Math Editor

PDE

Method of separation of variables - general approach

1. Assume the unknown function u(x,y)=X(x)Y(y)
2. Use the PDE to find two ODEs for X and Y
3. Solve the ODE for which boundary conditions are given
4. Insert the boundary conditions to reduce the number of constants
5. Solve the other ODE
6. Write down the general solution<br>
7. Insert the initial conditions to get the other constants
8. Substitute the constants in the general solution

Homogeneous Heat equation uxx​=k1​ut​⟺0<x<l

1. Assume that u=XT
2. Find the ODEs
   Substituting into the PDE X′′T=k1​XT′
   Separating the variables XX′′​=k1​TT′​=K
   Choosing negative signed separation constant K=−λ2 to get two ODEs
   T′=−λ2kT
   X′′=−λ2X
3. Solve the boundary conditioned ODE
   X=Acosλx+Bsinλx
4. 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​
5. Solve the other ODE
   T=Ce−λ2kt
6. The general solution is
   - Dirichlet: u(x,t)=Esinλx×e−λ2kt
   - Neumann: u(x,t)=X0​×T0​+Dcosλx×e−λ2kt
7. 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​
8. Finally we write the solution by substituting D,E,λ in the general solution
   - Dirichlet: u=n=1∑∞​bn​sinlnπx​×e−(nπ/l)2kt
   - Neumann: u=2a0​​+n=1∑∞​an​coslnπx​×e−(nπ/l)2kt

Non-Homogeneous Heat Equation ut​=uxx​+f(x)

1. 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)
2. 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
3. solve the homogeneous heat equation w(x,t)
   notice that the initial condition becomes w(x,0)=u(x,0)−v(x)
4. write the solution u(x,t)=w(x,t)+v(x)

Wave equation uxx​=c21​utt​⟺0<x<l

1. Assume that u=X⁢⁢⁢T
2. Find the ODEs
   Substituting into the PDE X′′T=c21​XT′′
   Separating the variables XX′′​=c21​TT′′​=K
   Choosing negative signed separation constant K=−λ2 to get two ODEs
   T′′=−λ2c2T
   X′′=−λ2X
3. Solve the boundary conditioned ODE
   X=Acosλx+Bsinλx
4. 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​
5. Solve the other ODE
   T=Dcosλct+Ecosλct
6. The general solution is
   u(x,t)=(Fcosλct+Gsinλct)sinλx
7. 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​​​
8. Finally we write the solution by substituting F,G,λ in
   u(x,t)=(n=1∑∞​bn​coslnπ​ct+n=1∑∞​λcbn​​sinlnπ​ct)sinlnπx​

Laplace’s equation uxx​+uyy​=0⟺0<x<l&0<y<h

1. Assume that u=XY
2. 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
3. Solve the homogeneous boundary conditioned ODE
   - x homo:
   X=Acosλx+Bsinλx
   -y homo
   Y=Acosλy+Bsinλy​
4. 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​
5. Solve the other ODE
   - x homo:
   Y=Dcoshλy+Esinhλy
   - y homo:
   X=Dcoshλx+Esinhλx
6. The general solution is
   - x homo: u(x,y)=(Fcoshλy+Gsinhλysinλx
   - y homo: u(x,y)=(Fcoshλx+Gsinhλxsinλy​
7. 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​​
8. Finally we write the solution by substituting F,G,λ
   - x homo: u(x,y)=n=1∑∞​sinhnπh/lbn​​sinhlnπy​×sinlnπx​
   - y homo: u(x,y)=n=1∑∞​sinhnπl/hbn​​sinhhnπx​×sinhnπy​