Exercise \(\mathbf{1}\)

Question

Consider the following statement. Let \(g: \mathbb{R} \rightarrow \mathbb{R}\) denote the function given, for \(x \in \mathbb{R}\), by \(g(x)=3 x-n\), where \(n\) is the unique integer sucht that \(x \in[n / 3,(n+1) / 3)\). The Fourier series of \(g\) converges pointwise to \(g\).
Determine if the statement is true or false and give a justification for your answer.

Solution

First we draw the graph of \(g\) where \(x\in [-1,1]\).

By the Dini's Theorem, we can observe that

\[\frac{g(0^-)+g(0^+)}{2} = 0.5 \neq g(0) =0 \]

Which means the value of Fourier's series is not the same as the funtion \(g\) when \(x = 0\) .

Hence the Fourier series of \(g\) does not converge pointwise to \(g\).

Exercise 2

question

Suppose that \(s, t\) and \(z: \mathbb{R}^{2} \rightarrow \mathbb{R}\) are functions of two indepedent variables \(x\) and \(y\), with \(s\) and \(t\) given explicitly by \(s(x, y)=x^{2} y\) and \(t(x, y)=x^{2} y+x y^{2}\), and with \(z\) given implictly by the partial differential equation

\[2 y \frac{\partial z}{\partial y}-x \frac{\partial z}{\partial x}=3 x y^{2} \]

with bounday condition \((\partial z / \partial x)(0, y)=(\partial z / \partial y)(1, y)\) and initial condition \(z(0,0)=42\). By using a change of variables, from \(x\) and \(y\) to \(s\) and \(t\), find the general solution for \(z\) in terms of \(x\) and \(y\). For the purposes of this question, you may assume that there exists a function \(v: \mathbb{R}^{2} \rightarrow \mathbb{R}\) such that \(v(s(x, y), t(x, y))=z(x, y)\) for all \((x, y) \in \mathbb{R}^{2}\).

Solution

Assume that \(z(x, y) =v(s(x, y), t(x, y))\) for all \((x, y) \in \mathbb{R}^{2}\).

then

\[\frac{\partial z}{\partial y} = \frac{\partial v}{\partial s}\frac{\partial s}{\partial y} + \frac{\partial v}{\partial t}\frac{\partial t}{\partial y} \\ \frac{\partial z}{\partial x} = \frac{\partial v}{\partial s}\frac{\partial s}{\partial x} + \frac{\partial v}{\partial t}\frac{\partial t}{\partial x} \]

Since \(s(x,y)=x^2y\quad t(x,y)=x^2y+xy^2\), we get

\[\frac{\partial s}{\partial x} = 2xy \\ \frac{\partial s}{\partial y} = x^2 \\ \frac{\partial t}{\partial x} = 2xy + y^2 \\ \frac{\partial t}{\partial y} = x^2 + 2xy \]

then

\[\frac{\partial z}{\partial y} = \frac{\partial v}{\partial s}x^2 + \frac{\partial v}{\partial t}(x^2 + 2xy)\quad \tag \dagger \\ \frac{\partial z}{\partial x} = \frac{\partial v}{\partial s}2xy + \frac{\partial v}{\partial t} (2xy + y^2) \]

Take the equations above into \(2 y \frac{\partial z}{\partial y}-x \frac{\partial z}{\partial x}=3 x y^{2}\), we get

\[2 y (\frac{\partial v}{\partial s}x^2 + \frac{\partial v}{\partial t}(x^2 + 2xy))\\ - x (\frac{\partial v}{\partial s}2xy + \frac{\partial v}{\partial t} (2xy + y^2))=3 x y^{2}\\ \Rightarrow \frac{\partial v}{\partial t} 3xy^2 = 3xy^2\\ \]

Suppose \(x\neq 0\) and \(y \neq 0\), then

\[\frac{\partial v}{\partial t} = 1 \]

Integrate both sides,

\[v = t + m(s) \]

Where \(m\) is an arbitrary function. Substitute \(v\) by \(z\) ,

\[z(x,y) = x^2y + xy^2 + m(x^2y) \]

From the bounday condition \(\frac{\partial z}{\partial x}(0,y) = \frac{\partial z}{\partial y}(1,y)\),

\[\frac{\partial z}{\partial x} = 2xy + y^2 + 2xy \cdot m'(x^2y) \\ \frac{\partial z}{\partial y} = x^2 + 2xy + x^2 \cdot m'(x^2y) \\ \Rightarrow y^2 = 1 + 2y + m'(y)\\ \Rightarrow m'(y) = y^2- 2y -1 \\ \Rightarrow m(y) = y^3/3 - y^2 -y + C \]

because \(z(0,0)=42\),

\[z(0,0) = 0 + m(0) = 42 \\ \Rightarrow C = 42 \\ \Rightarrow m(y) = y^3/3 - y^2 -y + 42\\ \Rightarrow m(x^2y) = (x^2y)^3/3 - (x^2y)^2 - x^2y + 42 \]

Hence

\[z(x,y) = x^2y + xy^2 + (x^2y)^3/3 - (x^2y)^2 - x^2y + 42 \\ (Simplify) \quad z(x,y) = 42 + xy^2 - x^4 y^2 + (x^6 y^3)/3 \quad (x\neq 0, y \neq 0) \]

Consider \(x\neq 0, y = 0\), in this case \(z = 42\) can be easily calculated. Similarly, when \(x= 0, y \neq 0\), \(z= 42\).

Therefore, the answer is

\[z(x,y) = 42 + xy^2 - x^4 y^2 + (x^6 y^3)/3 \]

标签：mathbb,frac,42,2y,2xy,MVA,partial,Exercise
来源： https://www.cnblogs.com/kion/p/15975497.html