Temperature Distribution Solution for Rectangular HIJK with Laplace Equation

[matt222]

1. Homework Statement [/b

A rectangular of HIJK sides is bounded by the lines x=0, y=0, x=4, y=2.whatis the Temperature distribution T(x,y) over the rectangle by using the Laplace equation, boundary conditions are:
T(0,y)=0, T(4,y)=0 , T(x,2)=0, T(x,0)=x(4-x)

Homework Equations

d^2T/dx^2 + d^2T/dy^2 =0

The Attempt at a Solution

I started solving by using the separation of variable and find for
X=c1cos(npix/2)+c2sin(npix/2)
Y=c3cosh(npiy/2)+c4sinh(npiy/2)
c1 should go to zero so,
X=c2sin(npix/2)

so the final look for T(x,y)=c2sin(npix/2)[c3cosh(npiy/2)+c4sinh(npiy/2)]
the first two BC worked well, the 3rd BC got 0=c2sin(npix/2)c4sinh(npiy/2)
the 4th BC got c2c4sin(npix/2)=x(4-x), I subitute the 4th BC in the final T(x,y) and got

T(x,y)=x(4-x)cosh(npiy/2)+c4sinh(npiy/2)c2sin(npix/2)
to satisfy the BC, C3=0,
T(x,y)=x(4-x)cosh(npiy/2)
i have problem now with the 3rd BC at T(x,2), it will not satisfy the BC? what is my mistake

 

[vela]

Let's back up for a second. When you initially find the solution using separation of variables, there are no restrictions on the separation constant k². At that point, the best you can say is

$$\begin{align*} X(x) &= c_1 \cos kx + c_2 \sin kx \\ Y(y) &= c_3 \cosh ky + c_4 \sinh ky \end{align*}$$

Explain what applying the first boundary condition does to the solution. Then explain what applying the second one does.

 

[LCKurtz]

1. Homework Statement [/b

A rectangular of HIJK sides is bounded by the lines x=0, y=0, x=4, y=2.whatis the Temperature distribution T(x,y) over the rectangle by using the Laplace equation, boundary conditions are:
T(0,y)=0, T(4,y)=0 , T(x,2)=0, T(x,0)=x(4-x)

Homework Equations

d^2T/dx^2 + d^2T/dy^2 =0

The Attempt at a Solution

I started solving by using the separation of variable and find for
X=c1cos(npix/2)+c2sin(npix/2)

Recheck that, I think you should get

$$X_n=\sin(\frac{n\pi}{4}x)$$

Y=c3cosh(npiy/2)+c4sinh(npiy/2)

You have left off some squares inside the cosh terms. But more important, you don't have to stick with sinh and cosh; you can use any linearly independent pair. You will get much easier expressions to work with if you try for the Y solution an expression of the form

$$Y(y) = A\sinh(\mu y) + B\sinh(\mu(2-y))$$

Try it. And remember lamda and mu aren't equal to each other.