Need help with solving PDE

  • Thread starter hbl2012
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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=a1eμx+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''-n22Y=0.
The general solution is Y=b1en∏y+b2e-n∏y.
Then since Y(1)=0, b1en∏+b2e-n∏=0.


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

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

Answers and Replies

  • #2
pasmith
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[...]

Then we consider Y''-λY=0, i.e. Y''-n22Y=0.
The general solution is Y=b1en∏y+b2e-n∏y.
Then since Y(1)=0, b1en∏+b2e-n∏=0.
Now, here is my question.
Why b1+b2e-2n∏=0?
Divide both sides of the previous equation by [itex]e^{n\pi}[/itex].

and How it change in the black font? Yh(y)=-b2e-2n∏en∏y+b2e-n∏y
=-b2e-n∏e-n∏en∏y+b2e-n∏en∏e-n∏y
=-b2'e-n∏(1-y)+b2'en∏(1-y)
=b2'(en∏(1-y)-e-n∏(1-y))
=2b2''sin(hn∏(1-y))
I think that's a misprint for [itex]2b_2'\sinh(n\pi(1-y))[/itex].

Remember that
[tex]\sinh(x) \equiv \frac{e^x - e^{-x}}{2}.[/tex]
 
  • #3
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Divide both sides of the previous equation by [itex]e^{n\pi}[/itex].



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

Remember that
[tex]\sinh(x) \equiv \frac{e^x - e^{-x}}{2}.[/tex]
Got it! Thank you very much~~!!!
 

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