Method of Characteristics: How to solve PDE

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Recommends Matthew Cassell · 2015-03 · Mathematics Stack Exchange · on “Can't understand how equation was arrived at in method of characteristics?”

An explanation in the comments section wouldn't do it justice and I'm about to go to sleep. See here [this video] for a good video explanation.

Recommends hbp · 2015-12 · Mathematics Stack Exchange · on “difficulty in solving first order PDE: $ (y+xz)z_x + (x+yz)z_y = z^2 - 1$”

Just to expand d-e [LINK] 's comment. Some instructive tutorials can be found here (I do not know the author) Dr. Chris Tisdell, Method of Characteristics: How to solve PDE [this video] . Dr. Chris Tisdell, How to solve quasi-linear PDE [LINK] . We will follow the method described there. The characteristic equation is $$ \frac{dx}{y+xz} = \frac{dy}{x+yz} = \frac{dz}{z^2 - 1}. $$ Using the fact $$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow \frac{a+ c}{b+ d} \Longleftrightarrow \frac{a- c}{b- d}. \qquad (1) $$ we get $$ \frac{d(x+y)}{(x+y)(z+1)} = \frac{dz}{(z+1)(z-1)}. $$ which is equivalent to $$ \frac{d(x+y)}{x+y} = \frac{d(z-1)}{z - 1}. $$ Integrating this yields the first constant of integr

Recommends Pedro Delfino · 2018-11 · Mathematics Stack Exchange · on “Solve this Semi-Linear PDE (Partial Differential Equation) with the Characteristic Method”

I need to solve this linear PDE: $3u_x - 4u_y = y^2$ The initial condition provided is: $ u (0,y)= sin(y)$ I need to use the Characteristic Method . I learned the method from this video [this video] . I have reached an answer. However, I am not sure if it is wright. My intermediate steps are: First constant: $c_1= y + \frac{4}{3}x $ Second constant: $c_2= \frac{y^3}{3} + 4u $ Using an arbitrary function G to make the relation between both constants, $c_2 =G(c_1) $ , we have that: $\frac{y^3}{3} + 4u = G(y + \frac{4}{3}x) $ With the initial condition we have: $G(y) = \frac{y^3}{3} +4sin(y)$ After the definition of $G(y)$ above , I inputed the value of $c_1$ , having: $G(y + \frac{4}{3}x) = \frac{(

Cites Pedro Delfino · 2018-11 · Mathematics Stack Exchange · on “Nonhomogeneous Semi-Linear PDE with the Characteristic Method”

I need to solve this semi-linear PDE: $u_x - 3u_y = \sin (y) + \cos (x)$ The initial condition provided is: $ u (t,t)= t^2$ I need to use the Characteristic Method . I learned the method from this video [this video] . I have reached an answer. However, I am not sure if it is wright. I am insecure in two different moments of my solution. My intermediate steps are: First constant: $c_1= y + 3x $ Second constant: $c_2= x\cos y -\cos x-u$ The second constant above is my first source of insecurity . I needed to solve this equality as an ODE: $\frac{dx}{1}= \frac{du}{\sin x + \cos y}$ I did so with this approach: $\int (\sin x + \cos y) dx = \int 1 du $ Using integral properties: $(\int \sin x dx) + (\

Cites Axion004 · 2019-01 · Mathematics Stack Exchange · on “Solve quasi-linear PDE $uu_x+yu_y=x$ through the method of characteristics”

Related Question: PDE characteristic equation [LINK] I am doing a PDE problem from a course that I will take next semester. I watched this [this video] video in advance. It appears that I am doing something wrong with the initial data. Consider the following first order PDE \begin{cases} uu_x+yu_y=x, \\ u(x,1)=2x. \end{cases} State the condition which guarantees that the initial surface $\Gamma$ is not characteristic. Use the method of characteristics to find a solution of the PDE and discuss for which $(x,y)\in\mathbb R^2$ the solution exists. For the second part, the general form of the method of characteristics is $au_x+bu_y=f$ . Therefore, we have that $$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{f}