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Would it be possible for someone to guide me through this problem.

The PDE is, $\Delta u =0$

We have the following boundary solutions: where a and b are real numbers u(0,y)=c1, u(a,y)=c1, u(x,0)=g(x), $\frac{\delta u}{\delta v}$(x,b)=c2 where c1 and c2 are real numbers.

I have to solve this for x in [0,a] and y in [0,b]

Could someone help me out for this problem. I'm beginner at PDE. I was able to break up the problem to $V^{II}=\lambda V$ and $W^{II}=-\lambda W$. I know that I have to break it up and solve 4 PDEs.

For u(0,y)=c1, and all the other boundary's zero, I found $\lambda > 0$. For u(a,y)=c1, and all the other boundary's zero, I also found $\lambda > 0$. For the last 2 solutions I found $\lambda<0$.

I think the v in $\frac{\delta u}{\delta v}$(x,b)=c2 is the outward pointing normal vector and I believe it should equal y except I can't figure out why.

I've spent hours on the problem trying to work out the solutions and I can't seem to get anywhere. Any help would be greatly appreciated!

Thank you for your time!

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  • $\begingroup$ tutorial.math.lamar.edu/Classes/DE/LaplacesEqn.aspx $\endgroup$
    – user3417
    Oct 18, 2018 at 21:33
  • $\begingroup$ @RyanHowe, thank you! I've actually already looked through his notes and I can't seem to find anything about an outward pointing normal vector as part of the boundaries! $\endgroup$
    – Shaun
    Oct 18, 2018 at 21:35
  • $\begingroup$ one moment...i'll walk through it. $\endgroup$
    – user3417
    Oct 18, 2018 at 21:36
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    $\begingroup$ Subtract $c_1$ from $u$ to obtain $v=u-c_1$ where $v(0,y)=0$, $v(a,y)=0$ and $\Delta v = 0$. $\endgroup$
    – Disintegrating By Parts
    Oct 18, 2018 at 21:49
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    $\begingroup$ Separation of variables problems result in eigenfunction problems only in the variables where you have two homogeneous endpoint conditions. The last direction is solved in a different way. $\endgroup$
    – Disintegrating By Parts
    Oct 18, 2018 at 22:18

1 Answer 1

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Let $v=u-c_1$. The problem for $v$ is $$ \Delta v = 0 \\ v(0,y)=0,\;\; v(a,y)=0\\ v(x,0)=g(x)-c_1,\;\; v_y(x,b)=c_2. $$ This problem is the sum of solutions of the following two problems $$ \Delta v = 0 \\ v(0,y)=0,\;\; v(a,y)=0\\ v(x,0)=g(x)-c_1,\;\; v_y(x,b)=0. $$ $$ \Delta v = 0 \\ v(0,y)=0,\;\; v(a,y)=0\\ v(x,0)=0,\;\; v_y(x,b)=c_2. $$ Then $u=v+c_1$ is the desired solution.

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