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I've currently started reading 'Lectures on Partial Differential Equations' by Faris.

Page 44 he states the following forward equation: $$J=a(y)p-\frac{1}{2}\frac{\partial \sigma(y)^2p}{\partial y} = 0$$

I understand how to solve this and obtain the following: $$I(y) = \int_{y_0}^{y}\frac{2}{\sigma(z)^2}a(z)dz$$ i.e. integrate once and separate the variables. He then states solving for $p$ produces: $$p(y)=C\frac{2}{\sigma(y)^2}exp(I(y))$$ I don't understand where the $\frac{2}{\sigma(y)^2}$ comes from. Can someone guide me in the right direction please?

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1 Answer 1

It's been four days and you've managed to solve your own question. I'll now share it with you.

Note $J=0$ and $\sigma(y)$ is a function on of variable (i.e. there is no time variable).

  1. Rearrange the Forward Equation and use standard differential operators: $$a(y)p = \frac{1}{2}\frac{d(\sigma(y)^2p)}{dy}$$

  2. Multiply through by 2 and use the product rule: $$2a(y)p=\frac{d(\sigma(y)^2)}{dy}p+\frac{d(p)}{dy}\sigma(y)^2$$

  3. Now divide through by p and divide through by $\sigma(y)^2$: $$\frac{2a(y)}{\sigma(y)^2}=\frac{1}{\sigma(y)^2}\frac{d(\sigma(y)^2)}{dy}+\frac{1}{p}\frac{d(p)}{dy}$$

  4. Use an integration 'trick': $$\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy=\int_{}{}\frac{1}{\sigma(y)^2}\frac{d(\sigma(y)^2)}{dy}dy+\int_{}{}\frac{1}{p}\frac{d(p)}{dy}dy$$

  5. Simplify! $$\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy=\int_{}{}\frac{d(\sigma(y)^2)}{\sigma(y)^2}+\int_{}{}\frac{d(p)}{p}$$

  6. Perform rhs integration: $$\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy=\ln \sigma(y)^2 +\ln P +\ln C$$

  7. Exponential and rearrange: $$p(y) = C\frac{1}{\sigma(y)^2}exp(\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy)$$

Note: I am not sure where Mr Faris obtained the '2' in the final solution. Perhaps someone more knowledgable will be able to explain. I believe it is a mistake and the constant of integration should be correct.

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