ODE direction field doesn’t match solution

Question

For the ODE $y’=sqrt y$ and the initial condition $y(0)=0$ the solution is $y=frac{1}{4}x^2$. This doesn’t match the direction field for the ODE. I wondered why this was happening here? I feel I am missing/overlooking something very basic

Arthur · Accepted Answer

Below I have a plot of the slope field and of $y(x) = frac14x^2$.

We see that the right half fits the slope field perfectly, but the left half doesn't. So clearly something is wrong with the left half.
Not really. Or, it depends. Note that $frac14x^2$ perfectly fulfills $(y')^2 = y$. So one could say that the non-fit is a result of choosing the "wrong sign" of the square root to the left of the $x$-intersection.
But also note that $y(x) = 0$ is a perfectly valid solution to your differential equation. And for any $ain Bbb R$, setting $y(x) = 0$ for $xleq a$ and $y(x) = frac14(x-a)^2$ for $x>a$, is also a valid solution.
So with your initial condition, and with the original differential equation $y' = sqrt y$, all we can really say is that $ageq 0$, possibly "$a = infty$" (i.e. $y(x) = 0$ everywhere). With an initial condition like $y(1) = frac14$, on the other hand, we fix $a = 0$.

Yiorgos S. Smyrlis · Answer

This IVP
$$
y'=sqrt{y}, quad y(0)=0
$$
does not enjoy uniqueness. (This is because the flux $sqrt{y}$ is not sufficiently smooth.)
It possesses infinitely many solutions:
$$
y_c(x)=maxleft{0,frac{1}{4}(x-c)^2right}, quad c ge 0.
$$

Shubham Johri · Answer

On solving the differential equation by variable separable method, we get$$intfrac{y'}{sqrt y}=int dximplies x-2sqrt y=C$$The initial condition gives $C=0$. Note that in the solution $x=2sqrt y$, both $x,yge0$. But when you square it to obtain $y=x^2/4$, it becomes defined for negative $x$. The curve over $x<0$ does not actually satisfy the ODE:$$y=x^2/4,x<0implies sqrt y=|x/2|=-x/2$$ giving $y'=-sqrt y$ which is not the original ODE. You are probably seeing the mismatch between direction field and the curve tangents only over $x<0$ (look at the graph).

ODE direction field doesn’t match solution

3 Answers

Add your own answers!

Ask a Question