Question

Let u(x) and v(x) be two continuous functions satisfying the differential equations $\frac{du}{dx}+p(x)u=f(x)$ and $\frac{dv}{dx}+p(x)v=g(x)$, respectively. If $u(x_1)>v(x_1)$ for some $x_1$ and $f(x)>g(x)$ for all $x>x_1$, prove that any point $(x, y)$, where $x>x_1$, does not satisfy the equations $y=u(x)$ and $y=v(x)$ simultaneously.

Answer 1

Proof

Answer 2

To prove that any point $(x, y)$, where $x>x_1$, does not satisfy the equations $y=u(x)$ and $y=v(x)$ simultaneously, we need to show that $u(x) \neq v(x)$ for all $x > x_1$.

Let the given differential equations be:

1. $\frac{du}{dx}+p(x)u=f(x)$

2. $\frac{dv}{dx}+p(x)v=g(x)$

Subtract equation (2) from equation (1):

$\left(\frac{du}{dx}+p(x)u\right) - \left(\frac{dv}{dx}+p(x)v\right) = f(x) - g(x)$

$\frac{d}{dx}(u-v) + p(x)(u-v) = f(x) - g(x)$

Let $w(x) = u(x) - v(x)$. Substituting this into the equation, we get a first-order linear differential equation for $w(x)$:

$\frac{dw}{dx} + p(x)w = f(x) - g(x)$

This equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)=p(x)$ and $Q(x)=f(x)-g(x)$. The integrating factor (IF) is $I(x) = e^{\int p(x) dx}$. Note that $I(x)$ is always positive, i.e., $I(x) > 0$ for all $x$.

Multiply the differential equation for $w(x)$ by the integrating factor $I(x)$:

$I(x)\frac{dw}{dx} + I(x)p(x)w = I(x)(f(x) - g(x))$

The left side is the derivative of the product $I(x)w(x)$:

$\frac{d}{dx}(I(x)w(x)) = I(x)(f(x) - g(x))$

Now, integrate both sides with respect to $x$ from $x_1$ to $x$ (where $x > x_1$):

$\int_{x_1}^{x} \frac{d}{dt}(I(t)w(t)) dt = \int_{x_1}^{x} I(t)(f(t) - g(t)) dt$

Applying the Fundamental Theorem of Calculus to the left side:

$[I(t)w(t)]_{x_1}^{x} = \int_{x_1}^{x} I(t)(f(t) - g(t)) dt$

$I(x)w(x) - I(x_1)w(x_1) = \int_{x_1}^{x} I(t)(f(t) - g(t)) dt$

Rearrange the equation to solve for $w(x)$:

$I(x)w(x) = I(x_1)w(x_1) + \int_{x_1}^{x} I(t)(f(t) - g(t)) dt$

$w(x) = \frac{I(x_1)}{I(x)}w(x_1) + \frac{1}{I(x)}\int_{x_1}^{x} I(t)(f(t) - g(t)) dt$

Now, let's use the given conditions:

1. $u(x_1) > v(x_1) \implies w(x_1) = u(x_1) - v(x_1) > 0$.

2. $f(x) > g(x)$ for all $x > x_1 \implies f(x) - g(x) > 0$ for $x > x_1$.

3. As established, $I(x) > 0$ for all $x$.

Let's analyze each term in the expression for $w(x)$:

• First term: $\frac{I(x_1)}{I(x)}w(x_1)$

  Since $I(x_1) > 0$, $I(x) > 0$, and $w(x_1) > 0$, this term is strictly positive.

• Second term: $\frac{1}{I(x)}\int_{x_1}^{x} I(t)(f(t) - g(t)) dt$

  For any $t$ in the integration interval $(x_1, x)$, we have $t > x_1$. Therefore, $I(t) > 0$ and $f(t) - g(t) > 0$. This means the integrand $I(t)(f(t) - g(t))$ is strictly positive for $t \in (x_1, x)$. Since $x > x_1$, the length of the integration interval is positive. An integral of a positive function over a positive interval is positive. So, $\int_{x_1}^{x} I(t)(f(t) - g(t)) dt > 0$. Since $I(x) > 0$, the entire second term is also strictly positive.

Since both terms in the expression for $w(x)$ are strictly positive, their sum must also be strictly positive:

$w(x) > 0$ for all $x > x_1$.

Substituting back $w(x) = u(x) - v(x)$:

$u(x) - v(x) > 0 \implies u(x) > v(x)$ for all $x > x_1$.

Since $u(x) > v(x)$ for all $x > x_1$, it implies that $u(x)$ can never be equal to $v(x)$ for any $x > x_1$. Therefore, for any point $(x, y)$ where $x > x_1$, it is impossible for $y=u(x)$ and $y=v(x)$ to be satisfied simultaneously, as that would require $u(x)=v(x)$.

Explanation:

1. Define $w(x) = u(x) - v(x)$.

2. Subtract the second differential equation from the first to obtain a new differential equation for $w(x)$: $\frac{dw}{dx} + p(x)w = f(x) - g(x)$.

3. Solve this linear first-order differential equation for $w(x)$ using an integrating factor $I(x) = e^{\int p(x) dx}$.

4. The solution for $w(x)$ is $w(x) = \frac{I(x_1)}{I(x)}w(x_1) + \frac{1}{I(x)}\int_{x_1}^{x} I(t)(f(t) - g(t)) dt$.

5. Use the given conditions: $w(x_1) = u(x_1) - v(x_1) > 0$ and $f(t) - g(t) > 0$ for $t > x_1$. Also, $I(x) > 0$.

6. Show that both terms in the expression for $w(x)$ are positive for $x > x_1$.

7. Conclude that $w(x) > 0$, which means $u(x) > v(x)$ for all $x > x_1$.

8. Since $u(x) > v(x)$, $u(x)$ can never equal $v(x)$, thus $y=u(x)$ and $y=v(x)$ cannot be satisfied simultaneously for $x > x_1$.