How do I solve the Laplace equation using Robbin's Boundary Conditions?

[shreddinglicks]

Homework Statement

Equation

(d^2 u)/(dx^2 )+(d^2 u)/(dy^2 )=0

boundary conditions

u(0,y)=0
u(a,y)=50
u(x,0)=0
u^' (x,b)=-h[u(x,b)-70]

Relevant Equations

(d^2 u)/(dx^2 )+(d^2 u)/(dy^2 )=0

I've tried a few things. I did one method to try to accomplish the removal of the -70 in the derivative boundary condition. It came out as below. When plotting it however it gave a solution that didn't make sense.

 

[pasmith]

What is the boundary condition at x = a? What do you mean by the notation u', since u is a function of 2 varaibles? Do you mean \frac{\partial u}{\partial x}? Is h a constant?

If that is correct, then I would set <br /> u(x,y) = \frac{70h}{1 + ah}x + v(x,y) where v satisfies \begin{split}<br /> v(0,y) = 0, \\<br /> \frac{\partial v}{\partial x}(a,y) + hv(a,y) &amp;= 0, \\<br /> v(x,0) &amp;= -\frac{70h}{1 + ah}x, \\<br /> v(x,b) &amp;= 50 - \frac{70h}{1 + ah}x. \end{split}

 

[shreddinglicks]

What is the boundary condition at x = a? What do you mean by the notation u&#039;, since u is a function of 2 varaibles? Do you mean \frac{\partial u}{\partial x}? Is h a constant?

If that is correct, then I would set <br /> u(x,y) = \frac{70h}{1 + ah}x + v(x,y) where v satisfies \begin{split}<br /> v(0,y) = 0, \\<br /> \frac{\partial v}{\partial x}(a,y) + hv(a,y) &amp;= 0, \\<br /> v(x,0) &amp;= -\frac{70h}{1 + ah}x, \\<br /> v(x,b) &amp;= 50 - \frac{70h}{1 + ah}x. \end{split}

Yes, exactly as you said. du/dx and h is a constant.

I realized I made an error.

The conditions are actually

I'll fix it also on the original post.

It's what you have but y instead of x.

The problem is, this gives an answer where for all y when u(0,y)=0 when x should be zero, x does not equal zero.

The funny thing is if I manipulate the equation by multiplying the first term by x/a. I can plot a solution that makes sense.

 

[pasmith]

You're only satisfying the boundary condition at x = a. No wonder you don't get the right value at x = 0.

Setting u(x,y) = \frac{70hy}{1 + bh} + v(x,y) gives \begin{split} v(0,y) &amp;= -\frac{70h}{1 + bh}y \\<br /> v(a,y) &amp;= 50 - \frac{70h}{1 + bh}y \\<br /> v(x,0) &amp;= 0 \\<br /> \left.\frac{\partial v}{\partial y}\right|_{(x,b)} + hv(x,b) &amp;= 0. \end{split} This gives you <br /> u(x,y) = \frac{70h}{1 + bh}y + \sum_{n=1}^\infty X_n(x) \sin k_n y. Now X_n is a linear combination of \cosh k_n x and \sinh k_n x. But instead of taking X_n(0) = 0, you must take \sum_{n=1}^\infty X_n(0) \sin k_n y + \frac{70h}{1 + bh}y = 0. You can't just ignore the \cosh k_n x solution, you need it to satisfy this condition.
It is, however, more convenient to take advantage of the fact that \sinh k_n(a- x) \equiv \sinh k_n a \cosh k_n x - \cosh k_n a \sinh k_n x is a linear combination of \cosh k_n x and \sinh k_n x and set <br /> X_n(x) = \frac{A_n \sinh k_n(a-x) + B_n \sinh k_n x}{\sinh k_n a} so that \begin{split}<br /> 0 &amp;= u(0,y) = \frac{70h}{1 + bh}y + \sum_{n=1}^\infty A_n \sin k_n y \\<br /> 50 &amp;= u(a,y) = \frac{70h}{1 + bh}y + \sum_{n=1}^\infty B_n \sin k_n y \end{split}