I'm trying to solve the partial differential equation $$2u_x + u_t = x,\qquad u(x, 0) = f(x)$$ by the method of characteristics while not retaining an extra parameter. I'm able to find the characteristic lines fairly easily, but for whatever reason, I cannot find a final solution for $u(x, t)$ that satisfies both the initial condition and the PDE itself.

Here's my work so far:

Using $dx/2 = dt/1$, I've found that the characteristic curves are $x - 2t = c$, for arbitrary $c$.

Using $du / x = dx / 2$, I've found that $u = 1/4 x^2 + k$, for arbitrary $k$.

My next logic here was to say that $k = g(c) = g(x - 2t)$, for arbitrary function $g$.

This is where it begins to break, though. Normally, I'd substitute the initial condition $u(x, 0) = f(x)$ into the equation and find how $g$ relates to $f$, and then finally give an answer for $u$. However, the initial condition here is completely violated no matter what $g$ equals, since there is no way to get $1/4x^4$ to equal $x$ by changing $g$, so clearly there's a big gap in my logic. How would I be able to solve this problem using this general method but without this flaw? My professor said it is possible to solve this without introducing a parameter.

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    $\begingroup$ I get that $$u = \frac{x^{2}}{4} + g(x-2t)$$ and hence \begin{align} u(x,0) &= \frac{x^{2}}{4} + g(x) \\ &= f(x) \\ \implies g(x) &= f(x)-\frac{x^{2}}{4} \\ \implies g(x-2t) &= f(x-2t)-\frac{(x-2t)^{2}}{4} \end{align} $\endgroup$ – mattos Nov 11 '19 at 3:43

You correctly found two equations of characteristic curves : $$x - 2t = c$$ $$u = \frac14 x^2 + k$$ Then you correctly wrote $k = g(c)$ but you made a mistake at next step. You wrote $k= g(x - ct)$ which is not correct since $c=x-2t$. The correct writting is : $$k=g(x-2t)$$ $$u=\frac14 x^2+g(x-2t)$$ Then the condition $u(x,0)=f(x)=\frac14 x^2+g(x)$ gives $$g(x)=f(x)-\frac14 x^2$$

Now the function $g$ is determined, any variable $X$ : $$g(X)=f(X)-\frac14 X^2$$ We put this function into the general solution $u=\frac14 x^2+g(x-2t)$ where $X=x-2t$ : $$u=\frac14 x^2+f(x-2t)-\frac14 (x-2t)^2$$


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