This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: MATH 220: DUHAMEL’S PRINCIPLE Although we have solved only the homogeneous heat equation on R n , the same method employed there also solves the inhomogeneous PDE. As an application of these methods, let’s solve the heat equation on (0 , ∞ ) t × R n x : (1) u t- k Δ u = f, u (0 , x ) = φ ( x ) , with f ∈ C ∞ ([0 , ∞ ) t ; S ( R n x )), φ ∈ S ( R n ) (say) given. Namely, taking the partial Fourier transform in x , and writing F x u ( t, ξ ) = ˆ u ( t, ξ ), gives ∂ ˆ u ∂t ( t, ξ ) + k | ξ | 2 ˆ u ( t, ξ ) = ˆ f ( t, ξ ) , ˆ u (0 , ξ ) = ( F φ )( ξ ) . We again solve the ODE for each fixed ξ . To do so, we multiply through by e k | ξ | 2 t : e k | ξ | 2 t ∂ ˆ u ∂t ( t, ξ ) + e k | ξ | 2 t k | ξ | 2 ˆ u ( t, ξ ) = e k | ξ | 2 t ˆ f ( t, ξ ) , and realize that the left hand side is d dt parenleftBig e k | ξ | 2 t ˆ u ( t, ξ ) parenrightBig . Thus, integrating from t = 0 and using the fundamental theorem of calculus yields e k | ξ | 2 t ˆ u ( t, ξ )- ˆ u (0 , ξ ) = integraldisplay t e k | ξ | 2 s ˆ f ( s, ξ ) ds, so ˆ u ( t, ξ ) = e − k | ξ | 2 t ( F φ )( ξ ) + integraldisplay t e − k | ξ | 2 ( t − s ) ˆ f ( s, ξ ) ds. Finally, u ( t, x ) = F − 1 ξ parenleftbigg e − k | ξ | 2 t ( F φ )( ξ ) + integraldisplay t e − k | ξ | 2 ( t − s ) ˆ f ( s, ξ ) ds parenrightbigg = F − 1 ξ parenleftBig e − k | ξ | 2 t ( F φ )( ξ ) parenrightBig + integraldisplay t F − 1 parenleftBig e − k | ξ | 2 ( t − s ) ˆ f ( s, ξ ) parenrightBig ds = (4 πkt ) − n/ 2 integraldisplay R n e −| x − y | 2 / 4 kt φ ( y ) dy + integraldisplay t (4 πk ( t- s )) − n/ 2 integraldisplay R n e −| x − y | 2 / 4 k ( t − s ) f ( s, y ) dy ds. Note that the parts of the solution formula corresponding to the initial condition and the forcing term are very similar. In fact, let the solution operator of the homogeneous PDE u t- k Δ u = 0 , u (0 , x ) = φ ( x ) , be denoted by S ( t ), so ( S ( t ) φ )( x ) = u ( t, x ) . Then (4 πk ( t- s )) − n/ 2 integraldisplay R n e −| x − y | 2 / 4 k ( t − s ) f ( s, y ) dy = ( S ( t- s ) f s )( x ) , 1 2 where we let f s ( x ) = f ( s, x ) be the restriction of f to the time s slice. Thus, the solution formula for the inhomogeneous PDE, (1), takes the form (2) u ( t, x ) = ( S ( t ) φ )( x ) + integraldisplay t ( S ( t- s ) f s )( x ) ds. The fact that we can write the solution of the inhomogeneous PDE in terms of the solution of the Cauchy problem for the homogeneous PDE is called Duhamel’s principle . Physically one may think of (2) as follows. The expression S ( t- s ) f s is the solution of the heat equation at time t with initial condition f ( s, x ) imposed at time s . Thus, we think of the forcing as a superposition (namely, integral) of initial conditions given at times s between 0 (when the actual initial condition is imposed) and time t (when the solution is evaluated). Conversely, one could say that the initial condition amounts to a delta-distributional forcing,...
View Full Document
- Fall '10
- Boundary value problem, Partial differential equation, Boundary conditions