Partial DIfferential Equations

[stunner5000pt]

Solve using separation of variables and find particular solution
\frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} - u = 0 \ for\ 0 <x<1, t >0
\frac{\partial u}{\partial t} (x,0) = 0
0,t) = u(1,t) = 0
to assume u(x,t) = X(x) T(t)
then \frac{X''}{X} = \frac{T'' - T}{T} = \lambda
solving for X(x) yields X(x) = C_{1} \cos{\sqrt{\lambda} x} + C_{2} \sin{\sqrt{\lambda} x}
and i get C1 = 0 and C2 assumed to be 1 and \lambda = \sqrt{n \pi}
now for T(t),
T'' - (1 + n^2 pi^2) T = 0
now here is where i am stuck...
is the answer just the
C_{1} \cos{\sqrt{1 + n^2 \pi^2} t}+ C_{2} \sin{\sqrt{1 + n^2 \pi^2} t}
i'm not quite sure what to do after this
the answer in my book is \cos{\sqrt{n^2 \pi^2 -1}} t} \sin (n \pi x)
i got the x part right... but what about the t part

 

[Physics Monkey]

Hi stunner5000pt,

Sorry, I got all confused there, and I read your differential equation backwards. Your problem is simply that \lambda < 0 in order to obtain sines and cosines and meet the boundary condition for the x part. In reality, you should have - \lambda = n^2 \pi^2 and not \lambda = n^2 \pi^2.

Once you have solved the t part, apply the boundary condition you haven't used yet.

 

[stunner5000pt]

the DE is
\frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} - u = 0

2 negative signs

what is wrong with the equation?

 

[Physics Monkey]

Yeah, my fault, I read it as
<br /> \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial t^2} - u = 0,<br />
and then realized my mistake as I was thinking about the solution.

 

[stunner5000pt]

uh.. the t part was the part at which i was stuck.

the condition given is \frac{\partial u}{\partial t} (x,0) = 0
im not quite sure howto relate this to what i have already...

 

[Physics Monkey]

First, as I said, your general solution to the t part in your first post is incorrect, it should be
<br /> T(t) = C_1 \sin{(\sqrt{n^2 \pi^2 -1} \,t )} + C_2 \cos{(\sqrt{n^2 \pi^2 - 1}\, t)}<br />
Now, regarding the boundary condition, what do you mean you don't know how to relate it to what you have? Just take the derivative of u(x,t) = X(x)T(t) with respect to t and set it equal to zero for t=0.

 

[stunner5000pt]

why should the generla solution for T be that?

 

[Physics Monkey]

Read my post 2 above, and try to figure it out.

 

[stunner5000pt]

so why should - \lambda = n^2 \pi^2
i m not sure about the 'conditions on the x part' to which you are referring. And wouldn't the solution become imaginary if that was the case?

 

[Physics Monkey]

What is the solution to
<br /> \frac{1}{X}\frac{d^2 X}{dx^2} = \lambda,<br />
isn't it
<br /> X = C_1 e^{\sqrt{\lambda} x} + C_2 e^{-\sqrt{\lambda} x}.<br />
you get real exponentials and not real sines and cosines if \lambda &gt; 0. If you try applying the boundary conditions with real exponentials then you find C_1 = C_2 = 0. Take it from here.

 

[stunner5000pt]

i get it now lambda mustb e imaginary iof you are going to use Eler's identity and expand E into cos and sin. ANd thus carries onto the T(t) part where C1-0 and C2 =1 and the solution is what appears in the book

thank you for your help! I appreciate it!