# Heat equation and help with BC

#### piotrekkk888

##### New member
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
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.

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.

• topsquark

#### piotrekkk888

##### New member
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
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.

• topsquark