Mathematical solution
In mathematics , and specifically partial differential equations (PDEs), d´Alembert's formula is the general solution to the one-dimensional wave equation :
u t t − c 2 u x x = 0 , u ( x , 0 ) = g ( x ) , u t ( x , 0 ) = h ( x ) , {\displaystyle u_{tt}-c^{2}u_{xx}=0,\,u(x,0)=g(x),\,u_{t}(x,0)=h(x),} for − ∞ < x < ∞ , t > 0 {\displaystyle -\infty <x<\infty ,\,\,t>0}
It is named after the mathematician Jean le Rond d'Alembert , who derived it in 1747 as a solution to the problem of a vibrating string .
Details The characteristics of the PDE are x ± c t = c o n s t {\displaystyle x\pm ct=\mathrm {const} } (where ± {\displaystyle \pm } sign states the two solutions to quadratic equation), so we can use the change of variables μ = x + c t {\displaystyle \mu =x+ct} (for the positive solution) and η = x − c t {\displaystyle \eta =x-ct} (for the negative solution) to transform the PDE to u μ η = 0 {\displaystyle u_{\mu \eta }=0} . The general solution of this PDE is u ( μ , η ) = F ( μ ) + G ( η ) {\displaystyle u(\mu ,\eta )=F(\mu )+G(\eta )} where F {\displaystyle F} and G {\displaystyle G} are C 1 {\displaystyle C^{1}} functions. Back in x , t {\displaystyle x,t} coordinates,
u ( x , t ) = F ( x + c t ) + G ( x − c t ) {\displaystyle u(x,t)=F(x+ct)+G(x-ct)} u {\displaystyle u} is C 2 {\displaystyle C^{2}} if F {\displaystyle F} and G {\displaystyle G} are C 2 {\displaystyle C^{2}} .This solution u {\displaystyle u} can be interpreted as two waves with constant velocity c {\displaystyle c} moving in opposite directions along the x-axis.
Now consider this solution with the Cauchy data u ( x , 0 ) = g ( x ) , u t ( x , 0 ) = h ( x ) {\displaystyle u(x,0)=g(x),u_{t}(x,0)=h(x)} .
Using u ( x , 0 ) = g ( x ) {\displaystyle u(x,0)=g(x)} we get F ( x ) + G ( x ) = g ( x ) {\displaystyle F(x)+G(x)=g(x)} .
Using u t ( x , 0 ) = h ( x ) {\displaystyle u_{t}(x,0)=h(x)} we get c F ′ ( x ) − c G ′ ( x ) = h ( x ) {\displaystyle cF'(x)-cG'(x)=h(x)} .
We can integrate the last equation to get c F ( x ) − c G ( x ) = ∫ − ∞ x h ( ξ ) d ξ + c 1 . {\displaystyle cF(x)-cG(x)=\int _{-\infty }^{x}h(\xi )\,d\xi +c_{1}.}
Now we can solve this system of equations to get F ( x ) = − 1 2 c ( − c g ( x ) − ( ∫ − ∞ x h ( ξ ) d ξ + c 1 ) ) {\displaystyle F(x)={\frac {-1}{2c}}\left(-cg(x)-\left(\int _{-\infty }^{x}h(\xi )\,d\xi +c_{1}\right)\right)} G ( x ) = − 1 2 c ( − c g ( x ) + ( ∫ − ∞ x h ( ξ ) d ξ + c 1 ) ) . {\displaystyle G(x)={\frac {-1}{2c}}\left(-cg(x)+\left(\int _{-\infty }^{x}h(\xi )d\xi +c_{1}\right)\right).}
Now, using u ( x , t ) = F ( x + c t ) + G ( x − c t ) {\displaystyle u(x,t)=F(x+ct)+G(x-ct)}
d'Alembert's formula becomes: u ( x , t ) = 1 2 [ g ( x − c t ) + g ( x + c t ) ] + 1 2 c ∫ x − c t x + c t h ( ξ ) d ξ . {\displaystyle u(x,t)={\frac {1}{2}}\left[g(x-ct)+g(x+ct)\right]+{\frac {1}{2c}}\int _{x-ct}^{x+ct}h(\xi )\,d\xi .}
Generalization for inhomogeneous canonical hyperbolic differential equations The general form of an inhomogeneous canonical hyperbolic type differential equation takes the form of: u t t − c 2 u x x = f ( x , t ) , u ( x , 0 ) = g ( x ) , u t ( x , 0 ) = h ( x ) , {\displaystyle u_{tt}-c^{2}u_{xx}=f(x,t),\,u(x,0)=g(x),\,u_{t}(x,0)=h(x),} for − ∞ < x < ∞ , t > 0 , f ∈ C 2 ( R 2 , R ) {\displaystyle -\infty <x<\infty ,\,\,t>0,f\in C^{2}(\mathbb {R} ^{2},\mathbb {R} )} .
All second order differential equations with constant coefficients can be transformed into their respective canonic forms . This equation is one of these three cases: Elliptic partial differential equation , Parabolic partial differential equation and Hyperbolic partial differential equation .
The only difference between a homogeneous and an inhomogeneous (partial) differential equation is that in the homogeneous form we only allow 0 to stand on the right side (f ( x , t ) = 0 {\displaystyle f(x,t)=0} ), while the inhomogeneous one is much more general, as in f ( x , t ) {\displaystyle f(x,t)} could be any function as long as it's continuous and can be continuously differentiated twice.
The solution of the above equation is given by the formula: u ( x , t ) = 1 2 ( g ( x + c t ) + g ( x − c t ) ) + 1 2 c ∫ x − c t x + c t h ( s ) d s + 1 2 c ∫ 0 t ∫ x − c ( t − τ ) x + c ( t − τ ) f ( s , τ ) d s d τ . {\displaystyle u(x,t)={\frac {1}{2}}{\bigl (}g(x+ct)+g(x-ct){\bigr )}+{\frac {1}{2c}}\int _{x-ct}^{x+ct}h(s)\,ds+{\frac {1}{2c}}\int _{0}^{t}\int _{x-c(t-\tau )}^{x+c(t-\tau )}f(s,\tau )\,ds\,d\tau .}
If g ( x ) = 0 {\displaystyle g(x)=0} , the first part disappears, if h ( x ) = 0 {\displaystyle h(x)=0} , the second part disappears, and if f ( x ) = 0 {\displaystyle f(x)=0} , the third part disappears from the solution, since integrating the 0-function between any two bounds always results in 0.
See also
Notes