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Homework Help: Explaining a physical model (differential equation)

  1. May 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Check on labb41.jpeg (I am not used to formatting my mathematical equations on the web, I can only format in my local text-editing programs).

    2. Relevant equations
    Again, check on labb41.jpeg

    3. The attempt at a solution
    I have plot the solution to U(I) as according to the assignment. Please check solution1.jpeg to se my solution plot. I am very sure that my solution is correct.

    The only question I have about this assignment is why the temperatures in the real model (I assumed that it is meant about the given differential equation) cannot be higher than the temperatures in the equation to be solved with Newton-Raphson's method and then plot U(I) for I=0.1,0.2,...,10 in the first assignment. I am not sure about how to explain this, so a little help here would be worth gold to me now.

    Thanks in advance for helping me with this problem!

    P.S. If this thread is posted in wrong forum, please tell me were I should ask these questions then so I will know it for further times.

    Attached Files:

  2. jcsd
  3. May 24, 2015 #2


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    Have you done the first part, set up the numerical model and solved the equation numerically? Doing that and considering the assumptions made in setting up the Newton-Raphson formula should give you some ideas.
  4. May 24, 2015 #3
    I have solved the latter equation by the help of Newton-Raphson's method for a every given I and I also plotted the solution just like shown in solution1.jpeg. I have also solved the differential equation with finite differences method and the highest temperature for a given I is somewhat smaller than the temperature for the same given I that is plotted according to solution1.jpeg after solving the equation with Newton-Raphson's method. So it is true that the temperatures in the differential equation for given I's are not bigger than the temperatures for same I's.

    But I still not feeling that it gives me any ideas about why the temperatures in the equation (to be solved with Newton-Raphson's method) are bigger than the temperatures received from solving the differentil equation with finite differences method. I can see that when I solve the equation f(u)=0 with Newton-Raphson's method, then I do omit the thermal conductivity. What significance does the thermal conductivity have to the temperatures in this case? I'm unfortunately lost in this problem :/
  5. May 24, 2015 #4
    Make a graph for αu4 and βΙ2σ and check the sign.
  6. May 24, 2015 #5
    Shall I make a graph for those terms from the simplified equation according to the assignment or from the given differential equation?
  7. May 24, 2015 #6
    To see the sing of $$\frac{d^2u}{dx^2}$$
    If is positive (and is positive for u>u0(some value)) then have the shape that you make for solution for various values for I.
  8. May 24, 2015 #7
    And in which way would it exactly explain that the temperatures gained in the solution of the equation when solving with Newton-Raphson's method for U(I) are higher than the temperatures you gain when solving the differential equation?
  9. May 24, 2015 #8
    Sorry for my mistake. If this is positive then the shape is opposite that you calculate, so it cannot satisfy boundary condition. For larger temperatures u(x) have cyrtosis to down.
  10. May 24, 2015 #9
    When I am graphing both the terms together, they are like parallell to each other on the graph (my visual view). The second term in the differential equation is not drawn with a negative sign as in the differential equation. The only observation I can make is that the first term is constant since there is no I value included over there while the second term grow very slightly since it is depending on the I value. So for 0.1<=I<=10, then the alpha*u^4 will be bigger than beta*(I^2)*sigma(u) and therefore will the sign on d^2u/dx^2 be positive. Have I done something wrong here...?

    I'm really sorry, I'm totally lost in this... :/
  11. May 24, 2015 #10
    for your data I take this diagram for I=10A that say for temperature larger than 150 K there is no solution that can satisfy the u'(L/2)=0 condition.
    graph here
    Equilibrium state is on the cross from x=L/2.
    Last edited: May 24, 2015
  12. May 24, 2015 #11
    Having problems with opening the graph (if it supposed to open the graph through your link). But can you answer about how this graphing term for term in the differential equation and sign change of d^2u/dx^2 will answer on the question why temperatures in the real model (the differential equation) cannot be higher than temperatures in the simplified model to be solved with Newton-Raphson's method? I need to get a better orientation of how you are thinking with these graphing and calculations.
  13. May 24, 2015 #12
    I fix it on my post. Your calculation is correct. My diagram is for I not for I2 than give maximum temperature 322K
    Last edited: May 24, 2015
  14. May 24, 2015 #13
    What exactly does your graph describe? As far as I understand, the x-axis is the temperature you use as an indenpendent variable. But how about with y-axis? And again, how is your graph (at I=10) relevant for the question I want to answer on as I have asked from the beginning in this thread? How can I, using your reasoning, answer why temperatures are lower in the differential equation than the simplified equation that is supposed to be calculated with Newton-Raphson's method? You understand what I mean?
  15. May 24, 2015 #14
    The correct graph
    This say that with temp larger than 322-324 K and I=10A, the u(L/2) have negative slope, so u'(L/2)=0 cannot satisfied. For t<322K the slope is positive so u(x) is monotonic rise. Boundary temperature on (L/2) have the maximum value.
    y-axis is nothing. the two terms of differential equation separately. The sum is the 2nd derivation.
    Last edited: May 24, 2015
  16. May 24, 2015 #15
    And the slope you are talking about can be received by substracting a function value for the green line with corresponding function value for the red line at the same x point? Just want to make sure about this.

    And again, how do you put this drawn graph in relation to the question that I want to get answer on? I'm sorry if I keep talk about the answer on the question I'm struggling with, but I really want to see some relation between your graph and the answer on that question.
  17. May 24, 2015 #16
    Sorry for my English. I understand that you want to prove that there is no higher temperature for this that you calculate by Newton-Raphson method. The answer for this is that the function u(x) is monotonic rise, left bound fix at 10K end right bound have the maximum value.
    This graphic is the graphical solution for the same problem that you solve by Newton-Raphson method.
  18. May 24, 2015 #17
    Okay, but do you rely on some specific theory that the graph you are solving actually has the temperatures that the differential equation would not be able to get such temperatures (just lower) in relation to the temperatures from the differential equation? I mean, you should somehow compare the temperature values between these two mentioned models (the diff equation and the simplified equation) somehow and strengthen your arguments if someone would ask you about it. Would like to hear more opinions from you about this :)

    By the way, thank you for your time effort to this problem. Really appreciateble, but I just want an answer black on white so I can feel me sure about this question I was not able to answer, so everything will be clarifying to me.

    Again, thank you very much for your time effort!

    P.S. Never mind, I'm not better at English myself.
  19. May 24, 2015 #18
    see this
    Your problem is about equilibrium, the final state, not a time depended prosses. You have not initial conditions, only boundary.
    u(x) on the right hand is not the indipended variable. The total equation steel is a Laplace equation.
  20. May 24, 2015 #19


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    It looks as if you can get quite a simple first integral of it and then you can use a relatively simple leap frog method to get a solution.
  21. May 24, 2015 #20
    $$ \frac{\partial{u(x)}}{\partial{x}} = \int \left(\alpha u^4(x) - \beta I^2\sigma(u(x))\right) dx $$
  22. May 24, 2015 #21
    What would the solution of the integral represent? How it is related to the answer on my question?
  23. May 24, 2015 #22


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    This is wrong. you should integrate w.r.t. u and you have a square root coming in.
  24. May 24, 2015 #23


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    The solution you obtain may be more accurate.
  25. May 24, 2015 #24
    Let me get this right:

    I am solving a equation where d^2u/dx^2=0 for every given I. And I think this is what you mean when you say "the final state"...? The conditions will not matter, since you will get the same solutions anyway. Well, I don't know so much about Laplace equations, I have not gone through it yet.

    But let's take a little example: I have uploaded an image named pic1.jpg. It represents the solution curve to the differential equation when I=10. Compare the maximal temperature in pic1.jpg with solution1.jpg that was uploaded in the first post of this thread. As you can see, the maximal temperature in pic1.jpg is smaller than the maximal temperature for u when I=10 in solution.jpg. Can you see the small differences? How shall I explain why this thing happens. What is the reason to the fact that the maximal temperature in pic1.jpg is smaller than the temperature at I=10 in solution1.jpg? How would your answer somehow explain that? Want now put everything in relation to each other.

    Attached Files:

    • pic1.jpg
      File size:
      11.6 KB
  26. May 24, 2015 #25
    Perhaps, but what is the point with integrating the whole differential equation? How will it help me to answer the question I am struggling with?
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