Register to reply 
Solve PDE Method of Characteristics help 
Share this thread: 
#1
Sep2011, 05:54 PM

P: 20

The problem: Solve for u(x,y,z) such that
[tex] xu_x+2yu_y+u_z=3u\; \;\;\;\;u(x,y,0)=g(x,y) [/tex] So I write [itex]\frac{du}{ds}=3u \implies \frac{dx}{ds}=x,\; \frac{dy}{ds}=2y\;\frac{dz}{ds}=1 .[/itex] Thus [tex]u=u_0e^{3s},\;\;x=x_0e^{s}\;\;y=y_0e^{2s}\;\;z=s+z_0 [/tex] but from here I can't figure out what to do, there are several ways I can write s....I have only done the method of characteristics before with two variables, and those are pretty much the only examples I can find. Can someone help please? 


#2
Sep2011, 06:19 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682

If [itex]x= x_0e^s[/itex] and [itex]y= y_0e^{2s}[/itex], you have [itex]s= ln(x/x_0)= (1/2)ln(y/y_0)[/itex] so that ln(x/x_0) ln(y^{1/2}/y_0^{1/2})= ln((y_0^{1/2}/x_0)(x/y^{1/2}= 0 which then gives [itex]x/y^{1/2}= constant[/itex]. Each curve with different constant is a characteristic. With [itex]z= s+ z_0[/itex] that gives also [itex]s= zz_0= ln(x/x_0)[/itex], have [itex]x/x_0= e^{zz_0}= e^{z_0}e^z[/itex] so [itex]x= (constant)e^z[/itex] or [itex]xe^{z}= constant[/itex]. Putting [itex]s= zz_0= ln(y^{1/2}/y_0^{1/2})[/itex], we get [itex]y^{1/2}e^{z}= constant[/itex]. We want to use those "characteristics" or "characteristic curves" as axes: let [itex]u= x/y^{1/2}[/itex], [itex]v= xe^{z}[/itex], and [itex]w= y^{1/2}e^{z}[/itex]. 


Register to reply 
Related Discussions  
PDE problem, Solve using Method of Characteristics  Calculus & Beyond Homework  7  
Method of characteristics  Calculus & Beyond Homework  1  
Method of characteristics  Calculus & Beyond Homework  0  
Method of Characteristics  Calculus & Beyond Homework  0  
Solve equation using the Method of Characteristics  Calculus & Beyond Homework  0 