Steady state heat flow with source
(10 pts)
Heat flow with a constant internal heat source is governed by
$$u_t = Du_{xx} + f, u(0,t)=a, u(1,t)=b.$$
Your task is to find the steady state temperature distribution. You can find your personal problem on this webpage.
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This is very much problem 6 on page 37 of our text.
Your answer should include a statement of the problem as given and your typed up solution including some description of how you found the solution.
Comments
Heat flow with a constant internal heat source is governed by
$$u_t = 2u_{xx} + 8, u(0,t) = 1, u(1,t) = 7.$$
Find the steady state temperature distribution.
The steady state distribution is when $u_t = 0.$ I solved the given equation for $u_t = 0,$ which gives me the value of $u_{xx} = u''(x) = -4.$ I then antidifferentiated this expression twice to get a general solution for $u,$ and I plugged the boundary conditions into this general solution to get the steady state solution of $u(x) = -2x^2 + 8x + 1.$
Heat flow with a constant internal heat source is governed by $u_t = 3u_{xx} +3$, U(0,t) = -4, U(1,t) = 5. Find the steady sate temp distribution.
Setting Ut = 0, I ended up with Uxx = -1. I then integrated twice to end up with
$$u(x,t) = -\frac{1}{2}x^2+ax+b.$$
Plugging in the conditions U(0,t) = -4 and U(1,t) = 5, I was able to solve for the constants a and b thus getting a steady state solution of U(x,t) = -1/2x^2 + 9.5x -4
Looking for the temperature distribution in a steady state problem (i.e. the time derivative is not relevant once the system reaches steady state)
My given equation and boundary values were
$$ u_t = 3u_{xx} + 8, u(0, t) = -4, u(1, t) = 6 $$
when $ u_t = 0 $ the sytem is steady state, then solving for $u_{xx} $gives $ u_{xx} = \frac{8}{3} $
anti-differentiating twice gives the general solution $ u(x) = \frac{8}{3}x^2 + ax + b $
Then, using the boundary values:
$$ u(x) = \frac{4}{3}x^2 + \frac{26}{3}x - 4 $$
@hadley
I think your math is good. It's hard to be certain because an expression like 1/2x^2 isn't as clear as say
$$\frac{1}{2}x^2 \: \text{ or } \: \frac{1}{2x^2}.$$
I guess you mean one of those. That's why we like mathematical typesetting.
As described in this Meta post, our mathematical typesetting tools are based on LaTeX snippets. You can read more about that in the post and in the cheat sheet linked from there. I went ahead and edited your post slightly. I bet you can finish the rest based on what you see there.
Heat flow with a constant internal heat source is governed by
$$u_t = 2u_{xx} + 6, u(0,t) = -1, u(1,t) = 4.$$
For a steady state temperature distribution, $u(x,t)$ does not change with respect to time, so $u_t = 0$. Hence, $2u_{xx} + 6 = 0$. Solving this equation for $u_{xx}$ yields $u_{xx} = -3$, and then integrating twice gives the solution
$$u(x,t) = -\frac{3}{2}x^2 +Cx + D$$
where C and D are constants. By setting $u(0,t) = -1$ and $u(1,t) = 4$, I was able to solve for C and D, and get a final solution of
$$u(x,t) = -\frac{3}{2}x^2 +\frac{13}{2}x - 1.$$
Heat flow with a constant internal heat source is governed by:
$$u_t=1u_{xx}+3, u(0,t)=1, u(1,t)=2.$$
Since this is a steady state temperature distribution, $u(t)=0$ since $u(x,t)$ does not change with time. Thus,
$$1u_{xx}+3=0.$$
Solving for $u_{xx}$ yields $u_{xx}=-3$. Integrating this twice with provide the result,
$$u(x,t)=-\frac{3}{2}x^2+Ax+B,$$
where A and B are two constants. Solving for A and B using the initial conditions of $u(0,t)=1, u(1,t)=2$ produces the steady state distribution of
$$u(x,t)=-\frac{3}{2}x^2+\frac{5}{2}x+1.$$
The heat flow equation with a constant internal heat source I was given is
$$u_t=u_{xx}+5.$$
My boundary conditions are
$$u(0,t)=-4\mbox{ and }u(1,t)=5.$$
To find the steady state temperature distribution we set the partial derivative of the function $u(x,t)$ with respect to time equal to zero, that is $u_t=0$, and then integrate in succession, yielding
$$u(x,t)=-\frac{5}{2}x^2+c_1x+c_2.$$
Using the given boundary conditions, $u(0,t)=-4$, $u(1,t)=5$, we determine $c_1=\frac{23}{2}$ and $c_2=-4$. Thus the steady state temperature distribution is represented by
$$u(x,t)=-\frac{5}{2}x^2+\frac{23}{2}x-4.$$
Heat flow with a constant internal heat source is governed by:
$$u_t = 1u_{xx} + 2, u(0,t) = -3, u(1,t) = 5.$$
When at the steady state temperature distribution, $u_t = 0$ therefore:
$$0 = u_{xx} + 2$$
This can be rearranged to give $u_{xx} = -2$. Next this function was integrated with respect to $x$ twice giving $$u(x) = -1x^2 + \alpha*x + \beta$$.
Plugging in the conditions $u(0,t) = -3$ and $u(1,t) = 5$ allowed for finding of $\alpha = 9$ and $\beta = -3$.
Therefore, the steady state temperature distribution is:
$$u(x) = -1x^2 + 9x - 3$$
Heat flow with a constant internal heat source is governed by u_t = 1u_{xx}+7, u(0,t)=-2, u(1,t)=4. This is how I found the steady state temperature distribution.
For the steady state temperature distribution, $u(x,t)$ does not change with respect to time so $u_t=0$, Then for my equation $$0=u_{xx}+7,$$ $$u_{xx}=-7.$$
Then I will need to integrate to find $u(x)$. We start with $$u_{xx}=-7.$$ Then after the first integral we get $$u_{x}=-7x+c_1.$$ The second integral is $$u(x)=\frac{-7}{2}x^2+c_1x+c_2.$$
With the conditions where $u(0,t)=-2$ and $u(1,t)=4$, we are able to find that $c_1=\frac{19}{2}$ and $c_2=-2$.
Therefore, the steady state temperature distribution is $$u(x)=\frac{-7}{2}x^2+\frac{19}{2}x-2.$$
$u_t=u_x{}_x+8 ,u(0,t)=-4 ,u(1,t)=4$ since we are looking for a steady state we are looking for when $$u_t=0$$ because our equation should not change with respect to time. By looking at our first equation we integrate up $$u_x{}_x$$ and net $$u=-4x^2+Ax+B$$ with A and B being some constants. We then solve for A and B using our initial conditions and get B=-4 and A=12 so our steady state looks like $$u(x,t)=-4x^2+12x-4$$
My heat flow equation is
$$ u_t=4u_{xx}+4$$
And my boundary conditions are
$$u(0,t)=-2$$ and $$u(1,t)=5$$ when $u_t=0$, solving for $u_{xx}$ gives us $u_{xx}=-1$.
Integrating twice, we get
$$u(x) = -x^2+ax+b$$
By using our initial conditions we get
$$u(x)=-x^2+(17/2)x-(7/2)$$ which is our steady state temperature distribution
Heat flow with a constant internal heat source is governed by
$$u_t = 2u_{xx} + 3, u(0,t)=-4, u(1,t)=3.$$
Find the steady state temperature distribution.
Steady state occurs when $u_t=0$. Solving the above equation for $u_{xx}$ given $u_t=0$, yields $u_{xx}=-3/2$. Integrating this twice gives $u(x)=\frac{-3}{4}x^2+cx+d.$ Using the boundary condition to solve for c and d gives you the steady state equation of $$u(x)=\frac{-3}{4}x^2+\frac{31}{4}x-4$$
My heat flow problem where I found the steady state temperature distribution: $$u_t=3u_{xx}+1$$
My boundary conditions were: $$u(0,t)=-3$$ and $$u(1,t)=4$$
Since $u_t=0$ when the bar is in steady state I used this to solve for $u_{xx}$, which I got was $u_{xx}=\frac{-1}{3}$. Then integrating twice: $$u(x)=\frac{-1}{6}x^2+Cx+D$$
I found these constants by using my boundary conditions: $C=\frac{43}{6}$ and $D=-3$
Finally I got: $$u(x)=\frac{-1}{6}x^2+\frac{43}{6}x-3$$
Heat flow with a constant internal heat source is governed by:
$u_t=2u_{xx}+7$, $u(0,t)=−2$, $u(1,t)=4$.
Because this is a steady state temperature distribution, $u_t=0$ since $u(x,t)$ does not change with time. So,
$0=2u_{xx}+7$
Solving for $u_{xx}$ gives $u_{xx}=\frac{-7}{2}$. By integrating this equation twice we get:
$u(x,t)=\frac{-7}{4}x^2+c_1x+c_2$,
where $c_1$ and $c_2$ are our two constants. Solving for $c_1$ and $c_2$ using the initial conditions of $u(0,t)=-2$, $u(1,t)=4$ produces the steady state distribution of
$u(x,t)=\frac{-7}{4}x^2+\frac{3}{2}x+\frac{19}{4}$
Heat flow with a constant internal heat source is governed by $u_t = 3u_{xx} + 3$, $u(0,t) = -4$, $u(1,t) = 5$ Find the steady state temp distribution.
Set $u_t = 0$ which yields $u_{xx} = -1$ ; Integrating twice I get :
$u(x,t) = \frac{-1}{2}x^2 + \alpha(t)x + \beta(t)$
where $\alpha$ and $\beta$ are the two constants. Solving for our initial conditions, the steady state temp distribution function is
$u(x,t) = \frac{-1}{2}x^2 + 9.5x - 4$
Heat flow with a constant internal heat source is governed by the following model:
$$u_t = 4u_{xx} + 5$$
Find the steady-state temperature distribution $u(x,t)$ if the boundary conditions are:
$$u( 0, t) = -2; u( 1, t) = 4$$
Steady-state implies that nothing changes with time, therefore:
$$\frac{du}{dt} = 0$$
Plugging the result into the model and rearranging the terms:
$$u_{xx} = - \frac{5}{4}$$
Integrate twice:
$$u_x = - \frac{5}{4} x + \phi (t)$$
$$u = - \frac{5}{8} x^2 + \phi (t)x + \psi (t)$$
Use boundary conditions to determine the unknown functions $\phi (t)$ and $\psi (t)$
$$u(0, t) = \psi (t) = -2$$
$$u(1, t) = -\frac{5}{8} + \phi (t) + (-2) = 4 \rightarrow \phi(t) = \frac{53}{8}$$
Therefore, the steady state temperature distribution is:
$$\boxed{u(x,t) = - 0.625 x^2 + 9.6 x -2}$$
We will look at heat flow with a constant internal heat source. This will be governed by, $$u_t=4{u_xx}+8.$$
Such that we have the initial conditions, $u(0,t)=-3$ and $u(1,t)=3.$
Given that we are looking for a steady-state distribution, we can set $u_t=0,$ such that $u(x,t)$ does not change with respect to time, $t.$
Therefore, we have the equation $$0=4{u_xx}+8.$$
Simplifying we get:
$$4{u_xx}=-8$$
$${u_xx}=-2.$$
Integrating twice we get
$${u_x}=-2x+{c_1}$$
$$u=-x^{2}+x{c_1}+{c_2}.$$
Using our initial conditions, we will solve for $c_1$ and $c_2$:
We get ${c_1}=7$ and ${c_2}=-3.$
Thus, our steady-state distribution function is:
$$u=-x^2+7x-3.$$
Heat flow with a constant internal heat source is governed by
$$u_t = 3u_{xx} + 4\ ; \ u(0,t) = 1, \ u(1,t) = 2$$
Find the steady state temperature distribution.
Steady state: $u_t = 3u_{xx} + 4 = 0 \ \longrightarrow \ u_{xx} = -\frac{4}{3}$
$$u = \int\int -\frac{4}{3}\ dx = -\frac{2}{3}x^2 + c_{1}x + c_2$$
Initial conditions:
$u(0,t) = 1 \ \longrightarrow \ c_2 = 1$
$u(1,t) = 2 \ \longrightarrow \ -\frac{2}{3} + c_1 + 1 = 2 \ \longrightarrow \ c_1 = \frac{5}{3}$
So, $\boxed{u(x,t) = -\frac{2}{3}x^2 + x + \frac{5}{3}}$
The equation for heat flow in my problem is
$$u_t=4u_{xx}+3$$
with boundary conditions $u(0,t)=-3$ and $u(1,t)=2$.
To find the steady state solution, we set $u_t=0$. Then,
$$0=4u_{xx}+3$$
$$\Rightarrow u_{xx}=-\frac{4}{3}$$
$$\Rightarrow u_{x}=-\frac{4}{3}x+\alpha$$
$$\Rightarrow u(x)=-\frac{2}{3}x^2+\alpha x+\beta.$$
Inserting the boundary conditions, we get
$$u(0)=\beta=-3$$
and
$$u(1)=-\frac{2}{3}+\alpha-3=2$$
$$\Rightarrow \alpha=\frac{17}{3}.$$
Therefore, the steady state solution is
$$\boxed{u(x,t)=-\frac{2}{3}x^2+\frac{17}{3}x-3}$$
My heat flow equation was
$$u_t=1u_{xx}+2$$
with boundary conditions of
$$u(0,t) = 0$$
and
$$u(1,t) = 7.$$
When $u_t = 0$ we can solve for $u_{xx}$ giving us $u_{xx}=-2$. Integrating this twice gives us
$$u(x,t) = -x^2 + c_1x + c_2.$$
We can then solve for these constants using our boundary conditions mentioned above, giving us the solution
$$u(x,t) = -x^2 + 8x.$$