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Solve the pde

$$(t+1)u_t=u_{xx}$$

subject to

$$\rm{B.C.}:u(0,t)=u(\pi,t)=0,\quad \rm{I.C.}: u(x,0)=\sin x+7\sin 6x.$$

I've used separation of variables with $k=(t+1)$ but I did not get the right answer for $u(x,t)$. I got the right answer for the previous problems, but that was when k equalled a constant, like 5 or 1. What should I do differently?

I used Paul's formula for the answer here.

Martin Sleziak
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Myles
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  • I think so. The right side of the first equation is the double derivative of u with respect to x and the left side of the equation is (t+1) multiplied by the derivative of u with respect to t. – Myles Sep 11 '13 at 04:36
  • Change variables $\displaystyle{\large t \to \tau}$ as follows: $\displaystyle{\large% t \equiv \exp\left(\tau\right) - 1}$. It switchs to a Difussion Equation. – Felix Marin Sep 11 '13 at 04:48
  • t equals e to the what now? Why? – Myles Sep 11 '13 at 04:49
  • @Myles if you change to a new variable $\tau$ like Felix states, observe $\tau=\log(t+1)$ hence $\dfrac{d\tau}{dt}=\dfrac1{t+1}$ – obataku Sep 11 '13 at 04:50
  • I don't understand how that can lead to the answer? – Myles Sep 11 '13 at 04:53

3 Answers3

5

$$(t+1)u_t=u_{xx}$$Assuming a separable solution, we let $u(x,t)=F(x)G(t)$ so $u_t=F(x)G'(t),u_{xx}=F''(x)G(t)$ thus:$$(t+1)F(x)G'(t)=F''(x)G(t)\\(t+1)\frac{G'(t)}{G(t)}=\frac{F''(x)}{F(x)}=-\lambda^2\\G'(t)+\frac{\lambda^2}{t+1}G(t)=0\\F''(x)+\lambda^2 F(x)=0$$Clearly $F(x)=C_1\cos\lambda x+C_2\sin\lambda x$. To determine $G'(t)$ observe we can multiply by $\mu=\exp\left(\lambda^2\int\dfrac1{t+1}\,dt\right)=(t+1)^{\lambda^2}$ giving us:$$G(t)=\frac{C_3}{(t+1)^{\lambda^2}}$$Given our boundary conditions:$$u(0,t)=u(\pi,t)=0\\u(x,0)=\sin x+7\sin6x$$hence $F(0)=0$ and therefore $C_1=0$ i.e. $F(x)=C_2\sin\lambda x$. The key is to consider that since $F(\pi)=0\Leftrightarrow C_2\sin\lambda x=0$ we have either $C_2=0\Leftrightarrow u(x,t)=0$ or $\sin\lambda\pi=0$ hence $\lambda\in\mathbb{Z}$. Given the infinite number of eigenvalues $\lambda$, we write:$$F(x)=\sum_{\lambda=0}^\infty b_\lambda\sin\lambda x$$and so it follows with $a_\lambda=C_2C_3b_\lambda$ $$u(x,t)=F(x)G(t)=\sum_{\lambda=0}^\infty a_{\lambda}\frac1{(t+1)^{\lambda^2}}\sin\lambda x\\u(x,0)=\sum_{\lambda=0}^\infty a_\lambda\sin\lambda x\\\sin x+7\sin6x=\sum_{\lambda=0}^\infty a_\lambda\sin\lambda x$$By the orthogonality of $\sin\lambda x$ we have $a_1=1,a_6=7$ and other $a_\lambda=0$. Thus our solution is given by$$u(x,t)=\frac1{t+1}\sin x+\frac7{(t+1)^{36}}\sin 6x$$

Wolfram agrees. Apologies if I left anything unclear -- let me know and I will try to explain further. I suggest reading up on elementary Sturm-Liouville theory.

obataku
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  • Your general solution for u(x,t) is right when I plugged it into my homework. I still am confused about how you got that general solution for u(x,t) though and why it's so different than when there's a constant in front of the u_t in the original question. – Myles Sep 11 '13 at 05:29
  • @Myles hmm? The general solution follows from letting $F(x)=\sum_{\lambda=0}^\infty b_\lambda\sin\lambda x$ – obataku Sep 11 '13 at 05:32
  • @Myles: $k$ used to be constant. But now there is $t$ which is also one of the variables and hence, is not constant. It the same difference as in the ODEs $y'=y$ and $ty' = y$. – Dirk Sep 11 '13 at 09:28
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Solutions which satisfies ${\rm u}\left(0,t\right) = {\rm u}\left(\pi,t\right) = 0$ are of the form $$ {\rm u}\left(x,t\right) = \sum_{n = 0}^{\infty}{\rm A}_{n}\left(t\right)\sin\left(n\pi x\right) $$ where $$ \left(t + 1\right)\,{{\rm d}{\rm A}_{n}\left(t\right) \over {\rm d}t} = -\left(n\pi\right)^{2}{\rm A}_{n}\left(t\right) \quad\Longrightarrow\quad {{\rm d}% \left\lbrack \left(t + 1\right)^{\left(n\pi\right)^{2}}{\rm A}_{n}\left(t\right) \right\rbrack \over {\rm d}t} = 0 $$

with ${\rm A}_{n}\left(0\right) = \delta_{n,1} + 7\delta_{n,6}$. Then

$$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad{\rm u}\left(x,t\right) = {\sin\left(\pi x\right) \over \left(t + 1\right)^{\pi^{2}}} + 7\,{\sin\left(6\pi x\right) \over \left(t + 1\right)^{36\pi^{2}}}\quad} \\ \\ \hline \end{array} $$

Felix Marin
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2

You can use separation of variables technique to solve the pde. So, assuming

$$ u(x,t)=F(x)G(t) $$

results in the two ode's

$$ F''(x)=\lambda F(x),\quad G'(t)=\frac{\lambda}{1+t}G(t). $$

Now, solve the first ode using the boundary conditions $u(0)=u(\pi)=0$. You can follow the technique in my answer for the heat equation.

Community
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Mhenni Benghorbal
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  • What would I use for the k value in the formula at the end of your answer? – Myles Sep 11 '13 at 05:13
  • @Myles: There is no $k$ here. Just follow the steps I gave you. If you plug $u=FG$ in the pde you will get the two ode's just like in the heat equation. – Mhenni Benghorbal Sep 11 '13 at 05:16