Does the PDE $$u_{tt} = u_{xxx}$$ with $x \in [0,1]$ and $t \in [0,T]$ with initial conditions $$u(x,0)=\sin(2 \pi x),\, \partial_t u(x,0)=0$$ and boundary condition $$u(0,t)=u(1,t),\, \partial_x u(0,t)=\partial_x u(1,t),\, \partial_{xx} u(0,t)=\partial_{xx} u(1,t)$$ have a analytical solution.

I've tried separation of variable and Fourier transformations, both failed for me.

  • $\begingroup$ Are the boundary conditions really on $\partial_{t}u$ and $\partial_{tt}u$ and not the spatial derivatives? $\endgroup$ – in_mathematica_we_trust Mar 28 '13 at 10:07
  • $\begingroup$ Are you sure that it is $u_{xxx}$ and not $u_{xx}$? $\endgroup$ – gerw Mar 28 '13 at 10:08
  • $\begingroup$ @ in_wolfram_we_trust: my mistake $\endgroup$ – Maikel Mar 28 '13 at 10:15
  • $\begingroup$ @ grew yes $u_{xxx}$ is correct $\endgroup$ – Maikel Mar 28 '13 at 10:16
  • $\begingroup$ And the initial condition is not $\sin(2\pi x)$? At the moment the initial and boundary conditions do not agree at the corner where they meet. $\endgroup$ – in_mathematica_we_trust Mar 28 '13 at 10:30

The entire system is periodic. So it makes sense to assume that the solution will be, too. We could perform a Fourier transform on $x$, but that would be unnecessary. Let's just assume the form we're looking for. That is,

$$ u(x,t) = \sum_{n=0}^\infty (A_n(t) \cos (2n\pi x)+B_n(t)\sin(2n\pi x)) $$ Now obviously, $A_n(0)=0$ and $B_n(0)=0$ except for $B_1(0)=1$. Similarly, $A_n'(0)=B_n'(0)=0$. Note that our $\cos$ and $\sin$ terms already satisfy the spatial boundary conditions. For the PDE itself,

$$ u_{tt} = \sum_{n=0}^\infty (A_n''(t)\cos(2n\pi x) + B_n''(t)\sin(2n\pi x)) $$ and

$$ u_{xxx} = \sum_{n=1}^\infty 8n^3\pi^3\left(A_n(t)\sin(2n\pi x) - B_n(t)\cos(2n\pi x)\right) $$ Therefore, we have that

$$ A_n''(t) = -8n^3\pi^3B_n(t)\\ B_n''(t) = 8n^3\pi^3A_n(t) $$ Substituting, we find

$$ B_n^{(4)}(t) = -64n^6\pi^6B_n(t)\\ B_n''(0) = B_n^{(3)}(0) = 0 $$ And so we have the general solution form

$$ B_n(t) = b_1e^{2(1+i)n^{3/2}\pi^{3/2}t}+b_2e^{2(1-i)n^{3/2}\pi^{3/2}t}+b_3e^{2(-1+i)n^{3/2}\pi^{3/2}t}+b_4e^{-2(1+i)n^{3/2}\pi^{3/2}t}\\ B_n(0) = b_1+b_2+b_3+b_4\\ B_n'(0) = C((1+i)b_1+(1-i)b_2-(1-i)b_3-(1+i)b_4)\\ B_n''(0) = 2iC^2(-b_1+b_2+b_3-b_4)\\ B_n^{(3)}(0) = 2iC^3(-(1+i)b_1+(1-i)b_2-(1-i)b_3+(1+i)b_4) $$ where $C=2n^{3/2}\pi^{3/2}$. Now, for $n\neq 1$, we have trivially $b_1=b_2=b_3=b_4=0$. But $B_1(0)=1$.

So our system of equations is

$$ \left(\begin{matrix}1&1&1&1\\1+i&1-i&-(1-i)&-(1+i)\\-1&1&1&-1\\-(1+i)&1-i&-(1-i)&1+i\end{matrix}\right)\left(\begin{matrix}b_1\\b_2\\b_3\\b_4\end{matrix}\right)=\left(\begin{matrix}1\\0\\0\\0\end{matrix}\right)\\ $$ Which we can solve to get $$ b_1=b_2=b_3=b_4=\frac{1}{4} $$ So we have

$$\begin{align} B_1(t) &= \frac{e^{2(1+i)\pi^{3/2}t}+e^{2(1-i)\pi^{3/2}t}+e^{2(-1+i)\pi^{3/2}t}+e^{-2(1+i)\pi^{3/2}t}}{4}\\ &=\frac{(e^{2\pi^{3/2}t}+e^{-2\pi^{3/2}t})(e^{2i\pi^{3/2}t}+e^{2i\pi^{3/2}t})}{4}\\ &=\cosh(2\pi^{3/2}t)\cos(2\pi^{3/2}t) \end{align}$$ And so,

$$\begin{align} A_1(t) &= \frac{B_1''(t)}{8\pi^3}\\ &= - \sinh(2\pi^{3/2} t) \sin(2\pi^{3/2} t) \end{align}$$ So finally, we get $$ u(x,t) = \cosh(2\pi^{3/2}t)\cos(2\pi^{3/2}t)\sin(2\pi x)\\- \sinh(2\pi^{3/2} t) \sin(2\pi^{3/2} t)\cos(2\pi x) $$

This can be simplified, though.

$$ u(x,t) = \frac{1}{2}(e^{-2\pi^{3/2}t}\sin(2\pi x+2\pi^{3/2}t)+e^{2\pi^{3/2}t}\sin(2\pi x-2\pi^{3/2}t)) $$ Sanity check: $$ u(x,0) = \frac{1}{2}(e^0\sin(2\pi x+0)+e^0\sin(2\pi x-0)) = \sin(2\pi x) $$

  • $\begingroup$ Why does the index in $u_{xxx}$ starts at 1? $\endgroup$ – Maikel Mar 28 '13 at 21:54
  • 1
    $\begingroup$ Because the $n=0$ term looks like this: $A_0(t)\cos(0)+B_0(t)\sin(0)=A_0(t)$, and so it does not depend on $x$ at all. So even the first derivative of that term is zero. $\endgroup$ – Glen O Mar 29 '13 at 1:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.