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Is there an ordinary differential equation with no solutions? Precisely, I mean

1) not in the IVP context, just a simple ODE without IVP conditions,

2) not an ODE with complex solutions but no real ones (such as $(y') ^2+1=0$),

3) not of the form $y^{(n)}=y^{(n)}+c$.

Thanks

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Of course you can set up a differential equation $$y'=f(x,y)$$ with $f(x,y)=1$ if at least one of $x$, $y$ is irrational and $=0$ otherwise. Such a differential equation will have no solution, I guess. But as soon as there is a small disk with center $(x_0,y_0)$ on which $f$ is continuous Peano's existence theorem guarantees a solution $x\mapsto y(x)$ on some small $x$-interval $[x_0-h,x_0+h]$ satisfying the initial condition $y(x_0)=y_0$.

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For sure, many ODEs do not have any analytic solution but I'm sure that's not what you meant. However assuming you cannot find any analytic solution, you will have to turn to your computer and write some code like Euler's Method which is the simplest. Approximate $y(x_{n+1}) = y(x_n) + hDy(x_n)$ where $x_{n+1} = x_n + h$.

Whatever your ODE, as long as you can make it a system of first order ODE (eg : $D^2y = -y$ is the same as $DY_1 = Y_2$ and $DY_2 = -Y1$ with $Y_1 = y$ and $Y_2 = Dy$), then you can find the next value using Euler's method or any numerical ODE solver (there are quite a few).

More generally, whenever you can write an ODE as $\frac{d\underline Y(x)}{dx} = \underline F(x, \underline Y(x))$ where $\underline Y = (Y_1, Y_2, ..., Y_n)^T$ is a vector, you can calculate an approximation to the solution by iterating : $$\underline Y(x+h) = \underline Y(x) + h\underline F(x, \underline Y(x))$$

However, you might bump into some ODE which is highly non-linear and you cannot write it as the above. In that case I still believe that there is an answer unless it falls in one of your $3$ conditions.

My reasoning is that if you convert you ODE into an algebraic equation where $D^ny$, your highest derivative, is the unknown then you have two choices :

1) Either it will have at least one real solution in which case you apply my above reasoning although the solution might not be explicit: $(D^3y)^5 - 5D^3y + D^2y = 0$ can be written algebraically as $x^5 - 5x + a = 0$ which does not have any closed form solution.

2) Or it will only have complex solutions and so it falls into your condition 3).

So for me, there is always a solution but it is not always possible to find it, at least not with today's knowledge. Perhaps one day the theory of non-linear ODE will have improved far enough to answer this question in a nicer way.

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