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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.

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  • $\begingroup$ 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
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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: $$ u(x,0) = \frac{1}{2}(e^0\sin(2\pi x+0)+e^0\sin(2\pi x-0)) = \sin(2\pi x) $$

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  • $\begingroup$ Why does the index in $u_{xxx}$ starts at 1? $\endgroup$ – Maikel Mar 28 '13 at 21:54
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    $\begingroup$ 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

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