Method of separation of variables - general approach
Assume the unknown function u(x,y)=X(x)Y(y)
Use the PDE to find two ODEs for X and Y
Solve the ODE for which boundary conditions are given
Insert the boundary conditions to reduce the number of constants
Solve the other ODE
Write down the general solution<br>
Insert the initial conditions to get the other constants
Substitute the constants in the general solution
Homogeneous Heat equation uxx=k1ut⟺0<x<l
Assume that u=XT
Find the ODEs Substituting into the PDE X′′T=k1XT′ Separating the variables XX′′=k1TT′=K Choosing negative signed separation constant K=−λ2 to get two ODEs T′=−λ2kT X′′=−λ2X
Solve the boundary conditioned ODE X=Acosλx+Bsinλx
Insert the boundary conditions - Dirichlet: The boundary condition u(0,t)=0 gives A=0 - Neumann: The boundary condition ux(0,t)=0 gives B=0 The other boundary condition u(l,t)=0 or ux(l,t)=0 gives λ=lnπ
Solve the other ODE T=Ce−λ2kt
The general solution is - Dirichlet: u(x,t)=Esinλx×e−λ2kt - Neumann: u(x,t)=X0×T0+Dcosλx×e−λ2kt
Insert the initial condition u(x,0)=f(x) you'll need to expand f(x) into half-range Fourier series: - Dirichlet: Fourier sine which gives E=n=1∑∞bn - Neumann: Fourier cosine which gives X0=2a0 and D=n=1∑∞an
Finally we write the solution by substituting D,E,λ in the general solution - Dirichlet: u=n=1∑∞bnsinlnπx×e−(nπ/l)2kt - Neumann: u=2a0+n=1∑∞ancoslnπx×e−(nπ/l)2kt
Non-Homogeneous Heat Equation ut=uxx+f(x)
Assume that the solution consists of two parts: a non-homogeneous steady state time independent solution v(x) a homogeneous transient solution w(x,t)
find the steady state equation by substituting 0=v′′+f(x) and solve the exact equation by integrating twice calculate the two constants using the given boundary conditions
solve the homogeneous heat equation w(x,t) notice that the initial condition becomes w(x,0)=u(x,0)−v(x)
write the solution u(x,t)=w(x,t)+v(x)
Wave equation uxx=c21utt⟺0<x<l
Assume that u=XT
Find the ODEs Substituting into the PDE X′′T=c21XT′′ Separating the variables XX′′=c21TT′′=K Choosing negative signed separation constant K=−λ2 to get two ODEs T′′=−λ2c2T X′′=−λ2X
Solve the boundary conditioned ODE X=Acosλx+Bsinλx
Insert the boundary conditions The boundary condition u(0,t)=0 gives A=0 The other boundary condition u(l,t)=0 gives λ=lnπ
Solve the other ODE T=Dcosλct+Ecosλct
The general solution is u(x,t)=(Fcosλct+Gsinλct)sinλx
Insert the initial conditions - u(x,0)=f(x) and expand f(x) into Fourier sine which gives F=n=1∑∞bn - ut(x,0)=g(x) differentiate u(x,t) partially with respect to t and expand g(x) into Fourier sine which gives G=n=1∑∞λcbn
Finally we write the solution by substituting F,G,λ in u(x,t)=(n=1∑∞bncoslnπct+n=1∑∞λcbnsinlnπct)sinlnπx
Laplace’s equation uxx+uyy=0⟺0<x<l&0<y<h
Assume that u=XY
Find the ODEs Substituting into the PDE X′′Y+XY′′=0 Separating the variables - x homogeneous boundary conditions: XX′′=−YY′′=K - y homogeneous boundary conditions: YY′′=−XX′′=K Choosing negative signed separation constant K=−λ2 to get two ODEs - x homo: X′′=−λ2X Y′′=λ2Y - y homo Y′′=−λ2Y X′′=λ2X
Solve the homogeneous boundary conditioned ODE - x homo: X=Acosλx+Bsinλx -y homo Y=Acosλy+Bsinλy
Insert the homogeneous boundary conditions - x homo: The boundary condition u(0,y)=0 gives A=0 The other boundary condition u(l,y)=0 gives λ=lnπ - y homo: The boundary condition u(x,0)=0 gives A=0 The other boundary conditionu(x,h)=0 gives λ=hnπ
Solve the other ODE - x homo: Y=Dcoshλy+Esinhλy - y homo: X=Dcoshλx+Esinhλx
The general solution is - x homo: u(x,y)=(Fcoshλy+Gsinhλysinλx - y homo: u(x,y)=(Fcoshλx+Gsinhλxsinλy
Insert the non-homogeneous boundary conditions - x homo: u(x,0)=0 gives F=0 so u becomes u(x,y)=Gsinhλy×sinλx u(x,h)=g(x) gives G=n=1∑∞sinhλhbn - y homo: u(0,y)=0 gives F=0 so u becomes u(x,y)=Gsinhλx×sinλy u(l,0)=g(x) gives G=n=1∑∞sinhλlbn
Finally we write the solution by substituting F,G,λ - x homo: u(x,y)=n=1∑∞sinhnπh/lbnsinhlnπy×sinlnπx - y homo: u(x,y)=n=1∑∞sinhnπl/hbnsinhhnπx×sinhnπy
