# Why integral curves cannot be tangent to each other?

I am watching Lecture 1 of the differential equation course on MIT OpenCourseWare. The teacher said, for $$y'= f(x,y)$$, if $$f(x,y)$$ is continuous around a point $$(x0,y0)$$, it would guarantee at least one solution, and if $$\frac{\partial f(x,y)}{\partial {y}}$$ is continuous, it would ensure uniqueness of the solution, i.e. there would be only one integral curve passing through $$(x0,y0)$$, meaning two integral curves could not be tangent to each other at $$(x0,y0)$$.

Can anyone explain why there would be only one integral curve passing through $$(x0,y0)$$, i.e. two integral curves could not be tangent to each other at $$(x0,y0)$$, if $$\frac{\partial f(x,y)}{\partial {y}}$$ is continuous at $$(x0,y0)$$? If 2 integral curves are tangent to each other at $$(x0,y0)$$, then they should have the same derivative at $$(x0,y0)$$, which is equal to $$f(x0,y0)$$, right? How would this make $$\frac{\partial f(x,y)}{\partial {y}}$$ discontinuous at $$(x0,y0)$$?

Thanks!

• From a point with an imposed derivative, you cannot "fork". – Yves Daoust Nov 16 '18 at 19:34
• The teacher said if you forked, then there would be 2 derivatives. But if the two integral curves were tangent to each other, there would still be only 1 derivative at this point. That's partly why I don't understand why two integral curves can not be tangent to each other. – fyang Nov 16 '18 at 19:39
• By what magic could the curves diverge ? – Yves Daoust Nov 16 '18 at 19:41
• What is the example ? – Yves Daoust Nov 16 '18 at 20:03
• Doesn't the teacher just say its hard to guess that the partial derivative being continuous is the required condition? – klirk Nov 16 '18 at 20:08

## 3 Answers

The teacher cites the Picard-Lindelöf theorem. It says that in a neighbourhood of the considered point, the ODE has a solution if $$f$$ is continuous and that the solution is unique if $$f(x,y)$$ is Lipschitz continuous in $$y$$.

By definition $$f(x,y)$$ is Lipschitz continuous in $$y$$, if there is a constant $$L\in \mathbb R$$ such that $$|f(x,y_1)- f(x,y_2)|\le L|y_1-y_2|$$ for all $$x,y_1,y_2$$.

I will sketch the proof for uniqueness:

Assume there are two solutions $$y_1(x), y_2(x)$$ which coincide at $$x=x_0$$.
Then $$|y_1(x)-y_2(x)|= |\int_{x_0}^x (y_1'(s)-y_2'(s))~ ds|\le\int_{x_0}^x |(f(s,y_1(s))-f(s,y_2(s)))~ds|\\ \le L \int_{x_0}^x |y_1(s)-y_2(s)|~ds.$$ So if we set $$z(x):=|y_1(x)-y_2(x)|$$, then $$z\le L \int_{x_0}^x z(s) ~ds$$.

This can be used to show $$z\le 0$$. (for example using the Grönwall inequality)
But obviously $$z\ge 0$$, so $$z=0$$, but this just means that $$y_1(x)=y_2(x)$$. Hence the solution is unique.

So why does the prof mention $$\frac{\partial f(x,y)}{\partial y}$$ should be continuous?

This is because the partial derivative being continuous in s neighbourhood implies that the function is Lipschitz continuous in this neighbourhood.

This is a consequence of the mean value theorem:

$$|f(x,y_1)- f(x,y_2)|\le \sup_{t\in [0,1]}|\frac{\partial f(x,y_1+t(y_2-y_1))}{\partial y}| \cdot |y_1-y_2|.$$ As we can choose the neighbourhood to be convex, the supremum is finite, (i.e $$<\infty$$) and takes the role of $$L$$.

Thus $$f$$ is Lipschitz continuous in the second variable.

For the differential equation $$y'=3|y|^{2/3}$$ you have (at least) $$4$$ solutions through the initial point $$(x_0,y_0):=(0,0)$$, namely $$y_1(x)\equiv0$$, $$\,y_2(x)=x^3$$, $$\,y_3(x)=\min\{0,x^3\}$$, and $$\,y_4(x)=\max\{0,x^3\}$$, the exact reason being that for the right hand side $$f(x,y):=|y|^{2/3}$$, while continuous at $$(0,0)$$, the partial derivative $${\partial f\over\partial y}$$ is not even defined there.

If

$$f_y(x, y) = \dfrac{\partial f(x, y)}{\partial y} \tag 1$$

is continuous in some neighborhood $$U$$of $$(x_0, y_0)$$, then $$f(x, y)$$ is Lipschitz continuous in $$y$$ on some, perhaps smaller, open ball $$B((x_0, y_0), \epsilon) \subset U$$ (we note $$(x_0, y_0) \in B((x_0, y_0), \epsilon)$$); this follows from the observation that, since the closure $$\bar B((x_0, y_0), \epsilon)$$ of $$B((x_0, y_0), \epsilon)$$ is compact and $$f_y(x, y)$$ is continuous, $$\vert f_y(x, y) \vert$$ is bounded by some real $$L > 0$$ on $$\bar B((x_0, y_0), \epsilon)$$; then for

$$(x, y_1), (x, y_2) \in B((x_0, y_0), \epsilon), \tag 2$$

we have

$$\vert f(x, y_2) - f(x, y_1) \vert = \left \vert \displaystyle \int_{y_1}^{y_2} f_s(x, s) \; ds \right \vert \le \left \vert \displaystyle \int_{y_1}^{y_2} \vert f_s(x, s) \vert \; ds \right \vert \le \left \vert \displaystyle \int_{y_1}^{y_2} L \; ds \right \vert = L\vert y_2 - y_1 \vert, \tag 3$$

which shows that $$f(x, y)$$ is Lipschitz on $$B((x_0, y_0), \epsilon)$$; thus we may invoke the Picard-Lindeloef theorem to affirm that there exists precisely one solution passing through any point in $$B((x_0, y_0), \epsilon)$$; in particular, there is exactly one integral curve $$y(x)$$ of

$$y' = f(x, y) \tag 4$$

with

$$y(x_0) = y_0. \tag 5$$

Thus, "two integral curves could not be tangent to each other at $$(x0,y0)$$" since there is in fact only one integral curve through this point!

Our colleague Christian Blatter has, in his answer, provided some good examples of how uniqueness may fail in the absence of Lipschitz continuity.