# Proving that a differential equation has unique solution.

Let $P,Q,f:[-1,1]\to\mathbb{R}$ continuous and $a,b\in\mathbb{R}$. Then I want to prove that the IVP

$$\begin{cases} u''(x)+P(x)u'(x)+Q(x)u(x)=f(x)\\u(0)=a,\qquad u'(0)=b\end{cases}$$

has a unique solution in a neighborhood of x=0.

To this end I have to show that the function $F(x,y,t)$ is lipschitz in the first variables and the constant independent of the parameter $t$.

(This comes when this problem is converted to solve the integral equation and invoke the condition that the operator $\Phi(x)(t)=x_0 + \int_0^tF(x(s),s)ds$ has only one fixed point and the use the Banach fixed point theorem)

So I chose $F(x,y,t)=f(x)-P(x)-Q(x)$ then:

$$|F(x,y,t)-F(u,v,t)|=|(f(x)-f(u))+(P(u)-P(x))+(Q(u)-Q(x))|$$

Now, by continuity of $f,P,Q$ in a compact set we have that they are uniformly continuous so we pick $\delta = \{ \delta_1,\delta_2, \delta_3 \}$ so if $|x-y|<\delta$ we get

$$|F(x,y,t)-F(u,v,t)|< \epsilon<\epsilon |x-y|$$

but the thing is that I don't know how to extend this property to all $x,y \in [-1,1]$

Can someone help me to fix this problem or provide another way to prove this Lipschitz condition ? because I don't know other way to proceed.

Thanks a lot in advance.

• So, do you have to reconstruct all steps of existence and uniqueness theorem or you can just use it? Because it's pretty easy to just use it here. – Evgeny Oct 7 '15 at 4:51
• I only have to use them :) so I only need to prove that the function I decided is Lipschitz in the first variables :). – user162343 Oct 7 '15 at 11:21
• Just observe that derivatives w.r.t. to $u$ and $u'$ ($Q(x)$ and $P(x)$ respectively) are continuous functions. And it's well known property that if function is continuously differentiable then it's locally Lipschitz :) and that's all that you really need here. – Evgeny Oct 7 '15 at 11:31
• Also, you should show that function $F(x, y, t) = f(t) - P(t) y - Q(t) x$ is Lipschitz in the variables $x$ and $y$, not the one that you've written. – Evgeny Oct 7 '15 at 11:33
• You can transform your second order ODE to the system of first order equations $\frac{dx}{dt} = y$, $\frac{dy}{dt} = f(t) - P(t)y - Q(t)x$. For this system existence and uniqueness theorem tells that for having unique solution it's enough for RHS to be continuous w.r.t. to $t$ and Lipschitz-continuous w.r.t. to $x$ and $y$. – Evgeny Oct 7 '15 at 12:44

Expanding the comment about transforming second order equation to the system:

$$\frac{dx}{dt} = y, \; \frac{dy}{dt} = f(t) - P(t)y- Q(t)x .$$

RHS is a vector now. Its first component is smooth w.r.t. all variables. The second component $F(x, y, t) = f(t) - P(t)y- Q(t)x$ is continuous w.r.t. $t$ and continuously differentiable in variables $x$ and $y$, because $\frac{\partial F}{\partial x} = -Q(t), \frac{\partial F}{\partial y} = -P(t)$ and both are continuous functions. Hence $F(x, y, t)$ is Lipschitz in variables $x$ and $y$. Therefore, we can apply existence and uniqueness theorem in this case.

If function $F(x)$ is continuously differentiable w.r.t. $x$, then it's Lipschitz continuous w.r.t. to them

Since $F(x)$ is continuously differentiable with, we can write $F(x_1) - F(x_2)$ as $(x_1 - x_2) \cdot F'_{x} (x_1 + \theta (x_2 - x_1))$ where $\theta \in \lbrack 0, 1 \rbrack$. From here we have an estimation: $$\vert F(x_1) - F(x_2)\vert = \vert (x_1 - x_2) \cdot F'_{x} (x_1 + \theta (x_2 - x_1)) \vert = \vert x_1 - x_2 \vert \cdot \vert F'_{x} (x_1 + \theta (x_2 - x_1)) \vert \leqslant C \cdot \vert x_1 - x_2 \vert,$$ where $C$ is the supremum of $F'_x (x)$ in some neighbourhood of point $x_1$ (it exists for compact neighbourhoods because $F'_x$ is continuous).

• I am back, what is your notation in $F'_x$? – user162343 Oct 9 '15 at 13:00
• Of course it's derivative of $F(x)$ w.r.t. to $x$. What else it could be in this context? – Evgeny Oct 9 '15 at 14:23
• Right :), and what do you think about my proof in the question math.stackexchange.com/questions/1471164/… ? How can I fix those bounds to get it of the form $L||w_1-w_2||$? – user162343 Oct 9 '15 at 14:25