Sol'n to PDE Integral: \frac{2b^2a}{3}

[Cyrus]

My PDE book does the following:

$$\int \phi_x^2 dx$$

Where,

$$\phi_x = b-\frac{b}{a} |x|$$

for $$|x|> a$$ and x=0 otherwise.

Strauss claims:

$$\int \phi_x^2 dx = ( \frac{b}{a} ) ^2 2a$$

However, I think there is a mistake. It can be shown that:

$$\frac{-3a}{b}(b- \frac{b|x|}{a})^3$$ is a Soln. Evaluate this between 0<x<a and you get:

$$\frac{b^2 a}{3}$$

Because the absolute value function is symmetric, its twice this value:

$$\frac{2b^2 a}{3}$$

Unless I goofed, I think the book is in error.

*Note: Intergration is over the whole real line.

 

[Dick]

I think you must mean |x|<a, right? Otherwise the integral isn't defined. I tried integrating (b-bx)^2 from 0 to a and I don't get anything close to either answer. Can you clarify?

 

[Cyrus]

O crap, its b-b/a|x| sorry. See above I fixed it.

 

[Dick]

Then I'm getting the same result as you.