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Let's consider heat equation $$\frac{\partial u}{\partial t} = \frac{\partial^2u}{\partial x^2}$$ or equivalently $$u_t = u_{xx}.$$ Of course $t > 0$ and $x \in [0, 1]$.

It's defined on a circle, thus

  1. $u(0, t) = u(1, t)$
  2. $u_x(0, t) = u_x(1, t)$

Moreover the initial condition is given by $$f(x) = u(x, 0)$$.

Using the separation of variables I managed to show that $$u(x,t) = X(x)T(t)$$ where $$X(x) = C_1 \sqrt\frac{1}{\lambda} \sin \big(\sqrt \lambda x\big) + C_2 \sqrt \frac{1}{\lambda} \cos \big(\sqrt \lambda x \big)$$ $$T(t) = Ae^{- \lambda t}$$

Using assumptions $(1)$ and $(2)$ we do get the following system of equations \begin{cases} C_2 = C_1 \sin(\sqrt \lambda) + C_2 \cos(\sqrt \lambda)\\C_1 = C_1 \cos(\sqrt \lambda) - C_2 \sin(\sqrt \lambda)\end{cases}

How can I solve the system above?

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1 Answer 1

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$$ C_2(1-\cos u) = C_1\sin u\\ C_1(1-\cos u) =-C_2\sin u $$

so $\cos u = 1$ and $\sin u = 0$ giving $\sqrt{\lambda} = 0+2k\pi,\ \ k = 0,1\cdots$

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