# Solve the initial value problem (PDE)

$$2u_t + x(1+t)u_x = u^2$$ $$u(x,0)=x$$

Tried to do this by method of characteristics.

So, using change of variables as follows $$\left\lbrace \begin{array}{c} p=x \\ q = 2\ln(x)-(t+t^2/2) \end{array}\right.$$

But I am stuck from this point onward. I seem to think, I might have done something wrong here.

Any hints? Thanks!

• Hint: Do the change $v=u^{-1}$. This simplifies the equation. – Pocho la pantera Sep 16 '13 at 17:45

$$2\frac{\partial u}{\partial t}+x(1+t)\frac{\partial u}{\partial x}=u^2\quad\text{with}\quad u(x,0)=x$$We will solve the above inhomogeneous Cauchy problem with variable coefficients using the method of characteristic curves. First, note this corresponds to $\dfrac{du}{ds}$$\frac{du}{ds}=\frac{\partial u}{\partial t}\frac{dt}{ds}+\frac{\partial u}{\partial x}\frac{dx}{ds}$$where clearly$\dfrac{du}{ds}=u^2,\dfrac{dt}{ds}=2,\dfrac{dx}{ds}=x(1+t)$. This yields$t(s)=2s\$ and so $$\frac{dx}{ds}=x(1+2s) \\ \frac{1}{x}\frac{dx}{ds}=1+2s \\ \int\frac1x\frac{dx}{ds}=\int(1+2s)~ds\\\log x=s+s^2\\ x(s)=\exp(s+s^2)$$Similarly,$$\frac{du}{ds}=u^2\\ \frac1{u^2}\frac{du}{ds}=1\\ \int\frac1{u^2}\frac{du}{ds}ds=\int ds \\ -\frac1u=s\\ u(s)=-\frac1{s}$$Can you take it from here?