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

simon
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  • There needs to be another condition at the boundary e.g., $\partial_x (0,t) = 0$ to fully specify the solution uniquely. – Ron Gordon Mar 29 '13 at 10:53
  • Can you explain. Because I think it's enough for a numerical solution using finite difference. – simon Mar 29 '13 at 11:37
  • 3rd degree in $x$. When you do separation of variables, you'll get a 3rd degree ODE, which requires 3 conditions to uniquely specify. – Ron Gordon Mar 29 '13 at 11:38
  • Does this also apply when it can't be solved by separation of variables? – simon Mar 29 '13 at 12:06
  • Yes, I think so, especially so if it is linear. – Ron Gordon Mar 29 '13 at 12:10
  • I tried to solve this equation with u(x,t)=X(x)T(t) but this didn't give me any solution that satisfied the initial conditions. – simon Mar 29 '13 at 12:12
  • I don't see what you mean. – simon Mar 29 '13 at 12:14
  • You'll need to elaborate on what you did. Your solution should resemble a Fourier series of some sort. – Ron Gordon Mar 29 '13 at 12:14
  • It's not quite true that you always require the same number of boundary conditions as derivatives. That said, it's quite unlikely that 2 boundary conditions are enough in this case. – Glen O Mar 29 '13 at 12:31
  • $u_{tt}=u_{xxx}, \quad$

    $u(x,t)=X(x)T(t), \quad$

    $u(x,0)=X(x)T(0)=\sin{\pi x}, \quad$

    $X(x)=\frac{\sin{\pi x}}{T(0)}, \quad$

    $u_{xxx}=X_{xxx}(x)T(t), \quad u_{tt}=X(x)T_{tt}(t), \quad$

    $\frac{T_{tt}(t)}{T(t)} = \frac{X_{xxx}(x)}{X(x)} = \lambda^2 = - \pi^2\frac{\cos{\pi x}}{\sin{\pi x}}$

     – simon Mar 29 '13 at 12:46
  • Since $\lambda^2$ is constant this can't be, right? – simon Mar 29 '13 at 12:50
  • You're not doing separation of variables correctly. $\sin \pi x$ is not supposed to be assumed to be the result's form. You're meant to solve the separated ODEs independently in terms of $\lambda$ to construct separated solutions, and then combine separated solutions (with multiplication by an unknown constant) to form the general solution. Then find the constants necessary to satisfy the initial conditions. – Glen O Mar 29 '13 at 12:58
  • Upon further investigation, it actually might be possible with the available boundary conditions. – Glen O Mar 29 '13 at 13:14
  • But not with separation of variables right. Since $\lambda=0$ gives a polynomial solution, so it can't satisfy the initial conditions. And $X_{xxx}=\pm X(x)$ gives $ c_1e^{\pm x}+ c_2 e^{\mp x/2} \cos{\frac{\sqrt{3} x}{2}}+c_3 e^{\mp x/2} \sin{\frac{\sqrt{3} x}{2}}$, which also can't satisfy the initial condition either. – simon Mar 29 '13 at 13:31
  • Once again, you're not doing separation of variables correctly. I'm writing up a "solution" right now with the separation of variables approach, although unfortunately it's a little too unwieldy for me to solve with any sort of ease. – Glen O Mar 29 '13 at 13:35
  • See here for the separation of variables.. – Mhenni Benghorbal Mar 29 '13 at 13:37

1 Answers1

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Using separation of variables, let

$$ \frac{T_{tt}}{T} = \frac{X_{xxx}}{X} = \lambda^6 $$ where we're assuming that $u=T(t)X(x)$. (Do you understand why I chose $\lambda^6$? Keep in mind that $\lambda$ need not be real) This allows us to find the general solutions

$$ T(t) = A_1e^{\lambda^3 t}+A_2e^{-\lambda^3 t}\\ X(x) = B_1e^{\lambda^2 x}+B_2e^{\alpha\lambda^2 x}+B_3e^{\alpha^2\lambda^2 x} $$ where $\alpha^3 = 1$ is one of the roots other than $\alpha=1$. We'll use $\alpha = -\frac{1}{2}+i\frac{\sqrt{3}}{2}$. Now, we'll find restrictions that satisfy our boundary conditions. We have $$ X(0)=B_1+B_2+B_3 = 0\\ X(1)=B_1e^{\lambda^2}+B_2e^{\alpha \lambda^2}+B_3e^{\alpha^2\lambda^2}=0\\ -(B_2+B_3)e^{\lambda^2}+B_2e^{\alpha \lambda^2}+B_3e^{\alpha^2 \lambda^2}=0\\ B_2(e^{\alpha \lambda^2}-e^{\lambda^2})+B_3(e^{\alpha^2 \lambda^2}-e^{\lambda^2})=0 $$ So let's write that

$$ X(x) = \hat C\left(\frac{e^{\alpha \lambda^2 x}-e^{\lambda^2 x}}{(e^{\alpha \lambda^2}-e^{\lambda^2})}-\frac{e^{\alpha^2 \lambda^2 x}-e^{\lambda^2 x}}{(e^{\alpha^2 \lambda^2}-e^{\lambda^2})}\right) $$

We also know that

$$ T'(0) = A_1\lambda^3-A_2\lambda^3 = 0 $$ So $A_1=A_2$ (except when $\lambda=0$), or $T(t) = \hat A \cosh \lambda^3 t$.

So now we have that

$$ u(x,t)=\int C(\lambda)\cosh (\lambda^3 t)\left(\frac{e^{\alpha \lambda^2 x}-e^{\lambda^2 x}}{(e^{\alpha \lambda^2}-e^{\lambda^2})}-\frac{e^{\alpha^2 \lambda^2 x}-e^{\lambda^2 x}}{(e^{\alpha^2 \lambda^2}-e^{\lambda^2})}\right)d\lambda $$

Now, $u(x,0)=\sin \pi x$, so

$$ \int C(\lambda)\left(\frac{e^{\alpha \lambda^2 x}-e^{\lambda^2 x}}{(e^{\alpha \lambda^2}-e^{\lambda^2})}-\frac{e^{\alpha^2 \lambda^2 x}-e^{\lambda^2 x}}{(e^{\alpha^2 \lambda^2}-e^{\lambda^2})}\right)d\lambda = \sin \pi x $$ This is how one does separation of variables.

Let's suppose that our solution will remain bounded - this requires that $\lambda^3$ is imaginary, and so $\lambda$ is also imaginary. So we have $\lambda=ik$. In this case, $\lambda^2=-k^2$, and our separable components look like

$$ T(t) = \hat A \cos (k^3 t) $$ and $$ X(x) = \hat C\left(\frac{e^{-\alpha k^2 x}-e^{-k^2 x}}{e^{-\alpha k^2}-e^{-k^2}}-\frac{e^{-\alpha^2 k^2 x}-e^{-k^2 x}}{e^{-\alpha^2 k^2}-e^{-k^2}}\right) $$ We could simplify this by converting $e^{-\alpha k^2 x}$ to $e^{k^2x/2}(\cos(\sqrt{3}k^2x/2)+i\sin(\sqrt{3}k^2x/2))$ and $e^{-\alpha^2 k^2 x}$ to $e^{k^2x/2}(\cos(\sqrt{3}k^2x/2)-i\sin(\sqrt{3}k^2x/2))$.

The problem, however, is that we cannot reasonably restrict the possible values of $k$ (for a finite domain, it is often restricted to a discrete set of numbers, allowing a sum formula), beyond noting that $k\geq 0$ is sufficient. Our integral defining $C(k)$ is

$$ \int_0^\infty C(k) \left(\frac{e^{-\alpha k^2 x}-e^{-k^2 x}}{e^{-\alpha k^2}-e^{-k^2}}-\frac{e^{-\alpha^2 k^2 x}-e^{-k^2 x}}{e^{-\alpha^2 k^2}-e^{-k^2}}\right)dk = \sin(\pi x) $$

Glen O
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