First order Linear PDE, Method of Characteristics

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SUMMARY

The discussion focuses on solving the first-order linear partial differential equation (PDE) given by \(x u_{x} + y u_{y} = 1 + y^2\) with the initial condition \(u(x,1) = 1 + x\). The method of characteristics is employed to derive solutions parametrically and in terms of \(x\) and \(y\). Two distinct solutions arise from the parametric and direct approaches, leading to confusion regarding the initial curve and the function \(G(y/x)\). The correct solution is confirmed as \(u = \frac{x}{y} + \ln y + \frac{y^2}{2} + \frac{1}{2}\), which satisfies both the PDE and the initial condition.

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  • Understanding of first-order linear partial differential equations (PDEs)
  • Familiarity with the method of characteristics
  • Knowledge of initial value problems in PDEs
  • Basic calculus, including differentiation and integration techniques
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Homework Statement


x*u_{x} + y*u_{y}= 1 + y^2
u(x,1) = 1+ x; -infinity < x < +infinity

Solve this parametrically and in terms of x and y

Homework Equations



We are supposed to solve this using the method of characteristics

The Attempt at a Solution



My problem is that solving the equation parametrically and with x and y give two different solutions.

I will first try to solve this parametrically, using variables s and t.

let x = s along the initial line y = 1 Is this assumption correct, or should the initial line be y = 0? I have only solved problems where the initial line as been explicitly stated. I chose y = 1 because it made the equations work out, and because the initial condition was u(x,1), so I am not sure if that is correct.

So,

dx/dt = x => x = s * e^t

dy/dt = y => y = Ce^t, since the initial line is y = 1, C= 1; y = e^t

du/dt = 1 + y^2 => u = t + \frac{e^{2t}}{2} + D

Now apply the initial condition:

u(x,1) = 1 + x = 0 + 1/2 + D => D = x + 1/2

So,

u = t + \frac{e^2t}{2} + x + 1/2

since y = e^t that means t = ln|y| (|| is absolute value... I would also like to know if y is always positive so I can remove the absolute value.)

So the solution for u is:

u = ln|y| + (y^2)/2 + x + 1/2

That is solving the equation parametrically, now I am a supposed so solve the equation treating x as a parameter variable and s as the other parameter variable:

Rewrite the equation to:

u_{x}+ (y/x)*u_{y} = (1 + y^2)/x (divided through by x)

so,

dy/dt = y/x => y = k(s)x (k is a function of s)

du/dt = (1+y^2)/x => u = (1+y^2)ln(x) + C(s) = (1+y^2)ln(x) + G(y/x)

Applying the initial condition:

u(x,1) = 1+x = 2ln(x) + G(1/x)

G(1/x) = 1+x - 2ln(x)

This is where I get confused:

First, the solutions I derived don't look at all like each other, and second for the later solution, I don't see how G(1/x) will tell me what G(y/x) is. I would appreciate any and all help, Thank You.
 
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I've done the entire problem a second time through and I get the same answer. My main concerns are:

What is the initial curve and why? y = 1?

When solving the problem in terms of x,y (the later solution), what is G(y/x)? I've tried applying the initial condition but I just get an equation for G(1/x). Again, thanks in advance.
 
The correct answer is:

u=x/y + ln y + y^2/2 + 1/2

You can check that this satisfies both the PDE and the initial condition:

u(x,1)=1+x

If you're still interested in this I'll try to write more. I haven't done this stuff for 30 years so I'm a little rusty.
 

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