# Simple Calculus with PDEs

Let

\begin{align} u(x,t)&=F(x+ct)+G(x-ct),\\ u(x,0)&=f(x),\\ u_t(x,0)&=g(x). \end{align}

How can I show that

\begin{align} F(x)&=\frac12f(x)+\frac1{2c}\int_{x_0}^xg(s)\,ds+C,\\ G(x)&=\frac12f(x)-\frac1{2c}\int_{x_0}^xg(s)\,ds+C\text{ ?} \end{align}

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Hint: From the first three equalities you have

$$f(x) = F(x) + G(x),$$

$$g(x) = c\,\Bigl(F'(x)-G'(x)\Bigr).$$

Substitute these into the two expressions you have at the end.

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That should be $f(x) = F(x) + G(x)$. – Robert Israel Dec 3 '12 at 8:32
@RobertIsrael thanks. – Antonio Vargas Dec 3 '12 at 8:33

I did the following:

\begin{align} u(x,0)&=F(x)+G(x)=f(x),\\ u_x(x,0)&=F'(x)+G'(x)=f'(x),\\\\ u_t(x,t)&=cF'(x+ct)-cG'(x-ct),\\ u_t(x,0)&=F'(x)-G'(x)=\frac{1}{c}g(x),\\\\ u_x(x,0)+u_t(x,0)&=2F'(x)=f'(x)+\frac{1}{c}g(x),\\ &\Longrightarrow F'(x)=\frac{1}{2}f'(x)+\frac{1}{2c}g(x),\\ \int_{x_0}^{x}F'(s)\,ds&=\frac{1}{2}\int_{x_0}^{x}f'(s)\,ds+\frac{1}{2c}\int_{x_0}^{x}g(s)\,ds,\\ F(x)&=\frac{1}{2}f(x)+\frac{1}{2c}\int_{x_0}^{x}g(s)\,ds+C. \end{align}

Equation $G(x)$ can be derived similarly.

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