non homogenous $c\frac{\partial^2 w}{\partial y^2} - \frac{\partial w}{\partial t}=-1-2y$

Question

I am doing this question As it can't be solved using separation of variables (my assumption according to what i did, after checking by substituting $w(y,t)=f(y)g(t)$ , and getting a term at last which is not depending on only single variable)

1. Did my assumption is right ?

2. So how to solve this equation, I am stuck ?

Answer 1

To get rid of the non-homogeneous part, you can substitute another variable and another function.

First, we can substitute:

$$z=2y+1 $$

Which gives us:

$$4c w_{zz}-w_t+z=0$$

Then we introduce a new function:

$$w=v(z,t)-\frac{z^3}{24 c}$$

Substituting, we get:

$$4c v_{zz}-v_t=0$$

The boundary and initial conditions change in an obvious way, and the result is a separable equation for $v(z,t)$ which is easy enough to solve.

As a further hint, the conditions change to (if I didn't make a mistake somewhere):

$$v(0,t)=\frac{1}{24c}, \qquad v(2L+1,t)=\frac{(2L+1)^3}{24c}+4 \\ v(z,0)=\frac{z^3}{24c}+\Phi \left(\frac{z-1}{2} \right)$$

Answer 2

Differentiate both sides with respect to $y$, to obtain $\displaystyle c \frac{\partial^4 w}{\partial y^4} = \frac{\partial^3 w}{\partial t \partial y^2}$. From here, we can swap order of differentiation and use the substitution $\displaystyle v = \frac{\partial^2 w}{\partial y^2}$ to obtain $\displaystyle \frac{\partial^2 v}{\partial y^2} = \frac{\partial v}{\partial t}$. Solving from here is not too difficult. For more information/help, see https://en.wikipedia.org/wiki/Partial_differential_equation - there is a list of solutions to well-known PDEs there.