Heat equation and help with BC

piotrekkk888

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Joined
Oct 14, 2020
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2
Hello I have heat equation:
dT/dt=d^2T/dx^2
And:
T(t,0)=T(t,l)=2
T(0,x)=0

What kind of type BC is that?
And I don't know what should I do in separating variables process and then what to do with series (cos or sin?).

Thank you very much,
Piotr
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
22,405
Hello I have heat equation:
dT/dt=d^2T/dx^2
And:
T(t,0)=T(t,l)=2
T(0,x)=0

What kind of type BC is that? And I don't know what should I do in separating variables process and then what to do with series (cos or sin?).Thank you very much,
Piotr
I do not understand your question: What kind of type BC is that?

Please tell us - What are the different types of BC that you have learned about.

To solve the problem - start with the assumption:

T(t,x) = f(x) * g(t)

Any university level "elementary heat transfer" book will have solution to this (or very similar) problem worked out.

Please do some research and tell us what you found.
 

piotrekkk888

New member
Joined
Oct 14, 2020
Messages
2
Thank You for answer.
I know about Neumann, Dirichlet and mixed BCs.

Dirichlet BC: u(0,t) = g1(t), u(1,t) = g2(t),
and g1(t)=g2(t)=2 here?
my first steps with BC:
a)T(t,0)-2=0
b)T(t,L)-2=0
cause we need zero in separating variables
Separating variables:
left side of heat equation: X(x)Y'(t) is dT/dt=d/dt[X(x)Y(t)]
right side: X''(x)Y(t) is d^2T/dx^2=d^2/dx^2[X(x)Y(t)]
Both sides are equal:
X(x)Y'(t)=X''(x)Y(t)
X''(x)/X(x)=Y'(t)/Y(t)=-lambda
so
X''(x)+lambdaX(x)=0 and Y'(t)+lambdaY(t)=0

If lambda is not 0, general solution is:
X(x)=Acos(sqrt(lambda)x)+Bsin(sqrt(lambda)x)
Y(t)=Ce^(-lambda*t)

And from BC:
a)T(t,0)-2=0
b)T(t,L)-2=0
for a)
X(0)-2=0
A*1-0=2, A=2
for b)
X(0)-2=0
Is it correct thinking and where to go...?
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
22,405
Thank You for answer.
I know about Neumann, Dirichlet and mixed BCs.

Dirichlet BC: u(0,t) = g1(t), u(1,t) = g2(t),
and g1(t)=g2(t)=2 here?
my first steps with BC:
a)T(t,0)-2=0
b)T(t,L)-2=0
cause we need zero in separating variables
Separating variables:
left side of heat equation: X(x)Y'(t) is dT/dt=d/dt[X(x)Y(t)]
right side: X''(x)Y(t) is d^2T/dx^2=d^2/dx^2[X(x)Y(t)]
Both sides are equal:
X(x)Y'(t)=X''(x)Y(t)
X''(x)/X(x)=Y'(t)/Y(t)=-lambda
so
X''(x)+lambdaX(x)=0 and Y'(t)+lambdaY(t)=0

If lambda is not 0, general solution is:
X(x)=Acos(sqrt(lambda)x)+Bsin(sqrt(lambda)x)
Y(t)=Ce^(-lambda*t)

And from BC:
a)T(t,0)-2=0
b)T(t,L)-2=0
for a)
X(0)-2=0
A*1-0=2, A=2
for b)
X(0)-2=0
Is it correct thinking and where to go...?
Yes thinking is correct. To be "more correct", please study:


page 17 onwards deal with time-dependent heat conduction (in your case α = 1). Comprehending this topic will take some effort and time.
 
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