wave equation

[little scientist]

Hello. can we help me in this problem?

 

[Dr.Peterson]

I don't know. It's all Greek to me.

Can you help us by translating the problem (including words you didn't even show in Greek), and telling us where you need help?

Read this, and then do what we ask: https://www.freemathhelp.com/forum/threads/guidelines-summary.112086/

 

[Romsek]

usually you start by letting \(\displaystyle u(x,y)=\rho(x)\psi(y)\) and then solve the resulting two eigenvalue problems.

Does any of that sound familiar?

 

[little scientist]

solve the problem
when c>0
and g,h are sufficiently smooth functions that satisfy appropriate conditions of compatibility at x = 0

 

[little scientist]

usually you start by letting \(\displaystyle u(x,y)=\rho(x)\psi(y)\) and then solve the resulting two eigenvalue problems.

Does any of that sound familiar?

we can't used this method because x>0

 

[Romsek]

[MATH]u(x,t) = \rho(x)\psi(t)\\ u_{xx}(x,t) = \rho''(x)\psi(t)\\ u_{tt}(x,t) = \rho(x) \psi''(t)[/MATH]
[MATH]\rho(x)\psi''(t) - c^2 \rho''(x) \psi(t)=0\\ c^2 \dfrac{\rho''(x)}{\rho(x)} = \dfrac{\psi''(t)}{\psi(t)}= \lambda^2[/MATH]
[MATH]\rho''(x) - \dfrac{\lambda^2}{c^2}\rho(x) = 0\\ \psi''(t) - \lambda^2 \psi(t) = 0[/MATH]
[MATH]\rho(x) = c_1 e^{\frac{\lambda}{c}x}+c_2 e^{-\frac{\lambda}{c}x}\\ \psi(t) = c_3 e^{\lambda t}+c_4 e^{-\lambda t}[/MATH]
Using these as a starting point you then find the partials and select the constants to satisfy the boundary and initial conditions.
You've never seen any of this?

 

[little scientist]

i now the method but i'm 99% sure this method can't be used on this problem because x>0. we need the method of D'Alembert. but i don't now what can i do with this side condition.

 

[Romsek]

i now the method but i'm 99% sure this method can't be used on this problem because x>0. we need the method of D'Alembert. but i don't now what can i do with this side condition.

Sounds like you know more about it than I do then w/o going back to my books. Good luck!

 

[HallsofIvy]

"D'Alembert's method" looks for solutions of the form u(x- ct)+ v(x+ ct) for unknown functions u and v (we can think of "u(x- ct)" as a wave in shape u moving to the right at speed c and "v(x+ ct)" as a wave in shape v moving to the left at speed c. D'Alembert's method gives the solution in terms of "traveling waves" where "separation of variables" gives it in terms of "standing waves".)

It is easy to show that y(x,t)= u(x- ct)+ v(x+ ct) satisfies the wave equation, \(\displaystyle y_{xx}= c^2y_{tt}\) for any twice-differentiable u and v. The problem is to determine u and v so that y satisfies the boundary and initial conditions.

Here y(x, 0)= u(x)+ v(x)= g(x) and y_t(x, 0)= -cu'(x)+ cv'(x)= f(x). -u'(x)+ v'(x)= f(x)/c. Differentiating the first equation, u'(x)+ v'(x)= g'(x). Adding. 2v'(x)= f'(x)/c+ g'(x) so that v'= (f'(x)+ cg'(x))/c and v is an anti-derivative of that. Then u(x)= g(x)- v(x).

(I see no reason why "x> 0" should mean you cannot use either method. Use "D'Alembert's method" or "separation of variables" ignoring x> 0, then restrict x in the statement of the solution.)