# Change variable to solve equation

• MHB
• evinda
In summary, the conversation discusses solving the equation $xu_t+uu_x=0$ with $u(x,0)=x$ using a hint to change variables. The first attempt using the hint leads to an equation that does not help in using the method suggested in a previous post. The conversation then considers the possibility of solving the equation without using the hint, leading to a solution of $u(x,t)=\frac{x^2-C}{2t}$ with a discussion on whether the solution holds for $x=0$. Eventually, the conversation concludes that the solution can be corrected by defining $u(0,t)=0$ or using the sign function.
evinda
Gold Member
MHB
Hello! (Wave)

I want to solve the equation $xu_t+uu_x=0$ with $u(x,0)=x$. There is a hint to change variables $x \mapsto x^2$.

I have tried the following.

Let $x=y^2$.

Then $u_x=\frac{du}{dy} \frac{dy}{dx}+\frac{du}{dt} \frac{dt}{dx}=u_y \frac{1}{\frac{dx}{dy}}=\frac{1}{2y} u_y$.

Thus, $xu_t+uu_x=0 \Leftrightarrow y^2 u_t+\frac{1}{2y}uu_y=0$. But the latter doesn't help us to use the method of my post [m]Solve equation[/m].

So do we have to make an other change of variables? (Thinking)

Hey evinda! (Smile)

Suppose we do not use the hint?
We can divide the ewuation by x to make it easier can't we? (Wondering)

I like Serena said:
Hey evinda! (Smile)

Suppose we do not use the hint?
We can divide the ewuation by x to make it easier can't we? (Wondering)

Then we have that $u_t+\frac{u}{x} u_x=0$.

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

$\frac{d}{dt}u(x(t),t)=0$.

Thus, $u(x(t),t)=c$.

The characteristic line that passes through the points $(x,t)$ and $(x_0,0)$ has slope

$\frac{x-x_0}{t-0}=\frac{dx}{dt}=\frac{u}{x}=\frac{x_0}{x}$$x-x_0=t \frac{x_0}{x} \Rightarrow x_0 \left( \frac{t}{x}+1\right)=x \Rightarrow x_0=\frac{x^2}{t+x}. So u(x,t)=\frac{x^2}{t+x}. Is everything right? evinda said: Then we have that u_t+\frac{u}{x} u_x=0. \frac{dx}{dt}=\frac{u}{x} \frac{d}{dt}u(x(t),t)=0. Thus, u(x(t),t)=c. The characteristic line that passes through the points (x,t) and (x_0,0) has slope \frac{x-x_0}{t-0}=\frac{dx}{dt}=\frac{u}{x}=\frac{x_0}{x} This time it's not a line is it? After all, \d xt isn't constant although u is. Shouln't we solve \d xt=\frac ux, and then make sure that it satisfies the boundary condition? (Wondering) I like Serena said: This time it's not a line is it? After all, \d xt isn't constant although u is. Shouln't we solve \d xt=\frac ux, and then make sure that it satisfies the boundary condition? (Wondering) So we substitute here \d xt=\frac ux that u(x(t),t)=x_0 ? If so, then is it as follows?$$\frac{dx}{dt}=\frac{u}{x} \Rightarrow \frac{dx}{dt}=\frac{x_0}{x} \Rightarrow xdx=x_0 dt \Rightarrow \frac{x^2}{2}=x_0 t+c \Rightarrow x^2=2x_0 t+C \Rightarrow 2x_0 t=x^2-C \Rightarrow x_0=\frac{x^2-C}{2t}$$Thus, u(x,t)=\frac{x^2-C}{2t}. We have that for t=0, u(x,0)=x. But can that hold? Is it maybe as follows? Since \frac{1}{2t} \to \infty as t \to 0, it has to hold C=x_0^2. (Thinking) Last edited: evinda said: So we substitute here \d xt=\frac ux that u(x(t),t)=x_0 ? If so, then is it as follows?$$\frac{dx}{dt}=\frac{u}{x} \Rightarrow \frac{dx}{dt}=\frac{x_0}{x} \Rightarrow xdx=x_0 dt \Rightarrow \frac{x^2}{2}=x_0 t+c \Rightarrow x^2=2x_0 t+C \Rightarrow 2x_0 t=x^2-C \Rightarrow x_0=\frac{x^2-C}{2t}$$Thus, u(x,t)=\frac{x^2-C}{2t}. We have that for t=0, u(x,0)=x. But can that hold? Is it maybe as follows? Since \frac{1}{2t} \to \infty as t \to 0, it has to hold C=x_0^2. Yes. It's a bit clearer if we substitute t=0 and x=x_0 in the earlier expression 2x_0 t=x^2-C. (Mmm) I like Serena said: Yes. It's a bit clearer if we substitute t=0 and x=x_0 in the earlier expression 2x_0 t=x^2-C. (Mmm) Ok, so then we get that x_0^2+2 x_0 t-x^2=0. x_0=-t \pm \sqrt{t^2+x^2}. Thus, u(x,t)=-t \pm \sqrt{t^2+x^2}. Since u(x,0)=x, it follows that u(x,t)=-t+\sqrt{t^2+x^2}. Is this right? Also, do we have to check seperately the case x=0 since we have divided at the initial equation by x ? (Thinking) evinda said: Ok, so then we get that x_0^2+2 x_0 t-x^2=0. x_0=-t \pm \sqrt{t^2+x^2}. Thus, u(x,t)=-t \pm \sqrt{t^2+x^2}. Since u(x,0)=x, it follows that u(x,t)=-t+\sqrt{t^2+x^2}. Wait! (Wait) Wouldn't we get that u(x,0)=|x|? That's not correct, is it? (Wondering) evinda said: Is this right? Also, do we have to check seperately the case x=0 since we have divided at the initial equation by x ? (Thinking) I believe it suffices to observe that u is well-defined for x=0. (Wink) I like Serena said: Wait! (Wait) Wouldn't we get that u(x,0)=|x|? That's not correct, is it? (Wondering) Oh yes, right... So have I done something wrong? (Thinking) I like Serena said: I believe it suffices to observe that u is well-defined for x=0. (Wink) Why does this suffice? And also how do we deduce it? evinda said: Oh yes, right... So have I done something wrong? It's just that when x changes sign, the solution changes sign as well. We can fix it with;$$u(x,t)=-t+x\sqrt{1+\frac{t^2}{x^2}}$$evinda said: Why does this suffice? And also how do we deduce it? All our functions are continuously differentiable. Since the solution is valid for x\ne 0, it will also be valid for x=0 as long as it is defined. (Thinking) I like Serena said: It's just that when x changes sign, the solution changes sign as well. We can fix it with;$$u(x,t)=-t+x\sqrt{1+\frac{t^2}{x^2}}$$Ok. I like Serena said: All our functions are continuously differentiable. Since the solution is valid for$x\ne 0$, it will also be valid for$x=0$as long as it is defined. (Thinking) The limit$\lim_{x \to 0} \left( -t+x\sqrt{1+\frac{t^2}{x^2}}\right)$does not exist. What does this mean? (Thinking) evinda said: The limit$\lim_{x \to 0} \left( -t+x\sqrt{1+\frac{t^2}{x^2}}\right)$does not exist. What does this mean? (Thinking) Ah yes. It means my solution is not entirely correct. (Blush) Btw, that limit does exist, doesn't it? It's$0$. Either way, we should either additionally define$u(0,t)=0$, or we should make it$u(x,t)=-t+\operatorname{sgn}(x)\sqrt{t^2+x^2}$, where$\operatorname{sgn}$is the sign function. (Thinking) We know that the solution is continuous and we have that$\lim_{x \to 0} \left( -t+ x \sqrt{\frac{t^2}{x^2}+1}\right)=0$, so the solution will be$u(x,t)=\left\{\begin{matrix}
-t+x \sqrt{\frac{t^2}{x^2}+1}, & x \neq 0\\
0,& x=0.
\end{matrix}\right.$Right? evinda said: We know that the solution is continuous and we have that$\lim_{x \to 0} \left( -t+ x \sqrt{\frac{t^2}{x^2}+1}\right)=0$, so the solution will be$u(x,t)=\left\{\begin{matrix}
-t+x \sqrt{\frac{t^2}{x^2}+1}, & x \neq 0\\
0,& x=0.
\end{matrix}\right.\$

Right?

Yep! (Nod)

I like Serena said:
Yep! (Nod)

Nice... Thank you! (Smile)

## 1. How do I know which variable to change when solving an equation?

When solving an equation, you should choose the variable that you are trying to solve for. This is typically the variable that is isolated on one side of the equation. For example, if the equation is 2x + 3 = 9, you would change the variable x to solve for its value.

## 2. Can I change more than one variable when solving an equation?

Yes, you can change more than one variable when solving an equation, but it may make the problem more complex. It is best to change one variable at a time and then use the resulting equation to solve for the remaining variables.

## 3. What is the purpose of changing variables when solving an equation?

Changing variables in an equation is a useful algebraic technique that makes it easier to isolate and solve for a specific variable. It allows you to manipulate the equation and simplify it, making it easier to find the solution.

## 4. How do I change the variable in an equation?

To change the variable in an equation, you can use algebraic techniques such as addition, subtraction, multiplication, and division. For example, if the equation is 2x + 3 = 9, you can subtract 3 from both sides to change the variable x.

## 5. Are there any rules for changing variables in equations?

Yes, there are a few rules to keep in mind when changing variables in equations. The most important rule is to perform the same operation on both sides of the equation to keep it balanced. Additionally, be sure to keep track of any negative signs and follow the order of operations when simplifying the equation.

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