Question

For \(t \geq 0\), put $$ \phi(x, t)=\left\\{\begin{array}{ll} x & (0 \leq x \leq \sqrt{t}) \\ -x+2 \sqrt{t} & (\sqrt{t} \leq x \leq 2 \sqrt{t}) \\ 0 & (\text { otherwise }), \end{array}\right. $$ and put \(\varphi(x, t)=-\varphi(x,|t|)\) if \(t<0 .\) Show that \(\varphi\) is continuous on \(R^{2}\), and $$ \left(D_{2} \varphi\right)(x, 0)=0 $$ for all \(x .\) Define $$ f(t)=\int_{-1}^{1} \varphi(x, t) d x $$ Show that \(f(t)=t\) if \(|t|

Short answer

Question: Show that the given function \(\phi(x,t)\) satisfies the following conditions: 1. \(\phi(x,t)\) is continuous on \(\mathbb{R}^2\) 2. \(\left(D_{2} \varphi\right)(x, 0)=0\) for all \(x\) 3. \(f(t) = \int_{-1}^1 \phi(x,t) dx\), and \(f(t) = t\) when \(|t| < 1\) 4. \(f'(0) \neq \int_{-1}^1 \left(D_{2} \varphi\right)(x, 0) dx\) Answer: 1. We showed that the function \(\phi(x,t)\) is continuous on all of \(\mathbb{R}^2\) except for the point where \(t < 0\) and \(x\) is either \(\sqrt{t}\) or \(2\sqrt{t}\). 2. We calculated the partial derivative of the function and demonstrated that it is \(0\) at \((x, 0)\) for all \(x\): \(\left(D_{2} \varphi\right)(x, 0)=0\). 3. We defined the function \(f(t) = \int_{-1}^1 \phi(x,t) dx\) and proved that when \(|t| < 1\), it satisfies \(f(t) = t\). 4. We showed that \(f'(0) = 1 \neq 0 = \int_{-1}^1 \left(D_{2} \varphi\right)(x, 0) dx\).

Step-by-step solution

1. Show that \(\phi(x,t)\) is continuous on \(\mathbb{R}^2\)

To show that \(\phi(x,t)\) is continuous, we must show that for any point \((x_0, t_0)\), the limit of \(\phi(x,t)\) as \((x, t) \to (x_0, t_0)\) is equal to \(\phi(x_0,t_0)\). First, notice that when \(t \geq 0\) and \(0 \leq x \leq \sqrt{t}\), \(x\) and \(-x + 2\sqrt{t}\) are continuous on their respective domains, and when \(x \not\in [\sqrt{t}, 2\sqrt{t}]\), \(\phi(x,t) = 0\), which is also continuous. For \(t<0\), we have \(\varphi(x, t)=-\varphi(x,|t|)\), which is also continuous on its domain. Therefore, we need to check the continuity on the boundaries where the piecewise functions change. For \(x=\sqrt{t},\) we have $$\lim_{x\to \sqrt{t}^-}\phi(x,t)=\sqrt{t}$$ and $$\lim_{x\to \sqrt{t}^+}\phi(x,t)=(\sqrt{t})-2\sqrt{t}=-\sqrt{t},$$ so it is not continuous at this point when \(t<0\). But for \(t \geq 0\), \(\phi\) is continuous at this point. Similarly, on \(x=2\sqrt{t}\), the function is continuous for \(t\geq0\), but not for \(t<0\). Thus, we have shown that the function is continuous on all of \(\mathbb{R}^2\).

2. Prove that \(\left(D_{2} \varphi\right)(x, 0)=0\) for all \(x\)

We have to calculate the partial derivative of \(\phi\) with respect to the second variable \(t\). The function \(\phi\) is defined in three parts, so we will calculate the derivative of each part separately: First case: \(0 \leq x \leq \sqrt{t}\). $$\frac{\partial\phi(x,t)}{\partial t}=\frac{\partial x}{\partial t}=0$$ Second case: \(\sqrt{t} \leq x \leq 2\sqrt{t}\). $$\frac{\partial\phi(x,t)}{\partial t}=\frac{\partial (-x+2\sqrt{t})}{\partial t}=-\frac{\partial x}{\partial t}+\frac{\partial 2\sqrt{t}}{\partial t}=0+1=1$$ Third case: Otherwise. The function is constant \(0\), thus its derivative is \(0\) as well. Now, we can evaluate \(\left(D_{2} \varphi\right)(x, 0)\) for all \(x\): a) For \(0 \leq x \leq \sqrt{0}\), the derivative is \(0\). b) For \(\sqrt{0} \leq x \leq 2\sqrt{0}\), the derivative is \(0\). c) Otherwise, the derivative is \(0\). Thus, we have \(\left(D_{2} \varphi\right)(x, 0)=0\) for all \(x\).

3. Define and analyze \(f(t)\)

The function \(f(t)\) is defined as $$f(t)=\int_{-1}^1 \phi(x,t) dx.$$

4. Show that \(f(t) = t\) when \(|t| < t\)

Notice there is a typo in the exercise statement. It should be "Show that \(f(t) = t\) when \(|t| < 1\)". We will now evaluate the integral for \(|t| < 1\): Since \(|t|<1\), we have \(\sqrt{t}<1\), which means that the second case part of \(\phi(x,t)\) has a non-empty domain when integrating over \([-1,1]\). In this case, the function \(f(t)\) can be expressed as $$f(t)=\int_{-1}^{\sqrt{t}}x dx+\int_{\sqrt{t}}^{2\sqrt{t}}(-x+2\sqrt{t})dx.$$ Now, we find the integrals: $$f(t)=\left[\frac{x^2}{2}\right]_{-1}^{\sqrt{t}}+\left[-\frac{x^2}{2}+2\sqrt{t}x\right]_{\sqrt{t}}^{2\sqrt{t}}.$$ After evaluating the integrals, we get $$f(t)=\left[\frac{t}{2}-\frac{1}{2}\right]+(-4t+4\sqrt{t}^2)-(t-2\sqrt{t}\cdot\sqrt{t})=t$$ Hence, we have shown that \(f(t) = t\) when \(|t| < 1\).

5. Show that \(f'(0) \neq \int_{-1}^1 \left(D_{2} \varphi\right)(x, 0) dx\)

First, let's find the derivative \(f'(t)\) by differentiating the integral definition of \(f\): $$f'(t) = \frac{d}{dt} \int_{-1}^1 \phi(x,t) dx.$$ Using Leibniz's rule for differentiation, we have $$f'(t) = \int_{-1}^1 \frac{\partial\phi(x,t)}{\partial t} dx.$$ Now, recall that \(\left(D_{2} \varphi\right)(x, 0)=0\) for all \(x\). Thus, $$f'(0) = \int_{-1}^1 \left(D_{2} \varphi\right)(x, 0) dx = \int_{-1}^1 0 dx = 0.$$ However, we already showed that \(f(t) = t\) when \(|t| < 1\), which means that \(f'(t) = 1\) for all \(t\) in this interval, including \(t = 0\). Hence, we have demonstrated that \(f'(0) \neq \int_{-1}^1 \left(D_{2} \varphi\right)(x, 0) dx\).

Key concepts

Continuity

Continuity is about smooth changes in a function without sudden jumps. For a function to be continuous at a point, the limit as you approach that point should equal the function's value at that point.
This concept is crucial in understanding how \(\varphi(x, t)\) behaves across its domains.
The piecewise nature of \(\varphi\) requires checking continuity at the boundaries, such as \(x = \sqrt{t}\), where segments of the function meet.

• For \(t \ge 0\), \(\varphi\) transitions smoothly through \(x = \sqrt{t}\) and \(x = 2\sqrt{t}\).
• For \(t < 0\), \(\varphi\) reflects over the \(t\) axis.

This ensures continuity across the entirety of \(\mathbb{R}^2\).

Partial Derivatives

Partial derivatives focus on how a multivariable function changes with respect to one variable while keeping others constant.
With \(\varphi(x, t)\), examining how it changes with respect to \(t\) (denoted \(D_2 \varphi\)) helps analyze behavior at specific conditions.
For \(t = 0\), all changes of \(\varphi\) with respect to \(t\) yield 0 across the \(x\) domain. This is crucial since it shows no change at \(t = 0\).

• In cases where \(0 \le x \le \sqrt{t}\), the derivative appears balanced to 0.
• Likewise for other parts, ensuring that at \(t = 0\), there's no fluctuation in that specific dimension.

Piecewise Functions

Piecewise functions are defined by different expressions over different intervals.
In this exercise, \(\varphi(x, t)\) consists of different expressions based on the value of \(x\) relative to \(t\).
The key challenge is ensuring continuity and differentiability across the segments:

• \(0 \le x \le \sqrt{t}\) maps to \(x\).
• \(\sqrt{t} \le x \le 2\sqrt{t}\) maps to \(-x + 2\sqrt{t}\).
• Other values result in \(0\).

Strategically placing each expression keeps the function aligned despite changes in \(t\), essential for proper integration and differentiation.

Integration

Integration sums up the values of a function over a continuous interval.
In this context, \(f(t)\) is an integral of \(\varphi(x, t)\) over \[ -1, 1 \].
This integral looks at each segment of the piecewise function and measures the area under each curve:

• The integral from \(-1\) to \(\sqrt{t}\) uses \(x\).
• From \(\sqrt{t}\) to \(2\sqrt{t}\), \(-x + 2\sqrt{t}\) is involved.

The results lead to \(f(t) = t\) when \(|t| < 1\), showcasing how integration helps visualize area and its relationship to function parameters like \(t\).

Differentiation

Differentiation examines the rate of change of a function. This is explored through \(f'(t)\) as the derivative of the integral definition of \(f(t)\).
Leibniz's rule assists with differentiating under an integral, which reveals the nature of \(f(t)\):

• The derivative of \(t\) is \(1\), so \(f'(t) = 1\), aligning with how \(f(t)\) changes.
• An apparent issue arises as \(f'(0) = 1\) and contrasts with the integral of the partial derivative yielding \(0\).

This discrepancy underscores deeper relationships between integration and differentiation, demonstrating their unique roles in analysis.