0
$\begingroup$

I'm in a problem involving the heat equation:

\begin{cases} u_{t}=\frac{1}{10}u_{xx}\\ u_{x}(0,t)=u_{x}(\pi,t)=0 \\ u(x,0)=3-4\cos(2x) \end{cases}

Using separation of variables, and using the first boundary condition, I found

$$u(x,t)=\sum_{n=1}^{\infty}a_{n}\exp\left[\frac{-(n+1)^{2}t}{10}\right]\sin\left[\left(n+\frac{1}{2}\right)x\right] $$

I just need to use the second boundary condition to have the value for each $a_{n}$, but I don't know how to do this, since

$$u(x,0)=3-4\cos(2x)=\sum_{n=1}^{\infty}a_{n}\sin\left[\left(n+\frac{1}{2}\right)x\right]$$

If the boundary condition was like $3-4\sin(2x)$ would be easier. How to deal with it?

$\endgroup$
  • $\begingroup$ Either the conditions are incorrect or the solution is incorrect. $\endgroup$ – Leucippus Aug 23 '18 at 19:04
1
$\begingroup$

From your boundary conditions it is clear that the solution has to be something like this $$u(x,t) = \sum_{n=0}^\infty b_n(t)\cos(n x)\tag1$$ because clearly $$u_x(x,t) = -\sum_{n=0}^\infty nb_n(t)\sin(nx)\\ u_x(0,t) = -\sum_{n=0}^\infty nb_n(t)\underbrace{\sin(n0)}_{\text{is zero even}\\ \text{for }x=\pi} = u_x(\pi,t) = 0\;\;\;\;\forall n\in\mathbb{N}$$ The solution $(1)$ it is clearly a good solution even judging on the initial condition which can be written as a sum of cosines $$u(x,0) = 3\cos(0)-4\cos(2x)$$

Now using the "Fourier method", we use the ansatz $(1)$ and substitute it in the equation to find $b_n(t)$, which gives us an ODE in the variable $b_n(t)$, mainly $$b'_n(t) = \frac{n^2}{10}b_n(t)\\ b_n(t) = c_n e^{\frac{n^2}{10}t}$$

To finde the solution to the PDE we just have to use the initial condition $$u(x,t) = \sum_{n=0}^\infty c_n e^{\frac{n^2}{10}t}\cos(nx)\\ u(x,0) = \sum_{n=0}^\infty c_n e^{\frac{n^2}{10}0}\cos(nx) = \sum_{n=0}^\infty c_n \cos(nx) = 3\cos(0)-4\cos(2x)$$ from which we get $$n=0\implies c_n=3 \\n=2 \implies c_n=-4$$

So the solution is $$u(x,t) = 3-4e^{\frac{4}{10}t}\cos(2x)=3-4e^{\frac{2}{5}t}\cos(2x)$$

$\endgroup$
  • 1
    $\begingroup$ You are right. I just changed the sin and cos terms at some steps before. Thank you! $\endgroup$ – Mateus Rocha Aug 23 '18 at 19:38

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.