About PDE of the form $u_{tt} - u_{xx} -u=0$

Question

I am trying to answer the following question:

Given the equation: $$ \left\{ \begin{array}{ll} u_{tt} - u_{xx} -u=0 , \mathbb{R}\times (0,\infty) \\ u(x,0)=x , u_t(x, 0)=-x \end{array} \right. $$ Find a solution as a power series expansion about the origin and identify this solution.

I'm kind confused with the written of the problem itself. Am I supposed to assume that there is a solution as a power series and find its form? I tried this way, but couldn't get any reasonable expression for it. Or how should I solve this kind of problem?

Thanks in advance, for any help.

Answer 1

If you are said to solve the problem using a series expansion, you should suppose that such a power series exists. After finding the closed form, you can find the disk of convergence over which the power series is convergent and valid.

Actually, this PDE is classified as linear. A general method for such equations can be substituting the variables or using the separation of variables as $$u(x,t)=f(x)g(t)$$and applying the initial condition.

Remark

A power series for a bivariate $u(x,t)$ can be written as $$u(x,t)=\sum_m\sum_n a_{mn}x^mt^n$$

Answer 2

Consider the Cauchy problem for second order PDE: $$ u_{tt}=F(t,x,u,u_{t},u_{tx},u_{xx}),~ u(0,x)=f(x),~u_{t}(0,x)=g(x). $$ Suppose that $f$ and $g$ are analytic in a neighborhood of origin of $\mathbb{R}$ and suppose that $F$ is analytic in a neighborhood of point $(0,0,f(0),g(0),f^{(1)}(0),g^{(1)}(0),f^{(2)}(0))$ of $\mathbb{R}^{7}$. Then Cauchy-Kovalevsky theorem assert that the given Cauchy problem has solution $u(t,x)$ which is analytic in neighborhood of origin of $\mathbb{R}^{2}$ and the solution is unique in class of analytic functions.

$\textbf{Edit:} ~$Under these assumptions, you can find Taylor series for $u(t,x)$ around origin. If you expand $u(t,x)$ around $(0,0)$, then $$ u(t,x)=x-tx+\frac{xt^{2}}{2!} -\frac{xt^{3}}{3!}+... $$ So, solution in closed form is $u(t,x)=x \text{e}^{-t}$