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"Given the differential equation $$x''(t)+x(t)=0$$

Is there more than 1 solution satisfying the initial value problem $$x(0)=3 \text{ and } x(\pi)=-3$$

Now, I thought I had heard that there are only ONE solution to a differential equation with an initial value problem. But now I'm told that the answer to the above question is yes.

Can someone explain why there are more solutions to this?

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  • $\begingroup$ @Alex5207: Are the ICs correct or did you mean for one of those to be $x'() = ...$? $\endgroup$ – Moo Dec 8 '17 at 22:49
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    $\begingroup$ Either there's a mistake in the question somewhere or else it doesn't even have solution, as shown in the answer below. $\endgroup$ – DonAntonio Dec 8 '17 at 23:23
  • $\begingroup$ Very sorry for the inconvenience! I accidently forgot - in front of 3 in $x(\pi)=-3$ - Corrected now! $\endgroup$ – Alex5207 Dec 9 '17 at 7:58
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Note that these are not initial conditions but rather boundary conditions, where you have information at 2 different points instead of the same point. Because these are not initial conditions, uniqueness is not guaranteed.

Looking at general solution $$ x(t) = A\cos t + B\sin t $$

we have $x(0) = A$ and $x(\pi) = -A$

I'd wager that there is a typo in the problem statement, as this gives us $A=-A=3$ which yields no solution.

But suppose that the B.C's were consistent, such as $x(\pi)=-3$ or $x(2\pi)=3$, then a solution does exist with $A=3$, then $$ x(t) = 3\cos t + B\sin t $$

As there is no constraints on the constant $B$, it is free to vary. Thus you have infinitely many solutions.

I suspect that this was the intent of the question, however the wording is faulty, as this isn't even an initial value problem to begin with.

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  • $\begingroup$ You are completely right about the ICs, I had a typo. $x(\pi)=-3$ and not 3. I corrected this now $\endgroup$ – Alex5207 Dec 9 '17 at 7:59
  • $\begingroup$ Good, does the answer make sense now? $\endgroup$ – Dylan Dec 9 '17 at 8:40
  • $\begingroup$ Yes indeed. The boundary vs. initial conditions makes good sense. So, by definition, are initial conditions conditions that specify x(0)=x0 and x'(0)=y0? $\endgroup$ – Alex5207 Dec 9 '17 at 8:43
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    $\begingroup$ Yes, you are correct. It can be defined on any point, so more like $x(t_0) = x_0,\ x'(t_0)=y_0$ $\endgroup$ – Dylan Dec 9 '17 at 8:57
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$$x''(t)+x(t)=0$$ The general solution is : $\quad x(t)=c_1\cos(t)+c_2\sin(t)$

First condition : $\quad x(0)=3=c_1\cos(0)+c_2\sin(0)=c_1 \quad\to\quad c_1=3$ $$x(t)=3\cos(t)+c_2\sin(t)$$

The second condition $\quad x(\pi)=-3=3\cos(\pi)+c_2\sin(\pi)=-3\quad$ is satisfied any value of $c_2$.

Thus the solution of the ODE with specified conditions is : $$x(t)=3\cos(t)+c_2\sin(t)\quad\text{with arbitrary constant } c_2 .$$

Conclusion : They are an infinity of solutions.

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Before we find the apply the initial value information we have:

$x = A\cos t + B\sin t$

Now we look at the additional information for guidance on the values of A,B.

Normally, two additional pieces of information would let you pin down two variables. But in this case.

$B\sin 0 = B\sin \pi$ so these coordinates give you no insight on $B.$

but $x(0) = x(\pi) = 3$ also presents a problem for you.

So, as I see it, there is no solution to this particular set of conditions.

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  • $\begingroup$ Just want to say I accidentally downvoted your answer. You are right that there is no solution. $\endgroup$ – Dylan Dec 9 '17 at 4:05

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