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I am just starting to study PDE and don't know how solve the Dirichlet problem on the line. I read PDE book, but they discuss in more than one dimension by using separation of variable. How does one do it in one dimension. I know you just get $\sin(nx)$, but how do you get that?

In fact, I am even more confused about the separation of variables technique on the plane.

d$u_{xx}+u_{yy}=0 $ Using separation of variables, one gets $u(x,y)=X(x)Y(y)$, which leads to the conclusion that $\frac{d^{2}X}{dx^{2}}+\lambda X=0$

and $ \frac{d^{2}Y}{dy^{2}}-\lambda Y=0 $.

Should it just be $\frac{d^{2}X}{dx^{2}}=0$ and analogously for $Y$?

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Why would that be? Given $\frac{d^2 X}{dx^2} + \lambda X = 0$, you know that the solution of this ODE is $$ X(x) = \begin{cases} a e^{\sqrt{-\lambda} x} + b e^{-\sqrt{-\lambda} x} & \text{ if } \lambda < 0 \\\ ax + b & \text{ if } \lambda = 0 \\\ a \cos( \sqrt{\lambda} x) + b \sin(\sqrt{\lambda}x) & \text{ if } \lambda > 0 \end{cases} $$ and given that $\frac{d^2 Y}{dy^2} - \lambda Y = 0$, you have $$ Y(x) = \begin{cases} a e^{\sqrt{\lambda} y} + b e^{-\sqrt{\lambda} y} & \text{ if } \lambda > 0 \\\ ay + b & \text{ if } \lambda = 0 \\\ a \cos( \sqrt{-\lambda} y) + b \sin(\sqrt{-\lambda}y) & \text{ if } \lambda < 0 \end{cases} $$ You can either find those by explicitly solving the differential equations (these are ODE techniques, not PDE, so assuming you have done an ODE course you should know how to do that) or just by substituting and knowing that two linearly independent solutions span the solutions of an homogeneous linear ODE of order $2$.

I am not sure the rest of your proof is right though. I'm not familiar with the Dirichlet problem you speak of (I am not a fan of differential equations, I'll admit) and I don't understand why you've written $du_{xx} + u_{yy} =0$, the "$d$" looks weird there. If your equation is $u_{xx} + u_{yy} = 0$, then everything I did works just fine.

Hope that helps,

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