Solving the 1D Heat Equation with Given Parameters

[jc2009]

Problem:IF there is heat radiation within the rod of length L , then the 1 dimensional heat equation might take the form
u_t = ku_xx + F(x,t)

Find u(x) if F = x , k = 1 , , u(0)=0 , u(L) = 0

the problem is that i am not sure what this is asking me , how can i find u(x) if i have only u(0)= 0 , k =1 , u(L) = 0 ,and F = x


this problem becomes just an ordinary differential equation but still i don't fully understand or how to proceed from there

any hints would be appreciated

 

[alle.fabbri]

have you tried green's function method?

 

[coomast]

Problem:IF there is heat radiation within the rod of length L , then the 1 dimensional heat equation might take the form
u_t = ku_xx + F(x,t)

Find u(x) if F = x , k = 1 , , u(0)=0 , u(L) = 0

the problem is that i am not sure what this is asking me , how can i find u(x) if i have only u(0)= 0 , k =1 , u(L) = 0 ,and F = x


this problem becomes just an ordinary differential equation but still i don't fully understand or how to proceed from there

any hints would be appreciated

I assume that a steady-state solution is required, therefore the time derivative vanishes in the pde and you get the following equation to solve:
[tex]\frac{d^2u}{dx^2}+x=0[/tex]
which has the solution:
[tex]u(x)=-\frac{x^3}{6}+Ax+B[/tex]
Using the boundary conditions, you get:
[tex]u(x)=\frac{x}{6}\cdot \left[L^2-x^2\right][/tex]
Hope this is what has been asked for.

coomast

 

[jc2009]

I assume that a steady-state solution is required, therefore the time derivative vanishes in the pde and you get the following equation to solve:
[tex]\frac{d^2u}{dx^2}+x=0[/tex]
which has the solution:
[tex]u(x)=-\frac{x^3}{6}+Ax+B[/tex]
Using the boundary conditions, you get:
[tex]u(x)=\frac{x}{6}\cdot \left[L^2-x^2\right][/tex]
Hope this is what has been asked for.

coomast

THank you so much