Modified heat equation steady state

[dp182]

Homework Statement

determine the steady state equation for the given heat equation and boundary conditions

Homework Equations

U_t=U_xx-4(U-T)
U(0,T)=T U(4,T)=0 U(x,0)=f(x)

The Attempt at a Solution

I put U_t=0
so 0=U_Inf''-4(U_inf-T)
then once I tried to integrate I ended up with a U_inf that's impossible to isolate is there a specific method used for this kind of heat equation?

 

[hunt_mat]

The steady state is usually when the system is time independent, so:
$$\frac{\partial U}{\partial t}=0$$
So this will turn your equation into and ODE, which I see you have already done and your ODE is:
$$\frac{\partial^{2}U}{\partial x^{2}}-4U+4T=0$$
I am not too sure what T is, is it a constant? A function?

 

[LCKurtz]

Homework Statement

determine the steady state equation for the given heat equation and boundary conditions

Homework Equations

U_t=U_xx-4(U-T)
U(0,T)=T U(4,T)=0 U(x,0)=f(x)

The Attempt at a Solution

I put U_t=0
so 0=U_Inf''-4(U_inf-T)
then once I tried to integrate I ended up with a U_inf that's impossible to isolate is there a specific method used for this kind of heat equation?

What do you mean by "impossible to isolate". Calling what you call U_inf u(x) for simplicity the equation you have written is just

u''(x) - 4u(x) = -4T
u(0) = u(T) = 0

A second order constant coefficient ODE.

 

[dp182]

I believe T is constant. so I tried using variation of parameters to solve (U_inf=X) X''-4X=-4T
and came up with a particular solution of
c₁e^2x+c₂e^-2x-T then solved for the constants using boundary conditions. is this method right as all the example I have seen involve just integrating it twice.

 

[LCKurtz]

I believe T is constant. so I tried using variation of parameters to solve (U_inf=X) X''-4X=-4T
and came up with a particular solution of
c₁e^2x+c₂e^-2x-T then solved for the constants using boundary conditions. is this method right as all the example I have seen involve just integrating it twice.

That should work, but I'm guessing it gets a bit messy solving for the constants. Here's a little trick you might like to learn. You probably know that instead of using the pair {e^2x,e^-2x} for the general solution of the homogeneous equation, you could have used {sinh(2x),cosh(2x)}, which simplifies things a bit when you calculate X(0) = 0. But an even better choice is the pair {sinh(x),sinh(T-x}. Try evaluating the constants when you write your solution

X(x) = Asinh(x)+Bsinh(T-x) + T

and you will see what I mean.