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Separating variables in a PDE with multiple constants

How to solve this PDE?Solving one dimensonal PDESolving the Laplace equation in terms of exponential of hyperbolic trigonometric functionsPDE with periodic boundary conditionsSolving a PDE through separation of variablesPDE using separation of variablesUsing Laplace Transforms to solve a PDETextbook Example Solving Heat Conduction PDE Using Separation of VariablesProblem with satisfying Boundary conditions for 1D heat PDESolving PDE using separation of variables (Heat diffusion)

0

$begingroup$

My question is:

How do you use separation of variables on a PDE that has more than one constant in it?

All the examples I can find in my book/online only have one constant in it, like $$ fracpartial upartial t=kfracpartial^2
upartial x^2+kfracpartial^2 upartial y^2$$

Which then means only one of the resultant ODEs has $k$ in it. If there is more than one constant, what do you do? Still use separation of variables? My attempt is below the question.

The problem I am working on is:

Solve

$displaystyle fracpartial upartial t=k_1fracpartial^2
upartial x^2+k_2fracpartial^2 upartial y^2$

on a rectangle $(0 lt x lt L, 0 lt y lt H)$ subject to

$displaystyle u(0,y,t)=0, fracpartial upartial y(x,0,t)=0$

$displaystyle u(0,y,0)=alpha(x,y)$

$displaystyle u(L,y,t)=0, fracpartial upartial y(x,H,t)=0$

What I have done so far:

Let $u(x,t)=X(x)Y(y)T(t)$. Then using separation of variables we have:

(1) $displaystyle fracd Td t=-lambda(k_1+k_2)T, text IC: T(0)=alpha(x,y)$

(2) $displaystyle fracd^2Xdx^2=-mu frac(k_1+k_2)Xk_1, text BCs: X(0)=0, X(L)=0$

(3) $displaystyle fracd^2Ydy^2=-(lambda - mu) frac(k_1+k_2)Yk_2, text BCs: Y'(0)=0, Y'(H)=0$

pde boundary-value-problem heat-equation

share|cite|improve this question

edited Mar 12 at 6:02

Dylan

13.9k31127

asked Mar 11 at 20:13

LovesPeanutButterLovesPeanutButter

836

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What if you now solve these ODE's? The first gives an exponential, the second and third gives sines and cosines. That gives you $T,X$ and $Y$. And then check if $XYT$ is a solution?
  $endgroup$
  – Jens Wagemaker
  Mar 11 at 20:56
  

  
  
  
  

add a comment | 

0

$begingroup$

My question is:

How do you use separation of variables on a PDE that has more than one constant in it?

All the examples I can find in my book/online only have one constant in it, like $$ fracpartial upartial t=kfracpartial^2
upartial x^2+kfracpartial^2 upartial y^2$$

Which then means only one of the resultant ODEs has $k$ in it. If there is more than one constant, what do you do? Still use separation of variables? My attempt is below the question.

The problem I am working on is:

Solve

$displaystyle fracpartial upartial t=k_1fracpartial^2
upartial x^2+k_2fracpartial^2 upartial y^2$

on a rectangle $(0 lt x lt L, 0 lt y lt H)$ subject to

$displaystyle u(0,y,t)=0, fracpartial upartial y(x,0,t)=0$

$displaystyle u(0,y,0)=alpha(x,y)$

$displaystyle u(L,y,t)=0, fracpartial upartial y(x,H,t)=0$

What I have done so far:

Let $u(x,t)=X(x)Y(y)T(t)$. Then using separation of variables we have:

(1) $displaystyle fracd Td t=-lambda(k_1+k_2)T, text IC: T(0)=alpha(x,y)$

(2) $displaystyle fracd^2Xdx^2=-mu frac(k_1+k_2)Xk_1, text BCs: X(0)=0, X(L)=0$

(3) $displaystyle fracd^2Ydy^2=-(lambda - mu) frac(k_1+k_2)Yk_2, text BCs: Y'(0)=0, Y'(H)=0$

pde boundary-value-problem heat-equation

share|cite|improve this question

edited Mar 12 at 6:02

Dylan

13.9k31127

asked Mar 11 at 20:13

LovesPeanutButterLovesPeanutButter

836

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What if you now solve these ODE's? The first gives an exponential, the second and third gives sines and cosines. That gives you $T,X$ and $Y$. And then check if $XYT$ is a solution?
  $endgroup$
  – Jens Wagemaker
  Mar 11 at 20:56
  

  
  
  
  

add a comment | 

$begingroup$

My question is:

How do you use separation of variables on a PDE that has more than one constant in it?

All the examples I can find in my book/online only have one constant in it, like $$ fracpartial upartial t=kfracpartial^2
upartial x^2+kfracpartial^2 upartial y^2$$

Which then means only one of the resultant ODEs has $k$ in it. If there is more than one constant, what do you do? Still use separation of variables? My attempt is below the question.

The problem I am working on is:

Solve

$displaystyle fracpartial upartial t=k_1fracpartial^2
upartial x^2+k_2fracpartial^2 upartial y^2$

on a rectangle $(0 lt x lt L, 0 lt y lt H)$ subject to

$displaystyle u(0,y,t)=0, fracpartial upartial y(x,0,t)=0$

$displaystyle u(0,y,0)=alpha(x,y)$

$displaystyle u(L,y,t)=0, fracpartial upartial y(x,H,t)=0$

What I have done so far:

Let $u(x,t)=X(x)Y(y)T(t)$. Then using separation of variables we have:

(1) $displaystyle fracd Td t=-lambda(k_1+k_2)T, text IC: T(0)=alpha(x,y)$

(2) $displaystyle fracd^2Xdx^2=-mu frac(k_1+k_2)Xk_1, text BCs: X(0)=0, X(L)=0$

(3) $displaystyle fracd^2Ydy^2=-(lambda - mu) frac(k_1+k_2)Yk_2, text BCs: Y'(0)=0, Y'(H)=0$

pde boundary-value-problem heat-equation

share|cite|improve this question

edited Mar 12 at 6:02

Dylan

13.9k31127

asked Mar 11 at 20:13

LovesPeanutButterLovesPeanutButter

836

$endgroup$

share|cite|improve this question

edited Mar 12 at 6:02

Dylan

13.9k31127

asked Mar 11 at 20:13

LovesPeanutButterLovesPeanutButter

836

share|cite|improve this question

edited Mar 12 at 6:02

Dylan

13.9k31127

asked Mar 11 at 20:13

LovesPeanutButterLovesPeanutButter

836

edited Mar 12 at 6:02

Dylan

13.9k31127

edited Mar 12 at 6:02

Dylan

13.9k31127

asked Mar 11 at 20:13

LovesPeanutButterLovesPeanutButter

836

asked Mar 11 at 20:13

LovesPeanutButterLovesPeanutButter

836

1 Answer
1

active

oldest

votes

2

$begingroup$

You could also change the order of the assignment of variables. In
$$
fracT'(t)T(t)=k_1fracX''(x)X(x)+k_2fracY''(y)Y(y)
$$
all terms must be constants, so that you can set $fracX''(x)X(x)=-λ$, $fracY''(y)Y(y)=-μ$ and thus in consequence $fracT'(t)T(t)=-(k_1λ+k_2μ)$. I think the expressions for the constants in this form are less complex.

share|cite|improve this answer

answered Mar 11 at 21:08

LutzLLutzL

59.6k42057

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So using this method, the three ODEs to solve become: (1) $displaystyle X''+lambda X=0, text BCs: X(0)=0, X(L)=0$ (2) $displaystyle Y''+mu Y=0, text BCs: Y'(0)=0, Y'(H)=0$ (3) $displaystyle T'+(k_1 lambda +k_2 mu)T=0, text IC: T(0)=alpha(x,y)$ ?
  $endgroup$
  – LovesPeanutButter
  Mar 11 at 21:48
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes up to the last condition. The IC for $T$ is the Fourier coefficient corresponding to $(λ,μ)$ in the expansion of the initial function $α$ in the eigenbasis.
  $endgroup$
  – LutzL
  Mar 11 at 23:22
  

  
  
  
  

add a comment | 

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2

$begingroup$

You could also change the order of the assignment of variables. In
$$
fracT'(t)T(t)=k_1fracX''(x)X(x)+k_2fracY''(y)Y(y)
$$
all terms must be constants, so that you can set $fracX''(x)X(x)=-λ$, $fracY''(y)Y(y)=-μ$ and thus in consequence $fracT'(t)T(t)=-(k_1λ+k_2μ)$. I think the expressions for the constants in this form are less complex.

share|cite|improve this answer

answered Mar 11 at 21:08

LutzLLutzL

59.6k42057

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So using this method, the three ODEs to solve become: (1) $displaystyle X''+lambda X=0, text BCs: X(0)=0, X(L)=0$ (2) $displaystyle Y''+mu Y=0, text BCs: Y'(0)=0, Y'(H)=0$ (3) $displaystyle T'+(k_1 lambda +k_2 mu)T=0, text IC: T(0)=alpha(x,y)$ ?
  $endgroup$
  – LovesPeanutButter
  Mar 11 at 21:48
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes up to the last condition. The IC for $T$ is the Fourier coefficient corresponding to $(λ,μ)$ in the expansion of the initial function $α$ in the eigenbasis.
  $endgroup$
  – LutzL
  Mar 11 at 23:22
  

  
  
  
  

add a comment | 

2

$begingroup$

You could also change the order of the assignment of variables. In
$$
fracT'(t)T(t)=k_1fracX''(x)X(x)+k_2fracY''(y)Y(y)
$$
all terms must be constants, so that you can set $fracX''(x)X(x)=-λ$, $fracY''(y)Y(y)=-μ$ and thus in consequence $fracT'(t)T(t)=-(k_1λ+k_2μ)$. I think the expressions for the constants in this form are less complex.

share|cite|improve this answer

answered Mar 11 at 21:08

LutzLLutzL

59.6k42057

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So using this method, the three ODEs to solve become: (1) $displaystyle X''+lambda X=0, text BCs: X(0)=0, X(L)=0$ (2) $displaystyle Y''+mu Y=0, text BCs: Y'(0)=0, Y'(H)=0$ (3) $displaystyle T'+(k_1 lambda +k_2 mu)T=0, text IC: T(0)=alpha(x,y)$ ?
  $endgroup$
  – LovesPeanutButter
  Mar 11 at 21:48
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes up to the last condition. The IC for $T$ is the Fourier coefficient corresponding to $(λ,μ)$ in the expansion of the initial function $α$ in the eigenbasis.
  $endgroup$
  – LutzL
  Mar 11 at 23:22
  

  
  
  
  

add a comment | 

$begingroup$

You could also change the order of the assignment of variables. In
$$
fracT'(t)T(t)=k_1fracX''(x)X(x)+k_2fracY''(y)Y(y)
$$
all terms must be constants, so that you can set $fracX''(x)X(x)=-λ$, $fracY''(y)Y(y)=-μ$ and thus in consequence $fracT'(t)T(t)=-(k_1λ+k_2μ)$. I think the expressions for the constants in this form are less complex.

share|cite|improve this answer

answered Mar 11 at 21:08

LutzLLutzL

59.6k42057

$endgroup$

share|cite|improve this answer

answered Mar 11 at 21:08

LutzLLutzL

59.6k42057

answered Mar 11 at 21:08

LutzLLutzL

59.6k42057

answered Mar 11 at 21:08

LutzLLutzL

59.6k42057