How Do You Apply Boundary Conditions to Solve a Separable PDE in a Strip?

[MathematicalPhysics]

I need to find all the separated solns of

$$x^2 \frac{\partial^2 u}{\partial x^2} + x\frac{\partial u}{\partial x} + \frac{\partial^2 u}{\partial y^2} = 0$$

in the strip $${(x,y) : 0 < y < a, -\infty < x < \infty }$$
the separated solns must also satisfy u = 0 on both the edges, that is, on y=0 and y=a for all values of x.

Iv got the general solutions to be..

$$X(x) = Dlnx + C , (k = 0)$$
$$X(x) = Dx^{n} + Cx^{-n} , (k \neq 0)$$

and

$$Y(y) = A\cos{ky} + B\sin{ky} , (k \neq 0)$$
$$Y(y) = Ay + B , (k = 0)$$

where k is just the constant iv let the two bits equal when I separated the variables. (well -k^2 actually).

I just need help interpreting the conditions to sort out the constants..I think!

 

[Galileo]

I haven't checked your answer, but if it is correct then, since $u(x,y)=X(x)Y(y)$, the boundary conditions say:

$$u(x,0)=X(x)Y(0)=0$$
and
$$u(x,a)=X(x)Y(a)=0$$

So $Y(0)=Y(a)=0$

For example: if k=0, then applying the boundary condition at y=0 gives:
$$Y(0)=B=0$$

 

[M98Ranger]

Firstly, it is important to note that the general solutions you have obtained are in the form of separable variables, where X(x) and Y(y) are functions of x and y respectively, and k is a constant. This is a common technique used to solve partial differential equations.

Now, to apply the initial conditions, we need to consider the given strip {(x,y) : 0 < y < a, -\infty < x < \infty } and the fact that the separated solutions must satisfy u=0 on both edges, y=0 and y=a.

For y=0, we have Y(0) = 0. This means that A=0, as the sine and cosine functions are not zero at y=0. So the general solution for Y(y) becomes Y(y) = Bsin(ky).

Similarly, for y=a, we have Y(a) = 0. This means that Bsin(ka) = 0. Since we are looking for non-trivial solutions, we can assume that B is non-zero. This implies that sin(ka) = 0, which gives us ka = nπ, where n is any integer. So, k = nπ/a.

Now, we can substitute this value of k into the general solution for X(x) to obtain the final separated solutions:

X(x) = Dlnx + C , for k=0
X(x) = Dx^{n} + Cx^{-n} , for k \neq 0

Y(y) = Bsin(nπy/a) , for k \neq 0

Finally, we can use the initial conditions to determine the values of the constants D, C, and B. Since we are given that u=0 on both edges, we have:

u(x,0) = X(x)Y(0) = 0, which implies that C = 0 for k=0, and D=0 for k\neq 0.

u(x,a) = X(x)Y(a) = 0, which implies that B=0 for k=0, and B is non-zero for k\neq 0.

So, the final separated solutions that satisfy the initial conditions are:

X(x) = C , for k=0
X(x) = 0 , for k \neq 0

Y(y) = Bsin(nπy