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Consider the differential equations and:

  1. $x'=x^2$ with initial condition $x(0)=x_0$≠0

  2. $x'=x^2-1$ with initial condition $x(0)=x_0$

  3. $x'=x^2+1$ with initial condition $x(0)=x_0$

a. Verify that the theorem on existence and uniqueness applies.

So my understanding is that in order for the theorem on existence and uniqueness to apply, the scalar differential equation $x'=f(x)$ and f(x) and f'(x) both have to be continuous. Is this correct or am I missing some requirements?

b. Solve for an explicit solution

So I am looking for the flow of the differential equation, $ϕ(t;x_0)$ right? My book says that the flow is the funciton which give the solution as a function of time and $x_0$. I guess I don't really understand how to compute the flow, from my book I got:

"For a linear system, $x'=Ax$, the flow $ϕ(t;x_0)=e^{At}x_0$"

But I can't find a worked out solution in the book, so I am a little confused. So I have to find the matrix A and then the flow is just e to the power of that matrix multiplied by t and then multiply everything by the initial condition? How can I find A from these equations above?

c. What is the maximal interval of definition?

I can't find a definition in the book for the "maximal interval of definition", so I am assuming it's just asking for what values is the function defined?

  1. $x'=x^2$ with initial condition $x(0)=x_0$≠0

a.$$x'=f(x)=x^2$$

$$f'(x)=2x$$

Since both f(x) and f'(x) are continuous everywhere, the theorem on existence and uniqueness applies.

b.

c.

  1. $x'=x^2-1$ with initial condition $x(0)=x_0$

a. $$x'=f(x)=x^2-1$$ $$f'(x)=2x$$

Since both f(x) and f'(x) are continuous everywhere, the theorem on existence and uniqueness applies.

b.

c.

  1. $x'=x^2+1$ with initial condition $x(0)=x_0$

$$x'=f(x)=x^2+1$$ $$f'(x)=2x$$

Since both f(x) and f'(x) are continuous everywhere, the theorem on existence and uniqueness applies.

b.

c.

Math Major
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1 Answers1

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a) $f$ and $f'$ continuous is sufficient for the existence and uniqueness theorem (in fact you only need weaker conditions but this is useless here, see Cauchy-Lipschitz theorem)

b) "For a linear system, $x′=Ax$, the flow $\varphi(t;x_0)=e^{At}x_0$" This holds only for linear differential equations. These equations are not linear, for example, suppose that $x(t)$ is a solution of the first equation ($x'(t)=x^2(t)$), let $y$ be defined by $y(t)=\lambda x(t)$ with $\lambda$ a real. Then $y'(t)=\lambda x'(t)=\lambda x^2(t)\neq y^2(t)$ so $y$ is not a solution to this equation.

Be careful because here $x$ is a function of $t$ and not a variable. You can solve the first equation this way :

$$x'(t) = x^2(t)$$ $$\frac{x'(t)}{x^2(t)} = 1 $$ You can do this on an interval around $0$ where $x(t)$ is not equal to $0$ $$\int^{t}_0 \frac{x'(u)}{x^2(u)} du=\int^{t}_0 du$$ $$\left[ -\frac{1}{x(u)} \right]_0^{t}=t$$ $$\frac{1}{x(0)}-\frac{1}{x(t)}=t$$ $$x(t)=\frac{1}{\frac{1}{x_0}-t}=\frac{x_0}{1-x_0t}$$ You can check that $x(0)=x_0$ and $x'(t)=-x_0\frac{-x_0}{(1-x_0t)^2}=x(t)^2$.

c) From the uniqueness part of the theorem you get that the previous function is the unique solution of the problem. It is defined on the interval $\left]-\infty,\frac{1}{x_0}\right[$ if $x_0$ is positive and $\left]\frac{1}{x_0},+\infty\right[$ if $x_0$ is negative. This is the maximal interval.

I let you do the second and third equations, the integration step is a little harder.

Mirajane
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