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Seperation of Variables (double check please :)
I have a final coming up, and I want to make sure I have this method down.
Q: For the second-order wave equation u_{tt}=u_{xx}, the substitution of u=A(x)B(t) will give second-order equations for A nd B when the x and t variables are seperated. From B''/B=A''/A=\omega^2, find all solutions of the form u=A(x)B(t)
Assume: u(x,t)=A(x)B(t)
\frac{A(x)B''(t)}{A(x)B(t)}=\frac{A''(x)B(t)}{A(x)B(t)}
\frac{B''}{B}=\frac{A''}{A}=-\omega^2 is a second order ODE of the form:
B''+\omega^2 B = 0
Solving yields:
B(t)=c_1 \cos \omega t + c_2 \sin \omega t with the assumption that \omega > 0
and:
A(x)=d_1 \cos \omega x + d_2 \sin \omega x
therefore:
u(x,t) = A(x)B(t)= (c_1 \cos \omega t + c_2 \sin \omega t)(d_1 \cos \omega x + d_2 \sin \omega x)
And this is simply all the solutions right? It seems really straightforward, but sometimes when I think it is... it totally isn't. Thanks
I have a final coming up, and I want to make sure I have this method down.
Q: For the second-order wave equation u_{tt}=u_{xx}, the substitution of u=A(x)B(t) will give second-order equations for A nd B when the x and t variables are seperated. From B''/B=A''/A=\omega^2, find all solutions of the form u=A(x)B(t)
Assume: u(x,t)=A(x)B(t)
\frac{A(x)B''(t)}{A(x)B(t)}=\frac{A''(x)B(t)}{A(x)B(t)}
\frac{B''}{B}=\frac{A''}{A}=-\omega^2 is a second order ODE of the form:
B''+\omega^2 B = 0
Solving yields:
B(t)=c_1 \cos \omega t + c_2 \sin \omega t with the assumption that \omega > 0
and:
A(x)=d_1 \cos \omega x + d_2 \sin \omega x
therefore:
u(x,t) = A(x)B(t)= (c_1 \cos \omega t + c_2 \sin \omega t)(d_1 \cos \omega x + d_2 \sin \omega x)
And this is simply all the solutions right? It seems really straightforward, but sometimes when I think it is... it totally isn't. Thanks
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