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Alyssa P.
MSc Mathematical Biology
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Linear Algebra
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Question:

Given the system: $$dx(t)/dt = a_{11}x + a_{12}y$$ $$dy(t)/dt = a_{21}x + a_{22}y$$ What is the equation for the eigenvalues of the system?

Alyssa P.
Answer:

The eigenvalues are found by solving for $$\lambda$$ in the formula $$det(A - \lambda I) = 0$$ where $$I$$ is the identity matrix. For the above problem, the system $$(A - \lambda I) = 0$$ takes the matrix form: $$\left[ \begin{array}{cc} a_{11} - \lambda & a_{12} \\ a_{21} & a_{22} - \lambda \end{array} \right] = 0$$ Finding the determinant requires taking the difference between the products of the diagonals: $$(a_{11}-\lambda)(a_{22} - \lambda) - a_{21}a_{12} = 0$$ Simplified, the equation for the eigenvalues reads: $$\lambda^{2} - (a_{11} + a_{22})\lambda - (a_{21}a_{12} - a_{11}a_{22}) = 0$$

Numerical Analysis
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Question:

Using Taylor's series, derive the second central divided difference of $$f$$ at a point $$x$$ with spacing $$h$$.

Alyssa P.
Answer:

Taylor's series takes the expanded form: $$f(x + h) \approx f(x) + hf'(x) + \frac{h^{2}}{2!}f''(x) + \frac{h^{3}}{3!}f'''(x) + \frac{h^{4}}{4!}f^{4}(x) + \cdots $$ Using the forward divided difference, expand Taylor's series to the 2nd derivative: $$f(x + h) = f(x) + hf'(x) + \frac{h^{2}}{2}f''(x)$$ Do the same using the backward divided difference: $$f(x-h) = f(x) - hf'(x) + \frac{h^{2}}{2}f''(x)$$ Then taking the sum: $$f(x+h) + f(x-h) = 2f(x) + h^{2}f''(x)$$ Solving for the second derivative results in the 2nd central divided difference: $$f''(x) = \frac{f(x+h) - 2f(x) + f(x-h)}{h^{2}}$$

Differential Equations
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Question:

Consider the PDE: $$(1 - M^{2})u_{xx} + u_{yy} = 0$$ What is the type of the above second order linear PDE for different values of M?

Alyssa P.
Answer:

A 2nd order linear PDE in 2 dimensions takes the general form: $$au_{xx} + 2bu_{xy} + cu_{yy} + du_{x} + eu_{y} + fu = g$$, for $$(x,y) \in \Omega \subset \Re^{2}$$ In order to determine the type of the PDE use the formula for the discriminant: $$D = b^{2} - ac$$ If $$D > 0$$, the PDE is hyperbolic If $$D = 0$$, the PDE is parabolic If $$D < 0$$, the PDE is elliptic For the above problem, $$a = (1 - M^{2})$$, $$b = 0$$, and $$c = 1$$ so $$D = M^{2} - 1$$. Therefore, when $$M = \pm 1$$ the PDE is parabolic; when $$M = 0$$ the PDE is elliptic; and for all other values of $$M$$ the PDE is hyperbolic.

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