Solving PDE Using Method of Separation of Variables: Complete Guide

[hbl2012]

I'm new here, hope it is the right place to ask the question.

The PDE question is ∂²u/∂x²+∂²u/∂y²=0
and u(x,0)=f(x), u(x,1)=0, u(0,y)=0, u(1,y)=0.

I use the method of separate the variables, with is let u(x,y)=X(x)Y(y) and get X''/X+Y''/Y=0.

Then let X''/X=-Y''/Y=-λ, i.e. X''+λX=0, and Y''-λY=0.

Since u(x,0)=f(x)=X(x)Y(0), u(x,1)=0=X(x)Y(1), u(0,y)=0=X(0)Y(y), u(1,y)=0=X(1)Y(y), we can conclude that X(0)=0, X(1)=0, Y(0)=f(x), Y(1)=0.

For X''+λX=0, we have 3 cases

1) λ=0, i.e. X''=0, i.e. X=ax+b, since X(0)=0, X(1)=0, a=b=0, which is X(x) always 0. not wavelike.

2) λ<0, Let λ=-μ², i.e. X''-μ²X=0, General solution is X=a₁e^μx+a₂e^-μx. Since X(0)=0, X(1)=0, a₁=a₂=0, which is X(x) always 0. not wavelike.

3) λ>0, Let λ=μ², i.e. X''+μ²X=0, General solution is X=a₁cos(μx)+a₂sin(μx). Since X(0)=0, X(1)=0, a₂sinμ=0.
In order to get a nontrivial solution, let μ=n∏, then X(x)=sin(n∏x).
Then we consider Y''-λY=0, i.e. Y''-n²∏²Y=0.
The general solution is Y=b₁e^n∏y+b₂e^-n∏y.
Then since Y(1)=0, b₁e^n∏+b₂e^-n∏=0.


Now, here is my question.
Why b₁+b₂e^-2n∏=0?

and How it change in the black font? Y_h(y)=-b₂e^-2n∏e^n∏y+b₂e^-n∏y
=-b₂e^-n∏e^-n∏e^n∏y+b₂e^-n∏e^n∏e^-n∏y
=-b₂'e^-n∏(1-y)+b₂'e^n∏(1-y)
=b₂'(e^n∏(1-y)-e^-n∏(1-y))
=2b₂''sin(hn∏(1-y))

 

[pasmith]

[...]

Then we consider Y''-λY=0, i.e. Y''-n²∏²Y=0.
The general solution is Y=b₁e^n∏y+b₂e^-n∏y.
Then since Y(1)=0, b₁e^n∏+b₂e^-n∏=0.
Now, here is my question.
Why b₁+b₂e^-2n∏=0?

Divide both sides of the previous equation by $e^{n\pi}$.

and How it change in the black font? Y_h(y)=-b₂e^-2n∏e^n∏y+b₂e^-n∏y
=-b₂e^-n∏e^-n∏e^n∏y+b₂e^-n∏e^n∏e^-n∏y
=-b₂'e^-n∏(1-y)+b₂'e^n∏(1-y)
=b₂'(e^n∏(1-y)-e^-n∏(1-y))
=2b₂''sin(hn∏(1-y))

I think that's a misprint for $2b_2'\sinh(n\pi(1-y))$.

Remember that
$$\sinh(x) \equiv \frac{e^x - e^{-x}}{2}.$$

 

[hbl2012]

Divide both sides of the previous equation by $e^{n\pi}$.



I think that's a misprint for $2b_2'\sinh(n\pi(1-y))$.

Remember that
$$\sinh(x) \equiv \frac{e^x - e^{-x}}{2}.$$

Got it! Thank you very much~~!