Solve the One Dimensional Eigenvalue Problem: $$u_{tt} - u_{xx} = m^2 u$$

I have learned how to solve PDEs of the form $u_{tt} - u_{xx} = f(x,t)$ using d'Alembert's solution. However, I have not yet studied PDEs of this form. How does one solve this PDE?

  • 2
    $\begingroup$ I think simple separation of variable is sufficient. $\endgroup$
    – 민찬홍
    Nov 13 '17 at 3:54
  • $\begingroup$ Why do you call this "one-dimensional" when you derive in two directions? And don't you have boundary conditions? $\endgroup$ Nov 13 '17 at 4:09
  • 1
    $\begingroup$ I added the "wave-equation" and the "hyperbolic-equations" tag to your post. $\endgroup$ Nov 13 '17 at 5:09

Suppose we try separation of variables, setting

$u(x, t) = f(t)g(x); \tag 1$


$u_{xx} = f(t) g''(x), \tag 2$

$u_{tt} = \ddot f(t) g(x), \tag 3$

whence the equation

$u_{tt} - u_{xx} = m^2 u \tag 4$


$\ddot f g - f g'' = m^2 fg; \tag 5$

if we assume for the moment that $f(t) \ne 0 \ne g(x)$, then we may divide (5) through by $fg$, yielding

$\dfrac{\ddot f}{f} - \dfrac{g''}{g} = m^2, \tag 6$


$\dfrac{\ddot f}{f} - m^2 = \dfrac{g''}{g}. \tag 7$

We can now argue in the typical manner that, since the left-hand side depends only on $t$ but the right-hand side only on $x$, they must each equal the same constant value which, for reasons which will hopefully soon become clear, we denote by $-\lambda^2$:

$\dfrac{\ddot f}{f} - m^2 = -\lambda^2 = \dfrac{g''}{g}, \tag 8$

which leads to

$\ddot f + (\lambda^2 - m^2) f = 0, \tag 9$


$g'' + \lambda^2 g = 0.\tag{10}$

(10) has solutions of the general form

$g_\lambda(x) = a_+(\lambda)e^{i\lambda x} + a_-(\lambda) e^{-i\lambda x}, \tag{11}$

where $\lambda$ and the $a_\pm(\lambda) \in \Bbb C$ are determined by boundary/initial conditions placed upon $g(x)$; we can't get much more specific at this point without such information. For each admissible $\lambda$, we then have a solution for $f(t)$:

$f_\lambda(t)= b_+(\lambda) e^{i\sqrt{\lambda^2 - m^2}t} + b_-(\lambda)e^{-i\sqrt{\lambda^2 - m^2}t}; \tag{12}$


$u_\lambda(x, t) = f_\lambda(t)g_\lambda(x)$ $= ( b_+(\lambda) e^{i\sqrt{\lambda^2 - m^2}t} + b_-(\lambda) e^{-i\sqrt{\lambda^2 - m^2}t})(a_+(\lambda)e^{i\lambda x} + a_-(\lambda) e^{-i\lambda x}); \tag{13}$

it is easy to check that

$u_{\lambda tt} - u_{\lambda xx} = m^2 u_\lambda. \tag{14}$

Solutions of the form (13) for different $\lambda$ may be composed by summation or integration over $\lambda$ to yield more complicated solutions.

Finally, it is of course worth observing that the temporal frequencies $\sqrt{\lambda^2 - m^2}$ may become imaginary if $\lambda^2 < m^2$; the we obtain temporally exponentially growing/decreasing solutions $u(x, t)$.

In the above, I have decided to associate the $m^2$ terms with $f(t)$, but we could just as well have taken

$g'' + (\lambda^2 + m^2)g = 0 \tag{15}$

and obtained a different, though similar, set of solutions to (4). In fact, if we set

$m^2 = m_t^2 - m_x^2, \tag{16}$

we can write

$\dfrac{\ddot f}{f} - \dfrac{g''}{g} = m_t^2 - m_x^2, \tag{17}$


$\dfrac{\ddot f}{f} - m_t^2 = -\lambda^2 = \dfrac{g''}{g} - m_x^2, \tag{18}$

which leads to a more symmetrical analysis and set of solutions.

This could go on forever, but I can't, so I'll leave it at that. I would like to say in closing, however, that all this stuff smells like the Special Theory of Relativity and Particle Physics to my somewhat knowlewdge-able knose.


Let $p=mx$ ,

Then $u_{tt}=m^2u_{pp}+m^2u$

Let $q=mt$ ,

Then $u_{qq}=u_{pp}+u$

Similar to Solve initial value problem for $u_{tt} - u_{xx} - u = 0$ using characteristics,

Consider $u(p,a)=f(p)$ and $u_q(p,a)=g(p)$ ,

Let $u(p,q)=\sum\limits_{n=0}^\infty\dfrac{(q-a)^n}{n!}\dfrac{\partial^nu(p,a)}{\partial q^n}$ ,

Then $u(p,q)=\sum\limits_{n=0}^\infty\dfrac{(q-a)^{2n}}{(2n)!}\dfrac{\partial^{2n}u(p,a)}{\partial q^{2n}}+\sum\limits_{n=0}^\infty\dfrac{(q-a)^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(p,a)}{\partial q^{2n+1}}$



Similarly, $\dfrac{\partial^{2n}u}{\partial q^{2n}}=\sum\limits_{k=0}^nC_k^n\dfrac{\partial^{2k}u}{\partial p^{2k}}$




Similarly, $\dfrac{\partial^{2n+1}u}{\partial q^{2n+1}}=\sum\limits_{k=0}^nC_k^n\dfrac{\partial^{2k+1}u}{\partial p^{2k}\partial q}$

$\therefore u(p,q)=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nf^{(2k)}(p)(q-a)^{2n}}{(2n)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^ng^{(2k)}(p)(q-a)^{2n+1}}{(2n+1)!}$


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