I am told that the following PDE is resonant:

$$U_{tt}=U_{xx}, 0\le x\le\pi, t>0$$ $$U(0,t)=0, U(\pi,t)=\pi \sin t, t\ge0$$ $$U(x,0)=0, U_t(x,0)=x ,0\le x\le\pi$$

I am unable to see this immediately from the equation. I know that if I make the Ansatz for a particular solution in the form of $A(x)\sin t+B(x)\sin t$ I obtain a contradiction and for this reason I must adapt my guess to: $$A(x)\sin t+B(x)\sin t+t(C(x)\sin t+D(x)\sin t)$$

Can anyone clarify this for me?

  • $\begingroup$ This makes sense thank you! But don't you mean $v=U-\pi \sin t$?@Winther $\endgroup$ – john melon Jan 7 '17 at 14:17
  • $\begingroup$ OK. Thank you @Winther $\endgroup$ – john melon Jan 7 '17 at 14:20

If we introduce the field redefinition $v = u - x\sin(t)$ then the PDE becomes $v_{tt} - v_{xx} = x\sin(t)$ with boundary conditions $v(0,t) = v(\pi,t) = v(x,0) = v_t(x,0) = 0$. The solution for $v$ can be expanded in a Fourier series

$$v(x,t) = \sum_{n=1}^\infty a_n(t)\sin(nx)$$

The boundary conditions leads to the constraints $a_n(0) = 0$ and $a_n'(0) = 0$. Using the expression above in the PDE we obtain

$$\sum_{n=1}^\infty \left[a_n''(t) + n^2a_n(t)\right]\sin(nx) = x\sin(t) = \sum_{n=1}^\infty \left[\frac{2(-1)^{n+1}}{n}\sin(t)\right]\sin(nx)$$

where I have expanded $x$ in a Fourier series. Comparing both sides leads to the ODE

$$a_n''(t) + n^2a_n(t) = \frac{2(-1)^{n+1}}{n}\sin(t),~~a_n(0) = a_n'(0) = 0$$

which is a driven harmonic oscillators which you can solve using the methods in the link provided. This harmonic oscillator have a resonance when $n=1$.

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