# To solve a PDE using Separation of Variables.

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!

• tutorial.math.lamar.edu/Classes/DE/LaplacesEqn.aspx
– user3417
Oct 18, 2018 at 21:33
• @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! Oct 18, 2018 at 21:35
• one moment...i'll walk through it.
– user3417
Oct 18, 2018 at 21:36
• Subtract $c_1$ from $u$ to obtain $v=u-c_1$ where $v(0,y)=0$, $v(a,y)=0$ and $\Delta v = 0$. Oct 18, 2018 at 21:49
• 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. Oct 18, 2018 at 22:18

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.