Find a nontrivial solution of $$u_{tt}=\Delta u-(x_1^2+4x_2^2)u$$ for $x=(x_1,x_2) \in \mathbb{R^2}$, $t>0$ such that $u_t$ is a nonzero function and $|u(x,t)|\leq C$ for $x=(x_1,x_2) \in \mathbb{R^2}$, $t>0$, where $C$ is a positve constant.

I’m trying to extend the method from Evan’s book for nonlinear first-order PDE, $$F(D^2u,u,x)=u_{tt}-\Delta u+(x_1^2+4x_2^2)u=0$$ Then $F(p,z,x)=b(x)p+c(x)z$, here $b(x)=(1,-1,-1)$, $c(x)=(x_1^2+4x_2^2)$ and so $$D_pF=b(x).$$ $$\dot{x}(s)=b(x(s)),$$ $$\dot{z}(s)=b(x(s))\cdot p(s)=-c(x(s))z(s)$$ For here we have $$\dot{x}^1=-1,\dot{x}^2=1,\dot{x}^3=1.$$

Edit: From the comment, I know that this is a linear problem not nonlinear. Thus, the method not work. While I try to separate it but I fail. How to solve this problem?

  • 1
    $\begingroup$ That PDE is linear. $\endgroup$
    – Gonçalo
    Nov 24, 2023 at 1:32
  • $\begingroup$ As it stands at the moment, this is a linear pde. I’d suggest you see what standard separation of variables leads to. $\endgroup$ Nov 24, 2023 at 1:52
  • $\begingroup$ It’s also a second order PDE not first order, so the method of chracteristics doesn’t apply $\endgroup$
    – JackT
    Nov 24, 2023 at 2:46
  • $\begingroup$ Oh, I see why it can't work. But I still stuck to solve it, I try separation of variables but I don't know how to separation while it has a variable $t$ not only $x.$ I rewrite it as $$u_{tt}-u_{x_1x_1}-u_{x_2x_2}+(x_1^2+4x_2^2)u=0.$$ But how to separate? $\endgroup$
    – Apple
    Nov 24, 2023 at 2:57
  • $\begingroup$ @A rural reader I still stuck to solve this problem. Do any can give more hint? or detail thoughts, thanks? $\endgroup$
    – Apple
    Nov 24, 2023 at 8:33

1 Answer 1


Since you are in fact dealing with a linear PDE and you are just looking for any nontrivial solution, seperation of variables should be your first try:

$$ u=T(t)X(x_1)Y(x_2)\\ \Rightarrow T''XY=TX''Y+TXY''-(x_1^2+4x_2^2)TXY\\ \Rightarrow\begin{cases} \frac{X''}{X}-x_1^2=\lambda\\ \frac{Y''}{Y}-4x_2^2=\mu\\ \frac{T''}{T}=\lambda+\mu \end{cases} $$ solving these equations for arbitrary $\lambda,\mu$ involves the parabolic cylinder function. $$ X(x_1) = c_1 D_{-\frac{\lambda+1}{2}}(\sqrt{2} x_1) + c_2 D_{\frac{\lambda-1}{2}}( \sqrt{2}i x_1)\\ Y(x_2) = c_3 D_{-\frac{\lambda+2}{4}}(2 x_2) + c_4 D_{\frac{\lambda-2}{4}}(2i x_2) $$ with potentially complex $c_i$. Now somebody with deep knowledge of parabolic cylinder functions would be able to determine for which choices of $\lambda,\mu$ there exists constants $c_i$ such that the resulting function is nontrivial, real and remains bounded.

However since you are only looking for one (!) solution we can make our lives easier by observing that $D_0(x)=e^{-x^2/4}$.

Therefore the special choice of $\lambda=-1,\mu=-2$ allows for a solution in terms of "standard" functions, namely

$$ X=e^{-x_1^2/2}\\ Y=e^{-x_2^2}\\ T=A\cos(\sqrt{3}t)+B\sin(\sqrt{3}t) $$

Please note that this is only one very specific class of solutions.

  • $\begingroup$ Why you choice $\lambda$ and $\mu$ like that? Is you choice arbitrary, then after choice we need to check the condition from question for here is bounded and $u_t$ nonzero function, right? $\endgroup$
    – Apple
    Nov 24, 2023 at 12:39
  • $\begingroup$ So for here solution are not unique since many possible pair of $(\lambda,\mu)$ can met the condition here. The choice you choose is a solution here. $\endgroup$
    – Apple
    Nov 24, 2023 at 13:42
  • 1
    $\begingroup$ I added some additional lines, hopefully its more clear now. $\endgroup$
    – maxmilgram
    Nov 24, 2023 at 14:03

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