# Solve quasi-linear PDE $uu_x+yu_y=x$ through the method of characteristics

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I am doing a PDE problem from a course that I will take next semester. I watched this video in advance. It appears that I am doing something wrong with the initial data.

Consider the following first order PDE

$$\begin{cases} uu_x+yu_y=x, \\ u(x,1)=2x. \end{cases}$$

1. State the condition which guarantees that the initial surface $$\Gamma$$ is not characteristic.
2. 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

$$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{f}$$ or $$\frac{dx}{u}=\frac{dy}{y}=\frac{du}{x}$$

In order to solve these equations, we need to find the value of two constants $$C_1$$ and $$C_2$$. First, let

$$\frac{dx}{u}=\frac{du}{x}$$

Then,

$$\frac{du}{dx}=\frac{x}{u}~~\Rightarrow~~~ udu = xdx ~~\Rightarrow~~~ \frac{u^2}{2}=\frac{x^2}{2}+C ~~\Rightarrow~~~ C_1=x^2-u^2$$

Next, let

$$\frac{dx}{u}=\frac{dy}{y}$$

Then,

$$\frac{dy}{dx}=\frac{y}{u} ~~\Rightarrow~~~ \frac{1}{y}dy = \frac{1}{u}dx ~~\Rightarrow~~~ \ln(y)=\frac{x}{u}+C ~~\Rightarrow~~~ C_2=y+e^{\frac{x}{u}}$$

So, we have that $$C_1=x^2-u^2$$ and $$C_2=y+e^{\frac{x}{u}}$$. We can combine the two constants such that $$C_2=F(C_1)$$ where $$F$$ is an arbitrary differentiable function. Then,

$$y+e^{\frac{x}{u}}=F(x^2-u^2)$$

We now need to apply our initial data. We are given that $$u(x,1)=2x$$. Therefore,

$$1+e^{\frac{1}{2}}=F(-3x^2)$$

This doesn't appear to help us find the solution. If instead we set $$C_1=F(C_2)$$ then

$$x^2-u^2=F(y+e^{\frac{x}{u}})$$

Applying the initial data produces

$$-3x^2=F(1+e^{\frac{1}{2}})$$

That also appears to be incorrect. I must have made a mistake in applying the initial data. How can we apply the initial data to solve for $$u$$?

$$\dfrac{dy}{dt}=y$$ , letting $$y(0)=1$$ , we have $$y=e^t$$

$$\begin{cases}\dfrac{dx}{dt}=u\\\dfrac{du}{dt}=x\end{cases}$$

$$\therefore\dfrac{d^2x}{dt^2}=x$$

$$x=C_1\sinh t+C_2\cosh t$$

$$\therefore u=C_1\cosh t+C_2\sinh t$$

Hence $$\begin{cases}x=C_1\sinh t+C_2\cosh t\\u=C_1\cosh t+C_2\sinh t\end{cases}$$

$$x(0)=x_0$$ , $$u(0)=F(x_0)$$ :

$$\begin{cases}C_1=F(x_0)\\C_2=x_0\end{cases}$$

$$\therefore\begin{cases}x=F(x_0)\sinh t+x_0\cosh t\\u=F(x_0)\cosh t+x_0\sinh t\end{cases}$$

$$\therefore\begin{cases}x_0=x\cosh t-u\sinh t=x\cosh\ln y-u\sinh\ln y\\F(x_0)=u\cosh t-x\sinh t=u\cosh\ln y-x\sinh\ln y\end{cases}$$

Hence $$u\cosh\ln y-x\sinh\ln y=F(x\cosh\ln y-u\sinh\ln y)$$

$$u(x,1)=2x$$ :

$$F(x)=2x$$

$$\therefore u\cosh\ln y-x\sinh\ln y=2x\cosh\ln y-2u\sinh\ln y$$

$$u(x,y)=\dfrac{x(\sinh\ln y+2\cosh\ln y)}{2\sinh\ln y+\cosh\ln y}$$

For the first question, on $$\Gamma$$ it is specified that $$u(x,y)=2x$$. Therefore, $$u_x=2$$ and $$u_y=0$$. Putting them into the PDE leads to $$uu_x+yu_y=4x$$ which is contradictory with $$uu_x+yu_y=x$$. So, the boundary condition is not on a characteristic.

For the second question, we should follow the method of characteristics. Lets first write the general form as $$au_x+bu_y=f$$. Therefore, we have that

$$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{f}$$ or $$\frac{dx}{u}=\frac{dy}{y}=\frac{du}{x}$$

In order to solve these equations, we need to find the value of two constants $$C_1$$ and $$C_2$$. First, let

$$\frac{dx}{u}=\frac{du}{x}$$

Then,

$$\frac{du}{dx}=\frac{x}{u}~~\Rightarrow~~~ udu = xdx ~~\Rightarrow~~~ \frac{u^2}{2}=\frac{x^2}{2}+C ~~\Rightarrow~~~ C_1=u^2-x^2$$

Next, observe that $$\frac{a}{b}=\frac{c}{d} \iff \frac{a+c}{b+d}=\frac{c}{d}$$. So,

$$\frac{dx}{u}=\frac{dy}{y}=\frac{du}{x}$$

can be written as

$$\frac{dx+du}{u+x}=\frac{dy}{y} ~~\Rightarrow~~~ \ln(x+u) = \ln(y)+C ~~\Rightarrow~~~ \ln\Big(\frac{x+u}{y}\Big)=C ~~\Rightarrow~~~ C_2=\frac{x+u}{y}$$

So, we have that $$C_1=u^2-x^2$$ and $$C_2=\dfrac{x+u}{y}$$. We can combine the two constants such that $$C_2=F(C_1)$$ where $$F$$ is an arbitrary differentiable function. Then,

$$\frac{x+u}{y} =F(u^2-x^2)$$

or

$$u=-x+yF(u^2-x^2)$$

We now need to apply our initial data. We are given that $$u(x,1)=2x$$. Therefore,

$$2x=-x+f((2x)^2-x^2) ~~\Rightarrow~~~ 3x=F(3x^2)$$

Letting $$w=3x^2$$, we see that

$$x^2=\frac{w}{3} ~~\Rightarrow~~~ x=\pm\sqrt{\frac{w}{3}}$$

So, $$F(w)=\pm3\sqrt{\frac{w}{3}}$$. Therefore,

$$u=-x+yF(u^2-x^2) = -x+y\pm3\sqrt{\frac{u^2-x^2}{3}} = -x+y\pm\sqrt{3(u^2-x^2)}$$

Hence,

$$u(x,y)=-x+y\pm\sqrt{3(u^2-x^2)}$$ $$u_x=-1\pm\frac{\sqrt{3}{x}}{\sqrt{u^2-x^2}}\quad\text{and}\quad u_y=1$$

Thus,

$$uu_x+yu_y=u\Big(-1\pm\frac{\sqrt{3}{x}}{\sqrt{u^2-x^2}})+y=...=x$$

No, it doesn't. I must have made a mistake somewhere.

• It appears that a mistake is made after setting $3x=F(3x^2)$. This produces $u=-x+yF(u^2-x^2) = -x+y\pm3\sqrt{\frac{u^2-x^2}{3}} = -x+y\pm\sqrt{3(u^2-x^2)}$, which is not a function of x and y. – Axion004 Jan 12 '19 at 19:58