3

I find on my textbook that PDE cannot be solved using the method of separation of variables if the boundary conditions are inhomogeneous. Why is that ?

Mohamed Mostafa
  • 1,228

1 Answers1

5

The idea of the method is to find simple (=separable) solutions and then combine them to obtain the solution you're really looking for.

Now say $u$ and $v$ solve your linear PDE with a boundary condition $u=v=A$ for $x=a$. Then their sum $f=u+v$ solves the same PDE, but $f=A+A$ for $x=a$. So $f$ satisfies the same boundary condition only if $A=A+A$, i.e., if $A=0$. So unless you have homogeneous boundary condition, you can't combine your simple solutions!

Hans Lundmark
  • 53,395
  • Your statements are irrefutable but I don't think they shed light on why separation of variables fails. Perhaps a little more explanation will help. – Amey Joshi Jul 30 '18 at 12:32
  • 1
    @AmeyJoshi: To me, this explains exactly why it fails, so I don't know what else to add, but feel free to add your own answer if you know something more illuminating that you want to share with the rest of us. – Hans Lundmark Jul 30 '18 at 12:57
  • I am struggling to complete the answer. We combine the simple solutions by multiplying them. At this stage, the non-homogeneity does not come in the way. It does not affect even when we find the constants of integration of one of the ODE's. But it does matter when we try to fit the initial condition using these constants. So I can describe what goes wrong when we use the method. I also understand how inhomogeneity places a role but somehow I am not able to see why it creates a problem. – Amey Joshi Jul 31 '18 at 13:48
  • Your response tells that a linear combination of solutions is not a solution if the BCs are inhomogeneous. But I am struggling to tie it with the method of separation of variables. It would be nice if you can help me connect the two. – Amey Joshi Jul 31 '18 at 13:51
  • 1
    @AmeyJoshi: No, we don't combine the simple solutions by multiplying! I'm talking about separated solutions of the PDE, that is, solutions of the form $u(x,t)=X(x)T(t)$ where you have already multiplied $X$ and $T$ (which in turn are solutions of certain ODEs). And by “combining them” I mean forming a linear combination $u(x,t)=\sum c_n X_n(x) T_n(t)$ in order to satisfy the initial condition, which is the only thing not taken into accunt yet. – Hans Lundmark Jul 31 '18 at 14:13
  • Let me take help of an example. Let us consider $u_t = u_{xx}$ with $u(0, t) = 1, u(1, t) = 1$ and $u(x, 0) = \phi(x)$. If we suppose that $u(x,t) = X(x)T(t)$ then the equation separates as $T^{\prime\prime}/T = -\lambda^2 = X^{\prime\prime}/X$. We then get $u(x,t) = e^{-\lambda^2 t}(c_1e^{i\lambda x} + c_2e^{-i\beta x})$. If we put this in the boundary and initial conditions we get (1) $c_1 + c_2 = e^{\lambda^2 t}$, (2) $c_1e^{i\beta} + c_2e^{-i\beta} = e^{\lambda^2 t}$ and (3) $c_1e^{i\beta x} + c_2e^{-i\beta x} = \phi(x)$. [Continued in next comment.] – Amey Joshi Jul 31 '18 at 14:38
  • Now, even if we solve equations (1) and (2) to get $c_1, c_2$ we are unable to guarantee that (3) is satisfied. Further, unlike the homogeneous case, where we get $c_1, c_2$ for a discrete range of $\beta$, over here we get them for all but a discrete set of $\beta$. So I am in trouble even before I consider a linear combination of solutions. – Amey Joshi Jul 31 '18 at 14:40
  • 1
    @AmeyJoshi: You're not supposed to impose the initial condition on the individual separated solutions $u_n=X_n T_n$; that doesn't work even in the homogeneous case! You only do once you have formed the linear combination $u=\sum c_n u_n$ (in order to determine the constants $c_n$). – Hans Lundmark Jul 31 '18 at 15:09
  • @AmeyJoshi: In your particular example, the boundary conditions applied to $u(x,t)=X(x)T(t)$ give $X(0)T(t)=X(1)T(t)=1$ for all $t$, so $T(t)$ must be constant, so you must have $\lambda = 0$. So you are right that there is a problem already at this stage, but it's not the problem that you think – instead you don't get sufficiently many separated solutions (but even if you did, you wouldn't be able to combine them anyway). – Hans Lundmark Jul 31 '18 at 15:21
  • Let us continue this discussion in chat. – Amey Joshi Jul 31 '18 at 15:32