1
$\begingroup$

I'm working out the Galerkin method for the heat equation $$\frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} = 0$$ subject to $u(0,t)=0,u_x(1,t)=v(t)$.

I want to use a Fourier basis to represent the solution, i.e. $u(x,t) = \sum_{j=1}^n c_j(t) \phi_j(x)$ where $\phi_j(x) = \sin j\pi x$.

If I try to formulate the weak form of this PDE using the basis functions above, the effect of the boundary condition $v(t)$ goes away, which is strange.

This is because if I compute $\int_0^1 \frac{\partial^2 u}{\partial x^2} \phi \,dx$ using integration by parts, I get $u_x(1,t)\phi(1) - u_x(0,t)\phi(0) - \int_0^1 \frac{\partial u}{\partial x} \frac{\partial \phi}{\partial x}\,dx$.

However, $\phi(1)=0$! Did I commit a mistake? What would be the proper way of doing this?

$\endgroup$
  • $\begingroup$ I am not an expert of Galerkin method, but aren't you already enforcing $u(1,t) = 0$ because of your choice of trial space? I don't think you can enforce $u(0,t) = 0$, $u(1,t) = 0$ and $u_x(1,t) = v(t)$ at the same time but I might be wrong. $\endgroup$ – QuantumApple May 1 at 21:01
  • $\begingroup$ @QuantumApple has this right: If you write your solution as $u(x,t)=\sum_j c_j(t) \phi_j(x)$ where all of the $\phi_j$ are zero at $x=1$, then necessarily $u(1,t)=0$. You simply can't represent a nonzero $v(t)$ by adding up zeros. If you want a $u$ that's nonzero at $x=1$, then you will also have throw a few functions into your basis that are nonzero there. $\endgroup$ – Wolfgang Bangerth May 1 at 21:55
  • $\begingroup$ Thanks for the replies. But my boundary condition at $x=1$ is not dirichlet, it's neumann. It's $u_x(1,t) = v(t)$. Just because $u(1,t)=0$ doesn't mean $u_x(1,t) = 0$, right? $\endgroup$ – user1237300 May 1 at 22:25
  • $\begingroup$ @QuantumApple. Well, $sin \pi x$ is a function that is 0 at $x=1$ but has a non-zero derivative there/ $\endgroup$ – user1237300 May 1 at 22:41
  • $\begingroup$ @user1237300 This is true, but I think that you cannot enforce 3 BC at the same time in 1D. This is simply not a mathematically well-defined problem. Another way to phrase it is that the real solution $u(x,t)$ is probably $\neq 0$ at $x = 1$. But your choice of trial functions will make it so that your solution will be $0$ there. So even if you tried somehow to enforce $u_x(1,t) = v(t)$ manually, my guess is that you would not converge to the right solution anyway (because it is a priori $\neq 0$ at $x=1$). $\endgroup$ – QuantumApple May 1 at 23:03
0
$\begingroup$

Inspired from this Computational Science SE post:

  • Transform your problem from one with inhomogeneous BC to homogeneous BC. This is done by substracting any function $B(x,t)$ with the right inhomogeneous BC from $u(x,t)$ to create a new function $h(x,t) = u(x,t) - B(x,t)$. For instance, take $B(x,t) = \frac{2v(t)\sin(\pi x/2)}{\pi}$. Your problem becomes $\frac{\partial h}{\partial t} - \frac{\partial^2 h}{\partial x^2} + \frac{\partial B}{\partial t} - \frac{\partial^2 B}{\partial x^2} = 0$, with BC $h(0,t) = 0$ and $h_x(1,t) = 0$.
  • Choose the right basis functions (that satisfy the correct BC). Here for instance, you can take $\phi_j(x) = \sin((2j+1)\pi x/2)$. Write $h_j(x,t_n) = \sum_j c_j(t_n) \phi_j(x)$.
  • If you take your space test to be equal to your trial space, enforce the following condition $$\int_0^1 \left( \frac{\partial h}{\partial t} - \frac{\partial^2 h}{\partial x^2} + \frac{\partial B}{\partial t} - \frac{\partial^2 B}{\partial x^2} \right) \phi_j(x,t) = 0, \quad \forall j, t_n$$

As I am not familiar with the Galerkin method, you should take this with a grain of salt. This might not be the correct or the most efficient way to deal with your BC. I've came with this after reading a bit about the Galerkin method online and going through the post linked at the top.

As for wether or not the $\phi_j$'s form an orthonormal basis of $L^2[0,1]$ (for the scalar product $\left\langle f, g \right\rangle = \int_0^1 f(x) g(x) dx$) when $j$ spans $\mathbb{N}$, again I am not an expert but I believe they do, since they are the solutions to the following Sturm-Liouville problem:

$$\frac{d^2 \phi}{dx^2} = -E \phi; \quad \phi(0) = 0; \quad \phi'(1) = 0.$$

| cite | improve this answer | |
$\endgroup$

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.