Deriving d'Alemberts solution - Boundary conditions

In summary, the solution to the given partial differential equation can be obtained using Fourier transform and incorporating the boundary conditions leads to the final solution u(x,t) = (f(x+t) + f(x-t))/2.
  • #1
divB
87
0
Hi,

I shall show (using Fourier transform) that the solution to

[tex]
\frac{\partial^2 u(x,t)}{\partial t^2} = \frac{\partial^2 u(x,t)}{\partial x^2}\\
u(x,0) = f(x) \\
u_t(x,0) = 0
[/tex]

is

[tex]
u(x,t) = (f(x+t) + f(x-t))/2
[/tex]

I got it almost: Taking the Fourier transform in the variable x (leaving t constant) and the derivative theorem gives:

[tex]
\hat{u}_{tt}(s,t) + 4\pi^2 s^2 \hat{u}(s,t) = 0
[/tex]

And the solution to this differential equation is:

[tex]
\hat{u}(s,t) = A e^{j2\pi s t} + B e^{-j2\pi s t}
[/tex]

However, now I am stuck in incorporating the boundary conditions and I do not see why A and B should become [itex]f(x)=u(x,0)[/itex].

My approach was: Since u(x,t) was transferred to Fourier domain via x rather than t, [itex]u(s,0) = \hat{u}(x,0)[/itex]. Derivating the solution for t and filling t=0 gives A=B. But I neither see the factor of 1/2 there, nore why A and B should be f(x). Can anyone give me a pointer?

Thanks!
 
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  • #2
Got it; finally!
It is so trivial from where I was: A=B was correct, it is a constant regarding t but not s, so A(s)=B(s).

The other "boundary" condition is u(x,0)=f(x), so when transforming back my solution: u(x,t) = A(x+t)+B(x-t) with t=0: u(x,0)=A(x)+B(x) which means that A(x)=B(x)=f(x)/2
 

Related to Deriving d'Alemberts solution - Boundary conditions

What is d'Alembert's solution?

D'Alembert's solution is a mathematical method used to find solutions to partial differential equations. It was developed by French mathematician Jean le Rond d'Alembert in the 18th century.

How is d'Alembert's solution derived?

D'Alembert's solution is derived by using the method of characteristics, which involves converting the partial differential equation into a set of ordinary differential equations. These equations are then solved to find the solution to the original partial differential equation.

What are boundary conditions?

Boundary conditions are constraints placed on the solution of a differential equation. They are used to determine a unique solution and can be specified at the boundaries of the domain or at specific points within the domain.

Why are boundary conditions important in deriving d'Alembert's solution?

Boundary conditions are important in deriving d'Alembert's solution because they help to determine the constants of integration that are needed to find the solution. Without boundary conditions, the solution would not be unique and may not accurately represent the physical system being studied.

What types of boundary conditions are commonly used in d'Alembert's solution?

The most commonly used boundary conditions in d'Alembert's solution are initial conditions and boundary value conditions. Initial conditions specify the solution at a given point in time, while boundary value conditions specify the solution at the boundaries of the domain.

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