Parabolic pde with additional term

[idraftscience]

Homework Statement

Solve
$\frac{\partial v}{\partial t} = k\frac{\partial^2 v}{\partial x^2} - v$

$0\leq x \leq L$

$t > 0$

Homework Equations

$v(x,0) = f(x), v(0,t)=0, \frac{\partial v}{\partial x}(L,t) = -v(L,t)$

The Attempt at a Solution

I've already attempted to solve this using separation of variables, but I'm not sure if that is a valid approach as the equation is not homogeneous (?) due to the v(x,0) = f(x) term. I've also attempted using green's theorem but I keep getting stuck, though this may be due to a math error. Can anyone give me any insight as to where I may be going wrong?

 

[haruspex]

If you can separate the variables, it will produce a valid solution. The question is whether all solutions can be achieved that way. The critical issue is whether a linear sum of two solutions produces a solution. If so, arbitrary solutions can be represented as an infinite sum of solutions obtained by separation of variables.
Does this equation have that property?

 

[pasmith]

Homework Statement

Solve
$\frac{\partial v}{\partial t} = k\frac{\partial^2 v}{\partial x^2} - v$

$0\leq x \leq L$

$t > 0$

Homework Equations

$v(x,0) = f(x), v(0,t)=0, \frac{\partial v}{\partial x}(L,t) = -v(L,t)$

The Attempt at a Solution

I've already attempted to solve this using separation of variables, but I'm not sure if that is a valid approach as the equation is not homogeneous (?) due to the v(x,0) = f(x) term. I've also attempted using green's theorem but I keep getting stuck, though this may be due to a math error. Can anyone give me any insight as to where I may be going wrong?

The equation is homogeneous and linear, so separation of variables should work. (Not being homogeneous is in any case not fatal to separation of variables, you just have to use a "particular integral plus complementary functions" approach; not being linear is fatal.)

The eigenfunctions are $$v(x,t) = T(t)X(x) = T(t)\sin(\lambda x)$$ where the possible values of $\lambda$ are determined by the condition that $$X(L) + X'(L) = 0,$$ and since you're given arbitrary initial data the fact that you can't find $\lambda$ analytically is not important. Sturm-Liouville theory tells you that the resulting eigenfunctions will be orthogonal with respect to the inner product $$\int_0^L f(x)g(x)\,dx$$ as usual. You can find $\|X\|^2 = \int_0^L X(x)^2\,dx$ in terms of $L$ and $\lambda$, but obviously if you can't find $\lambda$ analytically you can't find $\|X\|^2$ analytically.