Imposing boundary condition and normalization on an ODE

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I want to use DSolve to solve a differential equation while imposing a boundary condition and normalization.

How can I do that?

Let's take for example a simple heat-equation: $y''[x]=a y[x]$.
Easy to solve:

DSolve[y''[x] == a y[x], y[x], x]

I can also impose a no-flux condition at position L, $nabla rho|_L=0$

DSolve[y''[x] == a y[x], y'[L] == 0, y[x], x]

but I cannot (or don't know how to) impose a normalization condition, $int_0^Lrho(x)dx=1$

DSolve[y''[x] == a y[x], y'[L] == 0, Integrate[y[x], x, 0, L] == 1,

When I evaluate the above expression, I obtain the following error:

There are fewer dependent variables than equations, so the system is overdetermined"

But that is something I can do by hand. What is the problem? Is it a bug or am I misunderstanding something?

differential-equations boundary-conditions

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edited 1 hour ago

m_goldberg

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asked 1 hour ago

user134414

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up vote
4
down vote

favorite

I want to use DSolve to solve a differential equation while imposing a boundary condition and normalization.

How can I do that?

Let's take for example a simple heat-equation: $y''[x]=a y[x]$.
Easy to solve:

DSolve[y''[x] == a y[x], y[x], x]

I can also impose a no-flux condition at position L, $nabla rho|_L=0$

DSolve[y''[x] == a y[x], y'[L] == 0, y[x], x]

but I cannot (or don't know how to) impose a normalization condition, $int_0^Lrho(x)dx=1$

DSolve[y''[x] == a y[x], y'[L] == 0, Integrate[y[x], x, 0, L] == 1,

When I evaluate the above expression, I obtain the following error:

There are fewer dependent variables than equations, so the system is overdetermined"

But that is something I can do by hand. What is the problem? Is it a bug or am I misunderstanding something?

differential-equations boundary-conditions

share|improve this question

edited 1 hour ago

m_goldberg

83.2k870191

asked 1 hour ago

user134414

211

New contributor

user134414 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
4
down vote

favorite

up vote
4
down vote

favorite

I want to use DSolve to solve a differential equation while imposing a boundary condition and normalization.

How can I do that?

Let's take for example a simple heat-equation: $y''[x]=a y[x]$.
Easy to solve:

DSolve[y''[x] == a y[x], y[x], x]

I can also impose a no-flux condition at position L, $nabla rho|_L=0$

DSolve[y''[x] == a y[x], y'[L] == 0, y[x], x]

but I cannot (or don't know how to) impose a normalization condition, $int_0^Lrho(x)dx=1$

DSolve[y''[x] == a y[x], y'[L] == 0, Integrate[y[x], x, 0, L] == 1,

When I evaluate the above expression, I obtain the following error:

There are fewer dependent variables than equations, so the system is overdetermined"

But that is something I can do by hand. What is the problem? Is it a bug or am I misunderstanding something?

differential-equations boundary-conditions

share|improve this question

edited 1 hour ago

m_goldberg

83.2k870191

asked 1 hour ago

user134414

211

New contributor

user134414 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

I want to use DSolve to solve a differential equation while imposing a boundary condition and normalization.

How can I do that?

Let's take for example a simple heat-equation: $y''[x]=a y[x]$.
Easy to solve:

DSolve[y''[x] == a y[x], y[x], x]

I can also impose a no-flux condition at position L, $nabla rho|_L=0$

DSolve[y''[x] == a y[x], y'[L] == 0, y[x], x]

but I cannot (or don't know how to) impose a normalization condition, $int_0^Lrho(x)dx=1$

DSolve[y''[x] == a y[x], y'[L] == 0, Integrate[y[x], x, 0, L] == 1,

When I evaluate the above expression, I obtain the following error:

There are fewer dependent variables than equations, so the system is overdetermined"

But that is something I can do by hand. What is the problem? Is it a bug or am I misunderstanding something?

differential-equations boundary-conditions

differential-equations boundary-conditions

share|improve this question

edited 1 hour ago

m_goldberg

83.2k870191

asked 1 hour ago

user134414

211

New contributor

user134414 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 1 hour ago

m_goldberg

83.2k870191

asked 1 hour ago

user134414

211

New contributor

user134414 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

share|improve this question

edited 1 hour ago

m_goldberg

83.2k870191

edited 1 hour ago

m_goldberg

83.2k870191

edited 1 hour ago

m_goldberg

83.2k870191

83.2k870191

asked 1 hour ago

user134414

211

New contributor

user134414 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

user134414

211

asked 1 hour ago

user134414

211

211

New contributor

user134414 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

user134414 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

user134414 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
5
down vote

You can solve your problem by introducing a second ode (defining the antiderivative):

sol = 
 DSolve[y''[x] == a y[x], Y'[x] == y[x], y'[L] == 0, Y[L] == 1, y, Y, x][[1]]

y -> 
 Function[x, 
 (E^(-Sqrt[a] x) (E^(2 Sqrt[a] L) + E^(2 Sqrt[a] x)) C[1])/(1 + E^(2 Sqrt[a] L))], 
 Y -> 
 Function[x, 
 (E^(-Sqrt[a] x) 
 (Sqrt[a] E^(Sqrt[a] x) +Sqrt[a] E^(2 Sqrt[a] L + Sqrt[a] x) - 
 E^(2 Sqrt[a] L) C[1] + E^(2 Sqrt[a] x) C[1])) /
 (Sqrt[a] (1 + E^(2 Sqrt[a] L)))]

share|improve this answer

edited 26 mins ago

m_goldberg

83.2k870191

answered 1 hour ago

Ulrich Neumann

5,330413

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote

You can solve your problem by introducing a second ode (defining the antiderivative):

sol = 
 DSolve[y''[x] == a y[x], Y'[x] == y[x], y'[L] == 0, Y[L] == 1, y, Y, x][[1]]

y -> 
 Function[x, 
 (E^(-Sqrt[a] x) (E^(2 Sqrt[a] L) + E^(2 Sqrt[a] x)) C[1])/(1 + E^(2 Sqrt[a] L))], 
 Y -> 
 Function[x, 
 (E^(-Sqrt[a] x) 
 (Sqrt[a] E^(Sqrt[a] x) +Sqrt[a] E^(2 Sqrt[a] L + Sqrt[a] x) - 
 E^(2 Sqrt[a] L) C[1] + E^(2 Sqrt[a] x) C[1])) /
 (Sqrt[a] (1 + E^(2 Sqrt[a] L)))]

share|improve this answer

edited 26 mins ago

m_goldberg

83.2k870191

answered 1 hour ago

Ulrich Neumann

5,330413

add a comment | 

up vote
5
down vote

You can solve your problem by introducing a second ode (defining the antiderivative):

sol = 
 DSolve[y''[x] == a y[x], Y'[x] == y[x], y'[L] == 0, Y[L] == 1, y, Y, x][[1]]

y -> 
 Function[x, 
 (E^(-Sqrt[a] x) (E^(2 Sqrt[a] L) + E^(2 Sqrt[a] x)) C[1])/(1 + E^(2 Sqrt[a] L))], 
 Y -> 
 Function[x, 
 (E^(-Sqrt[a] x) 
 (Sqrt[a] E^(Sqrt[a] x) +Sqrt[a] E^(2 Sqrt[a] L + Sqrt[a] x) - 
 E^(2 Sqrt[a] L) C[1] + E^(2 Sqrt[a] x) C[1])) /
 (Sqrt[a] (1 + E^(2 Sqrt[a] L)))]

share|improve this answer

edited 26 mins ago

m_goldberg

83.2k870191

answered 1 hour ago

Ulrich Neumann

5,330413

add a comment | 

up vote
5
down vote

up vote
5
down vote

You can solve your problem by introducing a second ode (defining the antiderivative):

sol = 
 DSolve[y''[x] == a y[x], Y'[x] == y[x], y'[L] == 0, Y[L] == 1, y, Y, x][[1]]

y -> 
 Function[x, 
 (E^(-Sqrt[a] x) (E^(2 Sqrt[a] L) + E^(2 Sqrt[a] x)) C[1])/(1 + E^(2 Sqrt[a] L))], 
 Y -> 
 Function[x, 
 (E^(-Sqrt[a] x) 
 (Sqrt[a] E^(Sqrt[a] x) +Sqrt[a] E^(2 Sqrt[a] L + Sqrt[a] x) - 
 E^(2 Sqrt[a] L) C[1] + E^(2 Sqrt[a] x) C[1])) /
 (Sqrt[a] (1 + E^(2 Sqrt[a] L)))]

share|improve this answer

edited 26 mins ago

m_goldberg

83.2k870191

answered 1 hour ago

Ulrich Neumann

5,330413

You can solve your problem by introducing a second ode (defining the antiderivative):

sol = 
 DSolve[y''[x] == a y[x], Y'[x] == y[x], y'[L] == 0, Y[L] == 1, y, Y, x][[1]]

y -> 
 Function[x, 
 (E^(-Sqrt[a] x) (E^(2 Sqrt[a] L) + E^(2 Sqrt[a] x)) C[1])/(1 + E^(2 Sqrt[a] L))], 
 Y -> 
 Function[x, 
 (E^(-Sqrt[a] x) 
 (Sqrt[a] E^(Sqrt[a] x) +Sqrt[a] E^(2 Sqrt[a] L + Sqrt[a] x) - 
 E^(2 Sqrt[a] L) C[1] + E^(2 Sqrt[a] x) C[1])) /
 (Sqrt[a] (1 + E^(2 Sqrt[a] L)))]

share|improve this answer

edited 26 mins ago

m_goldberg

83.2k870191

answered 1 hour ago

Ulrich Neumann

5,330413

share|improve this answer

share|improve this answer

edited 26 mins ago

m_goldberg

83.2k870191

edited 26 mins ago

m_goldberg

83.2k870191

edited 26 mins ago

m_goldberg

83.2k870191

83.2k870191

answered 1 hour ago

Ulrich Neumann

5,330413

answered 1 hour ago

Ulrich Neumann

5,330413

answered 1 hour ago

Ulrich Neumann

5,330413

5,330413

add a comment | 

add a comment | 

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