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Tutor profile: Chelsea B.

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Chelsea B.
Assistant lecturer in partial differential equations
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Questions

Subject: Calculus

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Question:

a) What does it mean for a series to be absolutely convergent? b) Is the series $$\sum_{n=1}^{\infty} \frac{(-1)^n n^3}{3^n}$$ absolutely convergent?

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Chelsea B.
Answer:

a) A series $$\sum a_n$$ is absolutely convergent if the series of the absolute values $$\sum |a_n|$$ is convergent. b) We can use the ratio test for series to test absolute convergence as follows: Set $$a_n =\frac{(-1)^n n^3}{3^n}$$. Then, $$\frac{|a_{n+1}|}{|a_n|} = \frac{3^n}{n^3}\cdot\frac{(n+1)^3}{3^{n+1}} = \frac{1}{3}(1+\frac{1}{n})^3 \to \frac{1}{3}$$ as $$n\to \infty$$. Therefore, $$\lim_{n\to \infty} \frac{|a_{n+1}|}{|a_n|} = \frac{1}{3} <1$$. By the ratio test, the series $$\sum_{n=1}^{\infty} a_n$$ is absolutely convergent.

Subject: Differential Equations

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Question:

Consider $$x'(t)=x-\frac{1}{x^2}$$, $$x(0)=-1$$. How do you find a solution to this initial value problem?

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Chelsea B.
Answer:

Note that the differential equation may be written as $$\frac{dx}{dt}= \frac{x^3-1}{x^2}$$. Separating variables, we have $$\frac{x^2}{x^3-1}\frac{dx}{dt} =1$$. We can now solve by integrating both sides with respect to $$t$$ to obtain $$\int \frac{x^2}{x^3-1}\frac{dx}{dt} ~ dt=\int ~ dt$$, which is equivalent to $$\int \frac{x^2}{x^3-1} ~ dx= t+C$$, where $$C$$ is a constant. Setting $$g(x)=x^3-1$$ so that $$g'(x)=3x^2$$, we can write our equation as $$\int \frac{g'(x)}{g(x)} ~ dx= t+C$$. Evaluating the integral on the left gives $$\int \frac{g'(x)}{g(x)} ~ dx =\ln|g(x)|$$ and so $$\frac{1}{3} \ln|x^3-1| =t+C$$. Simplifying gives $$|x^3-1| =e^{3t+B}$$, for $$B$$ a constant. Since $$e^s>0$$ for all $$s\in\mathbb{R}$$, we may drop the absolute value to obtain $$x(t)=(Ae^{3t}+1)^{\frac{1}{3}}$$. Finally, using the initial condition, we find that $$x(t)=(1-2e^{3t})^{\frac{1}{3}}$$.

Subject: Partial Differential Equations

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Question:

Consider the following boundary value problem for the heat equation $$ \partial_tu=2\partial_x^2u, \ x\in(0,2),\ t>0 $$ $$u(0,t)=1, ~ u(2,t)=-1$$ $$u(x,0)=0 $$. How would you convert the above problem to a problem with homogeneous boundary conditions?

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Chelsea B.
Answer:

To convert the BVP to a problem with homogeneous boundary conditions, we consider its corresponding equilibrium solution, denoted $$u_E$$. (where $$\partial_tu_E=0$$) By definition, $$u_E$$ satisfies the differential equation $$u_E''(x) =0$$ with boundary conditions $$u_E(0)=1$$ and $$u_E(2)=-1$$. We use the boundary conditions to solve for $$u_E$$ to obtain $$u_E(x)=1-x$$. Let $$w=u-u_E$$. Since $$u$$ is a solution of the initial BVP and $$u_E$$ is the corresponding equilibrium solution, $$w$$ satisfies the problem with zero boundary conditions, given by $( \partial_tw =2\partial_x^2w\\ w(0,t)=0, ~ w(2,t)=0\\ w(x,0)=x-1 $)

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