t^2 u - \partial_t^2 u + u^3 = 0. \end{equation}
