# Classical solution to PDE using M.o.C

Mathematics Asked by xotix on October 27, 2020

Consider the transport equation
$$u_y+u^2u_x = 0 tag{1}$$
with initial condition
$$begin{cases}1 &, xleq 0\ 0 &,xgeq 1\sqrt{1-x} &, 0

Find the solution using the method of characteristics. Up to which time is the solution defined in a classical sense?

So we get

begin{align}x_t &= hat{u}^2 \ y_t &= 1 \ hat{u}_t &= 0end{align} tag{3}

with

begin{align}x_0(s) &= s \ y_0(s) &= 0 \ hat{u}_0(s) &= u(s,0)end{align} tag{4}

so we get

begin{align}x(t,s)&=hat{u}^2cdot t + s \ y(t,s)& = t \ hat{u}(t,s) &= u(s,0)end{align} tag{5}

We now invert the map $$(s,t) mapsto (x,y)$$:

$$underline{sleq0}:$$
$$x=y+s Rightarrow s=x-y text{with} xleq y tag{6}$$

$$underline{sgeq 1:}$$
$$x=s text{with} xgeq 1 tag{7}$$

$$underline{0
$$x=(1-s)cdot y + s = y-sy+s = y + s(1-y) quad Rightarrow quad s=frac{x-y}{1-y} tag{8}$$
with
$$yx>1 text{if} y>1 tag{9}$$

The characteristic are intersecting at the point $$(x,y)=(1,1)$$. Thus, the solution is defined up to $$y=1$$. We get:

$$u(x,y)=begin{cases} 1 &, xleq y \ 0 &, xgeq 1 \ sqrt{frac{1-x}{1-y}} &, y

Question: How exactly can I see that the lines intersect at $$(x,y)=(1,1)$$?

You should have written $$hat u(t,s) = u(0,s)$$. With this, the characteristics $$binom{x(t,s)}{y(t,s)}$$ are $$binom{x(t,s)}{y(t,s)} = begin{cases} binom{s + t}t & sle 0 \ binom{(1-s)t+s}t & sin(0,1) \ binom st & sge 1end{cases}$$

With the graph as a guide, It's clear that two lines of the first type $$(s,s'le 0)$$ don't intersect; similarly for $$s,s'ge 1$$; Two lines for $$s,s'in (0,1)$$ do intersect, and they do so at $$t=1$$:

$$binom{(1-s)t+s}t = binom{(1-s')t+s'}t iff (1-s)t+s = (1-s')t+s' iff s(t-1) = s'(t-1)$$ so if we have two different lines ($$sneq s'$$) then there is a solution at $$t=1$$. I'll leave the other cases to you, which should be easy to check.

Correct answer by Calvin Khor on October 27, 2020

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