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Math Help - Solving differential equations: Help explain the answers please

  1. #1
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    Solving differential equations: Help explain the answers please

    I have done most of the questions, I am just struggling on a few parts. The question is:

    a) Use the method of Separation of Variables to find the General Solution.
    b) Analyse the right hand side and determine for what pairs  (x_0,y_0) the initial value problems (IVP) would be guaranteed to have a solution by Peano's theorem, and a unique solution by Picard's theorem.
    c) Find solutions satisfying the specified initial conditions, and determine the intervals where they exists and where they are unique.

    1)  \frac{dy}{dx} = y^{\frac{1}{2}}. IC = y(1) = 0.


    Answers:

    a) I got:  y = (\frac{2}{3}(x + C) )^{\frac{3}{2}} , which was correct. But for some reason, they said this was the "typical solution" and I also needed a "special solution" where  G(y) = y^{\frac{3}{2}} = 0 \implies y = 0. My first question is: What does the special solution do?

    b) I know Picard's and Peano's theorem, so I checked where the RHS was continuous and everything and I got that the RHS of my IVP is continuous for all (x,y), and the partial derivative w.r.t 'y' is continuous everywhere except y = 0, so the two intervals I need are:

    The solutions to the IVP exist for any pair  (x_0,y_0) \in R (but for some reason the answers said it is contained in R multiplied by R. Why?), and the unique solution will exists for all values of  x_0 , but not when  y_0 = 0 and so the interval would be  (x_0,y_0) \in (-\infty,0) \cup (0,\infty), again, they however multiplied both intervals by some 'R'.

    I understand this, I just don't get why they multiplied everything by that 'R', it looks like an R for the set of Real numbers, but I don't see why you need to multiply it by this.

    c) This is the main one I am having trouble with. This is what it says in the answers:

    The initial condition is satisfied for the special solution y = 0 ( Again, what does this mean? What would happen if it didn't satisfy the special solution? ). For the typical solution, we have  0 = (\frac{2}{3}(0 + C))^{\frac{3}{2}} , hence C = 0 and  y = (\frac{2x}{3})^{\frac{3}{2}} . This general form is a "compound" solution (Whatever that means)of the form

    y = {  0 , x \in (\infty, B) or  (\frac{2}{3}(x - B))^{\frac{3}{2}}, x \in [B, \infty)

    depending on one arbitary parameter  B \geq 0 . Each of these solutions is defined for all  I = (-\infty, \infty), and they are NOT UNIQUE IN ANY INTERVAL, since B is not uniquely defined.


    Apart from the special solution bit, I get the bit up to C = 0 and your "new" equation, but I don't understand it after that. How do you know the solution is in that other form? How do I solve an equation like that?

    Thank you
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Solving differential equations: Help explain the answers please

    Quote Originally Posted by ironz View Post
    I have done most of the questions, I am just struggling on a few parts. The question is:

    a) Use the method of Separation of Variables to find the General Solution.
    b) Analyse the right hand side and determine for what pairs  (x_0,y_0) the initial value problems (IVP) would be guaranteed to have a solution by Peano's theorem, and a unique solution by Picard's theorem.
    c) Find solutions satisfying the specified initial conditions, and determine the intervals where they exists and where they are unique.

    1)  \frac{dy}{dx} = y^{\frac{1}{2}}. IC = y(1) = 0.


    Answers:

    a) I got:  y = (\frac{2}{3}(x + C) )^{\frac{3}{2}} , which was correct. But for some reason, they said this was the "typical solution" and I also needed a "special solution" where  G(y) = y^{\frac{3}{2}} = 0 \implies y = 0. My first question is: What does the special solution do?

    b) I know Picard's and Peano's theorem, so I checked where the RHS was continuous and everything and I got that the RHS of my IVP is continuous for all (x,y), and the partial derivative w.r.t 'y' is continuous everywhere except y = 0, so the two intervals I need are:

    The solutions to the IVP exist for any pair  (x_0,y_0) \in R (but for some reason the answers said it is contained in R multiplied by R. Why?), and the unique solution will exists for all values of  x_0 , but not when  y_0 = 0 and so the interval would be  (x_0,y_0) \in (-\infty,0) \cup (0,\infty), again, they however multiplied both intervals by some 'R'.

    I understand this, I just don't get why they multiplied everything by that 'R', it looks like an R for the set of Real numbers, but I don't see why you need to multiply it by this.

    c) This is the main one I am having trouble with. This is what it says in the answers:

    The initial condition is satisfied for the special solution y = 0 ( Again, what does this mean? What would happen if it didn't satisfy the special solution? ). For the typical solution, we have  0 = (\frac{2}{3}(0 + C))^{\frac{3}{2}} , hence C = 0 and  y = (\frac{2x}{3})^{\frac{3}{2}} . This general form is a "compound" solution (Whatever that means)of the form

    y = {  0 , x \in (\infty, B) or  (\frac{2}{3}(x - B))^{\frac{3}{2}}, x \in [B, \infty)

    depending on one arbitary parameter  B \geq 0 . Each of these solutions is defined for all  I = (-\infty, \infty), and they are NOT UNIQUE IN ANY INTERVAL, since B is not uniquely defined.


    Apart from the special solution bit, I get the bit up to C = 0 and your "new" equation, but I don't understand it after that. How do you know the solution is in that other form? How do I solve an equation like that?

    Thank you
    Looking at the DE...

    y^{'}=\sqrt{y} (1)

    ... You observe that, because the x is not in the DE, if \varphi(x) is solution of (1), then \varphi(x-\xi) with \xi arbitrary is also solution of (1). By simple inspection You find that...

    \varphi(x)=\begin{cases}\frac{x^{2}}{4} &\text{if}\ x \ge 0\\ 0 &\text{if}\ x<0\end{cases} (2)

    ... is solution of (1). Now the 'initial condition' is y(1)=0 and that means that any \varphi(x-\xi) with \xi \ge 1 is solution, so that it doesn't exist only one solution...




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    \chi \sigma
    Last edited by chisigma; December 24th 2011 at 04:51 AM.
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  3. #3
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    Re: Solving differential equations: Help explain the answers please

    Quote Originally Posted by ironz View Post
    I have done most of the questions, I am just struggling on a few parts. The question is:

    a) Use the method of Separation of Variables to find the General Solution.
    b) Analyse the right hand side and determine for what pairs  (x_0,y_0) the initial value problems (IVP) would be guaranteed to have a solution by Peano's theorem, and a unique solution by Picard's theorem.
    c) Find solutions satisfying the specified initial conditions, and determine the intervals where they exists and where they are unique.

    1)  \frac{dy}{dx} = y^{\frac{1}{2}}. IC = y(1) = 0.


    Answers:

    a) I got:  y = (\frac{2}{3}(x + C) )^{\frac{3}{2}} , which was correct. But for some reason, they said this was the "typical solution" and I also needed a "special solution" where  G(y) = y^{\frac{3}{2}} = 0 \implies y = 0. My first question is: What does the special solution do?

    b) I know Picard's and Peano's theorem, so I checked where the RHS was continuous and everything and I got that the RHS of my IVP is continuous for all (x,y), and the partial derivative w.r.t 'y' is continuous everywhere except y = 0, so the two intervals I need are:

    The solutions to the IVP exist for any pair  (x_0,y_0) \in R (but for some reason the answers said it is contained in R multiplied by R. Why?), and the unique solution will exists for all values of  x_0 , but not when  y_0 = 0 and so the interval would be  (x_0,y_0) \in (-\infty,0) \cup (0,\infty), again, they however multiplied both intervals by some 'R'.

    I understand this, I just don't get why they multiplied everything by that 'R', it looks like an R for the set of Real numbers, but I don't see why you need to multiply it by this.
    R\times R or R^2 is the "Cartesian product"- the set of ordered pairs of real numbers. Since each of x_0 and y_0 are real numbers, the pair (x_0, y_0) is in R\times R, not in R which contains numbers, not pairs of numbers.

    c) This is the main one I am having trouble with. This is what it says in the answers:

    The initial condition is satisfied for the special solution y = 0 ( Again, what does this mean? What would happen if it didn't satisfy the special solution? ). For the typical solution, we have  0 = (\frac{2}{3}(0 + C))^{\frac{3}{2}} , hence C = 0 and  y = (\frac{2x}{3})^{\frac{3}{2}} . This general form is a "compound" solution (Whatever that means)of the form

    y = {  0 , x \in (\infty, B) or  (\frac{2}{3}(x - B))^{\frac{3}{2}}, x \in [B, \infty)

    depending on one arbitary parameter  B \geq 0 . Each of these solutions is defined for all  I = (-\infty, \infty), and they are NOT UNIQUE IN ANY INTERVAL, since B is not uniquely defined.


    Apart from the special solution bit, I get the bit up to C = 0 and your "new" equation, but I don't understand it after that. How do you know the solution is in that other form? How do I solve an equation like that?

    Thank you
    Last edited by HallsofIvy; January 3rd 2012 at 08:13 AM.
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